System design and 8051

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The hardware, firmware, and software design of microprocessor systems
Recommended Tools:
1. HT-USB5130: USB Flash 8051uC Development Board
2. Keil C Compiler
3. FLIP Programmer for HT-USB5130
Systems Design with 8051 Microcontroller

Table of Contents
Title Page................................................................................... 1
Table of Contents...................................................................... 5
Section I – Foundations....................................................................4
Section II – Systems.........................................................................6
Section III – Applications...............................................................10
Section IV – Hardware...................................................................11
Section V – Architecture ................................................................12
Section VI – Communication .........................................................13
Section VII – Documentation ........................................................14
SECTION I - FOUNDATIONS 15
1. Introduction ............................................................................ 16
Why this book, now 16
Development environments 18
Book structure 18
Class structure 19
Credit where credit is due 20
2. Fundamental circuits.............................................................. 21
Fundamentals 21
Output 22
Input 23
Computer vs. microcontroller 24
History 101 24
Microcontroller 25
Microcontroller input/output 26
Sink or source 27
Propagation delay and power consumption 28
External input/output 29
3. Memory devices ...................................................................... 31
Where do you keep it 31

Table of Contents 5
Program 31
Data 32
Dual in-line package 32
Connections 33
How it works 34
Other uses 34
4. Project 0 - Memory.................................................................36
Project 0: Math using ROM 36
5. Micro primer...........................................................................38
Its all in the family 38
On-board data memory 39
Arithmetic 39
Other packages 40
Flash microcontroller 40
Program memory locks 40
Features 41
The extended family 42
6. Address and interface.............................................................43
What is the connection 43
Power 43
Clock 43
Reset 45
Ports 45
Port 0 46
Port 2 46
External memory 47
External program 47
External data 48
External 64K 48
Port 1 49
Port 3 49
7. Minimum system.....................................................................50
Minimalist 50
Project 51
Minimum software 51
Opcodes, mnemonics, comments 52
Classes of instructions 53
Schematic 53
8. Machine cycle time .................................................................55
First computer circuit 55

6 Systems Design and the 8051 Durham
Cycle time 56
Machine cycles 57
Long precise wait 58
In from out 58
Switch a bit 59
Circuit: led metronome 60
9. Project 1 – Output & time delay ........................................... 61
Project 1: Metronome 61
Program sample example 63
10. Software development............................................................ 66
The here and now 66
Instructions 67
Assembler directives 69
Step by step 71
Program with comments 71
Listing 73
Intel hex 75
Commentary 76
The top placement 77
The subs 78
Your comments, please 78
The bottom placement 79
Structure 79
11. Design practices ...................................................................... 81
Top down 81
Extreme programming (XP) 82
Steps for success 83
Process diagram 85
SECTION II - SYSTEMS 86
12. Switch, logic, and subs............................................................ 87
Switch hitter 87
Debounce 88
Bit manipulation 88
Masking logic 89
Rotate and exchange 90
Conditional branch 91
Subroutines 92
Stack 93
Circuit: led and switch 95

Table of Contents 7
13. Project 2 – Input & decisions.................................................96
Project 2: T-bird taillights 96
14. Register, timers, and interrupts ..........................................103
Timer registers 103
Timer 104
Interrupts 105
Counter & interrupt examples 106
Timer with interrupt examples 107
Circuit: interrupts 108
15. Project 3 – Clock & interrupt..............................................109
Project 3: Time to count 109
16. Board construction ...............................................................115
One step. Check! 115
Show and tell 115
Basics 116
Socket to me 117
What’s left 118
17. Project 4 - Development board............................................119
Project 4: Build from scratch 119
18. External memory ..................................................................122
Storage control lines 122
Address fetching 123
Timing sequence 124
Virtual memory 125
Wiring ROM or RAM 126
19. Bios.........................................................................................128
Definition 128
Bios main 129
Static memory test 130
Download 132
Downbyte 134
Checksum 135
ASCII to hex conversion 136
Memory switch 137
Use of low memory 139
20. Project 5 – Bios development tool .......................................140
Project 5: Develop operating system 140

21. Serial communications......................................................... 144
Background 144
Microcontroller 146
Generating baud rates 146
Mode 0 147
Mode 1 147
Mode 2 148
Mode 3 148
Timer/counter 2 baud rates 148
Timer baud table 149
Timer 1 and color burst 149
Serial initialization 150
Serial data protocol 151
Serial buffer 153
Circuit: serial 154
22. Project 6 – RS232 communications..................................... 155
Project 6: RS 232 to PC exchange 155
23. Expansion latches ................................................................. 162
I/O expansion port 162
I/O expansion memory 163
Latch in/out connection 163
Latch in/out code 165
24. Memory-mapped input and output..................................... 167
Accessing external data 167
The instruction 168
The setup 169
The hook-up 169
Latch in/out memory-mapped 171
25. Project 7 – I/O expansion..................................................... 173
Project 7: Unlimited I/O 173
26. Tables..................................................................................... 177
Data in code memory 177
Data byte 178
Characters available table 179
Movx vs. movc 181
Code messages 182
Enhanced serial messages 183

Table of Contents 9
27. Multiplexing ..........................................................................185
Perception 185
Multiplex 186
Circuit: displays 187
Code requirements 189
Code segment for port 190
Code segment for memory map 190
Binary to binary coded decimal 191
28. Project 8 - Seven-segment displays .....................................193
Project 8: Seeing what is not there 193
29. Matrix scanning ....................................................................200
Matrix inputs 200
Contact arrangement 201
Conflicts 203
Key debounce 203
Decipher 204
Complete solution 205
Connections 205
Test code 206
Decode flowchart 207
Keys procedure 210
Simple solution 211
Circuit: keypad 214
30. Project 9 - Keypad ................................................................215
Project 9: Debounce & matrix inputs 215
31. Liquid crystal display...........................................................226
Different display systems 226
LCD variations 227
Connections 227
Control 228
Control via port 229
Control via latch 230
Control via PLD 231
Command 232
Initialization 233
Cursor position 234
Message display 235
32. Project 10 - Text display ......................................................236

Project 10: Text message screens 236
SECTION III - APPLICATIONS 249
33. Infrared communications .................................................... 250
Local wireless 250
Philips protocol 251
Detected string 253
Connections 255
Circuit: infrared receiver 255
34. Project 11 - Wireless............................................................. 256
Project 11: Communicate with IR 256
35. Serial chips – IIC .................................................................. 263
Other chip interfaces 263
Inter integrated circuit 264
IIC details 265
IIC sequence 267
IIC bit bang 268
36. Serial chips – SPI.................................................................. 271
Serial peripheral interface 271
Analog to digital sensitivity 272
Analog to digital noise 273
LTC 1098 clocking 274
LTC 1098 operation 274
Program: LTC 1098 bit-bang 275
Onboard SPI control register 279
Program: EEPROM SPI register 280
TLC549 clocking 285
Circuit: SPI 286
37. Project 12 - A to D converter............................................... 287
Project 12: Analog / digital converter 287
38. Waveform synthesis.............................................................. 293
Real world output 293
Sensitivity 294
Circuit: digital to analog 295
Software 295
39. Project 13 – D to A converter .............................................. 297

Table of Contents 11
Project 13: Analog output 297
40. Project 14 - Photosensor.......................................................302
Project 14: Barcode reader 302
41. Project 15 – Analog control .................................................304
Project 15: Pulse width modulation 304
42. Project 16 - Digital feedback................................................308
Project 16: DC motor speed control 308
43. Math functions......................................................................311
Arithmetic 311
Extended precision 312
SECTION IV – HARDWARE 327
44. Parts and pin-outs.................................................................328
Watch your money 328
Proto then uC board 329
uC board only 329
uC board optional 330
Projects 330
uC board headers & jumpers 331
PLD / PEEL pin-out 331
Microprocessor pin-out 332
Buffer pin-out 333
RS232 & RS233 pin-out 334
7-Segment & LCD pin-out 335
A/D Converter pin-out 336
Memory pin-out 337
Cable pin-out, SPI & serial 338
45. Development board...............................................................340
Design 340
Options 340
HyperTerminal 341
Test Program 341
Schematic 341
Board specifications 343
46. In system programming .......................................................349
Serial downloading 349
Programming algorithm 350
Programming instruction 351

Programming schematic 352
Peripheral timing 352
Programming and printer 353
Connectors 354
SECTION V – ARCHITECTURE 356
47. Instruction set ....................................................................... 357
Microcontroller instruction set 357
Addressing modes 358
Data transfer 359
Arithmetic operations 360
Program branching 361
Logical operations 362
Bit manipulation 363
Instructions that affect flags 363
Instruction set 364
48. Memory organization........................................................... 366
Harvard vs. Princeton 366
Code addresses 367
External data addresses 368
Data memory expansion 368
Internal data addresses 369
Internal RAM low 371
Internal RAM high 372
Predefined bit addresses 373
Predefined bits port 3 374
49. Special function registers..................................................... 375
Reserved memory 375
Ports 376
Port 0 376
Port 1 377
Port 2 377
Port 3 377
PSW: Program status word 378
PCON: Power control register 379
Interrupts 380
IE: Interrupt enable register 381
IP: Interrupt priority register 381
Timer / counters 382
TCON: Timer/counter control register 383
TMOD: Timer/counter mode register 383

Table of Contents 13
Serial 385
SCON: Serial control register 385
50. SFR extended ........................................................................387
Enhanced registers 387
Timer/counter 2 387
T2CON: Timer/counter 2 control register 389
T2MOD: Timer 2 mode register 390
Timer 2 data registers 390
Serial peripheral interface 391
SPCR: SPI control register 392
SPSR: SPI status register 393
SPDR: SPI data register 394
WMCON: Watchdog 394
Using onboard EEPROM 395
SECTION VI – COMMUNICATION 399
51. Ascii........................................................................................400
What is it 400
ASCII-hex table 401
52. Rs 232.....................................................................................403
Connections 403
RS 232 pin outs 404
Schematic 405
Development board pin outs 406
53. Network connection..............................................................407
Network 407
Diagram – digital network 409
Diagram – analog telephone 409
Diagram – analog audio 410
54. Programmable logic device..................................................411
It is just logic 411
Combinational logic 412
First time user 413
Gated latch 414
OLMC and pin assigments 414
Registers 415
Combination output enable 416
Limitations 417
Program: combinational logic (*.psf) 417
Created Files 421

55. Circuit time & phase shift.................................................... 422
Background 422
Delay 423
Clock signals 424
Interaction 424
Ext program memory read cycle 426
Ext data memory read cycle 426
Ext data memory write cycle 427
SECTION VII – DOCUMENTATION 428
56. Extreme programming (XP) harmonization...................... 429
General guidelines 429
Program specifics 430
57. Documentation...................................................................... 442
Report 442
Computer aided design 444
58. End......................................................................................... 446
59. Author.................................................................................... 447
ž Ÿ

SECTION I - FOUNDATIONS
ž Ÿ

1
INTRODUCTION
Thought
Engineering is the tradeoff
between quality, time, and money
You can have two, but you cannot have all three.
MOD
Why this book, now ______________
Why is there a need for another book on system design or on the
8051 microcontroller? Actually, the answer is quite simple. There
are numerous books on systems and digital design. There are even
several on the 8051. However, we have found no reference that
treats the 8051 as the basis of a system.
In addition, most books present substantial theory before
introducing projects. The task of this book is to start building
projects immediately after discussion of the relevant topics. After
all, that is why most people want to be an engineer – it is to build
objects and see them work.
Most books and articles on the 8051 treat it as a simple device,
suitable for just a few tasks. Most authors have sample projects, but
they are not integrated. On the other hand, I have used the
microcontroller family as a basis for industrial and commercial
systems for over 20 years. I have also used it as the foundation for
an upper / graduate level university design class.

Chapter 1 Introduction 17
Why use the 8051 architecture? The core of the 8051 arguably is
used more than any other device. It is an expansion of the i8042
architecture, which is the fundamental processor used in keyboards
for personal computers. Phenomenally, the architecture has been
around for over 25 years. It is constantly being enhanced by various
vendors to perform tremendous tasks.
What is meant by a system design with the microcontroller? One
microcontroller forms the kernel for all the electronic and digital
functions associated with a system. The power of the 8051 devices
makes it very capable of filling the functions of a much larger
computer system.
The system approach is to integrate together in one microprocessor
every task simultaneously. These include digital in, digital out,
analog in, analog out, serial communications with other computers,
keypad, seven-segment display, liquid crystal display, local and
remote control, data logging, 32-bit math calculations with square
root, and all this in a real time environment. That is a system!
An earlier reference, Microcontrollers in Systems Design began the
task. It was very project oriented with an inclusive reference for the
8051. However, it had one shortcoming. It required the reader to
have a good grasp of computer systems and (s)he had to translate
the reference material into the projects.
This work, Systems Design and the 8051, takes a different tack. It is
based on engineering design principles that are elucidated as
required. Extreme programming (XP) approach to compatibility
allows each new project to be integrated as a separate, but
interactive module.
The book begins with the fundamentals and sequentially adds new
projects. This is accomplished by integrating a straightforward and
concise application explanation for all software commands. That is
followed by a circuit representation of the new components. The
project requirements are laid out. Finally, an exemplar program
shows a similar system that the designer can modify to complete the
project.

The procedure can obviously work. For many years, I have had
students step-by-step build a functioning, integrated computer
system around 10 projects. They start with a parts list, obtain the
parts, and build the projects into a working system. This is done in a
single semester. What an accomplishment!
Development environments ________
In most design environments, hardware and software are segmented.
Moreover, specialists typically handle the tasks separately. While
this may be beneficial for large systems development, it restricts
understanding of the overall big picture of computer applications.
This treatise will present the application of both hardware and
software to the solution of real problems. The most effective piece
of equipment to provide a simultaneous understanding of computer
architecture and programming is the microcontroller.
The microcontroller has a complete computer built on a chip. In
addition, much of the interface hardware to outside components has
been included on the chip. A working computer is obtained by
connecting switches to the input, digital displays to the output, and
adding a program to the memory.
The prerequisites for being able to successfully build the projects
are minimal. A reasonable understanding of TTL digital logic is
assumed. Proficiency in a high level programming language is also
assumed. Understanding of an assembly language would be very
beneficial, but is not necessary.
Book structure __________________
The structure of the treatise is to provide a working tool that can be
readily referenced. The first topics are general items and
fundamental connections. The next area is a group of projects that
can be constructed. Chapters that give the foundation material
precede each project. An exemplar program that can be modified to
complete the project follows the project description.

Chapter 1 Introduction 19
A parts list and development board are discussed. The instructions
and memory organization comes next. This is followed by
communications protocols. Then, a programmable logic device
(PLD) representation is used for the combinational logic. The final
chapters are documentation techniques. It is a good idea to look
over these section to have an idea of the available reference material
for use as the projects are developed.
The projects will begin with using a memory device as a table for
arithmetic functions. A number of projects will be constructed for
performing common control tasks. The epitome of control will
entail communications between two computers.
The projects are for demonstration of technology and gaining of
experience. Where necessary, a schematic and/or software are
provided. Often this is an illustration of a related idea. It is not
intended to be a solution to the project. It is intended to provide a
framework for tackling the project.
Class structure __________________
It is not necessary to perform all the projects or to do them all in
order to obtain a working knowledge of microprocessors.
A proven procedure is to use Project 1, 2, and 3. Project 4 can be
included if a development board is used. Then Project 6 and 7
should be completed. These will provide the foundation to do any
project. The remaining projects can be used in almost any order, if
desired.
Projects 8, 9, and 10 provide human interface devices.
Project 11, 12 and 13 give infrared communications and analog /
digital conversion exposure.
Within a semester, I typically use 1, 2, 3, 4, 6, 7, 8, 9, 10 and a
choice of 11, 12, or 13.

You can accomplish the tasks. The projects should prove both
challenging and enjoyable. Occasionally, you may be frustrated but
persevere. As a result, the completion of the task will be most
rewarding.
The material comes with a guarantee. If you complete all the tasks,
you will have the tools to design a computer system for any
purpose. Good Computing!
Credit where credit is due _________
Everything we know is developed from something we have read,
heard, or seen. Therefore, these other thoughts necessarily influence
what we write. To the best of our knowledge, we have given
specific credit where appropriate.
Rather than footnotes or references, we have listed the works that
have provided significant information in one way or another, since
this is often in concepts rather than quotes. Some of the information
is general and public domain, while some is device specific. The
generic information is used where possible.
Statements that are attributed to us are things we have used
commonly and do not recall seeing from someone else. Others
obviously have similar thoughts. If we have made an oversight in
any credits, we apologize and we would appreciate your comments.
Dr. Marcus O. Durham
ž Ÿ

2
FUNDAMENTAL CIRCUITS
Thought
Tell me the input, output, and
what to do in between, then
I can write a computer program.
Capt. Ed Fischel, USAF
Fundamentals ___________________
What are the fundamentals? Is it not input, output, and how these
are related? A mechanical switch represents every type of input
there is in a digital world. Similarly, a lamp or light emitting diode
(LED) is a surrogate for every possible output.
These two simple devices can effectively interface to any real world
circumstance by placing a buffer between the digital components
and the external items. Hence, a transistor or gate can substitute for
the switch and LED.
The power for most digital computer devices historically has been 5
volts DC referenced to ground. In many of the newer technologies,
this level is dropping below 2 volts. Regardless, the power voltage
is called VCC and has a digital value of 1 or True. Similarly, the
reference ground is zero volts, is called VSS, and has a digital value
of 0 or False.

The power is connected to the upper right corner of most chips.
Similarly, the ground is connected to the lower left corner.
Output _________________________
Since a light emitting diode (LED) is a logic device, it has polarity.
The positive connection is called the anode, it is the longest lead,
and is the arrow side of the diagram. The negative terminal is the
cathode, it has a flat spot on the edge of the LED, and is the bar in
the diagram.
A diode has a very low internal resistance. Hence, it has very
limited current handling capability. If it were connected directly
from power to ground, the current through the low resistance would
rise very high. The power dissipation would increase by the square
of the current. Then the diode would burn out or blow.
To restrict the current, common practice is to place a resistor in
series with the LED. In restricting the current, the brightness is
reduced. Therefore, for optimum brightness, the series resistor
should be calculated to give the optimum current.
VCC Ground
Various light emitting diodes have a voltage drop of 1.4 to 2.5
Volts. The maximum current for different LEDs is 20 up to 750
mA. However, the current must be restricted below the maximum
that can be supplied by the gate. To limit the current a resistor is
placed in series. Assume a current of 5 mA and 1.5 V drop.
R = V / I = 1.5 V/ 5 mA = 300 :
Typical values are 270 to 330 Ohms. If the lamp is not bright
enough, the specifics should be verified, and a different resistor
should be calculated.

Chapter 2 Fundamental Circuits 23
Input ___________________________
A switch must represent two states, both off and on. Therefore, it
must be connected to both power and ground. To prevent a short
circuit when the switch is closed, a resistor is placed between the
power and the switch. This is called a pull-up resistor. The other
side of the switch is connected to ground.
The output of the switch is taken between the resistor and the
switch. The other side of the output is ground.
VCC
Out
Ground
The resistor must be sized to limit the current in a short-circuit
condition. Many devices have a 10 mA limit on the amount of
current they can sink. When connecting a switch as a short circuit,
the current is often limited to 2.5 mA. Therefore, the resistor is sized
based on that value.
For a voltage source of 5 V and a current load of 2.5 mA, the
resistance is calculated from Georg Ohm’s rather famous law.
R = V / I = 5 V / 2.5 mA = 2000 Ohm
The common sized pull-up resistor is typically 2,000 to 2,200
Ohms.
Next consider the power rating of the resistor.
P = V*V / R = 5*5 / 2200 = 0.011 W
At this low quantity of power, virtually any resistor network has
adequate dissipation capability.

Computer vs. microcontroller ______
What is the difference in a computer, microprocessor, and
microcontroller? A computer is a device with an internal processor,
memory, and interface for input and output. The processor consists
of a central processing unit, a control unit, and an arithmetic
processing unit.
A computer is a complete system, while a microprocessor is
basically the core of the unit. The microprocessor will generally
require other chips to connect to the input / output interface.
A microcontroller is a microprocessor with additional interface
components as part of the chip. In essence, a microcontroller is
designed to connect directly to the input and output. In addition, it
has some memory as part of the chip. In some cases, these are also
called single chip computers.
A microcomputer is typically designed for general-purpose
applications. A microcontroller usually has a dedicated purpose as a
control or instrumentation device.
History 101 _____________________
The computer was originally developed during World War II. These
machines were physically very large, but had limited memory.
Large computers continued to develop until the very powerful
mainframe systems of the 1960’s. These machines still had discrete
magnetic core memory, transistors, and diodes.
At that time, the first integrated circuits (ICs) were being developed.
From the earliest days of the IC, microcomputers have been a major
component. A brief chronology of the microcomputer shows the
rapidity of the development.
1968 Intel formed from Fairchild Semiconductor

1971 Intel 4004 – 4 bit calculator
1972 Intel 8008 – eight bit
1974 Intel 8080 – eight bit TTL compatible
Motorola 6800
MOS Technology 6502
1978 Intel 8086 – sixteen bit basis of PC
1979 Intel 8051 – eight bit microcontroller
It is fascinating that the architecture and software of the 8051 has
continued for over 20 years. It is the core architecture for over half
of all microcontrollers.
Microcontroller _________________
The basic computer circuit is actually very simple. First power and
ground are connected to the upper right and lower left pins
respectively. Next, a crystal is connected to the XTAL pins to
provide stimulus for the internal oscillator. An 11.059 MHz crystal
will provide good performance and excellent communications.
Other choices are available and can be found in later sections. A
very small coupling capacitor of 10 – 40 pf is connected from each
side of the crystal to ground.
Third, pulling the RESET line high restarts the processor. The
RESET line can float low during normal operation. A power-on
reset circuit is often connected to the pin to automatically perform
the start function.
When using a flash microcontroller, a method of loading the
program is required. Program loading involves three lines, an input
for instructions, an output for data, and an input for clocking the
other two lines. An illustration is given in the figure below. Details
for implementing the in-system feature are provided in a chapter on
in-system programming in the reference section after the projects.
For microcontrollers without in-system programming, these lines
are not required.

Microcontroller input/output ______
The input / output interface is the defining feature of a
microcontroller. On the processor, there are four different ports for
this function. Each port represents one byte, which is connected in
parallel to pins of the chip.
The port is identified as a special function register within the
internal RAM. Because of the address of the ports, all are bit
addressable. The ports are bi-directional. In default operation, the
port will output a byte or bits of data.
Any bit of the port can be converted to input. Simply send a one to
the bit of the port. Then that bit can be read or input. If all ones are
sent to a port, then the entire byte becomes an input.
The internal circuitry for one bit of a port is shown. A latch holds
the output to the pin. A separate line provides the input.
For output, data is written to the port. The result is displayed on the
pin. For input, a one is sent to the port. This enables the line and
VCC
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset
Xtal2
Xtal1
Ground
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset

pulls the pin high. Therefore, with no other connection to the pin,
input from the port bit will be one. If the pin is pulled low by an
external switch, then the input is a zero.
VCC
port in
Sink or source ___________________
Connecting external devices to a gate can critically impact the chip.
A gate, whether in a microcontroller or a logic device, is limited to
the amount of current it can handle. If too much current is drawn by
an external connection, the chip may be damaged. Therefore, a
resistor or a buffer must limit the current.
A chip may provide VCC or Ground to the external devices.
Supplying VCC is the high state and is a source for current.
Supplying the ground connection is the low state and is a sink for
current.
Sink current is the amount of current that will flow into an output,
IOL, at low state. Source current is the amount of current that will
flow out of a gate, IOH, at the high state.
Total source (IOH) = input to load (IIH) * number of loads (m)
Total sink (IOL) = input to load (IIL) * number of loads (n)
The number of loads is called fan-out. The limiting fan-out is the
lower of the number of loads (m or n). The maximum current, and
resulting fan-out, must be observed to preclude overloading and
damaging a gate. When interfacing to different logic families, total
the load current for all devices.
Q
Latch
Q’
pin

Propagation delay and power consumption
Two other parameters determine the performance of a gate.
Propagation delay is the delay of low-to-high (tPLH)and high-to-low
(tPHL) transitions between the input and output of a gate.
tPLH tPHL
For TTL logic, power consumption is an average of power
consumption for high and low state output.
PD = VCC * (ICCH + ICCL) /2
The table below gives the performance for various logic families. A
minus sign indicates current is flowing out of the gate.
A later chapter on circuit time and phase shift addresses the timing
and propagation delay in more detail.
Description Range Unit Sym S LS AS ALS F CMOS
Propagation delay typical Ns tPD 3 9 1.7 4 3
P consumption per gate avg. mW PD 19 2 8 1.2 4
Speed*power (energy) pJ E 57 18 13.6 4.8 12
Low-level input voltage max V VIL 0.8 0.8 0.8 0.8 0.8 1.5
Low level output voltage max V VOL 0.5 0.5 0.5 0.5 0.5 0.05
High-level input voltage max V VIH 2.0 2.0 2.0 2.0 2.0 3.5
High-level output voltagemax V VOH 2.7 2.7 2.7 2.7 2.7 4.95
Low-level input current max mA IIL -2.0 -0.4 -0.5 -0.2 -0.6 -0.1 PA
Low-level output current max mA IIH 20 20 20 8 20 0.1 PA
High-level input current max mA IOL 50 8 20 20 20 0.88 PA
High-level output current max PA IOH 1000 -400 -2000 -400 -1000 -0.88 PA

External input/output _____________
An LED is shown connected with the chip as a sink. In this mode, a
zero “0” is sent to the port to illuminate the LED. This is the
preferred approach since chips sink more current than they can
source.
Vcc
port
This circuit is very similar to that used for a switch input.
Vcc
port
ground
In this schematic, an LED is shown connected with the chip as a
source. In this mode a one “1” is sent to the pin to illuminate the
LED. This is a poor design since the chip is limited in the amount of
current it can source.
Port Ground
One circuit of particular interest combines both an LED and a
switch on a single port pin. This permits both a display and an input.
Vcc
port

When the switch is closed, the lamp will be on. Therefore, the
switch must be open for the lamp to be controlled by the port. When
the port sends a zero, the lamp is on. When the port sends a one, the
lamp is off, and the switch can be read.
ž Ÿ

3
MEMORY DEVICES
Thought
Memory is the second thing to go.
He just does not remember the first.
Old quip
Where do you keep it _____________
Two fundamental types of memory are typically used in a computer
system. These are program memory and data memory. Program
memory is static and seldom changes. Data memory is dynamic and
changes with the conditions. These functions determine the
technology that can be used for each purpose.
Program ________________________
Program memory is typically completely changed when there is a
new program. Program memory can use one time technology. This
includes devices such as read only memory (ROM). The basic
version is programmable ROM or PROM. Enhanced versions allow
the device to be erased by ultraviolet light and programmed by
special hardware. These are called erasable programmable ROM or
EPROM.
As technology advanced, the chip could be erased and programmed
by changing the voltage. These are electrically erased PROM or

EEPROM. A variation of the technology is called flash memory.
The erasure and programming capability is built into the chip. This
is selected by using one of the pins.
Data ___________________________
Data memory requires the status to be changed frequently. Only a
part of the memory is typically involved. Therefore, random access
memory (RAM) is used. Not surprisingly, this technology is more
complex.
Some internal memory uses EEPROM technology to accomplish
this purpose. Because of the technology, the number of read/ write
cycles is often limited.
Dual in-line package _____________
A common version of separate memory chips uses dual inline
packaging (DIP). These are typically limited to prototyping or
limited quantity projects.
The quantity of memory in the package is constantly increasing as
technology improves. To handle the additional pins, other package
technologies are often used for production devices.
The pin-out for static RAM, PROM, and EEPROM is almost
identical. There are two basic differences. The programming pins
are called write pins and the output enable pins are called the read
pins.
A typical 32K address X 8-bit byte package is illustrated as a 256K
bit device in the following diagram.

Chapter 3 Memory Devices 33
Connections _____________________
Like most other DIP packages, power is supplied to the upper right
(pin 28), and ground is supplied to the lower left (pin14).
* V
A14 1 28 VCC
A12 2 27 /WE
A7 3 26 A13
A6 4 25 A8
A5 5 24 A9
A4 6 23 A11
A3 7 22 //OE
A2 8 21 A10
A1 9 20 //CE
A0 10 19 DQ7
DQ0 11 18 DQ6
DQ1 12 17 DQ5
DQ2 13 16 DQ4
Ground 14 15 DQ3
Memory 256K
In addition, the chip must be selected or enabled. The Chip Enable
not (/CE) line must be pulled low for the chip to function. If there
are multiple memory chips, this is done by an address decoding
chip. If there is only one chip, then the line is simply connected low.
There is one precaution about connecting address lines on the
memory chip. Avoid letting any pin float. Connect the line to an
address decoder. If the address line is unused, connect it to ground.
When sending data to the chip, the Write Enable not (/WE) line
must be pulled low. When obtaining data from the chip, the Output
Enable not (/OE) line must be pulled low. Obviously, the two lines
should not be pulled low simultaneously.

How it works ____________________
Since reading is common to both types of memory, an output or
read will be used as a description of how a memory chip works. The
combination of address lines is used to select a particular byte of
information stored in the memory. The information stored in that
location will be imposed on the output pins.
The number of addresses is a power of two function of the address
lines. If there are ‘m’ address lines, then the number of addresses
can be calculated easily.
Addresses = 2m
For eight (8) address connections, there are 256 addresses. The first
address is zero (0) and the last would be 255.
Ten addresses lines would access 1024 locations. In common usage,
this is rounded down and called a 1K device. Larger sizes are
proportionally rounded to give common sizes that are powers of
two. A few examples are 16K, 32K, 64K, and 512K.
Each address accesses ‘n’ bits of information stored in the memory
and displayed on the output. In microcontroller systems, the number
of bits is 8 and is called a byte.
The total number of bits is used to describe memory devices. The
number of addresses and the number of bits output determines the
size of the memory. For example, a chip with 8 address lines and 8
output lines would be a 2K device.
2m
* 8 = 28
* 8 = 2048
Other uses ______________________
Since the chip will have an output based on the status of the address
lines, it can be used to represent a conventional logic network.
There are ‘n’ outputs and each output can have 2m
maxterms.

Chapter 3 Memory Devices 35
The ROM has fixed values, which makes it well suited for projects
that require a table lookup. This is particularly appropriate for
mathematics problems that are repeated frequently.
Consider an example.
Given: y = 2 x2
– 1
Allowable range: 0 d x d 3
What is the address (input) variable? x
What is the data (output) variable? y
How many addresses are there? 4 (0, 1, 2, 3)
How many address lines are required? m = 2 o 2m
= 4
What is the largest value output? 17
How many output lines are required? n = 5 o 2n
= 32
Create a table of values to implement the function. Negative values
are represented by setting the most significant bit (MSB) to 1.
Decimal numbers can also be represented by setting another bit.
Input Address lines Memory Output lines
0 00 -1 100001
1 01 1 00001
2 10 7 00111
3 11 17 10001
By connecting switches to the address lines and LED’s to the output
lines, the special purpose calculator is realized.
ž Ÿ

4
PROJECT 0 - MEMORY
Thought
Skip this project,
if not using external memory.
Project 0: Math using ROM _______
Purpose: Skip this project if you are not using external memory.
To show the function of a ROM.
To program an EPROM.
Preamble:
Every microcomputer has a Read Only Memory (ROM) to store the
basic system program. In the case of a development system, a
programmable and erasable ROM is preferred. Erasable
Programmable Read Only Memory (EPROM) is the most common
example. EPROM has lower cost than most other alternatives.
An EPROM contains fuses. These are disconnected or left as is to
create a pattern of information called programs. An EPROM has
address lines usually called A1, A2...An, on the circuit diagram. The
EPROM data lines are accessed for the output. The lines usually are
called O1, O2, O3...On.
The function of address and data lines is exactly the same as the
ones on a ROM. Placing information in an EPROM is called

Chapter 4 Project 0 - Memory 37
programming. Exposing the EPROM to an ultraviolet (UV) light
source will clear all locations of their programmed contents. The
duration of the exposure depends on the UV intensity. An
alternative is electrically erasable memory (EEPROM). The
programmer / burner electrically erases this version of storage. The
EEPROM is much faster and easier to erase and program.
Plan:
Demonstrate a ROM is a program memory. Program the ROM to
act as a table for a certain function.
Preparation:
Prepare a breadboard with a +5V power supply. Observe the
EPROM data sheet. Then wire light emitting diodes on the EPROM
data lines. Wire five switches on the address lines.
Investigate the use of the EPROM burner program on the personal
computer (PC). Become very familiar with these tools since they
will be used in burning all programmable chips. Before burning an
EPROM, make sure that the device is blank.
Check the EPROM to ensure any lock bits are off.
Procedure:
Implement the function F = x3
on the EPROM. Use the address lines
as the inputs (x) and data line as the outputs(F). Only 5 switches are
required on the address lines.
Presentation:
Demonstrate the output for x = 0, 1, 2...5. Only selected numbers
will be requested for demonstration.
ž Ÿ

5
MICRO PRIMER
Thought
Prim and proper
is an old Southern saying for
‘You have got it together.’
Its all in the family _______________
The 8051 family is the dominant core among all microcontrollers.
Intel originally developed the architecture about 1979. Other
vendors took the basic device and developed enhanced versions.
Various reports are that versions of the device represent over half of
all microcontrollers.
The standard package is a 40-pin, dual-in-line package (DIP) with
600-mil spacing between the rows of pins. It is without program
memory and has only 128 bytes of data memory. Program and data
memory are external chips. The most popular variation has program
memory and data memory on board the processor chip.
In this family of computers, program memory is separate from data.
This is called Harvard architecture and it is the design used by Intel
related machines. Mixed memory is Von Neumann’s Princeton
architecture. It is the design used by Motorola related machines.
Reduced instruction set computers (RISC) have very few
instructions but these can access any register in the computer.

Chapter 5 Micro Primer 39
Complex instruction set computers (CISC) have many instructions
and each is used to access a limited register or memory. The 8051
has characteristics of both because of its memory and software
strategy. However, it is primarily a CISC device.
On-board data memory ___________
The on-board data memory is divided into three memory areas.
First, there are 128 bytes of general purpose RAM. A significant
portion of this can also be accessed as registers or as bits.
The next 128 bytes are segmented into 21 Special Function
Registers (SFR). These bytes are the only ones accessible from the
upper 128 locations on a standard machine. The twenty-one
memory locations accessed as special registers cover many areas.
A & B accumulators
Timer/counter mode and control
Serial port mode and control
Stack pointer
Data pointer
Program status word for flags
Interrupt enable and priority
Parallel ports 0, 1, 2, and 3 for data I/O
Many machines now make an additional 128 bytes available as an
extended internal RAM. Additional data memory is often available
in the form of EEPROM or similar technology.
Arithmetic ______________________
Arithmetic functions are unsigned, binary integers. Familiar
commands include add, add with carry, subtract, and subtract with
borrow.

There are hardware multiply and divide instructions. The A register
is operated on by the B register. The results are in A with B as
supplemental.
Logical operations include AND, OR, and EXCLUSIVE OR. These
operate on A with the results placed back into A.
Other packages __________________
In addition to the 40-pin wide DIP package, there are other
variations. Although there are surface mount designs, our discussion
is oriented to packages that can be used for construction of a small
number of units.
The programmable logic chip carrier (PLCC) is a square
configuration that can fit in some spaces more effectively.
Projects that do not require external memory and that can be
implemented with only two ports use a very cost and size effective
arrangement. This is a 20-pin slim package.
Flash microcontroller ____________
One of the variations of the 8051 which has almost everything but
the kitchen sink is an Atmel design. Other than part of this section,
the section on in-system programming, and the chapter on extended
special function registers, all the remaining information in this book
is generic and applicable to the entire family of microcontrollers.
Atmel’s AT89S8252 flash microcontroller has several significant
characteristics. The features are listed in the following table. The
standard or core features are shown in the last column of the table.
Program memory locks ___________

Chapter 5 Micro Primer 41
Some versions have three lock bits that can be left unprogrammed
(U) or can be programmed (P) to obtain security and prevent further
programming or fetching of the program. In prototyping, these are
left unprogrammed.
The lock bits may be inadvertently set if the burner program is
improperly configured. With some in-system programmers, the bits
may also be inadvertently set.
Once programmed, the lock bits can only be unprogrammed with
the Chip Erase operation in either the parallel or serial modes.
Alternately, the chip can be placed in a burner and the bits cleared.
Features ________________________
MCS51 Compatibility 89s8252 8051
Bytes of in-system memory 8k none
Reprogrammable downloadable flash
SPI serial interface for program downloading
Endurance: 1,000 write/erase cycles
Bytes EEPROM 2K none
Endurance: 100,000 write/erase cycles
Operating range: volts 4 - 6 4.9-5.1
Fully static operation frequency: MHz 0 - 24 3-50
Three-level program memory lock yes
Eight bit internal RAM: bytes 256 128
Programmable I/O lines 32 32
Sixteen bit timer/counters 3 2
Interrupt source addresses 6 5
Programmable UART serial channel yes yes
SPI serial interface yes none
Low-power idle and power-down modes yes none
Interrupt recovery from power-down yes none
Programmable watchdog timer yes none
Data pointer two one
Power-off flag yes none

The extended family ______________
The data sheet for the device provides very substantial design
information. Much of this is similar to generic controllers.
A primer provides all the details about how to use the features that
are special to this machine. Both these data sheets are available from
the manufacturers and on the web.
The diagram illustrates the majors features of the machine.
ADDRESS
ž Ÿ
SERIAL
PORT
INTERRURT
CONTROL
4K
ROM
4K
ROM
BUS
CONTROL
4 I/0 PORTS
CPU
OSC
EXTERNAL
INTERRUPTS
RD WR
TIMER 0
128
RAM
128
RAM
TIMER 2
TIMER 1
P0 P2 P1 P3 TXD RXD

6
ADDRESS AND INTERFACE
Thought
Education:
Learn - Do -Teach
What is the connection ____________
The standard 40-pin package is used for most discussions. It
contains all the external connections that will be used on any
variation of the microcontroller. Some designs do not need all the
features. These will be smaller and will not include some pins.
Power __________________________
Before anything can work, power must be applied. The upper right
(pin 40) is connected to VCC, which is usually 5 volts. The lower left
(pin 20) is connected to the ground reference.
Clock __________________________
The clock or crystal is connected to pins 18 (XTAL2) and 19
(XTAL1). This provides a frequency basis for all the micro
operations and timing. Each cycle is divided into 12 phases which
trigger internal circuits to create an instruction.

* V
P1.0 1 T2 40 VCC
P1.1 2 T2 EX AD0 39 P0.0
P1.2 3 AD1 38 P0.1
P1.3 4 AD2 37 P0.2
P1.4 5 SS/ AD3 36 P0.3
P1.5 6 MOSI AD4 35 P0.4
P1.6 7 MISO AD5 34 P0.5
P1.7 8 SCK AD6 33 P0.6
Reset 9 AD7 32 P0.7
P3.0 10 RXD VPP 31 /EA
P3.1 11 TXD PROG/ 30 ALE
P3.2 12 INT0 29 /PSEN
P3.3 13 INT1 A15 28 P2.7
P3.4 14 T0 A14 27 P2.6
P3.5 15 T1 A13 26 P2.5
P3.6 16 /WR A12 25 P2.4
P3.7 17 /RD A11 24 P2.3
Xtal2 18 A10 23 P2.2
Xtal1 19 A9 22 P2.1
Ground 20 A8 21 P2.0
MCS 51 / 8051
When using a crystal, a buffering capacitor of 10 to 40 pf is
connected from each pin to ground. The crystal frequency ranges
from 3 to over 50 MHz. However, the choice of frequency is often
determined by outside connections. For example, serial
communications requires rather unusual frequencies such as 11.059
MHz.
If an oscillator is used, XTAL1 is the input to the oscillator
amplifier and the input to the internal clock. XTAL2 is the output
from the inverting oscillator amplifier.

Chapter 6 Address & Interface 45
Reset ___________________________
Reset is connected to pin 9. The machine runs when the line is low.
Reset occurs if the pin is pulled high for two machine cycles. In the
simplest realization, a switch performs this function. The switch
should be open for normal operations and momentarily closed for
reset.
The most elegant circuit occurs with power on. This circuit will
automatically reset the computer whenever power is applied. The
circuit consists of a 10 microfarad capacitor connected to Vcc and a
10 K Ohm resistor connected to ground.
When using in-system programming, the pin may be activated by a
software command. A 1500 Ohm resistor should be connected
between the cable and the reset pin.
Ports ___________________________
Ports are parallel locations within the processor. Each port includes
eight bits that are individually addressable. Three different notations
are used to represent a bit on a port, as shown in the diagram. These
include pin number, port number followed by the bit number, and
function such as address / data.
Limit the number of devices and resulting current connected to any
port. The ability to sink or source current is very restricted. A buffer
such as a latch or 7406 is highly desirable to protect the port.
Current limiting resistors should be used for LEDs connected
directly to the port.
The maximum output currents that the chip can handle are shown in
the table.

Condition IOL
Max per port pin 10 mA
Max for eight bit, port 0 26 mA
Max for eight bit, port 1,2,3 15 mA
Max total for all output pins 71 mA
To determine the number of devices that can be connected, it is
necessary to identify the current required by each load.
Total source (IOH) = input to load (IIH) * number loads (m)
Total sink (IOL) = input from load (IIL) * number loads (n)
Use lower power devices such as ALS or CMOS technology.
Port 0 __________________________
Port 0 consists of eight pins (pins 32-39). As a true bi-directional
I/O port, it can be used for normal input and output. As an output, it
can sink eight TTL loads.
When 1s are written to the pins, they become high-impedance
inputs. Therefore, it requires a pull-up or pull-down resistor to
operate. Because of other features, it is probably the least used port
in normal programming. However, when used as a memory
connection, the port has internal pull-up resistors.
Port 2 __________________________
Port 2 is another 8-pin I/O connection (pins 21-28). It is less
frequently used in normal programming.
It is quasi bi-directional, has internal pull-ups, and can sink/source
four TTL devices. When 1s are written to the pins, they are pulled
high by the internal pull-ups and the pin can be used as an input. As
inputs, pins that are externally being pulled low will source current
because of the internal pull-ups.

External memory ________________
If external data or program is required, then port 0 takes on a
multiplexing task. It has eight bits of the address impressed on the
port. Then this is followed by the eight bits of data associated with
that address.
A latch is used to trap the address. An eight bit flip/flop such as a
74573 is commonly used. It contains the 8 bus latches connected to
the Data (D) or input pins of the latch. The output or Q pins are
connected to the low order address pins of the memory device.
Two pins control the data capture and transfer. The chip or latch
enable (pin 11) is asserted high, the gates receive and pass
information. When the pin is asserted low, the flip/flops trap and
hold the information. The output enable not (pin 10) is hardwired
low to permanently activate the output.
When the 8 address bits are on the port, the Address Latch Enable
line (pin 30) is toggled. This pin is connected to the chip enable pin
of the external latch. Therefore, when the address is superimposed,
the latch will capture and hold the location.
External program ________________
If external program memory is employed with the microprocessor,
then the External Address (pin 31) must be pulled low. This will
transfer all program code from external memory.
If there is both internal and external program memory, the internal
memory will be accessed when the /EA line is high or not
connected. The next address after the top of the internal memory,
will go to the external, regardless of the status of /EA.

When a code byte is requested from the program memory, the
program storage enable not (/PSEN) is pulled low by the processor.
The code will enter on port 0 pins.
The /PSEN line is connected to the external programmable read
only memory (PROM) device output enable not (/OE) pin 22. The
PROM chip select not (/CS) is connected to ground, if it is the only
device. If there are multiple devices, the chip select must be
decoded perhaps through a programmable logic device (PLD).
Do not allow input address pins on any memory device to float. Pull
the pin low if it is not connected to the processor.
External data ___________________
External data memory is in the form of static ram (SRAM),
electrically erasable prom (EEPROM) or similar devices. If these
are in a DIP package the pins will correspond with those of program
memory.
However, the connections to the processor are very different. Data is
transferred from the processor to memory when the write not (/WR)
pin 16 is pulled low by the processor. The pin is connected to the
write enable not (/WE) pin 27 on the memory device.
Data is read from external memory when the read not (/RD) pin 17
is pulled asserted low by the processor. This is connected to the read
not (/RD) or output enable not (/OE) pin 22 on the memory device.
External 64K ____________________
The eight bits on port 0 can address a maximum of 256 bytes of
memory.
Address = 28
– 1 = 255

By using port 2 as the upper eight bits of an address, 64K of
external memory can be accessed.
Address = 216
– 1 = 65,535
Port 2 is not multiplexed. Therefore, an address latch is not
required. The lines of port 2 are wired directly to the corresponding
high bit address lines of the memory device.
Since the port is bit accessible, any bits not used for addressing are
available for use as general purpose I/O. Often the upper bit (A15)
is used to select memory or memory-mapped input/output.
Port 1 __________________________
Port 1 is the most used general purpose I/O (pins 1-8) on the
processor. The internal pull-ups operate like port 2. Most devices
are connected through these pins.
If the processor has a serial peripheral interface (SPI), it is
connected to the upper bits of port 1 (pins 5-8). These pins are
connected to external chips. If the SPI is used for in-system
programming, these pins are connected to an external cable.
Port 3 __________________________
Port 3 is truly a multifunction byte (pins 11-17). The internal pull-
ups operate like port 2. Pins not used for other functions can be used
for general purpose I/O. The pins provide input for timers and
interrupts as well as a bi-directional serial UART. Two pins are
shared with the read not and write not functions for external data
memory.
ž Ÿ

7
MINIMUM SYSTEM
Thought
Elegantly simple
Minimalist ______________________
The minimum system for a computer is the ability to do input,
decisions, and output based on the decision. This will provide an
outline for every other project that can be built.
Although the software code is very simple for this type project, the
implementation will be in a structure that can support very complex
problems. As systems grow, more interaction is needed. A later
software illustration identifies issues of compatibility, particularly
for variable descriptions.
The hardware required covers several issues. First, appropriate
power must be supplied to ensure operation in range. Second, the
fundamental computer is wired. This includes the processor, crystal,
and reset circuit. Third, a method of transferring the code to
memory is required. In this implementation, in-system programming
is used. Finally, connections to the external world are made in the
form of switches for inputs and lamps for outputs.
When building a circuit for digital processing, several practices will
provide results that are more consistent. Lay out the components
adjacent to other components that will be interconnected. Use short

Chapter 7 Minimum System 51
wires. Red should be used for power. Black is preferred for ground.
A variety of other colors for inter-connections will help
construction. Keep the board neat, since this will aid
troubleshooting and improve your attitude about the project.
Project _________________________
The project is to apply one switch and two LED’s to the computer.
One LED is on while the other is off. Change the state of the switch
and the LEDs will change state.
Minimum software _______________
On reset, program control transfers to address 0000h. The number
of digits in the address is based on the maximum amount of memory
that is directly accessible. There are sixteen bits for addressing. A
hexadecimal number is made up of four bits. Sixteen bits then
would be divided into four hexadecimal numbers.
Addresses 0000 – 0002h are the power on reset vector and are
reserved to send the program to another location. Addresses 0003 –
002Ah are reserved for interrupt processing in a standard processor.
Many enhanced processors reserve the next 8 bytes for expansion
interrupts. Additional explanation is given in the architecture section
covering memory organization.
Therefore, the lowest code address should be 0033h. However,
programmer information and ownership is often embedded in the
next code addresses. For that reason, the first address used for
program code is usually 0080h.
For normal procedures, the first command at address 0000h is a
jump to address 0080h. Several jumps are available. These include
absolute short, absolute long, and relative. Any of these can be used
for the first command in the computer.

Opcodes, mnemonics, comments ___
The most basic level of programming is using machine language
and op codes. This can be done for relatively small routines and for
practice in debugging. However, it is not time effective for any
program of significance. A later chapter introduces higher level
language techniques. The opcode list is provided in the reference
chapter on instructions set.
When coding, it is very desirable to use precise spacing for the
instructions. This makes the program easier to read and keeps all the
information about each instruction on one line of a standard size
paper. The comments will be truncated on the narrow paper of this
book.
;add OPCODES LABEL: MNEM REGISTER ;COM
0000 020080 ljmp INITIAL ;go
0080 D2B5 INITIAL: setb 0B5h ;on
This is the complete program necessary to turn on an LED
connected to port 3, bit number 5.
The address is a hexadecimal value from 0000 – FFFFh. Each
program location represents an eight bit byte of data.
The op codes are hexadecimal values that are decoded and
processed by the internal central processing unit. A series of micro
operations are executed to create the correct response. Mnemonics
are used for convenience in recognizing the instruction. The
processor only understands the op codes.
In general, an instruction gets data from a source register, operates
on the data, and places the results at a destination register. By
convention on this particular machine, the destination is noted first.
A comma separator is used before the source value.

Chapter 7 Minimum System 53
Classes of instructions ____________
Only three classes of instruction are available to a computer. These
are data transfer, mathematics manipulation, and program control.
Any instruction is a variation of these classes. In only a few cases,
the instruction may be a combination.
Data transfer instructions change the location of data. The major
instruction is a move (mov) from a source to a destination. In
keeping with the leaning toward a RISC machine, this is the only
transfer instruction. Other computers use many instructions to
accomplish this same task, including load, store, and transfer.
Mathematics manipulation instructions include arithmetic
operations such as add, subtract, multiply and divide. More limited
operations are increment and decrement. Logical operations are
AND, OR, and EXCLUSIVE-OR as well as clear, set, and rotate.
Again, the instruction set is reduced by not having a shift command.
Program control instructions change the value of the program
counter under certain conditions. Some of these instructions are
jumps, calls, and returns.
Schematic _______________________
A project should be illustrated with a schematic. Numerous
programs can be used to create the drawing. The preferred systems
will provide a circuit that can be transferred to a printed circuit
board design if desired.
The illustration should show the circuit so that another builder can
construct it. This particular example also shows the in-system
programming connections, although that is not normally shown. It is
provided so the novice can start from scratch and build it.

Vcc
ž Ÿ
VCC
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset
Xtal2
Xtal1
Ground
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset

8
MACHINE CYCLE TIME
Thought
Engineers do not run the world,
But make the world run.
First computer circuit _____________
A microprocessor is the engine of a microcomputer system.
Hardware refers to the physical, electronic components. Software
includes the application programs that provide user functions.
Firmware is the program that is resident with the computer and is
not easily changed.
The microprocessor firmware is a basic system program that is
usually stored in a ROM. This system program is intended to do
basic input / output (I/O). The system program also includes a
simple power-on reset routine to run the system when it is started.
On a personal computer system, this program is usually called ROM
BIOS (Basic Input/Output System).
A microcontroller has the program as an integral part of the system.
This makes the program look like both firmware and software. The
program memory may be internal or external to the computer chip.
If the program memory is external, an address latch is needed. Since
the data and address lines are time-multiplexed, a latch is used to

hold the address while the data/program is fetched from the ROM.
If the program memory is internal, then the system is very simple.
In the latter case, a connection for programming the in-system
memory is needed.
Cycle time ______________________
In its simplest implementation, any computer program can be used
as a timer. First, determine the speed of each instruction, then add
all the instruction cycles.
Each instruction in a computer requires a definite amount of time to
execute. A machine cycle is the internal count for executing an
instruction. This is based on the oscillator frequency.
The fixed rate of an instruction time delay is calculated as follows.
Divide the number of states for the processor by the crystal
frequency. There are twelve states.
Time = (sec/11.059E6 cycle)(12 period/mach cy)
= 1.08509 E-6 sec / machine cycle
To obtain a time delay for an instruction, multiply the time per
machine cycle by the number of cycles for the instruction. Repeat
for all the instructions in the routine, to obtain the total time delay.
A delay with two nested loops is shown. Registers R2 & R3 are
used for counting the number of times through the loops. A smaller
number on the inside loop gives more precise counting.
mov R3,#02H ;Outer loop counter
ZDEL2: mov R2,#01H ;Nested loop counter
ZDEL1: nop ;Delay
djnz R2,ZDEL1 ;Nested loop, 256 x
djnz R3,ZDEL2 ;Outer loop, 256 x

Chapter 8 Machine Cycle Time 57
The maximum count is obtained by preloading the registers with
zero (0). The first pass through the djnz instruction will decrement
before it tests for the jump. Decrementing from zero gives 0FFh.
To obtain less time, remove the outer loop that includes register R3.
To obtain greater time, add another loop outside the shown code.
Machine cycles __________________
The machine cycles for each instruction are gathered from the
instruction table toward the back of the book.
mov = 1 cycle
nop = 1 cycle
djnz = 2 cycle
ret = 2 cycle
The time for each loop is calculated by counting the total number of
cycles in the loop. The inner loop includes just three instructions.
mov = 1
nop = 1* loop count
djnz = 2* loop count
The total cycles depend on going through the inner loop as many
times as the outer loop counter.
Inner loop = (3 * inner count) + 1
Outer loop = [(inner loop + 2) * outer count] + 1
Plus = 4 for call & return
Total cycles
Determine the total cycles for the example given above. The inner
count is one (1), while the outer loop is two (2).
Inner loop = (3 * 1) + 1 = 4
Outer loop = [(4 + 2) * 2] + 1 = 17
Plus = 4 for call & return = 4

Total cycles= 25
The elapsed time is the total cycles multiplied by the time for each
cycle.
Time delay = (1.08509 E-6 sec / machine cycle) * 21
= 27.28 microseconds
Long precise wait ________________
Some routines for display require a longer wait string than typically
used. These can be developed by multiple calls to a precise delay.
One-tenth second is a good base counter.
;-------------------------------------------------
WAIT:
;-------------------------------------------------
; Create a 0.1 second wait.
; Count = 0.1/1e-6 = 100000
; Loop = 192 = C0H
mov R3,#0C0H ;Outer loop counter
ZDEL1: mov R2,#00H ;Nested loop counter
ZDEL2: nop ;Delay
ret ;Return to call
In from out _____________________
The four ports are four bytes of the special function registers. These
bytes are used as the input and output connections to the computer.
Each of the ports is bit accessible. The operation of the ports was
explained in an earlier chapter. This discussion is how to program
the ports.
The output is quite simple. Place the value in the register that
represents the port. This triggers a latch, which holds the value on

Chapter 8 Machine Cycle Time 59
the external pins. A simple mov to the location will implement the
transfer of data to the external pins. No other action is required.
An input from a pin requires a two-step process. First, the latch
must be pulled high by sending a one (1) to the location. This is
exactly like an output. Then the value on the pin is transferred to
port register memory. In other words, if the pin is pulled low
externally, then the register bit will be a zero (0).
Ports 1, 2, and 3 have an internal pull-up triggered by the FET. So,
no external resistor is required. port 0 has a high impedance FET.
Therefore, an external resistor must be used to pull the pins high or
low.
;ILLUSTRATE OUT/IN
mov 90h,#0FFh ;set port1 for input
mov A,90h ;read port status
mov 0B0h,#01 ;output port3,bit 1
Switch a bit _____________________
Port pins can be manipulated as bits as well as bytes. This permits a
pin to be used as a switch. A single bit is set for the input. Then the
bit is tested for high or low.
;ILLUSTRATE OUT/IN
setb 0B0h ;set port3 for input
jb 0B0h,EXIT ;bit=1,then go end
;else do this instr.

Circuit: led metronome __________
Vcc
Port
ž Ÿ

9
PROJECT 1 – OUTPUT & TIME DELAY
Thought
You see what you seek.
Yakov Smirnoff
Project 1: Metronome ____________
Purpose: To build a basic computer system.
To use a microcontroller with program memory.
To write a simple program.
To use the output port.
Preamble:
The microcontroller has I/O ports, which are built-in. In order to
transfer a logic value to output ports, treat the ports as memory
locations. Only the mov command is needed.
Although the processor uses machine code to execute, the
information is converted to Intel Hex format for sending to the
memory burner. This is simply placing the addressing on the front
of each string with a checksum inserted at the end.
To make a light flash, require the computer to place a signal on the
output. Then the computer has an elapsed time delay by counting
instructions. Next the processor sends a signal to stop the output.

This makes one flash. The cycle is repeated for multiple flashes. A
cpl command is easily used to toggle a bit.
Plan:
For the project, build a basic microcomputer system with memory
for the program. If the chip has on-board memory, the program may
be stored there. Then, the microcontroller will be used to implement
a digital metronome.
Preparation:
Supply a crystal to the microcontroller. The chip already has an on
board oscillator circuit. Observe the circuit for a complete board
shown in a later chapter. The circuit contains more items than are
required. For this project, neglect unnecessary peripherals.
Simply wire the microcontroller, crystal, and power, if using a
machine with on-board memory. Select a crystal frequency that is
compatible with serial communications. A frequency of 11.059
MHz is a very popular selection. The finished wired circuit is a
complete microcomputer system. Add the power-on-reset circuit to
make start-up easier.
Procedure:
Check the system that has just been built. Implement a metronome
on the computer. Since ports 0 and 2 are used for memory
addressing, ports 1 and 3 are free for direct input/output. Put an
LED on one of the free I/O port bits. The LED must blink at a
specific rate. The cycling output implementation is a metronome.
Writing a 1 and 0 sequentially to a bit of the output port is the actual
task.
Test a switch to start or stop the display.

Chapter 9 Project 1 - Output & Time Delay 63
Presentation:
Implement the program using only machine code without any
assembler or high-level compiler. Developing this ability can aid in
debugging future programs.
Demonstrate your circuit. Write an additional short note about your
delay time. The delay duration should be equal to 10 divided by the
number of characters in your first name. This note should follow the
program listings.
ž Ÿ
Program sample example __________
The following code contains a delay subroutine. The project is
implemented by modifying the values that go into R2 and R3, the
loop counters, to fit the delay requirements.
Write the program with mnemonics and comments. Then translate
to the address and machine code. The op codes and the number of
bytes are obtained from the instruction tables in the reference
section. This information will then be placed in a file to be
programmed into the code memory.
The file information can be placed in an eprom burner program
editor. The burner program will convert the output to the
appropriate format for loading in the microprocessor program
memory.
ADDR OP CODE MNEMONICS & COMMENTS
;-----------------------------
;Program: MODDelay.asm
;Update: 29 January 2003
;Initial: 17 October 1991
;

;By: Dr. Marcus O. Durham, PhD, PE
; Tulsa, OK, USA
; mod@superb.org
; www.ThewayCorp.com
; Copyright (c)1991, 2002.
; All rights reserved
;
;Purpose:
; A routine to demonstrate software
; delay. One LED will come on.
; After a delay, another will.
;
;Processor: 8031 family
;PROM: 8k (2000H) onboard
;Crystal: 11.059 MHz
;#############################
; PROGRAM
;#############################
0000 org 00H
0000 020080 START: ljmp INITIAL
0033 org 0033H
0033 4D617263 db 'Marcus O. Durham, PhD, PE'
0037 7573204F
003B 2E204475
003F 7268616D
0043 2C205068
0047 442C2050
004B 45
;-----------------------------
0080 org 0080H
INITIAL:
MAIN:
;-----------------------------
0080 D2B4 setb 0B4h ;turn on
0082 12008A lcall DELAY ;wait
0085 D2B5 setb 0B5h ;turn on
0087 020080 MAN9: ljmp MAIN ;Repeat
;-----------------------------
DELAY:
;-----------------------------

Chapter 9 Project 1 - Output & Time Delay 65
008A 7B00 mov R3,#00H ;Outer
008C 7A00 ZDEL2: mov R2,#00H ;Nested
008E 00 ZDEL1: nop ;Delay
008F DAFD djnz R2,ZDEL1 ;Nested
0091 DBF9 djnz R3,ZDEL2 ;Outer
0093 22 ret ;go back
end
ž Ÿ

10
SOFTWARE DEVELOPMENT
Thought
God is in the details.
The here and now ________________
Earlier discussions introduced software in terms of machine
language and op codes. Although that is great for understanding
how the machine works, it is atrocious for effective use of time with
large projects.
There are three fundamental language levels of programming.
Machine language is the most basic. It involves use of zeroes and
ones to form op codes. The op codes are used directly by the
computer.
The intermediate level is assembly language, which uses symbols
and mnemonics to represent the basic op codes. This includes
Assembler. Although it has many capabilities and is somewhat
transportable, “C” is very similar to assembly because of its
terseness and difficulty in comprehending the meaning of the code.
High level languages are the most sophisticated and involve
conceptual statements similar to a spoken language. This includes
Basic and very advanced languages. Because of its advanced
features, most authors include “C” as a high level language. Both

Chapter 10 Software Development 67
Basic and “C” were developed in conjunction with operating
systems. Basic is to DOS as “C” is to Unix.
The software languages commonly used for programming the
microcontroller family are assembly, Basic, or C+. By far, the most
frequent is assembly language. Regardless of the source code, it
must be converted to an Intel Hex format for programming the
memory.
There is great flexibility in how program code is structured. There is
at least as one more way than the number of programmers.
Therefore, there is no ‘right’ way. Nevertheless, there are formats
that provide certain advantages.
The development of effective software using an assembler is
discussed in the next sections.
1. First comes the formatting of instructions.
2. Next is the description of directives.
3. A step-by-step process which goes from text to hex files
follows.
4. Then a program shows the source, assembler listing, and
hex code.
5. This is followed by practical suggestions for commenting.
6. Finally, structured programming is argued.
Substantially more information and illustrations are provided in the
extreme program harmonization example.
Instructions _____________________
The source program can be created using any word processor or
editor. Save the program unformatted in a flat text file. The
extension for the source depends on the particular software package.
*.txt and *.asm are the most common.
An earlier illustration is shown as an introduction to using an
assembler. This is a working program. It is the result of a listing
after the assembler completes working on the program.

ADDR OPCODES LABEL: MNEM REGISTER ;COM
0000 020080 ljmp INITIAL ;go
0080 D2B5 INITIAL: setb 0B5h ;on
The source code consists of the information on each line that starts
with the label. The assembler adds the address and op codes to each
line.
Label is an identifier for the address of that line of code. It is
restricted to 8 characters. Some assemblers are case sensitive;
therefore, it is a good idea to use the same capitalization in all
references. The label is followed by a colon (:). However, the
reference to the label does not have a colon. The assembler simply
translates the label to an address.
Comments must begin with a semicolon (;). Everything on a line
after the semicolon is ignored by the assembler. Neither does the
computer recognize or use the comments. Comments are for the
programmer. Comments should explain the purpose for the
instruction.
Mnemonics are abbreviations for an instruction. These are for
convenience rather than memorizing numbers for each instruction.
These are converted to op codes by the assembler. The op codes are
hexadecimal values that are decoded and processed by the internal
central processing unit.
The registers are listed as the destination with a comma separator
before the source. These registers are the information that is
manipulated by the instruction.
Names or variables starting with a letter are translated as a value.
Numbers must begin with 0 – 9. A hexadecimal number that begins
with a letter must be preceded by a 0. An example is 0Fh for the
number 15.

Numbers are identified by their base. Decimal numbers are followed
by a “d” or nothing. Hex numbers are followed by a “h”. Binary
numbers are followed by “b”.
When coding, it is very desirable to use precise spacing for the
instructions. The table lists the various components of a line of
code. It shows the preferred column for starting that code. This will
make the program easier to read and will assure all the information
fits on one page after the assembler processes the information.
Name Column Length
Label: 1 8 + colon
Mnemonic 11 5
Register 17 14
Comment 31 19 max
Heading comment 1 50
Assembler directives ______________
In addition to program code, the assembler must be informed about
certain operations. The directives are not comprehended by or
translated to the microcomputer.
Each assembler has it own format for the directives. There is no
standard among the various vendors. The ones illustrated are based
on the original Intel ASM51 assembler.
The first directive is org. This organizes the program to a specific
place in memory. The instruction simply gives the address for the
next instruction.
The second is equ. The equate directive provides a numeric
equivalence for a variable. This can simply be a number or a place
in memory.
The third is db and its related types. This defines byte values in code
memory. It often involves table values and ASCII messages. Include
a line with a copyright message. If it is not copyrighted, use

‘author’, then include your name. That provides a permanent record
of who did the programming.
Every program must terminate with the directive end.
Several assemblers are available for the 8051. The major differences
are in the syntax of the directives. Three commonly used formats are
compared.
TASM.EXE is a table lookup. It is accessed by the instruction:
tasm -51 -p -l Program2.asm
.org 0050h ;Next inst @ 0050H
Value .equ 12h ;predefined value
Table .byte 34h ;define code byte
.text "123" ;define ASCII
.end ;last thing
A51.EXE is the student version of PseudoCodes. It is accessed by
the instruction:
A51 -s Program2.asm
.org 0050h ;Next inst @ 0050H
.equ label,12h
Table: .db 34h ;define code byte
.db "123" ;define ASCII
ASM51.EXE is the Intel assembler with library. It is the original
assembler. It is very powerful. Unfortunately it is DOS based and
has not been maintained. It is accessed by the instruction:
ASM51 Program2.asm
org 0050h ;Next inst @ 0050H
Value equ 12h ;predefined value
Table: db 34h, '123' ;define code storage
end ;last thing
The major directives are listed in the table. Each assembler has a
different set, but these are the most common. There is an overlap

between assemblers. In addition there is an overlap in usage. As
illustrated above, the combination of equ and db will perform all the
other functions.
Dir Purpose
org Organize the program memory location
end End of the program for the assembler
equ Equate a variable to a value
db Define a byte of program memory to a value
dw Define two bytes of program memory to a value
ds Reserve a byte of program memory for a value
bit Equate a variable to a bit address
byte Define a byte of program memory to a value
data Equate a variable string to a number
text Equate a variable string to an ASCII value
Step by step _____________________
The program can be copied to or created in a text file. Then
assemble the file. This example has been successfully assembled
using Intel ASM51. The directives may need changing if using an
alternate assembler. If the assembler does not output a *.hex file,
convert the *.obj to an Intel Hex format.
Download the program in Intel Hex format to the program (code)
memory. Download can be either using a burner or the in-system
programming pins. When the processor is reset, the program will
run.
Program with comments __________
;-------------------------------------------------
;Program: MODIo.asm
;Update: 07 February 2003
;Initial: 01 December 1990
;
; Tulsa, OK, USA
; mod@superb.org

;Copyright (c)1990, 2003. All rights reserved
;
;Purpose:
; The LED on P3.4 will come on.
; The LED on P3.5 will go off.
; The LEDs will change state if P3.3 is at ground
; The program is very simple. Nevertheless,
; structured programming is used for illustration
;
;Assembler: Intel ASM51.exe
; ASM51 MODIo.asm
;#################################################
;
; ASSIGNMENTS
;
;#################################################
;PORT USE
P33 equ 0B3H ;Port3.3, switch input
P34 equ 0B4H ;Port3.4, LED
P35 equ 0B5H ;Port3.5, LED flashing
;#################################################
;
; PROGRAM
;
;#################################################
START:
;-------------------------------------------------
; When processor is reset, program control comes
; here. Jump to the first executable address
; after all interrupts reserved locations, etc.
org 00H
ljmp INITIAL

org 33H
db 'Marcus O. Durham, PhD, PE'
;-------------------------------------------------
org 80H
INITIAL:
;-------------------------------------------------
; Set the initial conditions. Output to LEDs and
; make P33 an input.
setb P34 ;turn on LED P34
clr P35 ;turn off LED P35
setb P33 ;make bit an input
;-------------------------------------------------
MAIN:
;-------------------------------------------------
; The main process loop will test the limit.
; If switch open, do nothing.
; If closed, change states.
; Since there is no delay & no debounce,
; the output is sensitive to when the switch is
; released.
jb P33,MAIN9 ;skip if off
cpl P34
cpl P35
MAIN9: sjmp MAIN ;play it again, Sam
;*************************************************
end
Listing _________________________
After the assembler runs, it produces several files. Perhaps one of
the most useful for debugging coding errors is the listing (*.LST).
This includes each line of the original code. The line is shifted to the
right. It is preceded by two very important pieces of information.

The first information is the memory location of the first byte of the
instruction. The next information is the machine code translation of
the instruction.
After the program, the listing shows the value of every variable in
the code.
An edited version of the listing is shown to save space and fit within
the page size. Comments have been truncated in addition to removal
of header information.
MCS-51 MACRO ASSEMBLER MODIO 01/28/:3 PAGE 2
LOC OBJ LINE SOURCE
0000 52 org 00H
0000 020080 53 ljmp INITIAL
54
55 ;-----------------------
56 ;AUTHOR
57 ;-----------------------
0033 58 org 33H
0033 4D617263 59 db 'Marcus O. Durham, PhD, PE'
0037 7573204F
003B 2E204475
003F 7268616D
0043 2C205068
0047 442C2050
004B 45
60
61 ;-------------------------------
0080 62 org 80H
63 INITIAL:
64 ;-------------------------------
65
0080 D2B4 69 setb P34
0082 C2B5 70 clr P35
0084 D2B3 71 setb P33
72
73 ;-------------------------------
74 MAIN:

75 ;-------------------------------
83
0086 20B304 84 jb P33,MAIN9
0089 B2B4 85 cpl P34
008B B2B5 86 cpl P35
87
008D 80F7 88 MAIN9: sjmp MAIN
89
90 ;*******************************
91 end
SYMBOL TABLE LISTING
------ ----- -------
N A M E T Y P E V A L U E A T T R I B U T E S
INITIAL. C ADDR 0080H A
MAIN . . C ADDR 0086H A
MAIN9. . C ADDR 008DH A
P33. . . NUMB 00B3H A
P34. . . NUMB 00B4H A
P35. . . NUMB 00B5H A
START. . C ADDR 0000H A
REGISTER BANK(S) USED: 0
ASSEMBLY COMPLETE, NO ERRORS FOUND
Intel hex _______________________
Data needs to be transferred between the assembled program and the
code memory. The machine code is converted to Intel hex format
for loading into program memory.
The file is shown in a table simply for explanation. It has been
divided into columns to indicate different parts of each data line.
Each line begins with a colon (:). This is followed by the number of
bytes on the line, the address of the first byte, a directive, then the
machine instructions in that block. The line concludes with the
checksum for the line.

Column 1 - Byte, number of byte/addresses on the line
Column 2 - 2 Bytes, hexadecimal address in memory
Column 3 - Byte, 00 if data, 01 if end-of-file
Column 4 - 16 Bytes, op code data for each address
Column 5 - Byte, checksum
: # Addr Dir Op Codes Data —
: 03 0000 00 020080 7B
: 10 0033 00 4D6172637573204F2E2044757268616D 34
: 09 0043 00 2C205068442C205045 8B
: 0F 0080 00 D2B4C2B5D2B320B404B2B5B2B680F7 --
: 00 0000 01 FF
The checksum is calculated from the data. Its purpose is to check
data transfers for loss of bits. The checksum is the two's
complement of all the other bytes on the line. It is calculated in the
following method.
1. Add all the bytes on the line.
2. Subtract the results from FF.
3. Add one to the result.
4. This is the checksum.
Commentary ____________________
Good software development is a skill, which is often neglected, and
which is best learned by example. Suggestions for good
documentation habits include comments, placement of code
information, and structure.
The first item that should be used is a header which gives the name
of the program and who wrote the document. Remember, there are
laws against plagiarism. The use of a version number of a program
(e.g. V1.3, V2.2) is strongly recommended. Update or increase the
number of the version every time you edit a file. This helps to keep
track of each change made to the files.

Following the header, there should be a list of major updates to the
program. Especially, note updates after release to the public.
The next major item that should follow is a short description of the
program. This should be long enough to give the user a general idea
of what the program does. The user should be able to use the
software without having to figure it out the hard way. Avoid going
into endless detail. Depending on the program, the description may
be anywhere from 2-3 paragraphs to about one page. A page and a
half may be required for BIG programs.
Any quirks of the program and any hardware dependencies about
which the user should be aware need to be included here. Examples
are software that supports a UART needing an 11.059 MHz crystal.
Assume that the user does not have your hardware available, but
that (s)he is intelligent enough to put together a system to run your
software. Therefore, state such things as where to hook-up port-pins
and what parts of the software to modify if a different crystal is
used. Similarly, the documentation should be good enough that the
user can modify the code or add to the program. (S)he should not
have to spend weeks trying to figure out whether the
additions/changes will affect the old code.
The top placement ________________
Place all the equate statements in one place such as at the top of the
file. Typical equates include ASCII codes, RAM locations, and
external locations in the memory map. This item can save you and
the user substantial time when digging for a particular item.
Remember to update these addresses when you change locations.
Often the equate list can save substantial amounts of time when
determining which items may adversely affect the program. This is
especially important if you are writing only one module of a large
program.

Document any test pin or port bit assignments. Note the uses for all
the registers dedicated to only one or two uses. If a register is used
as a general-purpose register, it can be so stated.
The subs _______________________
Documenting subroutines is another practice that will help writing
compatible procedures. Each routine should have a header, which
sets it apart from the rest of the program. The header should include
the following items as well as the subroutine name. Make a brief
description of what the routine does. List parameters that are passed
to the routine via which registers or memory locations. Note what
parameters are passed out of the routine and which subroutines are
called. In the optimum case, mention which registers, flags and
other storage variables are destroyed by the routine.
The destroyed variables may imply variables destroyed either by the
routine itself, or by the routine and all subroutines it calls. Noting all
these items requires a lot more work. However, it saves an
incredible amount of time when debugging, since it becomes
unnecessary to chase down each subroutine.
Your comments, please ___________
Furthermore, document the code itself. This should be done as you
write the program! It is not necessary to comment every line, but
one comment every 3 lines is not enough either. Seldom is a string
of code really that obvious.
Comments should describe the underlying process, not the actual
command that is being executed. For example, the comment for an
instruction like mov A, 45H should say why the value is being
moved. Discuss things such as the value should be in a certain
range.
If you do not write the comments until after you have finished
writing the program, often you will have forgotten what a particular

line of code does. Alternatively, you will run out of time and skip
the comments altogether. In that case, the client will not even bother
to look at the program.
The bottom placement ____________
Messages, tables and similar items, are often accessed by many
different subroutines. These should be placed in one area such as the
end of the file. If a message is only used by one subroutine, it is
acceptable to put the message at the end of the subroutine. Again,
consistency and find-ability are the goal.
Structure _______________________
Finally, it is worthwhile to stress the utility of structured
programming and extreme programming. Look at the main program
section of the code in the following pages. Notice the main program
consists almost entirely of CALL statements. Therefore, the
program is written in modular chunks.
First, this type of program is much easier to read and understand
than pages and pages of straight-line code with no subroutines.
Second, the structure forces you to think about what you are trying
to accomplish with the program. Hence, the program can be broken
down into distinct sub-tasks, which can be more easily tackled.
Often these subroutines can be used in several different parts of the
program, thus saving ROM space. If subroutines are well written,
they can also be used in other programs calling for similar
functions.
Third, it is easier to make changes in a modular program since the
effects of any modifications are more obvious.
Structured programming is logical. Its use reduces the likelihood of
becoming lost in reams of straight-line code. Long code segments
encourage forgetting what the program is trying to do. Structured

programming will not run quite as fast as straight-line code.
Nevertheless, it has the advantages of readability and it encourages
intelligent programming. The programs written for most projects
will not be time constrained.
ž Ÿ

11
DESIGN PRACTICES
Thought
One small step for man
One giant leap for mankind.
Commander Neal Armstrong
Top down _______________________
Engineering tasks can be divided into three functional descriptions.
1. Design involves developing a procedure or technique to do a
task that has not been done.
2. Application involves taking existing pieces or components and
combining them to perform a task.
3. Analysis involves observing an existing system and determining
how it works.
Since design must come first, it has top priority. However, it is
useless if it cannot be applied.
To be effective, the design must first consist of an overview or big
picture perspective. The broad project is then broken into individual
subsystems, which represent major components of the project. The
subsystems are loosely coupled. The subsystems are then segmented

into tasks that may be developed virtually independent of the other
tasks. This process is called top down design.
The procedure is called structured programming when applied to
computer code development. Niklaus Wirth developed Pascal as a
high level language that forces structure. He gave a definition:
“Structured programming is the formulation of programs as
hierarchal, nested structures of statements and objects of
computation.”
Projects should use a top down approach to the interfacing and
computer hardware as well as the program.
The project is the overall chore. It represents the complete program.
The subsystems are tied together with a main or control program. It
should primarily consist of calls to subroutines. An occasional
decision may be required to determine which subroutine to call.
Each subroutine should be independent of other subroutines. Data
arguments that must be transferred should be left in common
registers that can be accessed by any routine. The registers should
be defined prior to the main program. Then the main should call the
subroutines.
Extreme programming (XP) _______
Extreme programming (XP) is a practice used by large teams of
programmers. It is obviously based on top down design and
structured practices. It is primarily a method of communicating with
the design team. There are 12 core practices.
1. Customers define application features with user “stories”.
2. XP teams put small code released into production early.
3. XP teams use a common system of names and descriptions.

Chapter 11 Design Practices 83
4. Teams emphasize simply written, object-oriented code that
meets requirements.
5. Designers write automated unit tests upfront and run them
throughout the project.
6. XP teams frequently revise the overall code design, a process
called refactoring.
7. Programmers work side by side in pairs, continually seeing and
discussing each other’s code.
8. All programmers have collective ownership of the code and the
ability to change it.
9. XP teams integrate code and release it to repository every few
hours and never hold on to it longer than a day.
10. Programmers work at a sustainable pace, with no extended
overtime.
11. A customer representative remains onsite throughout the
development project.
12. Programmers must follow a common coding standard so all the
code in the system looks as if a single individual wrote it.
In keeping with extreme programming practices, a chapter on
software compatibility is included. This lists variable names,
locations, and formatting structure.
Steps for success _________________
A consistent process is used in the development of any project. A
seven-step pattern has been developed that works for software,
hardware, and all other technical procedures. This is a universal
model for engineering projects.

1. Entry
The point where the procedure starts arises from a reset or from
a decision.
2. Initialize
A sequence of events or conditions are involved in the planning
and set-up.
3. Input
The loop first must gather data from other sources.
4. Process
The core of the loop is manipulation of the data until a final
value is obtained.
5. Output
The defined results are displayed or become input to another
procedure.
6. Limit
A test or decision is performed to determine if the loop is
completed. The test may occur at three different places.
a. The test may be implemented before the input if there is a
predefined loop range. This is a If-Then-Else or For-Next
structure.
b. A decision is made after the input based on the input. This
is a data comparison structure.
c. A decision is made after the process is complete. This is a
Do-While some condition exists.
7. Exit
The point to leave the project causes transfer to another
procedure or task.

Chapter 11 Design Practices 85
Process diagram _________________
ž Ÿ
start
exit
processinput outputlimit?initialize

12
SWITCH, LOGIC, AND SUBS
Thought
People are where they are,
because of choices they make.
MOD
Switch hitter ____________________
A switch converts mechanical movement to electrical signals. The
normal circuit for wiring switches was offered in the chapter on
fundamental circuits.
One of the problems with mechanical switches is contact bounce.
When the switch is depressed, the contact will vibrate for some
time. Because of its speed, the microprocessor will count these as
multiple switch closures.
Hardware devices can catch the switch as a single bounce. This
could be a flip/flop or a one shot. However, software is much less
expensive in terms of real estate. The traditional software scheme
places a time delay on the switch input. Although this is effective, it
delays the processing and sampling rate of the system. A better
choice is to use a software solution that detects the first time the
switch is stable.

Debounce _______________________
Using a logical exclusive-or provides a very clever switch
debounce. The routine is entered multiple times until the last state of
the input is equal to the present state. This provides response the
first time the signal is stable.
;-------------------------------------------------
DEBOUNCE:
;-------------------------------------------------
;CHECK CHANGE BY xrl
mov B,A ;hold the input
xrl A,CharP ;exclusive or
jnz DEBN1 ;<>0, so a change
;KEEP DEBOUNCED
mov CharD,B ;0=no change,debounc
DEBN1: mov CharP,B ;not debounce
ret
The routine compares bytes. The value enters in the A register. The
previous value is kept in CharP. By using an EXCLUSIVE-OR, bits
that have changed are easily found. If bits have not changed, keep
the debounced value in CharD.
Bit manipulation _________________
The ability to manipulate bit addresses is a powerful feature that
separates this microprocessor from other computers. The internal
memory is byte addressable for both the data RAM and the special
function registers (SFR). Certain locations may be addressed as bits.
Any SFR whose address is divisible by 8 can be bit addressable.
This includes any hexadecimal address that ends with a 0 or 8. The
low order bit has the same address as the byte. For example, port 1
is located at address 90h. The least significant bit would also be bit
90h, while the most significant bit would be 97h.

Chapter 12 Switch, Logic & Subs 89
In addition the internal data locations 20 – 2Fh (32 – 47d) are bit
addressable. The bit address is sequential. Address 20h, bit 0 has a
bit address of 00h. Bit 7 has an address of 07h. Address 2Fh, bit 7
has a bit address of 7Fh.
A number of instructions are available to manipulate the bits. The
two instructions that change individual bits are setb to make a one
and clr to make a zero.
;bit MANIPULATION
CLEAR: clr 93h ;clear port1,bit3
SETCARRY: setb C ;1oCarry
Masking logic ___________________
Not all the values that exist in a byte are always necessary for
evaluation. If only some of the bits are used, the others can be
removed from the byte. The process of selectively modifying some
of the bits is called masking.
A mask is a register that is similar to a template. It has a defined
value that prevents other values from being observed. For example,
a template or mask that is placed over a paper permits only a certain
item to be painted.
Masking requires the use of logical operators. The three operators
used with a template are AND, OR, and EXCLUSIVE-OR. The
instructions are logical duals.
The logical AND (anl) of a bit with a 0 clears the bit, while using a
1 leaves the bit unchanged.
The logical OR (orl) of a bit with 1 sets the bit, while using a 0
leaves the bit unchanged.
The logical EXCLUSIVE-OR (xrl) of a bit with 1 inverts
(complements) the bit, while using a 0 leaves the bit unchanged.

The instruction takes the value of the destination register, performs
the logical operation with the source register, then places the results
back in the destination.
;ILLUSTRATE LOGIC
ZERO: anl A,#11110111b ;A bit3 becomes 0
ONE: orl A,#00001000b ;A bit3 becomes 1
INVERT: xrl A,#00001000b ;A bit3 complemented
Logical orl and anl should be used to selectively set bits on output
ports. This will change the status of bits without impacting the other
functions. If there are multiple functions on the port, avoid mov
instructions.
Rotate and exchange _____________
Data bits can be moved one position to an adjacent bit location. The
data must be in the accumulator or A register.
If the register is rotated to the right, the least significant bit (LSB) is
rotated to the most significant bit (MSB). Conversely, if the register
is rotated to the left, the MSB is rotated to the LSB. No data is lost
in the transaction.
The carry bit (C) can be considered to be bit 8. It is treated as if it
were the next bit after the MSB. A rotate right with carry will cause
the LSB to go to the C and the C will move to the MSB. A rotate
left with carry will move the MSB to C and the C will go to the
LSB.
The carry can be individually set or cleared from other operations.
Therefore, it can be used to load values serially into a register.
The machine does not have a shift instruction since it has hardware
multiply and divide. A shift is similar to a rotate, except a bit falls
off the end with each shift. If a shift is required the rotate with carry
can be used.

Several different commands are available for moving and changing
the values of bits. In order to exchange the upper four bits (nibble)
with the lower four bits (nibble) in a byte, the swap command is
available.
A very powerful instruction permits the exchange of a byte between
the Accumulator and a register or internal data location. The value
in A goes to the register and the value in the register goes to A. This
very quick operation replaces three mov instructions.
;ROTATE & EXCHANGE
RIGHT: RRA A ;bit 0o7, 7o6, 1o0
LEFT: RLA A ;bit 0o1, 1o2, 7o0
RC: rrc A ;bit 0oC, Co7, 1o0
LC: rlc A ;bit Co0, 0o1, 7oC
UPLOW: swap A ;0l4,1l5,2l6,3l7
EXCHANGE: xch A,Hold ;AlHold
Conditional branch _______________
Often it is necessary to make a decision in a program. Three
components are necessary for a decision: a value to evaluate, a
reference value, and the result of the comparison.
The process is called a conditional branch. All these functions are
included in a single instruction, compare and jump if not equal
(cjne). The conditional branch is much like the IF-THEN-ELSE
procedure.
The value to be evaluated is stored in a register. The register is
either the accumulator, register zero or register one. The reference
value for comparison may be an immediate number or the value
may be in a RAM location. If the two values are equal, the next
instruction is executed. If the two values do not match, the program
counter is changed to the specified code address. Neither the
evaluation nor the reference value is changed by the instruction.

The conditional branch is a single very efficient instruction. Most
computers use a compare or test instruction, which sets a zero flag
in the flag register. Then the next instruction does a branch based on
the value of the flag.
However, since all decisions are essentially a comparison for zero
results, the compare and branch are very effectively combined in
one instruction.
Another conditional branch is the DO-WHILE type. The decrement
and jump if not zero (djnz) instruction comes in this class.
A register is preset to the limit. It is decremented each time through
the loop. When the register is cleared, the loop is exited.
;CONDITIONAL BRANCH
BEFORE: ;an entry
COMPARE: cjne A,#02,BEFORE ;A<>2, then BEFORE
DECREASE: djnz D,BEFORE ;D=D-1, D<>0,then BE
;else, next command
Subroutines _____________________
Some routines of a program are used more than one time in a
sequence. Since several instructions are involved in each routine, it
is beneficial to reuse the same code if possible. In addition, it is
easier to read a program that is segmented, rather than straight line
code.
The instruction lcall invokes a subroutine any place in code
memory. It uses a sixteen bit absolute address to reach the entire
64K available memory.
The main program calls a subroutine located at a label such as SUB.
The program counter is pointing to the next instruction after the
lcall. The value of the program counter is pushed onto the stack.
Two bytes are stored. The program counter is then loaded with the
address of the subroutine SUB.

A return, ret, instruction is used to indicate the end of a subroutine.
Once the subroutine is completed, the ret pops the stack and then
places that address into the program counter. This is the next
instruction after the lcall.
;SUBROUTINE
MAIN: lcall SUB ;SUB=label
ONE:
SUB: mov A,#00 ;do something
ret ;go back to ONE
Only four bytes and four clock cycles are required for the two
instructions that create a subroutine. The lcall uses 3 bytes and 2
clock cycles, while ret uses 1 byte and 2 clock cycles.
A structured program is realized by making the main program a
sequence of call statements. If adequate room is provided, changes
require only the addition or moving of a CALL.
Each subroutine should be stand-alone. All subroutines should be as
consistent as possible with arguments passed and returned in
common registers. The preferred registers are A or R0 – R7. This
permits easy linking of multiple routines.
Stack ___________________________
One of the unique features of the machine is a stack. A stack allows
information to be pushed into a temporary location. The information
is removed in the reverse sequence from the order it was filled. This
is called a Last In, First Out (LIFO) sequence.
New data is added to memory on top of earlier data. The removal of
a data item makes the item below it appear to be the new top of the
stack.

A stack can be located anywhere in the internal data RAM. On reset,
the stack pointer is preset to data address 07h. This should be
changed since the stack will overwrite Bank 1.
Setting the address at which the stack is to begin initializes the
stack. The stack pointer (SP) contains the address for the top of the
stack. The stack grows up toward a higher address. The next byte is
stored at the address above the value placed in the stack pointer
register. Once the SP is initialized, it is seldom explicitly changed
again.
The subroutine instructions, lcall and ret, places and retrieves the
two program counter bytes on the stack.
There are two instructions that move a byte between direct internal
memory and the stack. push places the direct value into the next
memory address above the location in the stack pointer. pop moves
the value at the stack pointer location to the direct internal memory.
Although other instructions may be between them, always pair stack
operations.
;STACK
INITIAL: mov SP,#5Fh ;move stack to 5Fh
;other commands
SUB: push Direct ;direct to stack
;other commands
pop Direct ;stack to direct
The stack can use the entire internal data memory. The large stack
makes the machine capable of pseudo-reverse polish operations. In
many ways the machine appears to be a stack processor.

Circuit: led and switch ___________
Vcc
Port
ž Ÿ

13
PROJECT 2 – INPUT & DECISIONS
Thought
Sell to people that are buying.
Jim Stovall
Project 2: T-bird taillights ________
Purpose: To learn the basic I/O system.
To implement a T-Bird tail light system.
To make use of subroutine calls.
To use an assembler.
Preamble:
A microprocessor has several advantages over a hardwired
sequential logic circuit. This will become apparent in this project.
Other implementations of this project have been with discrete logic
and with programmable logic devices.
In order to communicate with the real world, a microprocessor must
have an input-output system. This is called an I/O port. The
computer treats the I/O port as a register and as internal data
memory. Therefore, data can be written to the port without
initialization. A one must be written to each bit before it is read.
The use of subroutine calls is preferable in large programs. Use of
subroutines will make debugging the program much easier. Even in

Chapter 13 Project 2 – Input & Decisions 97
a small size program, this method is still a winner when similar
chunks of codes are repeated several times in the program.
Plan I: Basic input-output
Build a microcomputer system. Then, the computer will input bits
from switches and output the bits to an LED.
Preparation I:
Connect the basic operating microcomputer, crystal and a power
supply. Observe how switches must be connected between power,
ground, and the computer port. Review the logic instructions.
Procedure I:
Input four bits from switches on port 1 or 3. Output the bits to
LED’s on the same port. When the switch is operated, the
appropriate LED should be energized. This should be a continuous
scan process.
Presentation I:
Demonstrate your circuit. Write a note about the differences in set
up for input and output. Discuss your procedure for moving (mov)
the values from one bit location to another.
Plan II: T-bird Taillight System
In this project, implement a display reminiscent of an old
Thunderbird tail light system.

Preparation II:
Wire 6 LEDs on one port and 2 switches on the other bits of the
port. The input port switches will represent brake, idle, turn left, and
turn right signals. A two-bit input will be sufficient. The LEDs will
represent the taillights. When the switch combination is selected, the
LEDs should have the following display:
Brake turns on all LEDs.
Idle turns off all LEDs.
Turn left turns on the right LED of the left bank of 3. Next turn
on the adjacent, then the left LED simultaneously.
Turn right. This will be the opposite of turn left. It will turn on
the left LED of the right bank of 3. Then it will turn on the
adjacent, then the next.
The sequence repeats until another kind of input is applied.
Procedure II:
There are no restrictions in choosing the port for your I/O system.
Remember to use proper software documentation. Then, the port
used will be obvious to the customer.
Use a delay routine to implement turn right and turn left.
Presentation II:
Explain your observations of how a microprocessor has advantages
over a hardwired logic circuit. Please write your explanation
following the program listing.
ž Ÿ

The exemplar program is similar to the project specifications.
However, it contains elements that should be modified to complete
the project as required.
;Program: ModTbird.asm
;Initial: 7 February 2003
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c)2003. All rights reserved
;Purpose:
; Check switches for status and display
; results in pattern of old T-Bird car
; P3.0-P3.1 used as direction input
; 00-idle, 01-right, 10-left, 11-brake
; P3.2-P3.4 left lights
; P3.5-P3.7 right lights
; P1.2 start switch
;
;Assembler: Intel ASM51
;
;#################################################
; ASSIGNMENTS
;#################################################
Pio equ 0B0h ;port for input output
Switch equ 092h ;switch input
;#################################################
; PROGRAM
;#################################################
org 00H

START: ljmp INITIAL
org 0033h
;-------------------------------------------------
org 0080H ;get past interrupt
INITIAL:
;-------------------------------------------------
setb Switch ;make switch input
mov Pio,#03h ;make bits 0&1 input
;-------------------------------------------------
MAIN:
;-------------------------------------------------
jb Switch,MAIN ;use as start,1=open
lcall INPUT ;get switch input
cjne A,#03h,TWO ;<>3 brake
lcall BRAKE ;else, =3, go brake
sjmp ZERO ;continue scan
TWO: cjne A,#02h,ONE ;<>2 left
lcall LEFT ;else, =2, go left
sjmp ZERO ;continue scan
ONE: cjne A,#01h,ZERO ;<>1 right
lcall RIGHT ;else, =1, go right
ZERO: lcall IDLE ;else, =0, idle
MAN9: ljmp MAIN ;Repeat
;-------------------------------------------------
INPUT:
;-------------------------------------------------
; Check bits for input.
mov A,Pio ;input port
anl A,#03h ;mask bits 2-7
ret ;exit
;-------------------------------------------------
OUT:
;-------------------------------------------------

; Show the results
mov Pio,A ;display results
ret ;exit
;-------------------------------------------------
DELAY:
;-------------------------------------------------
; Create a short time delay
mov R3,#00 ;Outer loop counter
ZDEL2: mov R2,#00 ;Nested loop counter
ZDEL1: nop ;Delay
ret ;exit
;-------------------------------------------------
RIGHT:
;-------------------------------------------------
; Flash right lights.
mov A,#10000000b ;first lamp
lcall OUT ;show it
lcall DELAY ;wait
mov A,#01000000b ;second lamp
lcall OUT ;show it
lcall DELAY ;wait
mov A,#00100000b ;third lamp
lcall OUT ;show it
lcall DELAY ;wait
ret ;exit
;-------------------------------------------------
LEFT:
;-------------------------------------------------
; Flash left lights.
ret ;exit
;-------------------------------------------------

BRAKE:
;-------------------------------------------------
; Turn all lights on.
; Ensure port input is not cleared.
ret ;exit
;-------------------------------------------------
IDLE:
;-------------------------------------------------
; Turn all lights off.
; Ensure port input is not cleared.
lcall DELAY ;
ret ;exit
;*************************************************
end ;Program end
ž Ÿ

14
REGISTER, TIMERS, AND INTERRUPTS
Thought
Everyone is given the same 24-hours.
What you do with it makes the difference.
MOD
Timer registers __________________
Special function registers (SFR) include tools to configure timers
and interrupts, as well as numerous other chores. The details of each
register are shown in a reference chapter. This discussion is how to
implement the register to perform a program task.
One of the more useful features is the built-in timer counters. These
can be used to count events or to count machine cycles. The
machine cycles then translate into time.
Timer 1 is used for serial communications control. So, for our
counting purposes, we will restrict operations to Timer 0. Many
chips have a third timer, but it will not be discussed since we want
to remain as generic as possible.
The timer control (TCON) register determines when the timer
operates. The timer is started by setting the run control bit (TR0),
and is stopped by clearing the bit under software control.

The timer mode (TMOD) register determines whether it is operating
as a timer or counter.
The counting register is sixteen bits. The low byte is TL0, while the
high byte is TH0.
During operation, the counting register increments on a 1 to 0
transition of the port3 pin (0B4h). This transition is checked every
machine cycle. During operation as a timer, the register is simply
incremented by the hardware. In essence, the timer is counting
machine cycles.
A preset value can be placed in the counting register. The counter
will increment from this value. A carry bit from the most significant
counter stage signals when the register has finished a complete
count.
The overflow causes a flag (TF0) to be set for the counter. By
monitoring the overflow flag, the program is aware of a completed
count cycle. The flag can be monitored by normal polling software.
Alternately, it can be used as an interrupt.
Timer __________________________
A timer can be used to indicate when a particular amount of time
has elapsed. This is accomplished by loading a preset value into the
timer register (TH0,TL0). The timer increments at each machine
cycle. Therefore, the preset is a negative value. The value is counted
backward (negative) from zero. As the count progresses, an
overflow occurs at count zero.
One machine cycle consists of six states. Each state lasts for two
oscillator periods. Hence one machine cycle takes 12 oscillator
periods.
The preset value is equal to the machine cycles subtracted from
zero.

Chapter 14 Timers & Interrupt 105
Preset = - time (seconds) * oscillator frequency (period/sec)
12 (oscillator periods / machine cycle)
Consider a time of 1/30 of a second from a crystal of 11.059 MHz.
Preset = - 1/30 (seconds) * 11,059,000 (period/sec) = -30719
12 (oscillator periods / machine cycle)
The decimal number must be converted to hexadecimal. This
permits the upper byte to be stuffed in TH0 and the lower byte to be
shoved into TL0.
Since the timer counts in machine cycles, the elapsed time in
seconds can be calculated from the value in TH0,TL0.
Time = TH0,TL0 * 12 (oscillator periods / machine cycle)
oscillator frequency (period/sec)
Interrupts _______________________
Some events are important enough that other tasks should be
suspended. These events are interrupts.
Processing interrupts is a common technique used for real-time
computer applications. The mechanism is usually found in data
acquisition systems, since timing is a critical problem.
The basic processor has five different sources of interrupts. These
are external 0, external 1, timer 0, timer 1, and serial.
Interrupts can arise on two pins of port3. These are INT0 (0B2h)
and INT1 (0B3h). The interrupts are initialized in the interrupt
enable (IE) register. The precedence can be set in the interrupt
priority (IP) register. Detailed explanations are provided in the
section on special function registers.

When an interrupt occurs, the next program counter address is
pushed to the stack. The program counter is set to the interrupt
handling address in low code memory. There are separate addresses
for INT0 (0003h) and INT1 (0013h). Each of these has 8 bytes
reserved to process the interrupt. If the interrupt service routine
requires more room, an ljmp can transfer to another location.
The interrupt processing routine is terminated by a reti instruction.
The instruction pops the stack for the location to reset the program
counter. The interrupt procedure is very similar to a subroutine that
has its own assigned memory location.
Counter & interrupt examples _____
The first task is to setup timer 0 to count. TH0,TL0 have results of
the count.
;-------------------------------------------------
COUNT0:
;-------------------------------------------------
setb 0B4h ;set T0 for input
mov TMOD,#00000101b ;T0,Mode1,counter
setb TR0 ;start the count
ret
The first illustration is to use counter 0 to register 5 events that
occur on the timer 0 pin of port 3 (0B4h).
;-------------------------------------------------
COUNTS:
;-------------------------------------------------
;COUNT 5 EVENTS
lcall COUNT0 ;set up timer
mov A,#5 ;preload the limit
HERE: cjne A,TL0,HERE ;T0 count low byte

Chapter 14 Timers & Interrupt 107
An interrupt can be used when count gets to a preset value stored in
TH0,TL0. For example, to interrupt after counting 5 events, preload
with -5 = 0FFFBh.
When count overflows to zero, the interrupt will transfer program
control to address 000Bh for processing. This requires the interrupt
to be enabled. If an interrupt is not desired, the overflow bit, TF0,
can be tested with software polling.
;-------------------------------------------------
COUNTINT:
;-------------------------------------------------
;INITIAL
setb 0B4h ;count 1 as input
mov TMOD,#00000101b ;counter=0, mode1
mov TH0,#0FFh ;minus leading ones
mov TL0,#0FBh ;-5
setb TR0 ;start counting
;OPTIONAL INTERRUPT
orl IE,#10000010B ;enable all & T0
;HERE: jnb TF0,HERE ;optional polling
ret ;get outta here
Timer with interrupt examples _____
A variation of the counter is to create a timer. The timer simply
counts machine cycles. The first routine will initialize timer 0 for
use as a clock. The mode will be sixteen bits and a software gate.
The routine preloads values and starts the timer.
If the optional interrupt is used, a transfer to address 000Bh is
generated when the time has elapsed. Alternately, if an interrupt is
not desired, the overflow bit, TF0, can be polled.
;-------------------------------------------------
TIMEINIT:
;-------------------------------------------------
; The preset time is calculated. The crystal osc.
; frequency is 11.059 MHz.

;
; Preset= (Sec * Osc freq)/12 osc/machine cy
;
; for 1 msec, THxTLx= -921.583= -400h = 0FC66h
; for 1/20 sec, THxTLx= -45954.167 = 4C7Eh
; for 1/30 sec, ThxTlx= -30719.444 = 8801h
; ;INITIAL
mov TMOD,#00000001b ;timer=0, mode=1
TIMERUN: mov TH0,#88h ;4C7Eh = 1/20 sec
mov TL0,#01H ;lo count
;OPTIONAL INTERRUPT
orl IE,#10000010b ;enable all & T0
ret ;MrButler to Atlanta
Circuit: interrupts _______________
Vcc
Port
ž Ÿ

15
PROJECT 3 – CLOCK & INTERRUPT
Thought
Objectivism
is rational self-interest.
Ayn Rand
Project 3: Time to count __________
Purpose: To use the internal registers.
To demonstrate external event triggers.
To calculate very precise times.
To implement interrupt processing.
Preamble:
The basic processor has 5 different sources of interrupts. These are
external 0, external 1, timer 0, timer 1, and serial. In this project,
only the external interrupts using INT0 or INT1 are required.
To receive an interrupt signal from the pins, make sure that the bits
corresponding to each pin are enabled. Set the bit to one, since the
ports are bit addressable.
When interrupts are used, an interrupt service routine is placed in
the program. The computer program jumps to an address that
corresponds to the interrupt.

Several registers are associated with the interrupt locations. Of
particular interest are the two counter/timer locations. These can be
used to count external events. Alternately they can count clock
cycles, which in essence, makes them a timer. The registers include
sixteen-bit data and two control registers.
Plan I: The External Interrupt System
Create a project to show that both interrupt pins work properly with
either a level or an edge-triggered signal.
Procedure I:
For example, illustrate the project by changing the LED displays
every time a trigger is sensed. This means the program is displaying
continuously but is interrupted by an external signal. Show at least
one interrupt source. The sources can be a combination of level-
triggered or edge-triggered with the INT0 / INT1 pins.
A more elegant design will count the number of times that the INTx
pin is triggered. Then, after a preset number of events, an LED will
illuminate.
Presentation I:
Demonstrate that the interrupt process is working properly.
Plan II: Digital Clock
Implement a real-time digital clock based on the timer register.
Timer 1 is used for serial communications; therefore, only Timer 0
should be used for these projects. Use an LED to show seconds and
other LEDs to show minutes.

Chapter 15 Clock & Interrupt 111
Procedure II:
Toggle the same LED every second. After each minute, illuminate
another LED. It is advisable to write the clock routine inside the
interrupt program and the display routine as the main program. This
will provide a more precise clock.
Presentation II:
Demonstrate the operation of the clock. Ensure all LEDs are off
before time start.
ž Ÿ
;-------------------------------------------------
;Program: MODTime.asm
;
; Tulsa, OK, USA
; mod@superb.org
;
;Purpose:
; A set of routines are provided to perform the
; basic functions of a clock.
;

;#################################################
;
; ASSIGNMENTS
;
;#################################################
TimPer equ 3CH ;time periods counter
TimFlag bit 20H ;time out
;#################################################
;
; PROGRAM
;
;#################################################
org 00H
START: ljmp INITIAL
;-------------------------------------------------
;INTERRUPT-Timer 0
;-------------------------------------------------
; The procedure provides direction when timer
; completes count.
org 0BH
ljmp TIMECAL ;interrupt processor
;-------------------------------------------------
org 0033H
;-------------------------------------------------
org 0080H
INITIAL:
;-------------------------------------------------
mov SP,#5Fh ;start stack @ 5f+1
lcall TIMEINIT ;start timer0
;-------------------------------------------------
MAIN:
;-------------------------------------------------
;PROCESS
jnb TimFlag,Man9 ;time cycle at end?

Chapter 15 Clock & Interrupt 113
clr TimFlag ;<> 1/2 sec
cpl P3.5 ;is 1/2, so flash
;-------------------------------------------------
TIMEINIT:
;-------------------------------------------------
;INTERRUPT
orl IE,#10000010b ;enable all & T0
; ;INITIAL
mov TMOD,#00000001b ;timer=0, mode=1
lcall TIMERUN ;preset & start T0
;-------------------------------------------------
TIMERUN:
;-------------------------------------------------
; The routine preloads values and starts the
; timer.
;
; The preset time is calculated. The crystal osc.
; frequency is 11.059 MHz.
;
; Preset= (Sec * Osc freq)/12 osc/machine cy
;
; for 1 msec, THxTLx= -921.583= -400h = 0FC66h
; for 1/20 sec, THxTLx= -45954.167 = 4C7Eh
; for 1/30 sec, ThxTlx= -30719.444 = 8801h
;INITIAL
clr TR0 ;stop counting
mov TH0,#88h ;8801h = 1/30 sec
mov TL0,#01H ;lo count
;-------------------------------------------------
TIMECAL:
;-------------------------------------------------
; This routine is called from the interrupt

; processor. First restart the timer.
;INITIAL
lcall TIMERUN ;preset & start T0
inc TimPer ;another time period
mov A,TimPer ;period count
cjne A,#15,TIMC9 ;periods in 1/2 sec
mov TimPer,#0 ;reset period
setb TimFlag ;end of cycle
TIMC9: reti ;leave Scarlet OHara
;*************************************************
end
ž Ÿ

16
BOARD CONSTRUCTION
Thought
Moral economics:
Maximizing personal benefit,
without compromising integrity.
Professor Durham
One step. Check! _________________
The procedures used by a board manufacturing or stuffing company
are considerably different from a single construction venture by
individuals that are building only one or a few projects.
These guidelines are oriented to the prototype construction of a
single board. The preferred procedure for populating a board is to do
one step at a time. Then check the work as you go. Troubleshooting
will be greatly reduced by this procedure.
Show and tell ____________________
Insert components carefully to ensure proper polarity. Confirm that
the notch on chips is aligned in the same direction as the notch on
the board. Other polarity sensitive devices include capacitors,
resistor packs, diodes, and transistors. If you are not sure, get help
or check data sheets. Incorrect polarity will assuredly destroy
semiconductors.

The board has many components that have to be soldered in a
relatively close space. Be cognizant of the difference in a pin hole
and a via. A via is simply a hole that connects circuit traces on
different layers of the board. Verify that your soldering does not
flow or connect to a via or another pin. The circuits will short and it
is very difficult to find later.
Improper soldering will keep the board from working and can be
nearly impossible to find. If you are uncomfortable with soldering,
ask for instructions or help. There are a few training videos that are
excellent tools. It is better to start correctly than to redo much of
your work.
Basics __________________________
Build the board the way you are learning the projects. Begin with
the basics, get it working, and then move to things that are more
complicated.
The first step is to get power to the board. If you do this first, you
can test the board without concern for damaging expensive
electronics. Install the voltage regulator, filtering capacitor, and the
power plug connector.
Then apply power. Verify that there is VCC (5V) and ground where
it is required on the board.
Next, install the minimum components necessary for the computer
to function. Mount the computer socket, the clock or crystal with
capacitors, and the reset resistor and capacitor. Carefully insert the
processor in the socket.
Then install the in-system programming circuit. This includes the
connection for the external cable. Refer to the section on parts and
pins to obtain cable pin-out information. The board can support both
serial peripheral programming and serial RS232 communications

Chapter 16 Board Construction 117
through similar jacks. Therefore, it is imperative that proper
configurations be used.
At this point, it is prudent to test the status of your work. Download
a simple program, like the metronome. Start the computer. Since
there are no I/O devices, use a meter or scope to verify the operation
of the appropriate pins.
Proceed only if everything is operating at this point. Otherwise,
troubleshooting and correcting become a major issue.
Socket to me _____________________
The board is designed for many functions and applications. It is not
necessary to stuff all the components. Only install those items that
are needed.
It is highly desirable to use sockets for some of the components. The
microprocessor should be in a socket. The programmable logic
device must be in a socket and memory devices must be in sockets.
It may be desirable to place in sockets other chips with many pins.
This is particularly true if your soldering skills are questionable.
The memory expansion has a special socket that will allow a 0.3” or
0.6” wide memory chip. To install the socket, cut the connecting
bars that hook the sides of the socket together. Install the 14-pin SIP
header in the center row of the socket.
The programming header (RJ45 or strip) must be installed on top of
the board. All other headers should be placed down. This allows the
board to be plugged onto a proto board.
Several jumpers are available to expand the flexibility of the board.
x External memory EA’: select 5V or Ground
x Seven-segment onboard: select ground or 7406
x Address Latch Enable: select ALE or PLD
x Memory Output Enable: select ground or PLD

x Port 0 resistor pack: select pull-up or pull-down or nothing
x LCD contrast: select auto circuit or jumper to Ground
Initially, connect EA’ to 5V, since external memory is not used. The
others can be connected if the components are added.
What’s left ______________________
The remaining items include latches for expansion, analog to digital
converter, liquid crystal display, infrared, seven-segment displays,
LED’s, pushbuttons, and resistor packs.
There is space for a coupling capacitor for each of the major
integrated circuits. If the speed is kept low, these generally are not
needed. Another alternative is for film capacitors that can be placed
under the chips.
For most prototype and project development, only a few of the
remaining items are required. It is strongly suggested that you test
each circuit system as you add components.
By careful placement and soldering, the board has been proven to
work trouble-free.
ž Ÿ

17
PROJECT 4 - DEVELOPMENT BOARD
Thought
A liberal selectively
acts the way he wants without regard for
accountability, absolutes, or consequences.
MOD
Project 4: Build from scratch ______
Purpose: To MOVe from proto-type to system integration.
To have a one-board development system.
Preamble:
Several chapters refer to board construction. One has a parts list and
pin-out diagrams. The board schematic and specifications are
included. Another has details for construction practices.
The numerous wires used on the proto-board are prone to loose
connections and improper operation. Once the fundamentals of
design and trouble-shooting are understood, a circuit board can be
used.
The board has every tool of a simple development system. The
software can be downloaded from a PC to the program memory. By
use of the expansion ports, any control circuit can be developed.

For the most basic computer, only the processor, crystal oscillator,
memory, and power supply are required. If the program memory is
internal, then the system is very simple.
A serial interface is added for troubleshooting. A PEEL is added for
combinational logic to decode addresses. Optional memory is
available. A liquid crystal display can be connected as well as the
circuits for automatic contrast control. An analog to digital
converter can be used.
Plan:
A custom-designed board has been developed for this purpose. The
board is commercially produced. The schematic is included in the
chapter on parts.
The board must be ordered in advance. Delivery time is typically
three weeks. The cost is much less if the boards are ordered in
quantity, since there is a substantial set-up fee for each order.
Preparation:
All projects to this point should be completed. The board requires
the components previously used on the proto-board. These will be
soldered on to the circuit board.
It is strongly recommended that sockets be used for the processor,
PEEL, and memory. Sockets may be used for other chips, but it runs
up the cost.
Procedure:
Once the board has been wired, software must be installed.
Continue to program the controller as before.

Chapter 17 Project 4 – Development Board 121
If using a processor with on-system memory, then connect the cable
to the appropriate connector.
If using external memory, program the EPROM. A download
program has been written for the development system. If the
download program is used, the hex code can be downloaded to the
SRAM. The program will transfer control to the microprocessor.
The download program has built-in test features. After the system
appears to be working, implement any previous project on the
completed board. This program is discussed in a later chapter.
Presentation:
Show the system works by downloading a program such as the
metronome. Then execute the program.
ž Ÿ

18
EXTERNAL MEMORY
Thought
A conservative
is constrained by
MOD
Storage control lines _____________
Harvard architecture separates program and data memory.
Moreover, memory can be internal or external. Both external data
and program memory are allocated as 64K byte pages. This size is
an inherent characteristic of sixteen bit addressing. The program
counter (PC) is a sixteen-bit register. Therefore, it can directly
access the entire page.
Since only one page of memory is available, any addresses that are
used for internal memory are not available in the external memory
chip.
Control lines are used to segregate data and program memory and to
isolate internal and external operation. On Reset, program control
transfers to program memory address 0000h.
If the external address not (/EA) line is low, the address points to
external memory. If /EA is high, the program counter points to

Chapter 18 External Memory 123
internal memory. Regardless of the setting, the next line executed
after the highest internal address is at the external device.
When the computer is accessing external program memory, the
program storage enable not (/PSEN) line is asserted low. This line is
connected to the output enable not (/OE) line on the memory chip.
The line is not invoked during internal memory fetches.
When the computer is accessing data memory, one of two lines is
asserted. Both are multipurpose lines on port 3. Read not (/RD) is
asserted low to get data from external memory. It is connected to
output enable not (/OE) on the memory chip. Write not (/WR) is
asserted low to send data to external memory. It is connected to the
write enable not (/WE) on the memory chip.
Address fetching _________________
The addressing is handled the same for both data and program
memory. The sixteen-bit address for external memory is placed on
two ports. Port 0 has the low byte and port 2 has the high address
byte. The eight bit data is transferred via port 0.
Port 0 multiplexes both address and memory. Therefore, an address
latch is required to catch the low memory. The address latch enable
(ALE) line is asserted high when addressing is on port 0. An
external eight-bit latch is required to trap the address information.
The data for the latch is setup when the latch enable (LE) pin is
driven high. When the pin is pulled low, the data is trapped by the
latch.
The ALE line is connected to the latch enable pin of the latch. When
the ALE line is asserted, the address is setup. The address is trapped
on the trailing (falling) edge of ALE. The latch holds the low
address. Port 2 holds its high address until the information byte
from memory is received.

Port 0 must be set for input to receive data. The CPU automatically
sets port 0 for input during a memory fetch. However, it does not
retain special function settings. Therefore, that information is
destroyed during a program fetch.
Since there is not input on port 2 during a fetch, it does not have to
be configured. Therefore, port 2 special function register data is
preserved during an external program fetch. The data reappears on
the port during cycles, which are not a part of external program
fetch.
The program counter (PC) contains the sixteen-bit address for
instructions that are fetched from program memory.
The data pointer register is also sixteen bits. The program must fill
the register. It will point to program memory when the movc
instruction is used. It will point to data memory when the movx
instruction is used. This is called indirect addressing
External data memory can be accessed with eight-bit or sixteen-bit
addressing. The data pointer low (DPL) contents are placed on port
0. If the address is sixteen bits, the data pointer high (DPH) contents
are placed on port 2. If the address is eight bits, the special function
register information is retained on port 2.
One other method can be used to address sixteen bits of data
memory without modifying the data pointer register. Port 2 is
preloaded with the high address byte, as if it were a special function
register. Then use eight-bit addressing with the registers R0 or R1.
Timing sequence _________________
Although this information is seldom necessary to construct projects,
it is of interest to some practitioners. A machine cycle is divided
into 6 segments (S1-S6). Each segment will provoke the internal
hardware of the CPU to perform a particular function.

Different types of instructions obviously have a different timing
sequence. The basic operation is to read a byte at S1 and read the
next byte at S4. What happens with each byte depends on the
number of bytes and the number of cycles required for the
instruction.
The one-byte, one-cycle instruction mov A, R1 operates with this
sequence. Read instruction at S1. Read next byte at S4 and discard.
Reread next instruction at next S1.
A two-byte instruction requiring one cycle is mov A,#. It reads byte
1 at S1, then reads byte 2 at S4. The next instruction is read at the
following S1.
There are a few one-byte, two-cycle instructions, such as ret and inc.
The sequence is read instruction at S1, read and discard next byte at
S4, S1, S4.
At S5 of the previous cycle, the address is valid when ALE drops
low. A byte is received after PSEN is asserted low during S6. PSEN
returns high during S1. Then the program counter is incremented to
prepare for the next instruction.
During an external fetch, /PSEN is asserted two times per cycle. An
exception is movx. It is not asserted in the second machine cycle.
The external timing diagrams and discussion are shown in a
reference chapter near the end of the book.
Virtual memory __________________
In a machine with Harvard architecture, data and program memory
are separated. The data is generally stored in random access
memory (RAM), while the program is stored in read only memory
(ROM). However, permanent or fixed data may be stored in the
program memory. Moreover, program code may be stored and
executed from a RAM chip.

These practices require hybrid procedures to create virtual memory.
Virtual memory is simply memory that appears in application to be
different from what it is designated.
Program store enable not (/PSEN) is actuated by the computer to
control the output enable not (/OE) on the ROM during a program
fetch and a movc instruction. Read not (/RD) is actuated by the
computer to control the output enable not (/OE) on the RAM during
movx.
If the program memory ROM is to be virtual data memory, either
the /PSEN or /RD control lines must be able to control the output
enable not (/OE). Since the signals are active low, an AND gate
must be used. If either input to the gate is low, the output will be
low. The AND is equivalent to an Invert-OR-Not gate.
The program store and the read signals are high-frequency.
Therefore, fast gates must be used. Type ALS or faster should be
employed.
Alternately a programmable logic device such as a 22CV10, can be
used rather than glue logic.
Wiring ROM or RAM ____________
The custom circuit board is discussed in detail in a reference
chapter. The board is implemented with an external memory socket.
The socket can be used for either ROM or RAM. Therefore, the
socket is a model of virtual memory. There are only a few
differences in connections. These involve writing information to
memory and reading the stored values.
This case will specifically discuss 32K x 8 memory, often referred
to as 256K.
ROM is programmed by VPP on pin 1. Address A14 is connected on
pin 27. ROM is read via output enable not (/OE) on pin 22. The
ROM read is activated by program store enable not (/PSEN).

RAM is written on write enable not (/WE) on pin 27. Address A14
is connected on pin 1. RAM is read via output enable not (/OE) on
pin 22. RAM read is activated by read not (/RD) and write not
(/WR).
A map clearly illustrates how the connections to the chip socket are
interrelated. A programmable logic device (22CV10) can be used as
the logic to switch the appropriate lines to the socket.
Direction Device Pin 1 Pin 27 Pin 22
to ROM VPP A14 /OE
from micro A14 /PSEN
to RAM A14 /WE /OE
from micro A14 /WR /RD
A Boolean equation is required for the line switching on each pin.
Mp1 = ucA14 & !ucA15 & (!WRn # !RDn)
Mp27 = (!PSENn & ucA14) # (WRn & !ucA15)
OEn = PSENn # !(!RDn & ucA15)
Address A15 is included in the equations to segment memory. A15
is used as a control line for input / output device expansion. The
expansion is mapped into high memory.
There is also a pin required for chip control. Functionally, since
there is only one memory device, the chip select can be left active
by connection to ground.
ž Ÿ

19
BIOS
Thought
An anarchist
is unconstrained by
MOD
Definition ______________________
This chapter and the associated project will only be used if your
design includes both an external EPROM and SRAM and the
program is loaded via the serial port. Otherwise, the topics can be
skipped with no loss of ability to build a workable system.
What is bios? Bios is an acronym for basic input output system. The
program begins when the computer is initialized. The procedure is
typically stored in read-only memory and cannot be changed.
Because of the permanent nature of the software, it is often called
firmware.
The firmware provides the interface to the keyboard and displays
until an operating system or application program takes over more
elegant implementation of the functions. It also provides for transfer
of control to the operating system.
The bios is a group of subroutines and procedures that are common
across the spectrum of projects. These are written so they can be

Chapter 19 Bios 129
called from applications programs. In some cases, they become the
model for other specialty routines.
The bios for the microprocessor system would include the firmware
implementation of the serial, seven-segment or liquid crystal
display, and keypad. That is what is commonly used as a backbone
for developing applications software.
The bios that will be discussed in this chapter is slightly different.
The firmware is loaded in the read-only memory that will be
accessed on initialization. However, the function of that code is to
use the serial port for loading an application program into static
ram. Program control is then transferred to run from the static ram.
The BIOS developed in this chapter is specifically for programming
an external EPROM, then using the serial port to load a program
into external SRAM. If you are using internal memory, this topic
will not apply to your projects. Nevertheless, you may find some of
the information beneficial.
Bios main _______________________
The firmware is the initial code if serial transfer is used to move a
program to external memory. It sets up RS232 loading of the
program from a personal computer (PC) to static ram. Then it
directs switching the program execution from ROM to RAM.
The code is for use with the most fundamental derivatives of the
Intel 8031. Many variations of the microprocessor have this or
similar capabilities as an integral part.
The program has several key functions:
1. Test the SRAM.
2. Download a new program from the serial port.
3. Save the program to SRAM
4. Send a control signal to switch from EPROM to SRAM.
5. Execute the next instruction from SRAM.
6. Display error messages.

A light emitting diode connected to port 3.5 (P3.5) is used for
messages. This very fundamental display can be implemented
without extensive code and additional hardware.
1. on-off = controller is operating.
2. on-off-on = serial download is complete.
3. long on-long off, repeated 32 times = serial fault.
4. fast on-fast off, repeated 128 times = SRAM fault.
5. longer on-fast off, repeated 10 times= not switch eprom
;-------------------------------------------------
BIOSMAIN:
;-------------------------------------------------
;INITIALIZE
lcall UART ;Config & start UART
;PROCESS
lcall RAMTEST ;Test RAM & interfac
mov DPTR,#SerGreet ;get start addr
lcall BIOSSER ;Send message on ser
lcall DOWNLOAD ;Serial DOWNLOAD Ram
lcall P35DONE ;Flash P35=done dnld
;TERMINATE
ljmp MEMSWIT ;Switch prog storage
;to RAM from EPROM
Static memory test _______________
This subroutine tests the microprocessor to static ram interface. That
includes the device, wiring, and control logic for the virtual memory
operation. Each RAM address is tested with two complete
write/read loops. Each loop involves writing an alternate bit pattern
to each address, then reading each address to confirm the value. For
the second loop, the starting bit pattern is the complement of the
starting bit pattern of the first loop.

Chapter 19 Bios 131
If any one of the values read does not equal the value written, then
the test fails. A failed test can result from a bad interface or a bad
RAM chip. The message LED on P3.5 cycles off/on 128 times.
The message LED at P3.5 turns on at the start of the test and turns
off at end of a successful test. The memory subroutine returns to the
main calling routine upon test completion.
;-------------------------------------------------
RAMTEST:
;-------------------------------------------------
;TURN ON LED & INIT
setb P3.5 ;Set LED on/uC Busy
mov TmpA,#02 ;two wr/rd passes
mov TmpC,#0AAH ;First wr=10101010
;LOOP
RAMT1: mov TmpB,TmpC ;Start write pass
mov DPTR,#0000H ;First RAM address
;WRITE
RAMT2: mov A,TmpB ;Value to write
movx @DPTR,A ;Write to RAM
cpl TmpB ;Flip bits
inc DPTR ;Increment address
mov A,DPL ;} Continue write
cjne A,#00H,RAMT2 ;} loop until each
mov A,DPH ;} RAM address has
cjne A,#80H,RAMT2 ;} been written to.
;READ & TEST
mov TmpB,TmpC ;Start read pass
mov DPTR,#0000H ;First RAM address
RAMT3: movx A,@DPTR ;Read value @ addrs
cjne A,TmpB,RAMT7 ;Comp to write value
cpl TmpB ;Flip bits
inc DPTR ;Increment address
mov A,DPL ;} Read each RAM
cjne A,#00H,RAMT3 ;} address until
mov A,DPH ;} comp fails or
cjne A,#80H,RAMT3 ;} all addrs read.
cpl TmpC ;Flip starting bits

djnz TmpA,RAMT1 ;Do second pass
;SUCCESSFUL TEST
clr P3.5 ;LED off/uC not Busy
sjmp RAMT9 ;Return to call
;FAILED TEST
RAMT7: mov TmpA,#0FFH ;Init loop counter
RAMT8: cpl P3.5 ;Toggle LED
lcall DELAY ;Delay to see LED
djnz TmpA,RAMT8 ;Loop 256 times
setb P3.5 ;Set LED on/uC Busy
RAMT9: ret ;Return to call
Download ______________________
The download routine is used to receive an ASCII file in Intel hex
format on the serial port. Each record begins with a colon (:) and
ends with a checksum.
The routine waits for a colon byte. Then the next byte determines
the number of hex values in one line or record. The stream next has
a two-byte value for the beginning program address where the data
will be stored. The following byte is to check end-of-file. Finally,
the stream of hex contents follows and is transferred to RAM.
;-------------------------------------------------
DOWNLOAD:
;-------------------------------------------------
;NO HANDSHAKE
clr P3.5 ;Hndshak-uP not BUSY
;RECEIVE READY
jnb RI,DOWNLOAD ;Wait for rec'd byte
lcall SERIN ;Input byte
cjne A,#3AH,DOWNLOAD ;Wait for : input
;COUNTER BYTES
DOWN1: lcall DOWNBYTE ;':' rec'd, get byte
mov R5,A ;Put in R5 for cntr
mov R4,A ;Init CHEKSUM in R4

Chapter 19 Bios 133
;ADDRESS
DOWN2: lcall DOWNBYTE ;Get next byte which
mov DPH,A ;is high order addrs
lcall CKSUMINC ;Update checksum
lcall DOWNBYTE ;Get next byte which
mov DPL,A ;is low order addrs
;CHECK FOR EOF
DOWN3: lcall DOWNBYTE ;Ck next byte, EOF?
cjne A,#0H,DOWN6 ;If so, exit, else
;NEXT byte
DOWN4: lcall DOWNBYTE ;Input HEX file
movx @DPTR,A ;Output to RAM
inc DPTR ;Next address
djnz R5,DOWN4 ;Repeat if more byte
;CHECK IF ERRORS
DOWN5: mov A,R4 ;Verify CKSM of recd
cpl A ;byte & HEX file.
inc A ;Complement, add 1 &
mov TmpA,A ;compare with in
lcall DOWNBYTE ;file. If OK,
cjne A,TmpA,DOWN7 ;continue, else err
ljmp DOWNLOAD ;Continue receiving
;end-OF-FILE
DOWN6: lcall DOWNBYTE ;Input file chksm.
cjne A,#0FFH,DOWN7 ;If not FF, error
ret ;Return to call
;FAILED
DOWN7: mov LoopC,#3CH ;Loop counter
DOWN8: cpl P3.5 ;Toggle LED/uP BUSY
djnz LoopC,DOWN8 ;Loop 60 times
setb P3.5 ;LED on/uP BUSY
ret ;Return to call

Downbyte ______________________
The downbyte routine is simply protocol for transferring serial
information into a byte.
This routine inputs two sequential ASCII values and converts them
to a corresponding binary value. The binary format is a nibble. The
low order and high order nibble are combined to yield the
hexadecimal Intel-hex format. Register A is used for input, R3 for
output, with R1 and R2 for processing.
The serial communications routines are discussed with substantial
detail in the next chapters.
;-------------------------------------------------
DOWNBYTE:
;-------------------------------------------------
;RECEIVE READY
jnb RI,DOWNBYTE ;Wait for full byte
lcall SERIN ;Input when rec'd
mov R1,A ;Put byte in R1
anl A,#40H ;Ck lead nibble, if
jz DOWB2 ;above 9, convert.
;
;HEX TO DECIMAL
DOWB1: mov A,#09D ;Conv ASCII A to F
add A,R1 ;by adding 9 then
sjmp DOWB3 ;adjust lead nibble
;
;DECIMAL
DOWB2: xch A,R1 ;Xchg ASCII value
;
;ADJUST
DOWB3: swap A ;Swap nibbles then
anl A,#0F0H ;strip unused nibble
mov R3,A ;Put hi nibble in R3
;
;GET NEXT byte
DOWB4: jnb RI,DOWB4 ;Wait for next byte

Chapter 19 Bios 135
mov R2,A ;Put ASCII byte n R2
anl A,#40H ;Strip lead nibble
jz DOWB6 ;If ASCII above 'A'
;
;HEX TO DECIMAL
;then adjust by + 9
DOWB5: xch A,R2 ;to stripped nibble
add A,#09D ;& save in R2 to
xch A,R2 ;combine w/hi nibble
;
;COMBINE
DOWB6: xch A,R2 ;Adj ASCII to A
anl A,#0FH ;Strip lead nibble
orl A,R3 ;combine nibbles, &
ret ;Output thru A
Checksum ______________________
The last item in the line record is a checksum. A checksum is
calculated and compared to the received check value. The checksum
is calculated by adding each byte received over the serial line into a
register that holds the total. Each byte added is already converted
into binary form.
If a serial communications checksum error is detected, the message
LED toggles off/on 30 times. The off/on cycle period is thrice that
of the RAM test error condition.
;-------------------------------------------------
CKSUMINC:
;-------------------------------------------------
add A,R4 ;add new byte to sum
xch A,R4 ;put sum into accum
ret
The procedure calls the serial input routine that was discussed in the
serial project. It also calls a routine to convert ASCII data to a
binary byte.

ASCII to hex conversion __________
The conversion routine inputs two sequential ASCII values and
converts them to a corresponding binary value. The binary format is
a nibble. The low order and high order nibble are combined to yield
the hexadecimal format. Register A is used for input, R3 for output
with R1 and R2 for processing.
;-------------------------------------------------
ASCIBYTE:
;-------------------------------------------------
;RECEIVE READY
jnb RI,ASCIBYTE ;Wait for full byte
mov R1,A ;Put byte in R1
anl A,#40H ;Ck lead nibble, if
jz ASCB2 ;above 9, convert.
;
;HEX TO DECIMAL
ASCB1: mov A,#09D ;Conv ASCII A to F
add A,R1 ;by adding 9 then
sjmp ASCB3 ;adjust lead nibble
;
;DECIMAL
ASCB2: xch A,R1 ;Xchg ASCII value
;
;ADJUST
ASCB3: swap A ;Swap nibbles then
anl A,#0F0H ;strip unused nibble
mov R3,A ;Put hi nibble in R3
;
;GET NEXT byte
ASCB4: jnb RI,ASCB4 ;Wait for next byte
mov R2,A ;Put ASCII byte n R2
anl A,#40H ;Strip lead nibble
jz ASCB6 ;If ASCII above 'A'
;
;HEX TO DECIMAL
;then adjust by + 9

Chapter 19 Bios 137
ASCB5: xch A,R2 ;to stripped nibble
add A,#09D ;& save in R2 to
xch A,R2 ;combine w/hi nibble
;
;COMBINE
ASCB6: xch A,R2 ;Adj ASCII to A
anl A,#0FH ;Strip lead nibble
orl A,R3 ;combine nibbles, &
ret ;Output thru A
Memory switch __________________
After the program download to the SRAM is complete, it is
necessary to switch control from the ROM to the RAM for program
execution. This routine does a memory-mapped output to change
the status of the mode latch in the programmable logic device.
Mode 0 is the default value and the program will run normally from
ROM. Mode 1 is sent as a signal to switch to SRAM
This is a switch of memory devices during program execution.
Therefore, the entry point into the program that will run from
SRAM will be at the address after the instruction movx @DPTR,A.
The routine is six bytes long. For that reason, back up from the
desired start location by six bytes to determine the org location for
the routine.
All register values retain their value during this switch since no reset
occurs. Therefore, the new program must initialize the register
values it uses.
The bios program can be in either internal or external memory. The
switch program will start at the next address after the switch.
Therefore, the SRAM program memory location can be at the same
address space as the original bios.
For example, consider the bios starts at address 0000h. It has the
requisite jump past the interrupts. It begins operation at address
0080h. Then the download procedure occupies the next few hundred

bytes. If the memory switch routine is org 007Ah (80h – 6h) in the
ROM, then the new program will also start at 0080h in the RAM.
;-------------------------------------------------
007A org 007AH ;After interrupt
MEMSWIT:
;-------------------------------------------------
007A 90FFFF mov DPTR,#0FFFFH ;Mode latch address
007D 7401 mov A,#1 ;Mode =1
007F F0 movx @DPTR,A ;Make switch to RAM
The process just described works fine if the ROM and RAM are
both external. However, if the ROM is internal, /EA will not be
configured for external operation. As a result, it is necessary to be a
little more creative with the software.
Assume there are 8K bytes of on board program memory. Then, the
top of the address space will be 1FFFh. The next executed address
will be 2000h, but it will be external memory. This is independent
of the /EA setting.
Consequently, the memory switch should be org 1FFAh, which is 6
bytes from the top.
;-------------------------------------------------
org 1FFAh ;After interrupt
MEMSWIT:
;-------------------------------------------------
Then the next program code line will execute at 2000h. In order for
the new program that has been downloaded to execute properly, it
must have the program organized to org 2000h.
;-------------------------------------------------
org 2000h ;start address
START:
;-------------------------------------------------

Chapter 19 Bios 139
Use of low memory _______________
One cost of this practice is the lower 8K of SRAM is not usable as
program memory. Nevertheless, if the control lines are kept correct
by the programmable logic device (PLD) program, there is nothing
to keep that section from being used as data memory. Although this
is slightly convoluted from a PLD program perspective, it is a true
use of virtual memory.
Suppose it is desirable to address a routine in the bios from the new
program in SRAM. The programs do not know each other exist. The
low memory SRAM overlays the same address space as the
program bios. This little feature gives a solution.
Create a routine that will display the location of all the routines that
may be called. Use the data to define the location to the SRAM
program. The SRAM procedure will require an equ for the name
and address of each routine that is called. Alternately, an org
directive can be set for the same location.
ž Ÿ

20
PROJECT 5 – BIOS DEVELOPMENT TOOL
Thought
Perfect practice makes perfect.
Proverb
Project 5: Develop operating system
Purpose: This project can be skipped if the program is loaded
into internal memory.
The design includes an EPROM for startup, and serial
communications to load the running program into
SRAM.
To download a new program from a PC computer.
Preamble:
Time required for software development by programming EPROM's
is too extensive for industrial applications. As a result, a
development system is used to speed-up the software process. A
development system usually consists of the target microprocessor,
memory, and a monitor program.
The monitor is a dedicated operating system that transfers program
code from an external computer to the project computer. In some
systems, the monitor is called a basic input / output system (BIOS).
The target microprocessor in the development system is sometimes
called an emulator.

Chapter 20 Project 5 – Bios Development Tool 141
In this project, a simple development system will be built for faster
software development and to minimize the number of times the
EPROM is burned.
Plan:
An external static RAM will be added to the microcontroller. The
static RAM will be used to store the program that is downloaded
from the host computer. The host will be a PC compatible computer.
First, the static RAM is used as data memory to receive a file,
which is program code from the host. Then the control lines will be
changed to make the RAM operate as program memory.
In order to download the program, the host computer has a driver or
file transfer program, such as HyperTerminal that transmits the data
on the serial port. The development system also has a program that
will monitor the received data. In this case, the microprocessor will
have a BIOS program reside in the ROM.
The interface used to download INTEL HEX files from the host to
the microcontroller system is simple RS-232 serial communication
without any control (handshake) lines.
Preparation:
Observe the comparison between RAM and ROM pins. For 32 K
byte devices, only two pins are different.
EPROM: Mp1=VPP, Mp27=A14
SRAM: Mp1=A14, Mp27=/WE
There are two other pins that are used for chip control. These are the
chip select and the output enable. Functionally, since there is only
one memory device, the chip select can be left active by connection
to ground. The output is determined by the function.

EPROM: /OE=/PSEN
SRAM: /OE=/RD
The RAM initially is used as data memory storage, then as program.
Therefore, both the EPROM and SRAM must have their control
lines switched. An address latch in the PLD (PEEL) is used to make
the change.
The indicator for the download process will use port P3.5. Connect
an LED with proper buffering to this port.
Finally, program the code memory with the BIOS program that was
discussed in the previous chapter.
Procedure:
The BIOS program will operate as a RAM tester. If P3.5 turns on,
then off, the right connections have been made. If the LED flashes
128 times, hardware debugging of the SRAM system is required.
The monitor program constantly scans for the incoming serial data.
The host PC computer will execute a download file transfer
program. This program transmits an INTEL HEX file from the host
to the development board. If P3.5 flashes 30 times, the download
process was not successful. Otherwise, the download was
completed.
The BIOS changes control to execute the program from RAM. The
application program should begin at an address above all interrupts
and other reserved spaces.
Using serial communications, download the new program hex code
to the SRAM. The program will transfer control to the
microprocessor.

Chapter 20 Project 5 – Bios Development Tool 143
Presentation:
Show the system works by downloading a program such as the
metronome. Then execute the program from the static RAM.
ž Ÿ

21
SERIAL COMMUNICATIONS
Thought
For lack of a nail…
the kingdom was lost.
Benjamin Franklin in
Poor Richard’s Almanac.
Background ____________________
Communications between systems takes many forms. For a number
of years, parallel communications was regarded as the preferred
mode since a full byte of data was available with each transfer.
However, the limitation is in the distance that the logic can drive the
circuit. Typically, this is less than 20 feet. In addition, many wires
are required.
The oldest form of electronic data communications is serial. In
1844, Samuel F. B. Morse developed the first serial
communications system with the practical implementation of the
telegraph. The first communications was Baltimore to Washington.
He developed a code for the on-off bits used to switch a remote
relay.
Unfortunately, it used uneven spacing. Shortly after Morse, Emile
Baudot invented the 5 digit code. Communications to this day
remember his name with Baud.

Chapter 21 Serial Communications 145
In 1902, Charles Krum developed an electro-mechanical system that
was the predecessor of the Teletype. This technique could literally
transmit data around the world.
His circuitry provided the basis of present day serial
communications using Electronic Industry Association (EIA)
standard RS232. The original standard was developed in 1960. The
standard was modified in 1969 and that one is still used in computer
systems. The system responds to a sequence of +25 and –25 volt
signals. Moreover, the current loop technique of the Teletype
spawned the structure of the 4-20 milliamp current loop used in
instrumentation systems.
With his system, Krum also developed a code for the sequence of on
and off bits. This code is the predecessor of the present day
American Standard for Information Interchange (ASCII) characters.
That standard was adopted in 1966.
Many different attempts have been made to eliminate RS232, but it
persists for several reasons. Just a few of these are noted. First,
because of its heritage, it is very simple to construct equipment.
Therefore, virtually any device or machine can include its features.
Second, because of its long use, there are innumerable devices that
are compatible. Therefore, it will persist for some time to come.
Third, the limited number of wires, simple configuration of wires,
and the long distance capabilities make it very economical.
The theoretical maximum speed of communications using the
system over a standard telephone circuitry is about 9600 baud. This
is because of the Nyquist frequency limit in the voice range.
However, clever compression algorithms have raised this value over
an order of magnitude. Nevertheless, simple systems still limit their
speed to 9600 baud.
Another major limitation is the number of items that can be
connected simultaneously on the network. The system is designed
for communications between only two devices on the network.

Microcontroller _________________
The microcontroller has serial communications as part of the basic
chip. A universal asynchronous receive transmit (UART) package is
included. The data exchange is on port3. Receive is bit zero (0B0h)
and transmit is bit one (0B1h).
Separate buffers are used for transmit and receive, therefore they
can occur simultaneously as true duplex communications. Since the
micro code is structured for communications, software
implementation is very simple. The speed is setup, a message is
transferred via the buffer, and a flag is checked for completion of
the message.
Integration only requires a method to convert the higher voltages of
serial RS232 to the VCC of the chip. Although a discrete circuit can
be built, commercial interface chips are very cost effective and
require little real estate.
Several registers are used in serial communications. These are serial
control, power control, timer mode, timer control, and timer 1 value.
The details of the bit patterns are illustrated in the chapter on special
function registers.
Generating baud rates ____________
The serial control register (SCON) determines the UART mode of
operation. This determines whether there is a fixed or variable baud
rate.
If the rate is variable, then the numeric value is placed in timer 1.
For serial communications, timer 1 would be used in the auto reload
mode. In this auto reload timer mode, the serial communications can
continuously run without additional software action from the
program. Auto reload is configured in the timer mode (TMOD)
register as mode 2 for timer 1.

The Baud rate calculation is dependent on the data protocol. The
mode may be as a shift register, eight bit UART, or 9-bit UART.
The Baud rate for each serial mode is shown in the following
equations.
Mode 0 _________________________
Mode 0 has a fixed baud rate, which is 1/12 of the oscillator
frequency. To run the serial port in this mode, none of the
Timer/Counters need to be set up. Only the SCON register needs to
be defined. This is simply a shift register.
Oscillator Frequency
Baud Rate = {--------------------}
12
Mode 1 _________________________
Mode 1 has a variable baud rate for eight bit data. Either Timer 1 or
Timer 2, if available, can generate the baud rate. The generic
procedure uses Timer/Counter 1.
K * Oscillator Frequency
Baud Rate = {------------------------}
32 * 12 * [256 – TH1]
A multiplier (K) is available to double the calculated baud rate. This
is the SMOD bit in the power control (PCON) register. If SMOD =
0, then K = 1. If SMOD = 1, then K = 2. Since the PCON register is
not bit addressable, one way to set the bit is logical ORing the
PCON register (orl PCON, # 80H). The address of PCON is 87H.
The reload time for the baud rate is placed in the high byte of Timer
1 (TH1). The equation to calculate TH1 can be written as follows.
K * Oscillator Frequency
TH1 = 256 – {------------------------}
32 * 12 * Baud Rate

TH1 must be an integer value. Rounding off TH1 to the nearest
integer may not produce the desired baud rate. In this case, it may
be necessary to choose another crystal frequency.
Mode 2 _________________________
In Mode 2, the baud rate is fixed for 9-bit data. It is 1/32 or 1/64 of
the oscillator frequency, depending on the value of the SMOD bit in
the PCON register. In this mode, none of the timers is used, and the
clock comes from the internal phase-2 clock.
SMOD = 1, Baud Rate = 1/32 Osc Freq.
SMOD = 0, Baud Rate = 1/64 Osc Freq.
To set the SMOD bit, use orl PCON,# 80H. The address of PCON is
87H.
Mode 3 _________________________
The baud rate in mode 3 is variable and sets up exactly the same as
in mode 1. This allows 9-bit data transfer.
Timer/counter 2 baud rates ________
Timer 2 can be used in the baud-rate generating mode. If Timer 2 is
clocked through pin T2 (P1.0) the baud rate is given by the
following equation.
Timer 2 Overflow Rate
Baud Rate = {---------------------}
16
If serial communications is being clocked internally, the baud rate is
given by the following equation.

Baud Rate = {------------------------------}
32 * [65536 – (RCAP2H,RCAP2L)]
To obtain the reload value for RCAP2H and RCAP2L the previous
equation can be rewritten as follows.
RCAP2H,RCAP2L = 65536 – {-------------------}
32 * Baud Rate
Timer baud table _________________
Several rates are commonly used and the choice of acceptable
crystal frequencies is rather limited. Therefore, a table can provide
the most commonly used values required for Timer 1.
TH1 7.3728
MHz
8.00
MHz
11.0592
MHz
11.0592
MHz
12.00
MHz
12.00
MHz
14.7456
MHz
22.1184
MHz
smod=0 smod=1 smod=0 smod=1 smod=1
E0 600 651 900 976 1,200
E6 -26 1,202
E8 -24 1,200 2,400 4,800
F0 -16 1,200 1,302 1,800 1,953 2,400
F3 -13 2,404
F4 -12 2,400 4,800 9,600
F8 -8 2,400 2,604 3,600 3,906 4,800
F9 -7 2,743 2,976 8,299 4,464 8,923 5,486
FA -6 3,200 3,472 4,800 9,600 5,208 6,400 19,200
FD -3 9,600 19,200 38,400
FF -1 19,200 20,833 28,800 57.6K 62,500 115.2K
Timer 1 and color burst ___________
Let us imagine that on one of his adventures, Mac MacGyver, from
the old television series, gets caught on a remote island. The only
communications is via an old instrumentation system operating at
300 Baud. He is challenged to make a computer interface. On
searching, he finds an old television that he can strip of its color

burst crystal. Although odd for computers, the frequency is the very
common 3.5795 MHz.
MacGyver asks you to determine what is the preload value that he
must load into the microcontroller?
1 * 3579500
TH1 = 256 – {--------------}
32 * 12 * 300
TH1 = -31.07
He must round this value to –31 or 0E1h. Since it is not precise,
there may be some loss of data. However, at this slow speed, that
should not be a problem.
Interestingly, this slow rate of 300 Baud is still frequently used with
long range radio communications such as amateur or “ham” radio.
Serial initialization _______________
The set-up for serial communication has many options that require
substantial details. Nevertheless, the actual code used is quite small.
A complete serial initialization contains a very limited number of
lines.
Register Timer 1 is used in the eight-bit, auto-reload mode.
Therefore, TMOD bit M1-1 is set and bits M0-1, C/T'-1 and GATE-
1 are clear. If the GATE were set, the external input on port3 would
control starting and stopping the timer.
GATE-1 C/T -1 M1 –1 M0 -1 GATE-0 C/T -0 M1 -0 M0 -0
Register PCON, bit SMOD can be used to double the baud rate. If
the bit is clear, the rate multiplier is K=1. If the bit is set, then the
multiplier is K=2. However, the speed multiplier is not used for
these examples.

SMOD — — — GF1 GF0 PD IDL
Register SCON is used to define serial communication as mode 1,
eight-bit UART, by setting bit SM1. Reception is enabled with bit
REN set.
SM0 SM1 SM2 REN TB8 RB8 TI RI
Set TI to indicate the previous serial transmission is complete.
Otherwise, the code that tests for a complete transmission will hang.
Register TCON, bit TR1, is set to enable the timer to begin
counting.
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Register TH1 contains the count for the baud rate.
TH1 = 256-(K * Osc Freq / 384 * baud rate)
For a 11.059 MHz oscillator, and 9600 baud operation, the value
placed in TH1 is 0FDHh (-3d).
;-----------------------------
UART:
;-----------------------------
mov TMOD,#00100000b ;Timer 1 Mode 2
mov TH1,#0FDH ;9600 BAUD @11.05MHz
mov SCON,#01010010b ;Set SM1, REN & TI
setb TR1 ;Start timer
ret ;Return to call
Serial data protocol _______________

Many options are available for data protocol. The SCON register is
configured for eight-bit or nine-bit data transfer. This will then
determine the hardware communications sequence. Eight bits are
normal communications. Nine are used when additional control bits
are required.
Eight-bit data is transferred in the SBUF register. If a ninth bit is
required it is placed in the SCON register.
Asynchronous operation is typical. This implies the device on both
ends of the serial communications line has its own clock.
Parity is a very simple method to determine data integrity. A parity
flag (P) is located at bit 0 in the Program Status Word (PSW)
register. The flag is set if there is an odd number of ones in the
accumulator. The flag is clear if there is an even number of ones in
the accumulator.
If parity is used, the bit is placed as the last bit in the transmission.
For eight-bit exchange, this would be the most significant bit in the
accumulator. However, because of the difficulty in managing eight
bits of data, plus a parity bit, the parity is often not used.
The standard data format for interchange with most serial devices is
ASCII. This is a 7-bit code for characters. The table is shown in the
reference chapter on ASCII.
However, it is not necessary to send ASCII. Any protocol structure
can be used. Nevertheless, both the receiving and transmitting
devices must be using the same structure.
The most common configuration is eight bits of data, no parity, and
one stop bit. The protocol is called 8-N-1.
The serial line is normally held high. When data is ready to be sent,
the line is pulled low for one bit time, the start bit. Then the eight bit
string is sent. This is followed by a high stop bit that allows the
receiver to process the character.

0 1 2 3 4 5 6 7
Wait Start Stop
Serial buffer _____________________
The exchange of data using the internal serial buffers is very simple.
First, verify that the last process is complete. Then clear the process
interrupt flag. Next, transfer the data between the accumulator and
the buffer.
Once the data is in the buffer, the program can go on to other things.
The hardware micro code will complete the exercise of shifting the
data between the port 3 pins and the buffer.
A simple scan and wait routine is most frequently used. Alternately,
a serial interrupt can be invoked. Since there is only one serial
interrupt, the handler first must decide if the interrupt was caused by
transmit or receive. If interrupt handling is used, the only change in
the processing routine would be removal of the first or polling line
from the next two routines.
These routines do not have hardware handshaking. If that is required
another port bit must be wired to the data cable.
Serial in brings data from an external device into the processor.
;-----------------------------
SERIN:
;-----------------------------
jnb RI,$ ;rcv busy so wait.
mov A,SBUF ;Get byte from SBUF
clr RI ;Clear RI
ret ;Return to call
Serial out exports data to an external device.
;-----------------------------
SEROUT:
;-----------------------------

jnb TI,$ ;xmit busy so wait.
clr TI ;clr for next xmit
mov SBUF,A ;mov byte to serial
ret
Circuit: serial ___________________
* V
1 20
2 T1IN 19
3 R1OUT 18
4 R1IN 17
5 T1OUT C2- 16
6 GND C2+ 15
7 Vcc 14
8 13
9 GND 12
10 C2- C2+ 11
P3.0 Rx P3.1 Tx
ž Ÿ

22
PROJECT 6 – RS232 COMMUNICATIONS
Thought
Serial is for communicating.
Cereal is for eating.
Project 6: RS 232 to PC exchange ___
Purpose: To use a serial interface system.
To implement a simple network system
To implement a door lock system with a dedicated uC.
Preamble:
Communication is one of the most important parts of a computer
system. Computer networks are commonly used for data
communications. Digital data communication uses one of two
modes.
Serial communications has several advantages over the parallel
type. A major benefit is that the number of wires is reduced
significantly. This reduces the cost of wiring in a long distance type
circuit. Another benefit is serial communications operate at a signal
level which can be transmitted much further without degradation.
Most major local area networks for computers use serial type
communications.

The processor has a built-in serial circuit that can be used for
communications. The computer has its own baud rate generator,
which uses a timer. Several bits in various registers must be set
before opening serial communications. The timer interrupt registers
are a necessary component, since Timer 1 determines baud rate. The
details of setting registers, interrupts, and the baud rate generator
can be observed from the reference section.
The EIA RS-232 is the most widely used standard in serial
communication. The primary signals are lines 2, 3, and 7. These are
receive, transmit, and ground signals. RS-232 has several handshake
lines that may be used. These other handshake lines are optional.
The RS-232 circuitry must be capable of handling +25 to –25 Volts.
The minimum voltage level of a line is +9 V for high and -9 V for
low. Therefore, the voltage level of the serial pin must be raised,
since it is operated in the TTL range of 0 and 5 volts. A voltage
doubler can be used.
The easiest way to interface RS-232 to the serial line is with a
special purpose IC such as MAX 232. The chip will receive an RS-
232 signal and send it to the controller at TTL level. The receive pin
(R1IN) will get data from the RS-232 transmit pin #2 (TXD) of the
PC. The MAX232 output pin (R1OT) will be connected to the
microprocessor pin #10 (RX). A ground (RS-232 pin# 7) must be
connected between the computers.
Plan:
Implement a microprocessor as a data communication controller that
is coupled to a PC.
Preparation:
Consult the data section for more details about serial
communications. The interface circuit for RS-232 communications
is given in the following pages. The software for the PC is typically

Chapter 22 Project 6 – RS232 Communications 157
the Windows based program HyperTerminal in the Accessories >
Communications folder. The cable from the PC can be constructed
using the diagram in the reference section.
Procedure:
The system works according to the following description. The
microprocessor will scan the switches for an input. When the
appropriate switch is set, an encoded message is selected. The
ASCII characters of the message will be sent serially to a host
personal computer. The host will evaluate the message. The PC will
send a certain code to the microcontroller. The microcontroller then
deciphers the returned code. A decision is made based on that
deciphered data.
If the code is valid, turn on or off an LED. This can represent a relay
that opens / closes a door. Alternately, it can be a signal to another
device.
More importantly, this procedure can also be used as a
troubleshooting tool to display messages on the PC at various stages
of a project.
Presentation:
Demonstrate the system. The PC with its serial communication
program is provided. Each one is encouraged to have a unique
implementation.
ž Ÿ

Program sample example _________
;-----------------------------
;Program: MODRs232.asm
;Update: 29 January 2003
;
; Tulsa, OK, USA
; mod@superb.org
;
;Purpose:
; A routine to demonstrate serial communication.
; A single character message will be displayed on
; the HyperTerminal or PC serial display.
;
;Baud: 9600
;#################################################
;
; PROGRAM
;
;#################################################
org 00H
START: ljmp INITIAL
org 0033H
;-----------------------------
org 0080H ;Addres past reserve

INITIAL:
;-----------------------------
;INITIALIZE
lcall UART ;config & start UART
;-----------------------------
MAIN:
;-----------------------------
;PROCESS
mov A,#3Fh ;ASCII ?
lcall SEROUT ;send character
;*************************************************
;
; SERIAL RS232
;
;*************************************************
UART:
;-----------------------------
;
; Initialize the registers that control the
; serial communications process.
;
; Timer 1 is used in eight bit auto-reload mode.
; So TMOD bit M1 is set and bits
; M1, C/T' and GATE are clear.
; With GATE set, INT1' & INT0' control the timer.
;
; Bit SMOD in register PCON can be used to double
; the baud rate when set. It's clear here, K=1.
; If set, then K=2.
;
; SCON is used to define serial communication
; mode 1, eight bit UART, by setting bit SM1,
; and enable reception with bit REN set.
;
; Bit TR1 in register TCON is set to enable
; timer.
;
; Set TI to indicate serial transmission is
; complete.

;
; Register TH1 contains the count for baud rate.
; TH1 = 256-(K * Osc Freq / 384 * baud rate)
;
; For 11.059 MHz oscillator,
; TH1 is 0FDH (-3D) for 9600 baud.
; TH1 is 0E8H (-24D) for 1200 baud.
mov TMOD,#20H ;Timer 1 Mode 2
; anl PCON,#7FH ;SMOD = 0, K=1
mov TH1,#0FDH ;9600 BAUD @11.05MHz
mov SCON,#50H ;Set SM1 & REN
;
setb TR1 ;Start timer
setb TI ;last xmission thru
;
ret ;Return to call
;-----------------------------
SERIN:
;-----------------------------
;
; General purpose serial receiver routine. It
; gets a byte from the serial buffer, SBUF,
; converts to the needed binary form by removing
; the leading bit, and clears the RI (receive
; interrupt) flag that is set by the uP after a
; full byte is received.
;
; Clearing RI allows next byte to be received.
; SBUF is the input and register A is the output.
jnb RI,$ ;xmit busy so wait.
mov A,SBUF ;Get byte from SBUF
clr RI ;Clear RI
; anl A,#7FH ;Clr MSB for ASCII
ret ;Return to call
;-----------------------------
SEROUT:
;-----------------------------
;
; Transfer one byte out to the serial port.
; TI is high while a transmission is happening.

; It goes lo, when a new byte can be sent.
jnb TI,$ ;xmit busy so wait.
clr TI ;clr for next xmit
mov SBUF,A ;mov byte to serial
ret
;*************************************************
;Program end
end
ž Ÿ

23
EXPANSION LATCHES
Thought
The process of success:
Dream – overcome – success.
MOD
I/O expansion port _______________
What do you do when you want to hang more devices on the
microprocessor? The standard chip is configured with four parallel
ports. However, these registers have many other functions.
Consequently, it is very easy to run out of pins for connections to
peripherals.
One choice is to place latches on one of the ports and select the
latches with control lines. This is very straightforward, but three
design constraints must be considered.
First, only four loads can be placed on a port because of fan-out
constraints. However, if they are not simultaneously selected, the
port sees only those that are active. Second, it takes additional port
pins for the control lines. The number of pins can be improved by
using a 2:4 or 3:8 decoder. That allows connection to more devices
with fewer lines. A similar function can be created in a
programmable logic device. Third, port expansion requires several
lines of code to invoke the control lines and the resulting latch.

Chapter 23 Expansion Latches 163
I/O expansion memory ____________
A seemingly infinite number of devices can be added with memory-
mapped input / output. In essence, a latch is placed on port 0 and is
addressed exactly as if it were data memory. Two design
configurations must be considered.
First, only a limited number of latches can be placed on port 0
because of fan-out considerations. Second, the address must be
decoded.
The most efficient application of memory-mapped input/output
would be to use low memory. However, that requires numerous
address lines. The more common approach is to use address A15 as
a control line for external input / output. This will give up to 32K
addresses. However, it restricts data memory to the lower 32K
addresses. Seldom is that a problem.
In addition to the address, a control line is required to activate the
latch. The read not (/RD) or write not (/WR) line must be ANDed
with the address. These lines are pulled low when the data is on the
port.
Latch in/out connection ___________
Two latches are inherent for an effective system. The key latch has
lines for output of column and input of row lines. The display-out
latch has eight data lines for display devices. The routines for these
latches are common to many procedures.
The latch is typically a D-type flip/flop. A common device is a
74573. The chip has eight transparent gates with two control lines.
This chip has the input pins on the left, and the output pins directly
across on the right side. The latch family should be ALS or CMOS
type, since they have less fan-out load.

When the latch enable (LE) is asserted high, the flip/flops are
transparent. The data is simply received by the flip/flops and
clocked through to the output. When the latch enable (LE) is
asserted low, the data on the input is trapped in the latch.
When the output enable not (/OE) is asserted low, the data on the
flip/flops is placed on the output terminals. When the /OE line is
asserted high, the output pins of the latches are high impedance.
Therefore, it does not interfere with other lines connected to the
same terminal pins.
When doing an output from the processor through the latch, the
output enable not (/OE) can be permanently wired low. When doing
an input from the latch to the processor, the latch enable (LE) line
can be permanently wired high.
The programmable logic must make the latch enable line high then
low. This may be done with hardware, but it is preferably done in
software using the select lines. The pins must be set high, then
cleared low for the latch to hold.
An input latch uses the output enable not (/OE) line. The line must
be asserted low by the processor before the value is read. So that
data is always available, the latch enable line is wired high, making
the latch transparent.
‘573
LE /OE
‘573
LE /OE

Chapter 23 Expansion Latches 165
Latch in/out code _________________
The latches can be connected as expansion registers on any one of
the ports. Consider using port 1 for the data. For latch select lines,
use 3 pins from Port 3 (0B3-0B2h).
Function Select
B4B3
unused 00
Keypad 01
Display out 10
ADC 11
The unused latch or an output latch is chosen as the default so the
latch enable line can be taken high then low. An input latch should
not be selected except at the time data is desired. Otherwise, it may
bias data on the port.
;-------------------------------------------------
LKEY:
;-------------------------------------------------
; Latch for keys. Select as #01 on B4,B3.
setb 0B3h ;key-in latch
clr 0B4h ;OE active
mov A,P1 ;get row in
ret
;-------------------------------------------------
LOUT:
;-------------------------------------------------
; Latch for output. Select as #00 on B4,B3.
mov P1,A ;set-up latch
clr 0B3h ; out latch
clr 0B4h ;LE hi
setb 0B4h ;LE lo
ret

;-------------------------------------------------
LDISPLAY:
;-------------------------------------------------
; Latch for display out. Select as #10 on B4,B3.
;byte OUT
mov P1,A ;data to display
clr 0B3h ;display-out latch
setb 0B4h ;LE hi
clr 0B4h ;LE lo
ret
An alternative addressing technique, memory mapping, is frequently
used to select latches. This is illustrated in the next chapter.
ž Ÿ

24
MEMORY-MAPPED INPUT AND OUTPUT
Thought
Doing the same thing and
expecting different results is ignorance.
Uncle Albert Einstein
Accessing external data ___________
The external memory can be up to 64K. The addressing is on port 0
for the low byte and port 2 for the upper byte. Two techniques can
be used to access the memory.
Eight-bit addressing can be use to access the lower 256 bytes. The
address is stored in R0 or R1. There are four banks selected by the
process status word (PSW) register. Therefore, up to eight banks
can have an address.
The sixteen-bit expansion of that technique simply uses the same
low byte bank registers, R0 and R1. The upper byte addressing is
placed on port 2 as if it were a special function register. This gives
many addresses that can be held without swapping data in the
registers.
The more common approach to sixteen-bit addressing is to use the
data pointer register (DPTR). The DPTR is the only two-byte latch
in the special function registers. It consists of a low byte (DPL) at
address 82h and a high byte (DPH) at 83h. These may be

individually loaded or the full address may be stuffed into the
DPTR.
The same register is employed for data memory and code memory.
Therefore, it must be changed frequently. That requires additional
temporary locations to hold the last values.
The instruction __________________
Only one type instruction is available to transfer data with external
memory. That is movx. The instruction may be executed with the
eight-bit R0 / R1 format or the sixteen-bit DPTR structure. The
instruction is only a single byte.
The mode of addressing is referred to as register-indirect addressing.
In essence, the address is placed in the register. Then the data at that
address is exchanged with the accumulator.
E0 movx A,@DPTR ;info @address to A
F0 movx @DPTR,A ;info @addres from A
E2 movx A,@R0 ;info @address to A
F3 movx @R1,A ;info @addres from A
The memory address to be accessed is stored in the register. On
execution of the instruction, the computer goes to the register, gets
an address for the instruction, and then transfers information at (@)
that address. In order to employ the indirect addressing mode, the
address must be stored in the register before the movx instruction is
executed.
The instruction activates either the read not (/RD) or the write not
(/WR) line. These lines are active low when the data is available at
the port.
When moving data to a register in a bank, the opcode is a one-byte
instruction that contains the register number. The first nibble is the
type instruction. The second nibble starts with 1 followed by the
binary equivalent of the register, xxxx1rrr.

Chapter 24 Memory-mapped I/O 169
The setup _______________________
When using the bank registers, the bank must be selected first. Next,
the address is placed in the register. Then the transfer is made with
the external location.
clr 0D4h ;PSW.4 bank 0
clr 0D3h ;PSW.3 bank 0
mov R1,#129 ;address= 129
; mov P2,#page ;optional paging
movx A,@R1 ;contents of 129 ĺA
The data pointer is considerably easier to setup. As a result, it is the
preferred technique.
mov DPTR,#8000h ;address= 8000h,A15
movx A,@DPTR ;contents of 8000hĺA
The generic microprocessor has a single DPTR. A very common
and useful enhancement to derivative machines is the addition of a
second data pointer register.
The hook-up ____________________
A latch or D-type flip/flop is commonly applied for the memory-
mapped input / output interface. A common device is a 74573. Pin-
out details are given in the reference section and a detailed operation
is exhibited in the expansion latch chapter. This chip has eight
transparent gates with two control lines.
When read or write lines are used for the latch enable (LE), the
process is very simple. The control line must be high during setup.
However, the read or write line is asserted low. Therefore, it must
be inverted before connecting to the latch enable. Then the latch

traps the data when the read or write line is released to high, which
makes the LE go low.
When doing an output to the latch, the output enable not (/OE) can
be permanently wired low. The latch enable (LE) is controlled by
the and of the address with the inverted write not (/WR) line.
D0 – D7
Address
/Wr
When doing an input from the latch, the process is slightly more
complex. The output of the latch is not permitted on the port except
when the port is ready to read. Otherwise, the load of the latch
would influence other inputs.
The latch output enable not is selected by the and-invert of the
address with the inverted read not (/RD) line. In addition, the latch
enable is wired high.
D0 – D7
Address
/Rd 5 V
The connection of the addresses with the read not or write not line is
accomplished in a programmable logic device (PLD). Therefore,
each of the two diagrams simply become an equation line of code in
firmware. The firmware latch is illustrated in the chapter on
programmable logic devices.
When the control is to the latch enable (LE) line, the device is self
latching simply by the transition of the read not (/RD) or write not
(/WR) line that is connected to the latch.
‘573
LE /OE
‘573
/OE LE

Chapter 24 Memory-mapped I/O 171
Latch in/out memory-mapped ______
The previous chapter illustrated latch selection by a port. A common
alternative is to connect external latches as memory-mapped
devices. Address A15 is frequently used as a simple decode address
select line. When ANDed with other address lines, 32K locations
can be used for expansion.
A couple of the common function latches are selected by these
addresses. Other addresses are used to drive particular lines high or
low, but these are not connected to drive a latch or flip/flop
Function Address
Memory map 8000h
Key Oen & Key LE 8001h
Display LE 8002h
8003h
8004h
The latch controls circuits can be glue logic. However, to save space
and increase flexibility, a programmable logic device is preferred.
The latches that require the data to be held can be locked or gated
with firmware in the PLD.
;-------------------------------------------------
LKEYIN:
;-------------------------------------------------
; Latch for key in.
mov DPTR,#8001h ;key-in latch
movx A,@DPTR ;get row in
ret
;-------------------------------------------------
LDISPLAY:
;-------------------------------------------------
; Latch for display out.
;byte OUT

mov DPTR,#8002h ;display out latch
movx @DPTR,A ;send column out
ret
It is very apparent that memory-mapped is simple and easily
implemented in the software. However, it does require slightly more
hardware or firmware to obtain the address combinations. This
arises because an address select line and read or write lines must be
ANDed with the lower address bits.
ž Ÿ

25
PROJECT 7 – I/O EXPANSION
Thought
Listening is
asking open ended questions.
MOD
Project 7: Unlimited I/O ___________
Purpose: To expand I/O system.
To decode an I/O port from memory space.
Preamble:
In a common microprocessor system, the I/O system provides the
interface to the real world. In many microcomputer systems, the I/O
port is accessed by special instructions such as IN and OUT.
However, an I/O port may be treated as external memory. This
technique is called memory-mapped I/O. The I/O device is accessed
like an external data memory location.
The I/O ports are already built within the chip. Moreover the ports
are an internal memory-mapped type.
To obtain additional external ports simply add a latch.

Plan:
For this project, implement the T-Bird Tail Light System at the
memory-mapped locations.
Preparation:
In this project, the I/O system will be changed, rather than using an
individual I/O port. Decode the input and output system as a
memory location on the same lines as the external SRAM. Memory
mapping is accomplished by wiring a latch to the address / data
lines.
The latch is enabled by an address line combination from the PEEL.
Connect the address / data lines of port 0 to the corresponding latch
data pins.
Decode the address through the PLD or discrete logic. Generally,
the address is selected from the higher order address (Port 2) lines.
AND the decoded address with the write not (/WR) line from the
processor. Remember that address lines are active high and the latch
enable is active low. Therefore, the output must be inverted.
Alternatively, a latch can be used for an input. The only change is
the address AND is with read not (/RD) rather than write not.
Some latches are used bi-directionally. These would simply have
the decoded address line connected, without either read or write.
Procedure:
Use the 74573 as a latch to hold the data for the display. Rewrite the
previous display program so that it will run at the new addresses.
Remember, the previous display was handled as a port location. If it
were written with appropriate subroutines, only two lines of code
will change.

Chapter 25 Project 7 - I/O Expansion 175
The new display address will be a memory location. Hence the
DPTR must be initialized every time anything is written to the
display system. The display should be LEDs connected to the output
of the latch.
Presentation:
Show that the memory-mapped display system is working properly.
Display various values on the relocated output LEDs.
ž Ÿ
;Program: MODmmio.ASM
;Update: 2 August 2004
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c)1991 - 2004. All rights reserved
;#################################################
; ASSIGNMENTS
;#################################################
AdSeven equ 8002h;MMIO address for Seven Seg
;#################################################
; PROGRAM

;#################################################
org 00H
START: ljmp INITIAL
org 0033h
db 25, 1,'Marcus O. Durham, PhD, PE'
;-------------------------------------------------
org 0080h ;get past interrupt
INITIAL:
;-------------------------------------------------
;-------------------------------------------------
MAIN:
;-------------------------------------------------
;DISPLAY
MAN1: mov A,#55h ;turn on
MAN2: lcall OUT ;output seven seg
mov A,#0AAh ;alternate segments
lcall OUT
sjmp MAIN
;-------------------------------------------------
OUT:
;-------------------------------------------------
; Create a loop to display latch.
mov DPTR,#AdSeven ;address
mov R2,#200 ;persistence loop
ZDEL2: movx @DPTR,A ;show it again
djnz R2,ZDEL2
ret ;Return to call
;*************************************************
end
ž Ÿ

26
TABLES
Thought
A project has four seasons-
plan, act, reap, reward
MOD
Data in code memory _____________
Some data is fixed. Therefore, the values can be permanently stored
in memory. This characteristic is more like program memory than
data memory. Often, fixed data is embedded in the program code.
What are these kinds of data? Message strings are one common
example. Tables are another. Tables can be used to quickly translate
values. For example, numbers shown on a seven-segment display
have a unique pattern. The pattern can be stored in the table to
correlate with a number.
Memory Pattern Base + Offset Number
200 3fh 200 + 0 0
201 06h 200 + 1 1
202 5bh 200 + 2 2
The pattern can be easily referenced if the first number is assigned
to a memory location called a base value. Then each other number is
an offset from the base. The base value can be stored in the data

pointer register (DPTR) or the program counter (PC). The index is
stored in the accumulator register. This is called base plus index
register addressing.
When information is transferred from data memory, the instruction
code is movx. The data table is stored in program memory rather
than data memory. Therefore, a unique transfer code is required to
load the data pattern.
The only instruction available for obtaining program code is movc.
It creates a code memory address by adding the base register, DPTR
or PC, and the accumulator as an index register. The contents of the
code address are moved to the accumulator.
movc A,@A+DPTR ;[A + DPTR] ĺA
movc A,@A+PC ;[A + PC] ĺA
Data byte _______________________
Fixed data will be stored as code in program memory. Data is stored
directly by setting the byte during machine language programming.
Rather than an instruction, directives tell the assembler that the
information is data. The directive type is Define (D). The directive
can reserve bytes (db), words(dw), or space(ds).
org 1000h ;start of table
RESERVE: db 3Fh ;insert list of byte
dw 1234h ;insert list of word
ds ;reserve space
In reality, the Define Byte (db) has become the directive that all
assemblers use and is the only one required for programming. By
placing a string of bytes, words are automatically defined. If
numbers are used with the byte, they are stored directly. If ASCII
characters are stored, they are included in a single quote.

Chapter 26 Tables 179
The table that is illustrated above can be placed directly into
memory. This is the list of values that represent bits of a seven-
segment display.
TABLE: db 3Fh ;a ‘0’
db 06h ;a ‘1’
db 5bh ;a ‘2’
To load the data, the corresponding number must be loaded into the
accumulator (A) register. The base determined by the org must be
loaded into the data pointer register (DPTR).
mov DPTR,#1000h ;base of table
mov A,#02h ;number seeking 7seg
movc A,@A+DPTR ;[1002]= 5bh ĺA
Be cautious that program instructions do not infringe into the table
area. Since the table is in program memory, the computer would try
to execute the byte as an instruction. The 3Fh would execute as
addc.
The /EA pin must be connected low to ground in order to access
external memory. If the program is strictly internal, then the pin is
left high.
Characters available table _________
A seven-segment display obviously has only a limited number of
elements. These are arranged in a square pattern. Therefore,
characters that require a diagonal are not possible. Furthermore, the
limited arrangement prevents creation of all possible letters. Some
letters can only be shown as lower and some as upper case.
Nevertheless, all numbers can be recognized.
The following table illustrates the available letters and numbers that
can be built.

Character Pattern
0 3F
1 06
2 5B
3 4F
4 66
5 6D
6 7D
7 07
8 7F
9 6F
A 77
C 39
E 79
F 71
H 76
I 06
J 1E
L 38
O 3F
P 73
S 6D
U 3E
Y 6E
b 7C
c 58
d 5E
g 6F
h 74
i 04
n 54
o 5C
r 50
t 78
u 1C
? 53
- 40
_ 08

The hexadecimal pattern corresponds to the bit pattern for the
seven-segment display. Segment a is the top bar on the display. The
details are shown in the reference section.
a
f b
g
e c
d
It is assumed that segment a is connected to bit 0 of the output port
latch.
bit 7 6 5 4 3 2 1 0
segment dp g f e d c b a
Movx vs. movc ___________________
Two transfers from external memory are available. They look
similar, but the registers are very different. movx moves information
to and from data memory. movc moves data from program memory.
movx is simply indirect addressing, while movc is index plus
indirect addressing.
The string of code provides a parallel comparison to the structure of
the instruction.
;EXTERNAL CODE
mov A,#02h ;number seeking 7seg
movc A,@A+DPTR ;[1002h]= 5bh ĺA
;EXTERNAL data
mov DPTR,#08000h ;memory-mapped I/O
movx A,@DPTR ;[8000h] ĺA

Code messages __________________
Another common use of code memory is to store a message string.
These messages may be for serial communications or displays. The
message is typically stored with ASCII characters rather than
hexadecimal numbers. Nevertheless, to the processor they are just
bits.
The structure for storing and retrieving the message string is
precisely the same as numbers. Since it is ASCII, the data is
contained in a single quote. One item that appears different is
simply a style issue. That is all bytes of the string are stored in one
line.
MESSAGE: db ‘Message to display’
The string is read as before. One additional requirement is the string
is read with a loop, since only one byte is obtained at a time.
mov A,#0 ;read initial byte
STRING2: movc A,@A+DPTR ;input byte
lcall SEROUT ;ship the byte
djnz LoopC,STRING2 ;go thru loop
The coded messages are a very powerful capability that permits
common language communications, rather than terse signals.
By using code memory to contain table for conversions and
messages for display, less data memory is required. These defined
byte data take very little space compared to the program memory.
Nevertheless, they do dramatically improve the interaction with the
processor.

Enhanced serial messages _________
By use of tables, messages can be significantly enhanced. This
allows communications to look very normal. The routines can be
added to any program for display.
;-------------------------------------------------
DISPLAY:
lcall UART ;initialize serial
mov DPTR,#ONLINE ;serial message
lcall BIOSSER ;send serial message
lcall SERCOMM ;send to serial
;-------------------------------------------------
BIOSSER:
;-------------------------------------------------
;Send a string to the serial port.
mov A,#0 ;read length string
movc A,@A+DPTR ;input byte
mov LoopC,A ;setup loop
BIOL2: mov A,#1 ;get 1st message
mov CharL,A ;character icd
lcall SEROUT ;send out byte
inc DPTR ;restore offset
djnz LoopC,BIOL2 ;go thru loop
ret
;-------------------------------------------------
SERCOMM:
;-------------------------------------------------
; Send byte on serial line.
; Convert byte to 2 ASCII characters.
;
;SAVE byte & SEGMENT
mov R0,A ;hold data
anl A,#11110000B ;keep high nibble

swap A ;move to low
mov DPTR,#TABASCII;ASCII table
movc A,@A+DPTR ;convert to ASCII
lcall SEROUT ;send first number
;GET LOW NIBBLE
mov A,R0 ;restore frm earlier
anl A,#00001111B ;keep low nibble
movc A,@A+DPTR ;convert to ASCII
lcall SEROUT ;send second number
mov A,#13 ;carriage return
lcall SEROUT
mov A,#10 ;line feed
lcall SEROUT
ret ;its all over
;#################################################
; CODE-STORED CONSTANTS
;#################################################
TABASCII:
;-------------------------------------------------
;look-up table for hex to ascii conversion:
db '0123456789ABCDEF'
;-------------------------------------------------
;Canned message to confirm serial port activation:
Online: db 6, 80h ,'Value=', CR, LF, 0
;#################################################
ž Ÿ

27
MULTIPLEXING
Thought
Our perception is an analog world,
the reality is discrete sampling.
Professor Durham
Perception ______________________
Perception is what you think. To you it is reality. However, in a
multiplexed digital world that is not true.
A phenomenal tool is the eye – brain sensory system. The eye sees
an object and translates the image to the brain. Actually, the eye is a
sampling system rather than a continuous analog network. The brain
processes the samples and gives the impression of continuous,
smooth information. Technology takes advantage of this visual
perception to reduce the quantity of information that must be
provided.
Consider some of the sampling to which the eye is exposed, but the
perception considers the data as stable. The tests are performed in
frames per second. The duration of the sampling is the reciprocal of
the frame rate.

Frame/sec Image
<18 Movie has flicker
18 Motion appears fluid
24 Movie picture rate
25 Television image rate
60 Fluorescent lamp
75 Computer monitor refresh
100 Cannot detect any flicker
220 Air Force pilots identify a plane
500 No detection, but sense something not as it should be
When there are small changes near 20 frames per second, the
changes appear to be fluid motion. Near 100 frames per second, the
eye cannot see any transitions. However, trained pilots can detect an
airplane when it is flashed before their eyes for 1/220 second. If an
image is flashed occasionally, in the order of 500 frames per second,
the mind never detects it. However, the brain does sense something
is not quite right.
A single white flash has a residue image on the brain for an
extended period of time. However, at 24 frames per second, black is
imperceptible. Subliminal imagery has been used in movies at rates
as low as 24 frames. These messages are overlaid on a black
background. The brain perceives the message, but it does not
register in the visual pathways.
Multiplex _______________________
What does the eye detection have to do with microprocessors? It has
everything to do with real world data. Virtually all continuous
analog signals that a person can detect operate at a rate less than 24
frames per second.
This phenomenon permits the microprocessor to sample input and
output at a rate much slower than the computer is operating.
Therefore, it can appear to be doing multiple things at one time.

Chapter 27 Multiplexing 187
Multiplexing is transmitting of several signals simultaneously on the
same circuit. Each message appears to be individual, but all the
signals are combined into one channel.
Consider a keyboard or keypad. How fast does it have to sample in
order to detect every keystroke? If a very fast typist can do 120
words per minute and a word consists of 5 letters, then the fastest
stroke rate is 10 strokes per second.
Similarly, displays that are updated more often than 25 times per
second will appear to be on continuously.
Our perception is an analog world, the reality is digital sampling.
The remaining projects will tap into this phenomenon.
Circuit: displays _________________
Seven-segment displays are commonly used to display limited data
in a large format. These consist of multiple individual elements that
all appear to be energized simultaneously. However, in a
microprocessor, that would require too many lines and connections.
Therefore, all the elements are in parallel on one port channel. Then
each display is progressively turned on then off. This digital
sampling creates a continuous imagery.
D0-D6
PLD
D0-D2
PLD
‘573
data
LE /OE
‘573
SELECT
LE /OE

Seven-segment displays are inexpensive and simple to implement.
Any number of displays can be used. The relatively large size makes
them effective for many situations. Data is output as a byte on a
port. Then other individual bits are used to select which display gets
the information. These are connected to the common pin.
Each segment of a display is an LED. Therefore, there are seven
LEDs in a display. There may also be another LED for a decimal
point or other similar symbol.
The current limiting resistors in series with the data seven-segments
are typically 270 to 330 Ohm, similar to any other LED.
The select line drivers are open collector inverters, such as 7406.
These require a pull-up resistor on the output. This permits the chip
to supply much more current. Alternately, field effect transistors
(FETs) can be used as drivers. When all seven-segments are on,
then substantial current is required through the common pin.
When a one is sent to the inverter, the output is low. If the seven-
segment is common cathode, this is proper. However, if the display
is common anode, the data to the inverter must be complimented.
Latches are shown on the data lines. The devices should be ALS or
CMOS type, since they are less fan-out load. These latches may be
connected as memory-mapped I/O. They may also be expansion
latches on a port. The only difference is the control logic in the
programmable logic device.
For the project, the data latch is connected to MMIO. However, a
latch is not used on the select lines. Each select line is connected
through a FET directly to port1.
The latches setup the output when the latch enable (LE) line is high.
When the line is then asserted low, the data is trapped in the latch.
That data is displayed as long as the output enable not pin is low.
Therefore, the programmable logic must make the latch enable line
high then low. This may be done with hardware in the case of

memory-mapped I/O controlled by the write not (/WR) line.
Alternately it is done in software if using port pins as select lines.
The pins must be set high, then cleared low for the latch to hold.
Code requirements________________
Once data is displayed, it must remain energized long enough for it
to be perceived by the eye. The period of time that it is maintained
energized is called persistence. The delay cycle necessary to hold
persistence is time dependent. Therefore, a faster machine will
require more delay loops.
The display is a routine that operates in three steps that are repeated
for each digit to be displayed.
1. Select the digit display line.
2. Send the data byte to the display.
3. Wait briefly for persistence of display.
The first routine (SEVEN) sets the bit that activates the common
line on one of the seven-segment displays. These lines of code
would be repeated for as many displays as shown. The output
routine gets the byte to display, shows it and waits. The information
to be displayed is stored at the location referenced by R0.
For the examples, the select lines are on port 1, but this is easily
changed in the equate directives. Display A is bit 0 (P1.0), display B
is bit 1 (P1.1), and display C is bit 2 (P1.2). The output is either on a
port or memory mapped.
;-------------------------------------------------
SegA equ 90h ;address
SegB equ 91h ;address
SegC equ 92h ;address
;-------------------------------------------------
SEVEN:
;-------------------------------------------------
;DIGIT 1
clr SegA ;turnoff select line

clr SegB ;turnoff select line
clr SegC ;turnoff select line
setb SegA ;Bit is hi, select A
lcall SEVOUT ;output & delay
clr SegA ;Bit is hi, select A
;REPEAT NEXT DIGITS
ret
Code segment for port ____________
The seven segment can be connected to a port or it can be memory
mapped. The first section has the entire code connected to ports.
;-------------------------------------------------
SEVOUT:
;-------------------------------------------------
;byte OUT
mov A,@R0 ;data byte
inc R0 ;next info to show
;PERSISTENCE WAIT
mov R2,#200 ;Nested loop counter
ZDEL1: mov Pio,A ;display digit
ret
Code segment for memory map ____
Alternately, memory mapping can be used. The only change is the
substituting the address information for the port. This is an excellent
example to see the similarities between the processes. The address is
used by the PLD code to select the latch. The PLD code causes the
latch to be held until the next write command.
;-------------------------------------------------
SEVOUT:
;-------------------------------------------------

;byte OUT
mov A,@R0 ;data byte
inc R0 ;next info to show
;PERSISTENCE WAIT
mov DPTR,#8002h ;addr for disp latch
mov R2,#200 ;Nested loop counter
ZDEL1: movx @DPTR,A ;display digit
djnz R2,ZDEL1 ;for persistence
ret
Binary to binary coded decimal _____
One routine often required is to convert a binary number into
individual digits so they can be displayed. The digits are displayed
as decimal values. The format is often called binary coded decimal
(BCD).
Because of the hardware divide command, the conversion of binary
to decimal is actually very straight forward, if the binary value is in
a single byte.
Simply do what your third grade teacher told you. Divide by the
place value to get the number for that place. The maximum number
is 255. So, first divide by 100 to get the number of hundreds. Take
the remainder and divide by the next place value of 10 to get the
number of tens for that place. Then remainder is the units.
;-------------------------------------------------
BINBCD:
;-------------------------------------------------
;byte BINARY TO BCD
mov A,GapD ;count 2
mov B,#100 ;divisor
div AB ;A/B, Quo= A, Rem= B
mov GapA,A ;# of 100's
mov A,B ;remainder
mov B,#10 ;divisor
div AB ;A/B, Quo= A, Rem= B
mov GapB,A ;# of 10's

mov GapC,B ;# of 1's
BINB9: ret ;out of here
ž Ÿ

28
PROJECT 8 - SEVEN-SEGMENT DISPLAYS
Thought
You are looking,
but are you seeing?
Dr. Eden Ryl
Project 8: Seeing what is not there __
Purpose: To design an interface which is a display system.
To fetch data from the program ROM.
Preamble:
Utilizing a software design for circuits is usually preferred to a
hardware design. A software design provides more flexibility than
does hardware. Software is very advantageous if the system is not
time constrained.
In this project build a 3-digit, 7-segment display. Only one byte may
be used to display LEDs. To display all the digits, use time
multiplexing.
The multiplex cycle must be very short, so that the human eyes will
not be able to see the blinking. The eye can discern any frequency
slower than 17 frames per second. Therefore, refresh the display
faster than 20 times a second. This task can be easily achieved, since
the uC is fast enough to process the output.

The software description of a digit can minimize the calculations
and simplify the system. The patterns of a letter for a 7-segment
display can be stored in a table. Use sequential characters (e.g. 0, 1,
2, 3, 4...a, b, c, etc.), so it will be easier to find any item. An easy
way to build a table is to program the data into the code memory.
Then the program has only to fetch the data from the table to display
a character.
The processor has capabilities of separate external memory for data
and for program code. To access code memory use the movc
command. This triggers the program storage enable not (/PSEN)
pin. To access data memory use the movx command. This triggers
the read not (/RD) or write not (/WR) pins. In both cases, the
address is stored in the data pointer (DPTR) register. The register
must be reinitialized every time it is used, if it has been modified
since the last access.
Plan:
In this project, implement a 3-digit, 7-segment time multiplexed
display system.
Preparation:
To drive the 7-segment displays, use a 74573 latch for the data
buffer. Connect the output of the latch to the seven pins of the
display. Repeat with a parallel connection to all other 7-segment
displays.
The common (cathode or anode) of each display must carry the
current of all the segments. Therefore, it should be connected to a
driver other than the microcontroller.
One option is to use an open collector driver. The 7406 hex inverter
is a popular open collector chip. The open collector must be
connected to 5 volts through an external pull-up resistor. Typically

Chapter 28 Project 8 - Seven-segment Displays 195
use 2.2 kOhm resistor with adequate power rating to handle the full
load current.
Alternatively, a field effect transistor (FET) can be used. These have
current capabilities in the 30 A range.
The driver input is a single bit that can come from any port or
memory-mapped I/O location. This becomes the digit select line.
Test your display system by illuminating simple characters such as
1 or 7.
Procedure:
Write a simple multiplex display program that has the following
abilities. First, turn off all digits. Second, select the first display.
Third, send data for the first display. Fourth, create persistence by a
very short delay of NOPs. Fifth, turn off all displays. Select the next
digit. Send data for the next display. Create persistence by a very
short delay of NOPs. Repeat steps five through eight for as many
displays as required. As the last step, turn off all displays.
Write an input routine that calculates a hexadecimal number in the
range of 00 – 0FFh, based on switch inputs. Caution: be sure that
stable data is on the input before starting the calculation. Several
techniques can be used. Software debounce is the easiest.
Separate the hex number into two bytes or digits. Do a table lookup
to convert the digits to 7-segment values. Send the digits to the
multiplex display program. Display the value for about two seconds.
Convert the hex number to a binary coded decimal value in the
range of 0 - 255. This will make separate bytes for each digit of the
decimal number. Do the same table lookup to convert the digits to
7-segment values. Send the digits to the multiplex display program.
Display the value until the input routine calculates a new number.

Presentation:
Show at least 4 hex-number conversions. The numbers may be
randomly chosen.
ž Ÿ
The latch key out and latch display routines are given in the chapter
on expansion latches.
;Program: ModSeven.ASM
; Tulsa, OK, USA
; mod@superb.org
;Purpose:
; Data is stored in GapA, GapB, GapC.
; Data is displayed on 3 7-segment displays.
; Display-out latch shows the data.
; The 3 displays are selected by port 3.2, 3.3, &
; 3.4
;
;#################################################
; ASSIGNMENTS
;#################################################
GapD equ 37H ;general purpose variables
GapC equ 36H
GapB equ 35H
GapA equ 34H
LoopC equ 07H ;loop counter

;R0 equ 00H ;destination indirect addr
P35 equ 0B5h ;switch input
SegA equ 90h ;Port 1.0
;#################################################
; PROGRAM
;#################################################
org 00h
START: ljmp INITIAL
org 0033h
;-------------------------------------------------
org 0080H ;get past interrupt
INITIAL:
;-------------------------------------------------
setb P35 ;make input
lcall HELLO ;start display
lcall SEVBCD ;BCD to 7 segment
;-------------------------------------------------
MAIN:
;-------------------------------------------------
lcall SEVSEG ;output seven segmen
MAN9: jnb P35,MAIN ;Repeat
;-------------------------------------------------
HELLO:
;-------------------------------------------------
; Preload a value into the general purpose bytes.
;PRELOAD
mov GapA,#0Fh ;data byte
mov GapB,#0Ah
mov GapC,#0Bh
ret

;-------------------------------------------------
SEVBCD:
;-------------------------------------------------
; Convert the binary coded decimal to seven seg.
; Use a table look up.
;INITIAL
mov DPTR,#TabSeven ;seven seg table
mov R0,#GapA ;base of digits
mov LoopC,#3 ;number of digits
;CONVERT
SBCD1: mov A,@R0 ;get BCD
movc A,@A+DPTR ;table offset to A
mov @R0,A ;replace w/ 7 seg
;NEXT DIGIT
inc R0 ;next
djnz LoopC,SBCD1 ;>=0, repeat process
ret ;else, exit
;-------------------------------------------------
SEVSEG:
;-------------------------------------------------
; Seven Seg is a routine that operates in 4 steps
; 1. Select the digit stored in GapS
; 2. Select the digit to turn on
; 3. Send the data
; 4. Wait briefly for persistence of the led.
; 5. Repeat
;INITIAL byte LOCATE
mov R0,#GapA ;least sig display
;SELECT LINES
clr SegA ;turnoff all selects
clr SegB
clr SegC
;DIGIT 1
setb SegA ;Bit is hi, select A

lcall SEVOUT ;output & delay
clr SegA
;DIGIT 2
;DIGIT 3
;TERMINATE
ret ;back to Hotlanta
;#################################################
; TABLES
;#################################################
TabSeven:
;-------------------------------------------------
; Seven-segment display
db 3Fh ;0
db 06h ;1
db 5Bh ;2
;complete the table
;*************************************************
end ;Program end
ž Ÿ

29
MATRIX SCANNING
Thought
Integrity is
recognizing someone’s short coming
and saying nothing about it to anyone.
Rosemary Durham
Matrix inputs ___________________
Although the processor is extremely powerful, one of the limits of
any computer is how many things can be connected. Several
techniques for expanding those options have been addressed.
An earlier chapter discussed expanding the number of devices by
using expansion memory and latches. Another chapter discussed the
challenges of getting more devices connected to the microprocessor
with the same number of pins. The concept is called multiplexing.
This chapter discusses another technique for obtaining more
information with limited connections. It employs a matrix network
to literally multiply the effect of pins.
Perhaps the most common method of data entry is a keyboard. A
personal computer (PC) has over 100 keys. To minimize the number
of wires, a microprocessor is embedded in the keyboard. This
converts all the keys to just a few lines.

Chapter 29 Matrix Scanning 201
Although the same concept is feasible for control systems, seldom is
that much data required. Another consideration is the system is
often constrained to a small size. Therefore, a much more common
arrangement is to use a keypad similar to a telephone. In some
cases, the keypad may only have three or four keys.
Regardless of the number of keys, it is necessary for the processor
to decipher which key is pushed. The simplest technique requires no
logic. One wire connects to each key and the electrical common
provides a return path.
# wires = # keys + 1
This technique would require way too many wires, a huge cable,
and an excessive number of pins on the microprocessor. An
alternative is to use a matrix. A wire is connected to each row and a
wire is connected to each column.
# wires = # rows + # columns
Compare the techniques for a 3x4 telephone keypad with three
columns and four rows for a total of 12 keys.
# wires = 12 +1 = 13 ;no logic network
# wires = 3 + 4 = 7 ;matrix network
A 4 x 4 keypad will add four more keys put will require only one
more wire. That is a huge gain.
The matrix requires rather extensive logic to decode the network.
Software is always preferred to hardware. Less real estate is
involved and the coding must only be done one time.
Contact arrangement _____________
The switch contacts for a matrix connect between a row and a
column. Whenever a particular switch is depressed, the row and

column become shorted together. There is only one combination of
row and column for any switch.
Since the keypad is a switch, it is connected exactly like any other
switch. A pull-up resistor is connected to the open side of the
switch. That contact becomes the input to the computer. By
convention, the rows will be used as inputs. As a result, a pull-up
resistor should be connected between each row and VCC. Therefore,
when a row is read, the value is ‘1’ if no key is pressed.
VCC Output
Input
The columns are applied as outputs. When a zero or ground is
placed on a column, then the switch looks like a traditional
connection.
The columns are configured to write through a latch. The output
enable not can be permanently active by connecting it to ground.
The latch is set up with data when the latch enable pin is asserted
high by the processor. Then, when the pin is asserted low, the latch
traps the data.
The rows are configured to read through a latch. Since this is an
input, the latch enable line can be pulled high. When the output
enable not (/OE) line is asserted low by the processor, the data will
be input to the computer.

Conflicts ________________________
If the project were only that easy, anyone could do it and there
would be no need for a design engineer. However, there are several
opportunities for improvement.
First, switches will bounce when they are exercised. This will cause
the impression of multiple key depressions. Hardware triggers can
be used, but they are expensive. Software is the best solution. The
logical exclusive-or instruction makes the problem almost trivial.
The chapter on switch inputs and logic gave an illustration of the
procedure.
Another decision process involves more than one key depressed at
the same time. Resolution of the keys can be made in at least four
ways.
1. n-key lockout on first. The first key depressed is recognized and
all others are locked out. A variation is to only recognize the
first key decoded.
2. n-key lockout on last. The last key released is recognized and all
others are locked out.
3. n-key rollover. All keys are recognized and placed in a first in-
first out (FIFO) memory until the computer can accept the data.
4. alternate key. The combination of keys are recognized as
another key. A common example is shift and alternate in
combination with the others.
Key debounce ___________________
The keys are typically a mechanical switch contact. As such, they
will vibrate when depressed. Because the computer is faster than the
bounce cycle time, the processor will detect multiple changes and
interpret this as multiple keys pressed. Hardware can be used to
filter the bouncing. However, software is much cheaper in terms of

real estate and investment. Although the topic was introduced with
switches, more detail is investigated now.
Many procedures can be used to eliminate the bounce. The simplest
is to use a logical EXCLUSIVE-OR to see if a bit has changed from
the last test. If it has changed, the bit is unstable, therefore, invalid.
If the bit has the same value on two passes, it is assumed it is stable
and has been debounced.
The routine is a very elegant procedure to see if an entire byte is
stable. This obviously could be modified for individual bytes.
However, for keypad decoding, we need a stable bit.
Three values are used – the input, previous character, and
debounced character. The previous character is updated with the
input on every pass through the test. However, the debounced
character is updated only on a stable input.
;-------------------------------------------------
DEBOUNCE:
;-------------------------------------------------
;CHECK CHANGE BY xrl
mov B,A ;hold the input
xrl A,CharP ;exclusive or
jnz DEBN1 ;<>0, so a change
;KEEP DEBOUNCED
mov CharD,B ;0=no change, CharD
DEBN1: mov CharP,B ;not debounce
ret
Decipher _______________________
Jethro Bodine on the old television program Beverly Hillbillies often
referred to ciphering. He was bragging about arithmetic calculations
where he took inputs and determined the results. The deciphering
procedure is the opposite. It is taking the results and determining the
inputs. The objective of this procedure is to decipher a key based on
the inputs.

The row lines are connected through a pull-up to give a one from
the 5 volts. Columns are used to output a zero from the processor.
These are not connected to a pull-up.
After a zero is sent to a column, the row is read. If the row is one,
then a key is not pressed. If the row is zero, then a key has been
pushed.
A single port can be used for a 4 x 4 keypad with sixteen keys.
Columns are connected to the high nibble and rows to the low
nibble. Even when a single port is not used, the same pattern is
employed.
Complete solution ________________
The process is a very sophisticated procedure that can be expanded
to any number of keys. It solves the problem of multiple keys and
uses highly developed decoding techniques. It is very powerful
code. Unfortunately, with more power, comes more complexity.
The first step in implementation of the keypad is obviously
construction. The next step is a very important tool to minimize
wiring problems. Run a KEYTEST program similar to the one
shown. The column output and the row input can be easily
modified, if memory mapped is not used as shown in the example.
The KEYTEST routine will check for proper wiring and
connections. After verification, implement the decipher routine in
steps, so it can be properly debugged.
Connections _____________________
The rows can be connected to ports or latches that are associated
with a port or memory-mapped. The columns can be similarly
connected. The only thing that changes with the connection is a few
lines of input/output code, which is used to select the control lines

for the latches. The remainder of the process remains the same,
regardless of the I/O connection.
For the particular process at hand, the row lines connect to a latch
that uses memory mapping. The technique has been discussed
extensively in other sections. Columns are connected to a separate
latch.
Memory mapped I/O shares space with the upper byte of memory
addressing. However, because of the connection configuration, there
is not a conflict. Address A15 selects the MMIO, so this line is not
connected to memory.
A15 A14 A13 A12 A11 A10 A9 A8
Mmio
The read and write address can be the same, since the address is
ANDed with the read not or write not line from the processor. The
addresses are selected using the data pointer register (DPTR). DPTR
is a sixteen-bit location that is comprised of a high (DPH) and low
(DPL) register. The high is associated with port 2 while the low is
allocated to port 0.
The programmable logic device (PLD) firmwire enables the key
row read latch at address 8001h. If an alternative is desired, the
firmware can be changed. Another technique is to use the MMIO
line from the PLD, then AND the line with an address that can be
decoded from the microprocessor lines.
Test code _______________________
The test code illustrates the basic concepts of deciphering a matrix.
The subroutine will determine that some key has been exercised.
However, it does not tell which one. That requires considerably
more logic and the topic will be addressed later. RUN KEYTEST
TO VERIFY THE KEYPAD WIRING!

;-----------------------------
KEYTEST:
;-----------------------------
; Output '0' to columns.
; Read rows, if any is '0', a key is pressed.
;COLUMN OUT
mov DPTR,#8001h ;keypad address
mov A,#0 ;key enable
movx @DPTR,A ;output columns
movx A,@DPTR ;read rows
;DECIPHER push
anl A,#0Fh ;mask hi nibble
cjne A,#0Fh,KEYT2 ;<>1, so key pushed
sjmp KEYTEST ;no key
KEYT2: cpl P35 ;LED change=key push
lcall WAIT ;delay multiple keys
sjmp KEYTEST
Decode flowchart ________________
Several steps are required to decode a keypad. The overview gives a
sequence of items required.
1. Read all rows simultaneously.
2. Debounce the row values into CharD.
3. Check CharD to see if any row is pushed.
4. If none pushed, clear hold flag, and exit with “None”.
5. If pushed, check if held down from a previous pass.
6. If held, exit.
7. Else, initialize variables.
8. Begin loop based on number of columns.
9. Write a zero to the first column.
10. Read rows.
11. If no row pushed, go to next column in loop at 8.
12. If pushed, decipher which row.
13. Check if multiple rows.
14. If multiple, then exit.
15. Calculate value of column, 0-3 and multiply by 4 for rows.

16. Add row number to get assigned number for key pushed.
17. Continue loop for next column.
18. Check if number is in legitimate range.
19. Convert key number to ASCII value, via table lookup.
20. Set flag for key is held.
21. Return
The list documents that the procedure is as complex as any routine
encountered. No specific item is difficult. However, the number of
items and the interaction makes a large number of operations.
The list is also shown in a flowchart format below. The information
is identical.

Read all rows
Debounce row values
Any row pushed? no
Clear key held flag
previous key held? yes
initialize variables
loop entry
write zero to first column
Read row
any row pushed? no
yes
decipher which row
multiple rows? yes
column value=col # * 4
add row number
decrement column number
last column?
yes
column number<0Fh?
CharK = None
convert key # to ASCII
Serial input
Set key held flag
return

Keys procedure __________________
The Keys routine directs the traffic for determining the key that is
pressed. The first task is to check if any key is pressed. To do so,
read the row and debounce the value. Enter the routine with the
value in the accumulator. Perform an exclusive-or with the previous
value, CharP. Exit the task with the debounced value in CharD.
Then use the procedure in the flow chart to decode the row and
column value. The routine waits until a key is pushed before
returning.
Row lines are connected to a pull-up to give 5V. The routine outputs
a zero to the columns. Then read the rows to determine if any key is
'0'. If so, it is pressed.
Rows are numbered 0,1,2,3. Columns are numbered 0,1,2,3.
However, columns are valued based on the number of rows.
Columns value = column number * number of rows
The normal configuration is four rows.
A = KeyCol * 4
Therefore, columns are valued 0,4,8,12. Then the row number is
added to give a location associated with each key. A table is used to
decode the value of each key location to ASCII.
For example, the key associated with the number “8” is pushed.
This is in column 1, row 2. The deciphering will return a column
location of 4 plus a row location of 2 for a key location of 6. The 6
is the offset pointer into a table location that will return andASCII 8.
If no key has been pressed, the procedure contains a serial routine at
the end. The routine checks if any external key has been pressed, in
lieu of the keypad.

The table look-up is dependent on the wiring of the columns and
rows. If an alternate hook-up is used, the table sequence will need to
be modified.
Simple solution __________________
A very simple sequence of code can be used when only a limited
number of keys are involved. This routine does not have a debounce
routine, keyheld detection, multiple key count, range check, or serial
check. These can obviously be added. However, at that point, the
routine begins to be very close to the full blown exemplar program
referenced in the project.
;#################################################
;CONSTANTS
;-------------------------------------------------
None equ 0FFh ;blank key
AdKey equ 8001h ;mmio latch
;-------------------------------------------------
;DEFINED VARIABLES
;-------------------------------------------------
;KEYS
KeyCol equ 17h ;present column number
;*************************************************
; KEYPAD
;*************************************************
KEYS:
;-------------------------------------------------
; Keypad is used to decode a matrix set of keys.
; A 4X3 keypad can be used.
;
; Columns connect to a latch, in the upper bits
; Rows connect to a latch, in the low bits.
;
; Column latch bits |7|6|5|4|3|2|1|0|
; Key column |0|1|2|3|-|-|-|-|
;

; Row latch bits |7|6|5|4|3|2|1|0|
; Key row |-|-|-|-|3|2|1|0|
;
;COL 0
mov A,#01111111b ;Column 0
mov KeyCol,#0 ;column location
lcall KEYROWRD ;input row
cjne A,#0Fh, KEYP7 ;<> F, so pushed
;COL 1
;COL 2
mov A,#None ;null key
sjmp KEYP9
;COLUMN VALUE
KEYP7: add A,KeyCol ;A=KeyCol*4 + KeyRow
mov DPTR,#TabKey ;start of table
movc A,@A+DPTR ;conversion
;TERMINATE
KEYP9: ret ;back to message
;-------------------------------------------------
KEYROWRD:
;-------------------------------------------------
; Output column, read row
; Determine row number
;COL OUT, ROW IN
KEYROW1: mov DPTR,#8001 ;pointer to latches
movx @DPTR,A ;read row

cjne A,#0Fh, KEYR9 ;<> F, so pushed
;INITIALIZE
clr C ;initialize test bit
mov B,#0 ;row number
KEYR4: rrc A ;row bit to C
jnc KEYR3 ;C=0,row down
inc B ;next row
sjmp KEYR4 ;next bit
KEYR3: mov A,B ;row number
;TERMINATE
KEYR9: ret ;back to message
;-------------------------------------------------
TabKey:
;-------------------------------------------------
; Three column keyboard
db ‘147*2580369#’

Circuit: keypad __________________
Column
Latch Q7
Col0
Q6
Col1
Q5
Col2
D0
Row0
D1
Row1
D2
Row2
D3
Row3
Row
Latch
ž Ÿ

30
PROJECT 9 - KEYPAD
Thought
The key to success?
The Golden Rule
Project 9: Debounce & matrix inputs
Purpose: To implement a matrix scanning mechanism.
To design a keyboard for a computer.
Preamble:
A keyboard is the most common input device for a computer
system. The common ASCII keyboard usually uses a dedicated
microcontroller to implement the task. The microcontroller reduces
significantly the number of ICs and the resulting cost. The decoded
value of each key is transferred serially to the main processor.
There are several problems with a mechanical keyboard that the
interface must resolve. For this project, the microcontroller is
expected to do all these tasks.
Any mechanical switch will bounce when it is pressed. A software
debounce may be implemented for most systems. Any actual real
time function is generally quite slow. For example, a very fast typist
will only type about 120 words per minute or 10 new keys per
second.

Second, the priority of the pressed keys must be established. In the
dedicated microprocessor system, the priority is determined by a
first-come, first-serve basis. Hence, the microprocessor will also act
as a buffer. In this project, priority buffering is not a problem, since
the system will not be dedicated to the keyboard. Nevertheless, a
simple buffering will be implemented for future use.
Plan:
Implement a 16-key keyboard system, using a 4 X 4 keyboard
matrix. Only one eight bit I/O bi-directional port is required, if using
ports. By simply masking one of the columns, a 3 X 4 keypad can
be used.
Preparation:
Use one of the microprocessor ports, if it is available. Alternatively
use a latch for the row and a latch for the column at a memory-
mapped I/O location.
Procedure:
Send a 4-bit output to the keyboard. Then read another 4-bits from
the keyboard to determine which key is pressed.
Implement the key debounce in software. This is easily done by
using an exclusive or with the previous key. Use a loop until a
stable value is achieved in two successive tests.
Once a stable value is detected, determine the column involved and
the row selected. Use the column with row information to decode
the key.
Perform a table look-up of the key value to determine the ASCII
equivalent. If no key was pushed, return a null value.

Chapter 30 Project 9 - Keypad 217
Presentation:
Demonstrate the keyboard operation by implementing a HEX
keypad (0 - F hex). Display a corresponding key value on the 7-
segment display or the serial port to the personal computer.
ž Ÿ
Most projects have been presented with an example code that had
additional items to be developed. In contrast, this section of code is
complete. It is necessary because of the involved process.
;-------------------------------------------------
;Program: MODkey.ASM
;Update: 26 July 2004
;
; Tulsa, OK, USA
; mod@superb.org
; www.DrMod.com
;
;Purpose:
; keypad input.
;
;Baud: 9600

;#################################################
;
; ASSIGNMENTS
;
;#################################################
;CONSTANTS
;-------------------------------------------------
;SYMBOLS
None equ 0FFH ;blank key
AdKey equ 8001h ;mmio latch
;-------------------------------------------------
;DEFINED VARIABLES
;-------------------------------------------------
;KEYS
KeyCol equ 17H ;present column number
KeyBit equ 16H ;byte moves a bit w/ column
KeyRow equ 15H ;row number pushed
KeyMul equ 14H ;multiple key count
CharK equ 13H ;key input character
CharD equ 12H ;debounced
;CHARACTERS, HOLD, COUNT
CharP equ 0FH ;undebounced previous input
CharL equ 0EH ;character to LCD & Serial
;
;R6 equ 06H ;size
;R5 equ 05H ;carry in multiply, GP
;-------------------------------------------------
;BITS ASSIGNMENTS
;-------------------------------------------------
;AT RAM byte 20H
FgKeyH bit 00H ;flag key held down
;#################################################
;
; PROGRAM
;

;#################################################
org 00H
START: ljmp INITIAL
org 0033h
;-------------------------------------------------
INITIAL:
;-------------------------------------------------
;INITIALIZE
lcall UART ;config & start UART
;-------------------------------------------------
MAIN:
;-------------------------------------------------
; The procedures are to input a key, send it on
; serial, and do a line return.
;PROCESS
lcall KEYS ;check keys
mov A,CharK
lcall SEROUT ;A =message value
;*************************************************
; KEYPAD
;*************************************************
KEYS:
;-------------------------------------------------
; Keypad is used to decode a matrix set of keys.
; Scan is a routine to check if any key is
; pressed. To do so, read the row and debounce.
; Exit with the debounced in CharD and the
; previous undebounced in CharP
;
; For a keypad, decode the row and column value.
;
; The routine waits until a key is pushed before
; returning.

;IS ANY KEY PUSHED?
lcall KEYALLRD ;all cols=0,read row
lcall DEBOUNCE ;CharD=debounced
lcall KEYPAD ;decipher key
;LAST KEY
mov A,CharK ;last key
cjne A,#None,KEYS9 ;<>none, have a key
sjmp KEYS ;=none, get a key
;TERMINATE
KEYS9: ret ;back to message
;-------------------------------------------------
KEYPAD:
;-------------------------------------------------
; KEYPAD is a routine to input buttons pressed on
; keyboard. A 4X4 keypad can be used.
;
; Columns connect to a latch, in the upper bits
; Rows connect to a latch, in the low bits.
;
; Column latch bits |7|6|5|4|3|2|1|0|
; Key column |0|1|2|3|-|-|-|-|
;
; Row latch bits |7|6|5|4|3|2|1|0|
; Key row |-|-|-|-|3|2|1|0|
;
; Row lines are connected to a pull-up to give 5V
; Output '0' to columns.
; Read rows, if any key is '0' it is pressed.
;
; A table is used to decode the value of
; each key to ASCII.
;
; If a key has not been selected, try for a
; serial input.
;ANY KEY BEEN push?
mov A,CharD ;CharD debounced key
cjne A,#0Fh,KEYP3 ;row<>1, one pushed
clr FgKeyH ;fg no key held down
sjmp KEYP8 ;return a null

;INITIALIZE
KEYP3: jb FgKeyH,KEYP8 ;key held, exit
lcall KEYINIT ;initialize
;SEND EACH COL A 0
KEYP4: lcall KEYROWRD ;read row
cjne A,#0Fh,KEYP5 ;0 in row pushed
sjmp KEYP6 ;no 0 in this row
KEYP5: lcall KEYROWQ ;query, KeyRow=row
mov A,KeyMul ;check multiple key
cjne A,#1,KEYP8 ;<>1, so null
;COLUMN VALUE
lcall KEYCOLQ ;query, KeyCol=col
add A,KeyRow ;A=KeyCol*4 + KeyRow
mov CharK,A ;upgrade the charac
;NEXT COLUMN
KEYP6: djnz KeyCol,KEYP4 ;remain col to write
;CHECK RANGE ERROR
mov A,#0Fh ;largest value
clr C
subb A,CharK ;key
jc KEYP8 ;invalid key
;CONVERT TO ASCII
mov A,CharK ;lookup last value
mov DPTR,#TabKey ;start of table
movc A,@A+DPTR ;conversion
mov CharK,A ;save key
setb FgKeyH ;flag key held down
sjmp KEYP9 ;exit
;NULL RESPONSE
KEYP8: mov CharK,#None ;nothing pushed
;SO CHECK SERIAL
jnb RI,KEYP9 ;no serial either
clr RI ;receive interrupt
mov CharK,SBUF ;serial input

;TERMINATE
KEYP9: ret ;back to message
;-------------------------------------------------
KEYINIT:
;-------------------------------------------------
; Initialize is a routine to set variables.
; A 4X4 keypad can be used with a single byte.
;
; KeyCol present column number
; KeyMul multiple key count
; KeyRow row number pushed
; KeyBit byte moves a 0 bit for the column move
;ANY KEY BEEN push?
mov KeyCol,#3 ;column number 0
mov KeyMul,#0 ;multiple key count
mov KeyRow,#0 ;row number 0
mov KeyBit,#01111111b;first column w/0
;TERMINATE
ret ;back home
;-------------------------------------------------
KEYALLRD:
;-------------------------------------------------
; Row Read is a routine that sends a 0 to columns
; It reads the row.
; For ALLRD, 0 is sent to all columns.
; For ROWRD, 0 is sent to each column
; successively
;
;SEND ALL 0
mov A,#00001111h ;all col=0
sjmp KEYROW1 ;COL OUT, ROW IN
;
;bit TO SEND
KEYROWRD: mov A,KeyBit ;0 bit is column
;SAVE FOR NEXT bit
rr A
mov KeyBit,A ;keep it
rl A ;use it

;COL OUT, ROW IN
KEYROW1: mov DPTR,#AdKey ;MMIO key address
movx @DPTR,A ;read row
;TERMINATE
ret ;back to message
;-------------------------------------------------
KEYROWQ:
;-------------------------------------------------
; Row Calculate determines the row number that is
; pressed. RowNum = 0,1,2,3
;
; A counter is set if multiple rows are pushed.
;
; The routine can handle 4 rows.
; If fewer rows are used, simply change the jump
; to the corresponding number of rows.
; If a 1 row pad is used, sjmp to ROW1.
;INITIALIZE
clr C ;initialize test bit
;; sjmp KEYR3 ;pad has only 3 rows
;IS THIS ROW=0
jc KEYR3 ;C=1,row not down
inc KeyMul ;multiple key count
mov KeyRow,#3 ;row number
;IS THIS ROW=0
;IS THIS ROW=0

;IS THIS ROW=0
sjmp KEYR9
;TERMINATE
KEYR9: ret ;back to message
;-------------------------------------------------
KEYCOLQ:
;-------------------------------------------------
; Column calculate determines the column that is
; pressed, KeyCol. Then it calculates the value
; for the key.
;
; Rows are numbered 0,1,2,3
; Columns are numbered 0,1,2,3
; Columns are valued 0,4,8,12.
; Columns value = column number * number of rows
; With four rows A = KeyCol * 4
;CALCULATE
mov A,KeyCol ;column w/ 0
dec A ;column # inc by 1
mov B,#4 ;qty of rows
mul AB ;A = column value
;TERMINATE
ret ;back to message
;+++++++++++++++++++++++++++++++++++++++++++++++++
;TABLE SETUP - KEYPAD CONVERSION
;-------------------------------------------------
; Tables are used to convert between formats.
; These include keypad & ASCII.
; The table pointer will show a decimal result
; that corresponds to a column/row location.
;-------------------------------------------------
TabKey:

;-------------------------------------------------
; Three column keyboard
db ‘147*2580369#’
;*************************************************
end ;Program end
ž Ÿ

31
LIQUID CRYSTAL DISPLAY
Thought
3C procedures:
Command, control, communications.
Different display systems __________
Numerous ways have been used to display information. The first
project started with a single light emitting diode (LED) for
information display. This is very basic and simple to use. As a
result, it has very simple information. Progressively, more LEDs
were added to give more info.
Serial communications were added. This permitted very clear,
detailed messages, However, it requires another computer or display
device.
Then seven-segment displays were implemented. The hardware and
software require multiplexing to get very many characters. These
characters are easy to read; however, the number of letters are
limited. All numbers can be seen, but some letters are not possible
and others can only be a lower or an upper case.
A complete text message and limited graphics can be shown using
liquid crystal displays. These are the most comprehensive imaging
devices for their size. As a result, they are very popular. One slight
drawback is the cost. The units are several times more expensive

Chapter 31 Liquid Crystal Display 227
than using a limited seven-segment display. As a result, very cost
competitive items will skip the displays.
High end systems will use a cathode ray tube (CRT) for imaging.
The size makes them very useful for showing large amounts of data.
However, the size and power consumption is a serious drawback for
many projects.
A compromise device for some specialty projects is a liquid crystal
display with a comparatively large screen. The layout is a large
matrix, therefore more imaging can be shown in addition to text.
These can be developed into quite sophisticated devices, so they are
infrequently used for simple control projects.
LCD variations __________________
Liquid crystal displays commonly come in one, two, or four line
versions. The length of line is typically sixteen or twenty characters.
Virtually all models are based on a common design with the same
microcontroller driver. As a result, the pin-outs are identical. A few
models simply renumber the pins from the opposite direction.
The instruction set to all liquid crystal displays is common between
manufacturers. The standard character set is built into the onboard
processor. Graphics can be created for the display, but these
typically require bit mapping.
Connections _____________________
Because of the control sequence, the LCD appears somewhat
tedious. There are three groups of routines.
1. LCD control lines.
2. LCD commands: initialize, instruction, data, and busy.
3. LCD communicate: display message.
The LCD has fourteen pins for connecting power, controls, and
data.

Pin Function
1 Vss, ground
2 Vdd, 5 V
3 Vo, power for contrast
4 Reg. Select (RS) 0= instruction, 1= data
5 RW, 0= Write, 1= Read
6 Enable
7-14 DB0-DB7 data bits
Connect contrast to ground for maximum viewing intensity. The
board has an elegant circuit for contrast control based on the
temperature sensed by a thermistor. For many applications, the
circuit can be bypassed with a grounding jumper.
Control ________________________
The liquid crystal display has three control lines- enable (pin 6),
read/write not (pin5), and register select (pin 7). It is wise to have
the enable line asserted low before the other control lines are
asserted. The register select and the read/write are activated by
software control. The commands are executed on the low to high
transition of the enable line. The high also allows data to be set-up.
Data is transferred with the enable line high.
Display of information requires initiating a control sequence. First
the enable line is pulled low. Then the control operation is
implemented. Then the enable line is pulled high. Then the data is
sent. Then the enable line is pulled low.
The timing diagram illustrates the relationship between the lines. A
detailed discussion of timing is included in a chapter of the
reference section.

Control via port __________________
The LCD has a microcontroller on board. It expects to communicate
with another computer. The simplest implementation uses port pins
for the control lines. This is very effective if there are available pins.
Any delay necessary is created in software.
D0-7
Port 0 Data
AD0-7
P1.0 4 RS
P1.1 5 RW
P1.2 6 En
microprocessor LCD
Regardless of the control line connection technique, the same
sequence must be exercised for the control to activate properly. The
timing diagram illustrates that the enable line must be taken through
multiple transitions for control and data transfer.
The enable line is first pulled low before the control lines are
asserted. Then the register select line is asserted high or low,
depending on the function for the LCD. Next the write line is pulled

low. On some versions, these two lines can be asserted
simultaneously.
The controls are activated when the enable line transitions to high.
The data is transferred between the microprocessor and the LCD on
the high to low transition of the enable line.
Control via latch _________________
The module is often physically wired through an expansion latch as
discussed in an earlier chapter. Therefore, this section will address
one procedure that has been used successfully. It will not be used
for project implementation.
The register has 8 lines that are configurable. The lines may be
shared with other output controls One arrangement that has been
used is illustrated. The upper four bits are for columns on a keypad.
The lower 3 bits are for the display select. Bit three is for RS485
select.
Flip/flop Function Control Bit
Q1 LCD 4 RS 0
Q2 LCD5 RW’ 1
Q3 LCD6 Enable 2
Q4 RS485 RE’/DE 3
Q5 Keypad Column 4
Q6 Keypad Column 5
Q7 Keypad Column 6
Q8 Keypad Column 7
Alternately, it can look like bit assignments.
[C C C C 3 E W S]
Other bits are on the latch (register) which will not be part of the
present operation. The value to the latch must be changed with
anl/orl or setb/clr to prevent these other bits from changing.

To ensure that other lines are not affected, the value placed on the
register can be maintained in temporary storage. However, since no
other output on the latch is processed while the LCD is being
driven, seldom is it necessary to hold the values. The exception
would be if other tasks were performed during the delays for the
LCD.
So look at a review of the process. The bits for the LCD control
lines are selected. The value may be saved in temporary storage.
The value is presented to the latch/register. The latch enable is made
active. The LCD then responds.
The next section illustrates the procedure that will be used for the
project implementation.
Control via PLD _________________
The LCD can be wired directly from a PLD a latch for the control
lines. The enable and register select lines are encoded in the PLD.
These are simply chosen based on a memory-mapped address.
D0-7
Port 0 Addr Data
AD0-7 A0-3
IO2 16 4 RS
1K
IO3 17 6 En
200pF
/Wr 17 10 I9 5 Wr
microprocessor latch PLD LCD
For most LCDs there must be a delay of at least 140 ns after the
controls are selected before the enable can be taken high. A RC
circuit is placed in the enable line to delay the response to the LCD
enable. Typical component values are 1 K Ohm and 200 pF.

Command ______________________
The LCD has a number of commands and instruction codes that are
required for display. The LCD can receive instructions for set-up,
send status, receive display data, or send the data currently in the
display back to the microcontroller. Two address lines (XX) select
the function.
The first table provides the instruction codes to the LCD.
Code Function
01 clear and home
02 home
04 cursor increment w/ data display
06 cursor decrement w/ data display
0A cursor on, flash off
0B cursor on, flash on
10 move cursor left
14 move cursor right
38 function set -8bit, 2line, 5x7dot
8X cursor position= DDRAM address
81 1st line, 1=home
C1 2nd line home
A1 3rd line home
E1 4th line home
The next table has the codes for the status response from the LCD to
the microprocessor.
Code Function
8X busy flag, X=current address
0X clear, ready to accept data
Two other commands are used for moving data.
Write Data: 8 bits of data to display

Read Data: 8 bits of data from display
All instructions for the LCD have a typical execution time of 40
microseconds in the module, except for two commands. Clear
Display and Return Home have a typical execution time of 1.64
milliseconds.
The LCD busy flag (BF) is clear when the module is ready to accept
another instruction. However, the busy flag (BF) cannot be read
until the first 3 LCD initialization bytes have been processed by the
module. Thus BF cannot be tested in the initialization subroutine.
The timing sequence of the module is slower than that of most
microprocessors. A memory-mapped interface limits the
microprocessor frequency to values less than about 9 MHz because
of timing. However, timing can be extended by using NOPs if a
faster clock is installed. Interfacing the LCD to a port allows the
processor to use any crystal frequency available, since additional
time is required for the instructions.
In addition, installation of different crystal frequencies dictates
changing the software counter used to create a 1 millisecond delay.
Insertion of additional NOPs may be necessary for crystal
frequencies above 40MHz.
Initialization ____________________
A rather extensive procedure is necessary to initialize the
microcontroller that is on-board the liquid crystal display module.
To take care of timing, several delays must be built into the
procedure. The busy flag cannot be used until after initialization.
Therefore, the delays must be created by software.
Moreover, as a practical matter, delay routines will be used in all
procedures rather than test for the busy flag. If the LCD module is
not plugged-in, then there is only a small delay. If busy were
checked, the process would hang, waiting for the response from the
LCD.

The initialization process is based on a Hitachi 44780 on-board
processor. Even if newer processors are used, the codes remain the
same to insure compatibility.
The references are Standish document A93531B page 2 and Hitachi
document CD-E613P 0591 page 90. The series of initialization
instructions are those required by the Philips data sheet.
First, after VCC is applied, allow 15ms for the module internal
initialization to be completed. The instruction mode is used for all
initializing. The instruction is sent, then there is a delay before the
next instruction.
Step Code Delay
1 wait 15 ms power on
2 30h 5 ms 8-bit
3 30h 100 us 8-bit
4 30h 100 us 8-bit
5 38h 200 us cursor
6 0Bh 100 us blink
7 01h 3 ms clear display
Cursor position __________________
The instruction codes contain items that are used for cursor
positioning. The first three bits of the message define the cursor line
that will be used. The remainder of the line contains the character
location on the line.
Line Hex Bits 765 43210
1 80 100 yxxxx
2 C0 110 yxxxx
3 A0 101 yxxxx
4 E0 111 yxxxx

The cursor position requires the most significant bit to be set. In
conjunction with the next two bits, the line is described. Then the
location is added to the line hex value.
The first position, 1, is home. Any location greater than fifteen
would require bit four to be set. When this is added to the line hex
value, it will be incremented by one. A few examples will clarify the
operations. Line one, home position would be 81h (80 + 1h). Line
one position 15 would be 8fh (80 + 0fh). Line one, position sixteen
would be 90h (80 +10h).
If using a 2 line display, send 80h to access the first space. Then
follow with the 16 or 20 characters for the length of the line. The
second line is accessed by sending 0C0h. Then the characters for the
length of the line.
A single line display is access exactly as a two line display. Send
80h to get to the first space. Follow that with the characters for one-
half of the line length. Then send 0C0h to access the second half of
the line. Follow that with the characters for the remaining half of the
line.
Message display __________________
The LCD is structured to display a string of characters. One
technique for sending a string is to define the characters with a
define byte ‘db’ directive. The procedure used in the project
includes 3 types of information with the line. First is the number of
characters to be sent. Second is the location for the starting
character, third is the string.
An alternative to counting the number of characters is to include a
termination character at the end of each string. Then test for the
termination character as the string is sent. An example is ‘0’. This
could be tested with a jz instruction.
ž Ÿ

32
PROJECT 10 - TEXT DISPLAY
Thought
Confidence is
the mental assurance
that something is true.
MOD
Project 10: Text message screens ___
Purpose: To display text in a friendly format.
To use a control register to expand functions.
Preamble:
A liquid crystal display (LCD) is the most common display
mechanism for simple text messages. The devices can have one,
two, or four lines of display. The length is typically 20 or 40
characters.
The device has an on-board processor and memory. Therefore, it has
extensive control capabilities. Data is sent to the display as an eight
bit byte. Instructions are also sent to control the machine. There are
three control lines - enable, read / write not, and data / instruction
not.
An extensive sequence of controls is required to initialize the
display. These need precise time delays before the next step.

Chapter 32 Project 10 – Text Display 237
Display of information requires initiating a control sequence. First
the enable line is pulled low. Then the control operation is
implemented. Then the enable line is pulled high. Then the data is
sent. Then the enable line is pulled low.
To prevent interference from other operations, interrupts are
typically disabled after the enable is high. The interrupts are enabled
after the enable is later pulled low.
The data pin outs and the control pin outs are standard among most
manufacturers. Some have a different power pin arrangement. There
are also lines for controlling the contrast. However, these are often
fixed. The board has an automatic circuit for contrast control. No
software is required for the contrast control. If the automatic feature
is not desired, it may be bypassed by grounding the contrast control
pin.
Plan:
Implement a multiple character liquid crystal display.
Preparation:
Connect the data lines through one latch. Connect the control lines
through another.
Procedure:
First, use an initialization routine to activate the LCD. Other
routines typically are clear screen, display data, advance cursor,
backup cursor, and flash cursor.
Next send data to the LCD. The information can come from the
keypad, serial line, or from internal tables.

Presentation:
Demonstrate the LCD operation by implementing a greeting
message. Then display a message from the keypad or from the serial
input.
ž Ÿ
;-------------------------------------------------
;Program: BiosLcd.ASM
;Update: 27 July 2004
;
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c)1991-2004. All rights reserved
;
;Purpose:
; A set of routines are provided to initialize
; and operate a liquid crystal display.
;
;Baud: 9600
;Handshake: not used at this speed
;
;#################################################
;
; ASSIGNMENTS

;
;#################################################
;CONSTANTS
;-------------------------------------------------
;
;R2 equ 02H ;loop, destination size,mat
Pio equ 0A0h ;Port 0
LcdRS equ 090h ;port 1.0
LcdRW equ 091h ;port 1.1, write not line
LcdEn equ 092h ;port 1.2
;#################################################
;
; PROGRAM
;
;#################################################
org 00H
START:
;-------------------------------------------------
ljmp INITIAL
;-------------------------------------------------
INITIAL:
;-------------------------------------------------
;
; The procedure initializes all common routines.
; It directs traffic for initiation messages.
;LCD INITIALIZE
lcall SCRINIT ;initialize screen
;-------------------------------------------------
MAIN:
;-------------------------------------------------
;
; The main procedure directs traffic.
; The main orchestrates execution of the
; subroutines.

;PROCESS
mov DPTR,#SerGreet ;get address
lcall BIOSLCD ;message headr RS232
lcall BIOSLCD ;second line
MAN9: ljmp MAN9 ;Halt
;-------------------------------------------------
BIOSLCD:
;-------------------------------------------------
;
; Send LCD message
; The first byte is the # of characters to send.
mov A,#0 ;read initial byte
mov LoopC,A ;set up loop
mov A,#1 ;length of message
movc A,@A+DPTR
mov CharL,A
lcall SCRINST
lcall DLYMIL
BIOL2: mov A,#2 ;get 1st message
mov CharL,A ;character lcd
;LCD data
lcall DLYMIC ;wait for next byte
lcall SCRDATA ;send out byte
inc DPTR ;restore offset
djnz LoopC,BIOL2 ;go thru loop
inc DPTR
inc DPTR
ret

;*************************************************
;
; LCD SETUP
;
;*************************************************
;
; Because of the control sequence, the LCD
; appears somewhat tedious.
;
; There are three groups of routines.
; 1. LCD control lines
; 2. LCD initialize, instruction, data, & busy
; 3. Message to display info.
;
; The LCD has 14 pins.
; 1 = Vss, ground
; 2 = Vdd, 5 V
; 3 = Vo, power for liquid crystal drive
; 4 = RS, 0= instruction, 1= data Reg. Select
; 5 = RW, 0= Write, 1= Read
; 6 = Enable
; 7-14 = DB0-DB7 data bits
;
; The LCD control lines are programmed as bits
; LcdRS = bit
; LcdRW = bit
; LcdEn = bit
;
;-------------------------------------------------
;SCREEN INSTRUCTION CODES
;-------------------------------------------------
; The following codes are required by the LCD
; display. The LCD can receive instructions for
; set-up, send status, receive display data, or
; send the data currently in the display back to
; the microcontroller.
; Two address lines (XX)select the function.
;
; Instruction Codes
; 01 ;clear and home
; 02 ;home
; 04 ;cursor increment w/ data display

; 06 ;cursor decrement w/ data display
; 0A ;cursor on, flash off
; 0B ;cursor on, flash on
; 10 ;move cursor left
; 14 ;move cursor right
; 38 ;function set-8bit, 2line, 5x7dot
; 8X ;cursor position= DDRAM address
; 81 ;each line has 40 chars, 1=home
; C1 ;41=2nd line home, 80+41=C1
;
; Status Response
; 8X ;busy flag, X=current address
; 0X ;clear, ready to accept data
;
; Write Data
; ;8 bits of data to display
;
; Read Data
; ;8 bits of data from display
;
; All instructions have a TYPICAL execution time
; of "40us" in the LCD module except instructions
; Clear Diplay and Return Home. These two
; have a TYPICAL execution time of 1.64ms.
;
; The LCD busy flag (BF) is clear when the LCD
; module is ready to accept another instruction.
;
; The LCD busy flag (BF) cannot be read until
; the first 3 LCD initialization bytes have been
; processed by the LCD module. Thus BF is not
; tested in this subroutine.
;
; Interfacing the LCD to port 1 allows the uC to
; use any crystal frequency available. A
; memory-mapped interface limits uC frequency to
; values less than about 9 MHz.
;
; Installation of different crystal frequencies
; dictates changing the software counter used to
; create a 1ms delay. Insertion of NOPs may be
; necessary crystal frequencies above 40MHz.
; See subroutines for control line values.

;-------------------------------------------------
SCRINIT:
;-------------------------------------------------
;--SUBS CALL -
; The routine initializes the Liquid Crystal
; Display. The busy flag stays busy until
; initialization is complete. The time is 15ms.
; Therefore delays must be built into the init
; routine before Busy can be used.
;
; Allow 15ms for Hitachi 44780 internal initial
; to complete after Vcc is applied.
; (Re: Standish document A93531B p. 2 &
; Hitachi document CD-E613P 0591 p. 90).
;
; The series of instructions are as required by
; Philips mfg data sheet.
;
; Instruction mode is used for all initializing.
;
; 232 must be on for LCD contrast control
; to get 10 volts.
;
; Delay is used rather than test for Busy line.
; If LCD is not plugged-in, then there is only
; a small delay. If Busy were checked, the
; process would hang.
;POWER ON RESET
mov LoopC,#20 ;20ms delay
SCRNI1: lcall DLYMIL ;1 ms routine
djnz LoopC,SCRNI1 ;loop
;CLEAR REGISTERS
mov CharL,#30H ;#1function set=8bit
lcall SCRINST ;send command
mov LoopC,#5 ;5ms delay
SCRNI2: lcall DLYMIL ;1 ms routine
djnz LoopC,SCRNI2 ;loop

lcall DLYMIC ;>100 microsec
;SET CURSOR
mov CharL,#38H ;#4funct set=8bit,2l
mov CharL,#0FH ;cursor on, blink
; mov CharL,#0CH ;cursor on,no blink
mov CharL,#01H ;display clear
lcall DLYMIL
lcall DLYMIL
lcall DLYMIL
;CHANGE MODE
mov CharL,#06H ;entry mode set
ret
;-------------------------------------------------
DLYMIL:
;-------------------------------------------------
; Delays shorter than the interrupt cycle are
; required. Since this is outside the normal
; program polling, hard calculated delays are
; used.
;
; The delay is based on clock cycles.
; mov=1, nop=1, djnz=2.
;
; The cycles in the loop are calculated.
; (1) for mov
; (2* count ) for both nop's
; (2* count-1) for djnz
; (2) for last jump

; = Total cycles
;
; Old crystal frequency was 7.3728Mhz
;
; Time = #states(12) * Total cycles/crys freq
; 11.059 MHz = 1.085 microsecond
;
; The inside loop has 925 cycles.
; At 11.059MHz this represents 1.003 ms
mov R2,#231 ;max loop=0
DLYM1: nop
nop
djnz R2,DLYM1 ;inside loop
ret
;-------------------------------------------------
DLYMIC:
;-------------------------------------------------
; Delay = 50 microsecond
;
; Delays shorter than the interrupt cycle are
; required. Since this is outside the normal
; program polling, hard calculated delays are
; used.
;
; The inside loop has 49 cycles.
; At 11.059Mhz,
; 1.085 microsec * this represents 53 microsec
mov R2,#12 ;max loop=0
DLYC1: nop
nop
djnz R2,DLYC1 ;inside loop
ret
;-------------------------------------------------
SCRINST:
;-------------------------------------------------
; The routine writes instructions to LCD.
; Write an instruction to the LCD requires
; control line RS= 0
; control line RW= 0
; Enable must be low before change R/W' or RS.

;
; CAUTION: A delay after data, before deselect
; can be interrupted and cause erratic data.
; disable interrupts until complete.
;CONTRL INSTRUCTION
clr LcdEn ;clr enable lo
clr LcdRW ;instruction& write
clr LcdRS
setb LcdEn ;set enable
;DISABLE INTERRUPTS
; anl IE,#7Fh ;disable main inter
;SEND INSTRUCTION
mov A,CharL ;select LCD data
mov DPTR,#8002h ;display latch
mov @DPTR,A ;send out
;CONTROL INSTRUCTION
lcall DLYMIL ;longer delay
;ENABLE INTERRUPTS
; orl IE,#80h ;enable all
ret
;-------------------------------------------------
SCRDATA:
;-------------------------------------------------
; The routine writes output data to LCD.
;
; Write data to the LCD requires
; control line RS= 1, data display on HD44780
; control line RW= 0, to write
; Enable must be low before change R/W' or RS.
;
; CAUTION: A delay after data, before deselect
; can be interrupted and cause erratic data.
;CONTRL INSTRUCTION

clr LcdRW ;instruction& write
setb LcdRS
setb LcdEn ;set enable
;DISABLE INTERRUPTS
; anl IE,#7Fh ;disable main inter
;SEND INSTRUCTION
mov A,CharL ;select LCD data
mov DPTR,#8002h ;display latch
mov @DPTR,A ;send out
;CONTROL INSTRUCTION
lcall DLYMIC ;short delay
;ENABLE INTERRUPTS
; orl IE,#80h ;enable all
ret
;#################################################
;
; TABLES
;
;#################################################
;
; Tables are used to convert between formats.
; These include kepad & ASCII.
;
; Tables are also used to carry display messages.
;
;
;+++++++++++++++++++++++++++++++++++++++++++++++++
;TABLE SETUP - MESSAGES
;-------------------------------------------------
; Predefined messages are in code memory.
; These are used in BIOSSER.
; Place at end of program code or org out of way
;
; The first byte is the number of characters.
; The second byte is the cursor location for LCD.

; The next bytes are the ASCII message.
;
; The format for intel assembler is illustrated.
;SerGreet: db 14, 0, 'uC BIOS 11V3- ';intel
;
; The format for Tasm is illustrated.
;SerGreet .byte 14
; .byte 0
; .text "uC BIOS 11V3- ";
;-------------------------------------------------
TABMESSG:
;-------------------------------------------------
SerGreet: db 8, 80h ,'uC Lcd--'
db 2,0C0h ,'hi'
;*************************************************
end
ž Ÿ

SECTION III - APPLICATIONS
ž Ÿ

33
INFRARED COMMUNICATIONS
Thought
Think excellent, do right,
obtain peace.
MOD
Local wireless ___________________
Wireless communications is highly desirable for many control
projects. Radio frequency communications is very effective, but
circuit design is tedious and requires additional technology.
Infrared communications is a very cost effective alternative.
Infrared is light that is slightly lower in frequency than the visible
range. It is a thermal energy that is easily generated and detected. It
does not suffer from electromagnetic interference. However, its
range is limited to line of sight. The receiver can experience
interference from other light sources. Pulsing light sources such as
fluorescent fixtures are a particular problem.
The technology is mature and widely used in consumer electronics.
The devices are used for video, audio, and control devices.
Transmitters are extensively available from numerous sources. As a
result, the processor project will be a receiver which can control
other devices.

Chapter 33 Infrared Communications 251
There are four basic technologies that are used. Sony uses a bit
width technology. The four - NEC, Apex, Hitachi, and Pioneer - all
use the same technology, which is another version of bit width. JVC
uses a complementary system which is a space width technology.
The major departure is the European model by Philips. It has a fixed
bit width and a fixed quantity of bits.
In its most basic implementation, an infrared signal is little different
from a red LED that is flashed on and off at some precise interval.
The receiver, then, is essentially a LED sensor that has a detector to
strip off extraneous carrier signals.
Philips protocol __________________
The discussion will focus on the Philips RC5 technology. The
circuit uses an infrared receiver with an integrated detector.
The infrared circuit consists of a carrier with a nominal 36-40 kHz
square wave. For a 40 kHz carrier, each cycle is 27 microseconds.
Then this is modulated with data bits. Therefore, a 40 kHz signal is
seen during on(0) and dc is seen during off (1). Because of the
variations in carrier frequency, it may be necessary to adjust the
cycle time slightly.
The protocol defines each bit as 1.728 ms long. Each IR string is 14
bits long. The bit stream is repeated every 130 ms, if the key is held.
This gives a long interval to resynchronize if there is a receive error.
Bit 1 is the most significant bit (MSB).
Bits Function Operation
1-2 AGC Automatic gate control, 10=good
3 check Bit flips with each new transmission
4-8 address 00= television
9-14 command Instructions out to latch

A data bit is both positive and negative. The first half cycle is setup,
then the second half is the state. On is 0 in the second half, while off
is 1 in the second half.
1 bit
RX data
0 bit
When the first falling edge is detected, this is used as the timer
frame. Wait 3/ 4 bit time to sync in the middle of the next bit.
The transmitter carrier frequency is turned on/off by the digital
signal. The receiver gets this modulated signal, which is a series of
carrier frequencies that are turned on/off. The detector removes the
carrier to leave the on/off pulses. The bit stream is fed into a single
bit input of the microprocessor.
bits modulated on carrier
bits with carrier removed
101 value of bits
The carrier frequency causes 32 pulses for each half cycle or 64
pulses per bit. The computer does not see these pulses, since they
are filtered by the detector.
This program has a wait cycle at the start. It is looking for the first
transition in the bit stream. A better approach would be to activate
the interrupt on INT1.
The first or start bit is always on (0). The second is also on (0).
These are the automatic gate control (AGC).
The third bit is a check (CHK). It toggles each time a new key is
pushed.

Next, follows 5 address and 6 command bits. The output should
remain on the latch for a short time to prevent glitches in data.
Check the first two bits for on. If not, wait more than 15 bit times to
begin a new sample.
Detected string __________________
The address is decoded to determine the equipment type. The
television remote is address 00. However, that is a code 10 at the
universal programmable remote, because of the detection circuit
waiting for a falling edge. There are addresses associated with other
devices as shown in the table.
Address Equipment
0 Television
2 Tele text
5 Video recorder
7 Experimental
16 Preamplifier
17 Receiver / tuner
18 Tape / cassette recorder
19 experimental
A command is associated with each piece of equipment. A table
lookup is used to convert the command to a value. The decode
command can be used to direct different operations. The table
shows a correlation for each number. It also gives an 8-bit value that
can be output on a latch.
Command bit 7 is 0 when a valid key is pressed. Otherwise it is a 1
for undefined values. That bit can be used to recognize values.
When key zero is pressed, all other bits are high. The protocol has
not defined commands that have FF as their description. Those
commands can be used for whatever task is required.
;-------------------------------------------------
IRRC5TAB:

;-------------------------------------------------
; The command code determines the function.
;
; VALUE TO P1 REMOTE KEY COMMAND
; ----------- ---------- -------
db 01111111b ; 0 ; 0
db 01111110b ; 1 ; 1
db 01111101b ; 2 ; 2
db 01111100b ; 3 ; 3
db 01111011b ; 4 ; 4
db 01111010b ; 5 ; 5
db 01111001b ; 6 ; 6
db 01111000b ; 7 ; 7
db 01110111b ; 8 ; 8
db 01110110b ; 9 ; 9
db 11111111b ; ; A
db 11111111b ; ; B
db 01110011b ; ON/OFF ; C
db 01110010b ; MUTE ; D
db 01110001b ; PP ; E
db 01110000b ; OSD ; F
db 01101111b ; Volume+ ; 10
db 01101110b ; Volume- ; 11
db 01101101b ; Bright+ ; 12
db 01101100b ; Bright- ; 13
db 01101011b ; Color+ ; 14
db 01111010b ; Color- ; 15
db 11111111b ; ; 16
db 11111111b ; ; 17
db 11111111b ; ; 18
db 11111111b ; ; 19
db 11111111b ; ; 1A
db 11111111b ; ; 1B
db 01100011b ; Contrast+ ; 1C
db 01100010b ; Contrast- ; 1D
db 11111111b ; ; 1E
db 11111111b ; ; 1F
db 01011111b ; Program+ ; 20
db 01011110b ; Program- ; 21
db 11111111b ; ; 22
db 11111111b ; ; 23
db 01011011b ; Timer ; 24
db 01010111b ; Special 1 ; 25
db 01000001b ; Special 2 ; 26

db 01000111b ; Special 3 ; 27
db 01001110b ; Special 4 ; 28
db 01000101b ; Special 5 ; 29
db 01010010b ; Special 6 ; 2A
Connections _____________________
Only three pins are required by the software. The data is input on
one pin.
Another pin is used to trigger an LED to show infrared is being
received.
The third pin shows a pulse generated by the software. This triggers
with each new bit. Therefore, a scope can look at the incoming
timing. This will ensure the code is sensing at mid bit.
---___---___------______---___---___--- IrInput
_|_____|_____|_____|_____|_____|_____|_ IrScope
After testing, this code can be removed with no loss of functionality.
Circuit: infrared receiver__________
Port3.3 13 1 IR
P3.4 14 2 Gnd
P3.5 15 3 Vcc
microprocessor infrared receiver
ž Ÿ

34
PROJECT 11 - WIRELESS
Thought
Peripheral vision or tunnel vision,
its your choice.
MOD
Project 11: Communicate with IR __
Purpose: To detect an infrared remote.
To control some device based on the remote command.
Preamble:
An infrared transmitter is used to control many consumer electronic
devices without physical wires. An oscillator of about 40 kHz
carrier is modulated with pulses for off and on states. When the
signal passes through a detector, the carrier is removed, leaving only
the pulses. These pulses represent digital data.
The transmit and receive procedures are typically performed by a
microcontroller. This is easier than making a custom chip.
For this project, connect an infrared photo detector. This is basically
a photo diode that will respond to the carrier. Connect the output of
the photo detector to an input port or mmio bit.

Chapter 34 Project 11 - Wireless 257
Basically three protocols are used by most remotes. The Philips,
Sony, and Eastern formats are most common. This project will use
the Philips protocol with the microcontroller operating as a
television. Therefore, any universal remote should be programmable
to operate with the microcontroller.
Plan:
Add an infrared detector to the microcontroller. Use Philips
protocol. Program the remote for a television.
Preparation:
Determine the address and command structure for the infrared
transmission. Create a table to decode the message.
Procedure:
First, write a subroutine to detect the pulses that are incoming.
Detect the first falling edge and use this as the start bit. Create
appropriate time delay before testing for the next pulse. Shift the
pulses in serially until all the data is received. If there is an error,
create an adequate delay to ensure the next message has not started.
If the data is valid, decode the pulses according to a table. Use the
results to initiate a control action. As a minimum, turn an LED on
and off, based on the infrared remote signal.
Presentation:
Demonstrate the infrared operation by displaying the results in
LED, seven-segment, or LCD format.
ž Ÿ

;-------------------------------------------------
;Program: MODInfra.ASM
;Date: 17 August 2002
;
; Tulsa, OK, USA
; mod@superb.org
;
;Purpose:
; A set of routines are provided to perform
; infrared receive functions. The code can be
; easily modified.
;
; This code has timing based on 11.059 MHz
; crystal. If another crystal is used, the delay
; routines must be modified.
;#################################################
; PROGRAM
;#################################################
org 00H
START: ljmp INITIAL
org 0033H ;Address past vectors
;*************************************************
; INITIAL & MAIN
;*************************************************
INITIAL:
;-------------------------------------------------

;INITIALIZE
;-------------------------------------------------
MAIN:
;-------------------------------------------------
;PROCESS
lcall IRRECV ;infrared:A=IrComd
lcall SEROUT ;display IrComd
;*************************************************
; INFRARED REMOTE
;*************************************************
; PHILIPS RC5 remote receiver.
; Adopted from code by Wagner Lipnharski 11/99
;-------------------------------------------------
;Ir Receiver Assignments
;-------------------------------------------------
IrOut equ 0B5h ;LED out
IrInp bit 0B3h ;IR serial input stream
IrScope bit 0B4h ;soft generate pulse for each bit
IrComd data 41h ;IR code received
;-------------------------------------------------
IRRECV:
;-------------------------------------------------
; Receive IR string of 14 bits.
;INITIAL
IRRE0: setb IrOut ;Turnoff IR Indicate
setb IrInp ;Input bit
;WAIT FOR FIRST bit
IRRE2: jb IrInp,IRRE0 ;Incoming bit is low
;-------------------------------------------------
; Interrupt entry on first falling edge.
; The code must be activated and the processor
; set up to jump to here.
; anl IE,#11111011b ;turn off ExtInt1
;-------------------------------------------------

;START BITS (AGC)
IRRE3: clr IrOut ;IR Indicator on 1st
lcall DLYPHI34 ;3/4 bittime=1.296ms
setb IrScope ;*scope pulse
clr IrScope ;*scope pulse
jb IrInp,IRRE31 ;2nd AGC,1st half hi
sjmp IRRE8 ;error, resync
;ADDRESS STREAM
IRRE31: lcall DLYPHI ;1 bit time delay
clr A ;IR Rx first low lev
mov B,#6 ;6 more bits
IRRE4: setb IrScope ;*scope pulse
mov C,IrInp ;IR state to Carry
rlc A ;Insert IR into A
lcall DLYPHI ;1 bit time delay
djnz B,IRRE4 ;Rotate 6 bits in A
;CHKbit & 5 ADDRESS
;DECODE ADDRESS
anl A,#00011111b ;mask CHK (flip) bit
cjne A,#0h,IRRE8 ;<> Address 00
;COMMAND STREAM
mov B,#6 ;6bit command stream
IRRE5: setb IrScope ;*scope pulse
mov C,IrInp ;IR state to Carry
rlc A ;Insert IR into A
lcall DLYPHI ;1 bit delay
djnz B,IRRE5 ;Rotate command to A
;SAVE COMMAND
mov IrComd,A ;Save Command
sjmp IRRE9 ;good stuff
;WAIT TO SYNC NEXT
IRRE8: lcall DLYPHI15 ;invalid, wait 15
ljmp IRRE0 ;bit time, restart

IRRE9: ret ;IRRE8 for Interrupt
;*************************************************
; DELAY
;*************************************************
; Delay routine using NOPS and a nested loop.
; Registers R2 & R3 are used for counting loops.
; Count the machine cycles for each instruction.
;
; This code has timing based on 11.059 MHz
; crystal. If another crystal is used, the delay
; routines must be modified.
;-------------------------------------------------
DLYPHI:
;-------------------------------------------------
; For time interval at 1 bit cycle, need
; time interval of 1.728 ms, so need
; .001728 * 11059000/12 = 1592.5 machine cycles
; Use 1.72, need 1585 machine cycles.
ZDLR2: mov R2,#10 ;Nested loop counter
ZDLR1: djnz R2,ZDLR1 ;Nested loop, 256 x
djnz R3,ZDLR2 ;Outer loop
ret ;Return to call
;-------------------------------------------------
DLYPHI34:
;-------------------------------------------------
; For interval at 3/ 4 of a bit cycle, need
; time interval of 1.728 * .75= 1.296 ms, so need
; .001296 * 11059000/12 = 1194.3 machine cycles
; Use .75 * 1.72 gives 1188 machine cycles.
nop
ret ;Return to call
;-------------------------------------------------

DLYPHI15:
;-------------------------------------------------
; For time interval at 15 bit cycle, need
; time interval of 1.728 * 15= 25.92 ms, so need
; .02592 * 11059000/12= 23,887.44 machine cycle
;
ret ;Return to call
;-------------------------------------------------
end ;Program end
ž Ÿ

35
SERIAL CHIPS – IIC
Thought
You can not serve two masters.
Jesus of Nazareth, ~AD 30
Other chip interfaces _____________
One of the limits of any computer is how many things can be
connected. Several techniques have been discussed for expanding
those options. These include expansion memory and latches,
multiplexing, and matrix networks. All these methods have used
parallel connections.
Because of real estate considerations, another method is frequently
used for accessing special purpose chips. Serial communications
with devices permits reducing the number of pins and the resulting
space. As with every engineering problem, there are trade-offs.
Because of the sequential bits, this method is slower and requires
considerably more software.
There are several competing technologies. The two dominant
protocols are serial peripheral interface (SPI) and inter integrated
circuit (IIC). These two have become de facto standards for
expansion of microprocessor capabilities.
The major differences are in the connections. SPI has a common bus
of three wires, but it requires a separate select line for each device.

IIC has only two wires for all communications. Although the wiring
is simpler for the IIC, the software is considerably more complex
than that for the SPI.
There are several other protocols that are very powerful and will be
mentioned only briefly. Since they are not used for on-board
connections, they will not be addressed in detail.
External serial communications such as Firewire and Ethernet are
accomplished using specialty chips or embedded technology in the
microprocessor. Universal serial bus (USB) is used for connection
to external computer peripherals. The controller area network
(CAN) is used in the automotive industry.
Inter integrated circuit ___________
During the 1980s Philips Semiconductor developed a bi-directional
serial bus for inter integrated circuit (IIC) communications. Two
common wires plus ground are used by all devices that are
connected on the bus. The common lines are serial data line (SDA)
and serial clock line (SCL).
Newer enhancements to the protocol allow 10-bit addressing and
communications speeds to 400 Kbits/second.
Every device connected to the lines necessarily has a unique
address. The chip can be a transmitter and/or receiver. A display is
only a receiver, while a memory chip is both.
The IIC is a multiple master bus. More than one integrated circuit
can initiate data transfer. The chip initiating the transfer is the bus
master, during which the others are bus slaves. A microprocessor is
typically the master.
The sequence of communications follows a set protocol.
1. The master sends a Start to get the attention of the slaves.

Chapter 35 Serial Chips - IIC 265
2. Then the master sends the Address of the slave along with a
read or write flag.
3. All slaves compare the address with their address. If the address
does not match, the slave waits for the Stop message.
4. If there is a match, the slave sends an Acknowledge.
5. When the master detects the acknowledge, it will write or read
the Data.
6. After completion, the master will send a Stop. This is a signal
the bus is released and a different transmission can take place.
Several versions of the microcontroller have IIC capability as an
internal design configuration. The pins are located on Port 1.
P1.7 P1.6
SDA SCL
Notice that these occupy the same space as the SPI protocol.
Therefore, both will not exist on the same chip. If both are required,
it is relatively easy to select another pin location for either protocol.
Then bit bang the pins for data transfer. Several bit bang examples
are provided in this and the next chapter.
IIC details ______________________
The interface is used as one de facto standard for serial
communications to peripherals. The system requires the following
two lines. SDA contains the I/O data. SCL is the clock and controls
data availability. SCL clocks data into the device on a rising edge. It
clocks out from a device on the falling edge.
The device addresses are listed. Since the microprocessor does not
have the capability for IIC internally, the data must be bit-banged.
IicSda bit 97H ;port serial I/O data bit
IicScl bit 96H ;port clock line

In the most common configuration, the system will be set up as
Master-Slave. The microprocessor is the master, so it controls the
clock. External devices are open collector, so the lines are high
when not used.
Data is sent MSB first. Any number of bits can be sent. The slave
address is a unique seven-bit number. This is followed by a
direction bit. Acknowledge (Ack) is sent by the receiver after each
byte.
The slave address has a device number as the four most significant
bits. Then three bits allow the address to select a particular device.
The final bit is 0 for write and 1 for read.
For example, an eprom slave device address is 0A0h. One device
has address of 0H. Follow with the direction bit. For greater than
256 bytes, the address is ORed with the device number.
Since all devices share a common line, one method of assuring the
bus is not busy is to check the clock. The problem is this causes a
hang-up until the lines are clear.
setb IicScl ;Scl=1
jnb IicScl,$ ;wait til not busy
A good practice is to leave the start and data transfer routines with
the clock low so the next transfer is ready.
There are numerous possible bit operations. These are identified in
the table. The device abbreviations are M=master, R=receiver,
T=transmitter

Operation Who Hold Change Clock
Free M/M Read-Scl=1 Read-Sda=1
Start M/M Scl=1 Sda=1 to 0
Slave Address M/M Scl=0 Sda=1 or 0 Scl=1
Direct Read M/M Scl=0 Sda=1 Scl=0
Direct Write M/M Scl=0 Sda=0 Scl=0
Acknowledge R/M Sda=0 Scl=0 to 1 Scl=0
NotAck R/M Sda=1 Scl=0 to 1 Scl=0
Data bits M/T Scl=0 Sda=1 or 0 Scl=1
Stop M/M Scl=1 Sda=0 to 1
IIC sequence ____________________
The sequence of subroutines has two layers of calls. The calls are
read, write, and acknowledge. Then these call other routines.
IICREAD:
lcall IICSTART ;start
lcall IICMSBOT ;MSB out
lcall IICMSBIN ;MSB in
lcall IICACK ;acknowledge
lcall IICNACK ;not acknowledge
lcall IICSTOP ;stop
IICWRITE:
lcall IICSTOP ;stop
IICDACK:
lcall IICSTOP ;stop
A short time delay is required for the IIC bus setup time. Four NOPs
are all that is required with 7 MHz crystal. Five are adequate with
the 11 MHz crystal.

IIC bit bang ____________________
The bit-banging routines are shown below. These include start, stop,
acknowledge, and not-acknowledge. These routines can be used
with any two bits on the processor, since they do not rely on the IIC
registers of special processors.
;-------------------------------------------------
IICSTART:
; Send IIC start sequence.
; Start M/M Hold-Scl=1 Change-Sda=1 to 0
;
; ;START SEQUENCE
setb IicSda ;Sda=1
setb IicScl ;Scl=1
%iicdly ;mac setup time wait
clr IicSda ;transition 1 to 0
clr IicScl ;end clock pulse
ret
;
;-------------------------------------------------
IICSTOP:
; Send IIC stop sequence.
; Stop M/M Hold-Scl=1 Change-Sda=0 to 1
; Leave with clock HI so the line is released.
;
; ;STOP SEQUENCE
clr IicSda ;Sda=0
setb IicScl ;Scl=1
setb IicSda ;transition 0 to 1
ret
;
;-------------------------------------------------
IICACK:
; Send IIC Acknowledge sequence.
; Acknow R/M Hold-Sda=0 Change-Scl=0 to 1
; Scl=0
; This is to follow each byte received.
;
; ;ACK SEQUENCE

clr IicSda ;Sda=0
setb IicScl ;start clock pulse
setb IicSda ;release sda
ret
;
;-------------------------------------------------
IICNAK:
; Send IIC not-acknowledge sequence.
; NotAck R/M Hold-Sda=1 Change-Scl=0 to 1
; Scl=0
; This is to follow each byte received.
;
; ;ACK SEQUENCE
setb IicSda ;Sda=1
setb IicScl ;start clock pulse
ret
;
;-------------------------------------------------
IICMSBIN:
; Data is input serially on the IIC
; with MSB arriving first.
; Data bit M/T Hold-Scl=0 Change-Sda=1 or 0
; Scl=1
;
; One byte is handled in the routine. This is
; rlc for C to go to the LSB of the byte.
; The data is transferred on a level clock.
; The SClk must be toggled.
; The data port bit must be set for input.
;
setb IicSda ;set port for input
mov B,#8 ;counter for 1 byte
; ;INPUT A byte
IICN1: setb IicScl ;strobe clock
mov C,IicSda ;MSB to ser data
clr IicScl ;
rlc A ;MSB to transmit
djnz B,IICN1 ;8 bits input

clr C ;error flag=clr
ret
;
;-------------------------------------------------
IICMSBOT:
; Data is output serially on the IIC
; with MSB first.
; Data bit M/T Hold-Scl=0 Change-Sda=1 or 0
; Scl=1
; One byte is handled in the routine. This is
; rlc for MSB to go to C.
; The data is transferred on a level clock.
;
; The SClk must be toggled.
; Leave with data/error code in C.
;
mov B,#8 ;counter for 1 byte
; ;SHIFT OUT byte
IICT1: rlc A ;MSB to transmit
mov IicSda,C ;MSB to ser data
; ;CLOCK
setb IicScl ;strobe clock
clr IicScl ;
djnz B,IICT1 ;8 bits input
; ;CHECK ACK FRM SLAVE
setb IicSda ;Sda=1, make input
setb IicScl ;strobe clock
mov C,IicSda ;Sda=0 is ack
IICT2: clr IicScl ;strobe clock
ret
ž Ÿ

36
SERIAL CHIPS – SPI
Thought
Don’t walk around with a chip
on your shoulder.
Grandpa
Serial peripheral interface _________
SPI was originally named and promoted by Motorola. It is also
called Microwire by National Semiconductor. Enhancements
include queued SPI.
SPI is a full duplex, synchronous, serial data transfer system. One
machine is selected as the master, the remainder are designated as
slaves. Three common wires plus ground are used by all devices
that are connected on the bus. The common lines are master out
slave in (MOSI), master in slave out (MISO), and clock (SCK). A
select line is unique to each slave device.
Communications speeds can exceed 1 Mbits/second.
The master creates the clock by asserting the strobe pin low and
high. The SPI standard allows either positive or negative clock
polarity. Two different protocols can be used for clocking 8-bit data.

The MOSI pin is the data output from the master, so it is the input to
all the slaves. Conversely, the MISO pin is the data input to the
master, so it is the data output from one of the slaves.
The slave select is a chip select for each slave. Therefore, several
lines are required for multiple slave integrated circuits. The slave
selected is the only one that responds to activity on the bus. The
others are high impedance, so they do not interfere.
If a device is trying to act as a master, it will assert the chip select
line. Therefore, this line can be read as an input by the master to
determine if there is a multiple master conflict. This will
automatically disable the outputs to prevent two masters.
Details for implementation are given in a sample program in the
next chapter.
Several versions of the microcontroller have SPI capability as an
internal design configuration. The pins are located on Port 1.
P1.7 P1.6 P1.5 P1.4
SCK MISO MOSI /SS
These lines are shared with the in system programming lines. To
assure there are no problems with the SPI chips, disconnect the in
system programming lines after the program is downloaded. This
can be accomplished by an isolation chip as illustrated on the board
schematic.
Analog to digital sensitivity ________
Analog to digital converters (ADC) are a very common type device
that are connected via the serial peripheral interface (SPI). Analog
values are converted to binary equivalent values and are input from
the converter to the microprocessor.
An analog to digital converter (ADC) chip senses continuous signals
in the range of 0-5 volt DC on its input. The device then samples the

Chapter 36 Serial Chips - SPI 273
analog data, holds the value, and converts it to a digital signal. The
number of bits used for the conversion determines the sensitivity.
Sensitivity is voltage range divided by digital range. An eight-bit
converter has a range of 0 to 255. Therefore, each bit has a
sensitivity of 5/256 = 0.0195 volts per bit.
If finer resolution is required, a larger number of bits must be used.
Common sizes are 8, 10, 12, 16, and 24 bit converters. Obviously,
the trade-off for more sensitivity performance is slower response
and more cost.
Analog to digital noise ____________
Noise on analog to digital converters can be a problem. This is
partially a characteristic of the chips. However, there are two bigger
issues. The wiring layout and the surrounding ambient noise tend to
be larger challenges. Analog ground and digital ground should be
totally isolated, except for a single-point connection. A ground
plane may be necessary below the analog section. Alternately a
shield may be necessary to cover the analog components.
Even with good design, often the low order bits of the conversion
results are noisy and unstable. On an eight bit converter, this may be
one or two bits. On twenty-four bit converters, it may be 4 bits.
These bits represent the noise threshold. Several techniques are used
to manage the unstable variations.
The simplest is to truncate the noise bits. Rotate the value right, then
left with zero fill for the number of unstable bits. The disadvantage
is this effectively reduces the sensitivity by a power of two.
Another techniques is using a smoothing routine. The running
average is the simplest. Add the previous value and the present
value. Then divide the result by two. This has the effective of
slowing down response time.

In the smoothing routine, weighting can be placed on either of the
values. Multiply the value by a factor before the addition. Then the
result is divided by the sum of the weighting factors.
LTC 1098 clocking _______________
The LTC1098 will be used as an application example. The
techniques can be easily adapted to other components.
The master triggers the clock bit. Data comes into a device on rising
edge and out from the device on a falling edge.
The sequence for reading the ADC is very dependent on the clock.
1. Start: Assert SCK high.
2. Select: Chip select high, chip select low.
3. Command: SCK low, assert data bit, wait, SCK high, wait,
repeat the cycle for each bit.
4. Ready: SCK low
5. Input: SCK high, SCK low, read data bit, repeat the cycle for
each bit.
6. Stop: SCK high
7. Deselect: Chip select high
LTC 1098 operation ______________
The LTC1098 is a two-channel, 12-bit analog-to-digital converter.
The master must configure the converter by writing a four bit
message.
1. MSB is the start bit, which is 1.
2. Bit1,2 is the single or differential ended mux address.
00 = differential channel 0 referenced to channel 1.
01 = differential channel 1 referenced to channel 0.
10 = single ended channel 0 referenced to ground.
11 = single ended channel 1 referenced to ground
3. Bit3 determines which bit is sent first.
1 = Most significant bit (MSB) is first with zero fill at end.
0 = Least significant bit (LSB) first.

After the four bits are output on the serial peripheral interface, the
master is configured to read twelve bits of data. If the system is
using a 10-bit or 8-bit device, the extra bits read will return a 0.
Data is written to the chip on a rising clock edge. Set the data on
the pin, then pulse the clock high followed by a low to complete the
cycle.
Read data from a chip on a trailing edge. Pulse the clock high
followed by a low, then read the data.
The LTC 1098 needs at least 10us after enabling, before the first
data bit is output. This is a relatively slow device for use as a
physical sensor. It has a 40Hz serial clock limit.
The twelve bits of data are read MSB first. When this is placed
directly in a register, it appears as sixteen bits with zero fill. This
appears to be a multiply by 24
or sixteen. Therefore, it is necessary
to shift the data to get the desired precision. This is a bit bang
procedure that can be used with any version of the processor.
Program: LTC 1098 bit-bang ______
;-------------------------------------------------
;Program: MODspi.ASM
;Update: 01 March, 2003
;Initial: May 23, 1994 @ Foxfire, MO
; Tulsa, OK, USA
; mod@superb.org
;
;Purpose:
; A set of routines are provided to read from
; a serial peripheral interface. The device is
; a 12-bit analog to digital converter.
;

;#################################################
; ASSIGNMENTS
;#################################################
;CONSTANTS
;-------------------------------------------------
;LATCH & SPI
Mosi equ 95h ;SPI Mosi from uC to slave
Miso equ 96h ;SPI Miso from slave to uC
Sck equ 97h ;SPI Clk
;-------------------------------------------------
;DEFINED VARIABLES
;-------------------------------------------------
QikB equ 19H ;Interrupt HEX value, msb
QikA equ 18H ;Interrupt HEX value, lsb
;#################################################
; PROGRAM
;#################################################
org 00h
START: ljmp INITIAL
org 0033h ;Address past vectors
;-------------------------------------------------
org 0080h ;Address past reserve
INITIAL:
MAIN:
;-------------------------------------------------
;PROCESS
lcall ADCIN ;read adc on spi
mov A,QikB ;MSB, channel 1
lcall SEROUT ;send to serial
mov A,QikA ;LSB, channel 1
lcall SEROUT ;send to serial
MAN9: ljmp MAN9 ;Halt
------------------------------------------------

ADCIN:
;-------------------------------------------------
; LTC1098, 2channel, 12-bit analog-digital convt
; Write 4 bit control message. Then do input.
;
; Write data to chip on rising edge:
; Set data, pulse hi, pulse low.
; Read data from chip on trailing edge:
; Pulse hi, pulse lo, read data.
;
; The LTC needs at least 10us after enable
; before first data bit is output.
; There is a 40Hz serial clock limit.
; The timing is for a 7.5 MHz crystal. So the
; delays will increase for 11 MHz.
;
;STOP CLOCK
setb SpiClk ;clock low for null
;ENABLE CHIP SELECT
; lcall ADCDES ;disable & shutdown
; lcall ADCSEL ;enabl falling edge
;READ 12 BITS, CH0
mov A,#0D0H ;wr cmd1101xxxx,ch0
; mov A,#0F0H ;wr cmd1111xxxx,ch1
mov B,#4 ;count bits shifted
lcall SPIMSBOT ;byte,MSB 1st
clr SpiClk ;clock low for null
mov B,#8 ;count bits shifted
lcall SPIMSBIN ;byte,MSB 1st
mov QIkB,A ;high byte in
mov B,#4 ;4 bits in
lcall SPIMSBIN ;byte,MSB 1st
mov QIkA,A ;byte w/ 0 fill LSB
;DESELECT
setb SpiClk ;hi before deselect
; lcall ADCDES ;disable & shutdown

;NEXT CHANNEL
; lcall ADCSEL ;enabl falling edge
ADCI9: ret
;-------------------------------------------------
SPIMSBIN:
;-------------------------------------------------
; Read serially from the SPI starting
; with the most significant bit (MSB). Data is
; clocked from the device on falling clock edge.
setb Miso ;Make Sdat an input
SPMI1: setb Sck ;Clk high
nop ;AD7714 t6 time
clr Sck ;Clock bit from ADC
nop ;AD7714 t5 time
mov C,Miso ;Sdat bit to C
rlc A ;Rotate bit to C
djnz B,SPMI1 ;Get all 8 bits
clr Miso ;undo Sdat input
ret
;-------------------------------------------------
SPIMSBOT:
;-------------------------------------------------
; Send B-bits. Most significant bit(MSB) first
; Data is clocked in device on rising clock edge.
;OUT & WAIT
SPMO1: rlc A ;Rotate MSB to C
clr Sck ;Clock bit to ADC
mov Miso,C ;Bit to port pin
nop ;Delay for ç & 40Hz
setb Sck ;CLK line high

djnz B,SPMO1 ;Send all bits
ret
;-------------------------------------------------
end
Onboard SPI control register ______
Many microprocessor version now have a serial peripheral interface
built onboard the chip. When using the internal connections, it is
unnecessary to develop some of the timing details for their
operation.
The operation is explained in detail in the section of special function
registers. The SPI control register is configured when using the
upper bits of port 1 for serial peripheral interface operations. For
typical operations the following bits should be selected.
SPIE SPE DORD MSTR CPOL CPHA SPR1 SPR0
0 1 0 1 0 1 0 1
This turns off interrupts (SPIE) and enables the SPI channel (SPE).
The order of bits is MSB first (DORD). The processor is the master
(MSTR). The serial clock polarity is low when idle (CPOL). The
slave may remain selected between samples (CPHA). The
microprocessor frequency is divide by 16 to provide a clock
frequency of less than 1 MHz.
Rather than using an interrupt, the transmission complete flag
(SPIF) is polled. The bit is set by the processor when transmission is
complete.

Program: EEPROM SPI register ___
Because of the internal SPI capabilities of some version of the
microcontroller, registers can be used to control SPI activity. This
example uses two-way communications with an external EEPROM.
;-------------------------------------------------
;Program: MODonspi.ASM
;Initial: July 28, 2003
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c) 2003. All rights reserved
; Original adapted from Atmel.
;
;Purpose:
; A set of routines are provided to write and
; read from a serial peripheral interface.
; The device is an eeprom.
;
;#################################################
; ASSIGNMENTS
;#################################################
;CONSTANTS
;-------------------------------------------------
;SPI REGISTERS
Spcr data 0d5h ;SPI control register
Spsr data 0aah ;SPI status register
Spdr data 86h ;SPI data register
Spif equ 10000000b ;interrupt flag
;SPI ASSIGNMENTS
Mosi bit 95h ;SPI Mosi from uC to slave
Miso bit 96h ;SPI Miso from slave to uC
Sck bit 97h ;SPI Clk
CSn bit 94h ;device slave select

;DEVICE COMMANDS
Rdsr equ 05h ;Read Status Register
Wrsr equ 01h ;Write Status Register
Read equ 03h ;Read Data from Memory
Write equ 02h ;Write Data to Memory
Wren equ 06h ;Write Enable
Wrdi equ 04h ;Write Disable
;bit DEFINITION
A8 bit acc.3 ;MSB of address
NRDY bit acc.0 ;hi= write cycle in progress
;#################################################
; PROGRAM
;#################################################
org 00h
START: ljmp INITIAL
;-------------------------------------------------
INITIAL:
;-------------------------------------------------
; SPI master mode initialization code.
; SPCR is setup as interrupt disable,
; pin enable, MSB first, polarity 0, phase 1
; clock rate /16
setb CSn ;deselect AT25040
setb Mosi ;initialize SPI pins
setb Miso
setb Sck
mov Spcr,#01010101b ;init SPI master
;-------------------------------------------------
MAIN:
;-------------------------------------------------
lcall SPIEEROM ;write/read eeprom
MAN9: sjmp MAIN

;-------------------------------------------------
SPIEEROM:
;-------------------------------------------------
; Write/Read AT25C040 EEPROM via the Serial
; Peripheral Interface (SPI).
; Completion of programming is checked by polling
; SPI interrupt is not used.
; Works w/ microcontroller clk of 24 MHz or less.
;
; Write one byte to AT25040 and verify
; (read and compare).
; Code to handle verification failure not shown.
; Needs timeout to prevent write error from
; causing an infinite loop.
;
; Information to write has value of Data.
; Address to write has 16-bit value of Address.
; if not valid write & read go to
lcall SWRENAB ;precede ea byte wr
mov A,#Data ;data
mov DPTR,#Address ;address
lcall SWRBYTE ;write
SPEE1: lcall SRDSTAT ;check write status
jb Nrdy,SPEE1 ;loop until done
mov DPTR,#Address ;address
lcall SRDBYTE ;read
cjne A,#Data,SPEE8 ;jump data chk fail
sjmp SPEE9
SPEE8: setb 0B5h :turn on led
SPEE9: ret ;valid write & read
;-------------------------------------------------
SRDSTAT:
;-------------------------------------------------
; Read device status. Returns status byte in A.
clr CSn ;select device
mov A,#Rdsr ;get command

lcall SPIIO ;send command
setb CSn ;deselect device
ret
;-------------------------------------------------
SWRENAB:
;-------------------------------------------------
; Enable write.
; Does not check for device ready before sending
; command. Returns nothing. Destroys A.
mov A,#Wren ;get command
lcall SPIIO ;send command
ret
;-------------------------------------------------
SRDBYTE:
;-------------------------------------------------
; Read one byte of data from specified address.
; Does not check for device ready before sending
; command. Called with address in DPTR.
; Returns data in A.
mov A,DPH ;get high byte addr
rrc A ;move LSB to carry
mov A,#Read ;get command
mov A8,C ;combine command
;& hi bit of addr
lcall SPIIO ;send com&hi bit add
mov A,DPL ;get low byte of add
lcall SPIIO ;send low byte addr
lcall SPIIO ;get data

ret
;-------------------------------------------------
SWRBYTE:
;-------------------------------------------------
; Write one byte of data to specified address.
; Does not check for device ready or write enable
; before sending command. Does not wait for write
; cycle to complete before returning.
; Called with address in DPTR, data in A.
; Returns nothing.
push Acc ;save data
mov A,DPH ;get high byte of ad
rrc A ;move LSB to carry
mov A,#Write ;get command
mov A8,C ;combine command &
;high bit of address
lcall SPIIO ;send com& hi bit ad
mov A,DPL ;get low byte of add
lcall SPIIO ;send low byte of
address
pop ACC ;restore data
lcall SPIIO ;send data
ret
;-------------------------------------------------
SPIIO:
;-------------------------------------------------
; Send/receive data through the SPI port.
; A byte is shifted in as a byte is shifted out,
; receiving and sending simultaneously.
; Waits for shift out/in complete before return.

; Expects slave already selected. Called with
; data to send in A. Returns data received in A.
mov Spdr,A ;write output data
SALL1: mov A,Spsr ;get status
anl A,#Spif ;check for done
jz SALL1 ;loop until done
mov A,Spdr ;read input data
ret
TLC549 clocking _________________
The TLC549 is a simple 8-pin a-to-d converters that can work on
the SPI bus. The device has an on-chip system clock that allows it to
have exception rates of conversions independent of the clock line
(SCK).
When the chip select line is asserted low, the most significant bit
(MSB) is placed on the data line. The next bits (A6-A0) are placed
on the data line with the falling edge of the clock line.
The hold function begins with the eighth clock cycle. After the eight
clock cycle, chip select must go high, or the clock line must remain
low for at least 36 internal system clock cycles to allow completion
of the conversion.
If the chip select is kept low for multiple conversion, the clock line
(SCK) must remain glitch free or the microprocessor and the device
will lose synchronization.
The clock line must maintain each state for more than 404 ns. The
conversion time takes a maximum of 17 microseconds. The
maximum clock frequency is 1.1 MHz.
The timing diagram illustrates the process

Circuit: SPI_____________________
Port1 1 Ref+ 8
MOSI 7 Analog In 2 Dout 7
SCK 8 3 Ref- 6
/SS 5 4 CS 5
microprocessor ADC
ž Ÿ

37
PROJECT 12 - A TO D CONVERTER
Thought
The whole world is an analog stage,
digital only plays bit parts.
Project 12: Analog / digital converter
Purpose: To develop peripheral expansion with serial devices.
To evaluate continuous or analog inputs.
Preamble:
Numerous peripheral devices are often desired for a computer
system. These may be memory, analog interfaces, or sensors. All
these devices require data input and output. If parallel
communications is used for the devices, they have a large footprint
that takes occupies a large amount of board space. An alternative is
to perform serial communications between the processor and the
peripheral device. This requires many fewer wires and typically the
device has a size of only 8 pins. As with every engineering design,
there is a tradeoff. The speed is slower, but the area is much smaller.
This is less costly and has better space performance.
Several protocols are used for serial interfacing. One of the simplest
to implement is the serial peripheral interface (SPI). It requires four
lines – a data out line, a data in line, a clock line, and a select line.

The data and clock lines can be common with other SPI devices.
However, each requires its own select.
An alternative is the inter-integrated circuit (IIC) format, which
requires only two lines – data and clock. The select is made by
addresses transferred on the data line. The tradeoff is the hardware
requirements are less, but the software to implement the addressing
is more complex.
Some microprocessors have SPI protocol as part of their system.
Others require it to be implemented in software through three or
four pins. For example, some versions use SPI to load the on-board
memory.
Many converters have the capability of sampling more than one
channel of analog input. A control must be sent to the device to
select which channel to convert. With an SPI chip, the control
would be an address sent to the ADC.
The ADC also requires a reference voltage as an input. The internal
circuitry is actually a comparator that evaluates the analog input in
relation to the reference voltage. Therefore, the converted input
often cannot quite reach the reference value.
The number of bits used for the conversion determines the
sensitivity. Sensitivity is voltage range divided by digital range. An
eight-bit converter has a range of 0 to 255. Therefore, each bit has a
sensitivity of 5/256 = 0.0195 volts per bit.
Disconnect the in-system programming cable to eliminate any noise
on the SPI lines.
Plan:
Add an analog to digital converter to the microcontroller. Use SPI
connections.

Chapter 37 Project 12 – A to D Converter 289
Preparation:
Use the power supply as the reference voltage. Connect the three
control lines to a port or memory addressed latch on the processor.
Connect a variable voltage source for the input signal.
Procedure:
First, write a subroutine to perform SPI communications. This will
include selecting the device, setting up data, and clocking the data
out of the chip.
Next, write a routine to select the ADC chip and the appropriate
channel. Call the SPI routine to perform the transfer.
Finally convert the data from a digital number to a number
corresponding to the voltage. Display the results on the serial line,
LCD, or LEDs.
Presentation:
Demonstrate the ADC operation by reading a DC voltage of a
known value. Display the digital and voltage values.
ž Ÿ
;-------------------------------------------------
;Program: MODadc.ASM
;Initial: 13 November 2003

; Tulsa, OK, USA
; mod@superb.org
;Copyright (c)2003. All rights reserved
;
; Modified code from original by Matt Olson.
;Purpose:
; A set of routines are provided to read from
; a serial peripheral interface. The device is
; an 8-bit analog to digital converter. This uses
; the SPI registers
;
;#################################################
; ASSIGNMENTS
;#################################################
;CONSTANTS
;-------------------------------------------------
;SPI
SpCr equ 0D5h ;SPI control register
SpDr equ 86h ;SPI data register
SpSr equ 0AAh ;SPI status register
AdCs equ 94h ;adc chip select
;#################################################
; PROGRAM
;#################################################
org 00h
START: ljmp INITIAL
org 0033h ;Address past vector
;-------------------------------------------------
INITIAL:
lcall SPIINIT ;init spi registers
;-------------------------------------------------
MAIN:

Chapter 37 Project 12 – A to D Converter 291
;-------------------------------------------------
;PROCESS
lcall SPIREAD ;read adc on spi
lcall DISPLAY ;show value someplace
MAN9: ljmp MAN9 ;repeat
;-------------------------------------------------
SPIINIT:
;-------------------------------------------------
; Setup SPI:
; Disable interrupts; enable spi; msb first;
; master; clk low when idle; cpha=1; f=osc/64
mov SPCR,#01010110B
ret
;-------------------------------------------------
SPIREAD:
;-------------------------------------------------
; Read byte from ADC.
;
;CHIP SELECT LOW
clr AdCs ;enable chip select
nop ;delay > 1.4usec
nop
nop
nop
nop
nop
; ;INIT data REGISTER
mov Spdr,#0AAH ;write anything
;CHECK STAUS
SPR1: mov A,Spsr ;read status reg
rlc A ;move spif to carry
jnc SPR1 ;<>1, so no data
; ;READ data REG
mov A,Spdr ;read data from reg
setb AdCs ;disable select line

;RESET STATUS REG
mov SPSR,#0 ;clear spsr
ret ;got byte, get out
;------------------------------------------------
end ;Program end
ž Ÿ

38
WAVEFORM SYNTHESIS
Thought
Quality = excellence.
Professor Durham
Real world output ________________
A computer operates in a discrete environment where everything is
either true or false, on or off, high or low, one or zero. The
perception is that the real world is an analog environment. In reality,
physical activities change very slowly and can be sampled easily.
Most natural physical phenomenon change slower than 30 times a
second.
The slowness of perception can be used to make an apparent analog
signal. A digital to analog converter (D/A) is a chip that takes a
binary input and converts it to a voltage.
A microprocessor can be used to generate a variety of waveforms.
This is accomplished by outputting a digital value for the wave. The
output is used as a signal for a digital to analog converter. The D/A
converter creates a continuous output from the previous digital
value.

Each wave can be described by a magnitude at some time. By
accessing of magnitudes at regular intervals of time, a representation
of the wave can be created.
The magnitudes can be held constant and the time interval between
outputs can be adjusted to vary the frequency. Alternately, the time
interval can be held constant and the table value multiplied by a gain
or scaling value to provide varying magnitude for the signal.
Multiple port latches or sequential bits can be sent for greater
precision. A full range capability of 8-bits is available on a port.
Therefore, the range of values in the table should be 0 to 255 to
obtain full sensitivity. The eight bits can include negative numbers.
The range of positive values is 0 to 127 (00 - 3fh). The negative
values are –1 to –128 (0FFh - 40h).
Negative numbers are acceptable for double-sided digital to analog
converters. However, a single-sided or monolithic device will
require all positive numbers.
By clever programming a single half-wave table can be used for
two-sided waves. The complement of the table is used for the
negative half-cycle. The values are directly output to a double-sided
converter.
To obtain values for a single-sided converter, 80h must be added
before the value is output. The 80h causes a positive shift in
magnitude so that all the values are greater than zero.
Sensitivity ______________________
The sensitivity is determined by the ratio of the volts to bits. When
using an 8-bit device, the range of values is 0 to 255. A common
power system is 5 volts. Therefore this device sensitivity is 5/256 =
0.0195 V/bit.

Chapter 38 Waveform Synthesis 295
Circuit: digital to analog __________
A typical 8-bit D/A converter has a relatively low level output
power. To drive a five-volt signal, an op-amp should be connected.
A 5-volt digital to analog converter with a parallel input is the
simplest design to implement with a microprocessor. The particular
layout shown is for a Max DAC5480. It is for reference only. Any
other design will work similarly.
Alternate designs use a serial input and many have multiple
converters on board.
D7-D0 VDD
VREF
VOUT
The converter circuit shows only three discrete elements. The
resistors are not required unless an adjustable gain is desired.
Element Value Location
R ref 2 K pot REF
R rfb 1 K RFB
C1 10 pF Out
Software ________________________
The software to implement the digital to analog converter is very
straightforward. It is simply a table lookup with the results display
on a port.
Options to the code are to complement negative numbers. Another
option is to add 80h for a range shift for a single sided DAC. Then
the data can be complemented if the waveform is mirrored.
‘573
LE /OE
4 D7 Vdd 14
Ref 15
11 D0 RFB 16
Out1 1
Out2 2
12 /CS
Gnd 3
13 /WR

By creating a series of waveforms, combining them, and shifting the
frequency of output, a wide variety of signals can be generated. An
amplifier on the op amp will provide audio tone levels.
The table that is created contains the wave value. Take the
waveform and divide into as many entries as necessary to
adequately describe the wave. Calculate the value at precise
intervals, then place these entries in the table.
Consider a sine wave. It can be describe by one-quarter wave. A six
table entry is likely adequate. Calculate the sine at 15 degree
intervals to get six entries in 90 degrees. For the second quarter
wave, the table is read in reverse. Then for the second half, the
values are complemented.
A sawtooth wave can be defined with just two points. Other wave
forms are similarly calculated.
ž Ÿ

39
PROJECT 13 – D TO A CONVERTER
Thought
Aphorism:
A concise formulation of a
principle, truth, or sentiment.
Definition
Project 13: Analog output _________
Purpose: To interface a D/A converter.
To use the system to create waveforms.
Preamble:
The D/A converter allows the output of a precise voltage level from
a digital signal. Coupling with a microprocessor allows the output of
this voltage to appear at a precise time. Thus the system allows the
output of voltage as a function of time.
The D/A converter used is similar to a DAC0806, which is
equivalent to the popular 1408. This is a device with eight-bit
inputs, which have a range of 0-255.
Use an op-amp on the output. Set the range of the op-amp to yield 0
volts when a digital 00 is applied. Adjust the output to +5 volts
when 255 is sent to the DAC. The +5 level can be adjusted by a
potentiometer. Use an oscilloscope to make voltage measurements.

The sensitivity is determined by the ratio of the volts to bits.
Therefore this device sensitivity is 5/256 = 0.0195 V/bit.
The integrated circuit (IC) will be interfaced to an I/O port or a
memory-mapped latch.
Plan:
Write programs to output saw-tooth and sine waveforms.
Procedure:
Write a routine to generate a saw-tooth waveform by continually
incrementing a register and sending the register contents to the D/A
converter. The frequency is controlled by a DELAY subroutine.
Write a routine to generate a sine waveform. To do this, determine a
table of sine values from 0 to 90 degree in 10-degree increments.
Convert the sine values to the appropriate HEX code for the D/A
converter. Store the values in a lookup table. Only store values
once. Use two's complement arithmetic to calculate the negative
values for quadrants 3 and 4. The frequency should be variable as in
the saw tooth program.
Notice that the values of the sine need only be known for one
quadrant of the waveform. The other segments may be formed
based on these values.
Presentation:
The routine should be such that it can be easily modified to vary the
frequency. Vary the frequency by changing the delay time between
successive outputs to the D/A converter.
ž Ÿ

Chapter 39 Project 13 – D to A Converter 299
;-------------------------------------------------
;Program: MODDac.ASM
;
; Tulsa, OK, USA
; mod@superb.org
;
;Purpose:
; Use digital to analog converter for waveform
; synthesis.
;
;#################################################
;CONSTANTS
;#################################################
;OUTPUT
Pio equ 90h ;port for i/o
;#################################################
; PROGRAM
;#################################################
org 00H
START: ljmp INITIAL
org 0033h

;-------------------------------------------------
INITIAL:
;-------------------------------------------------
;INITIALIZE
lcall DACINIT ;initialize counters
;-------------------------------------------------
MAIN:
;-------------------------------------------------
; Display a wave form.
;PROCESS
lcall DACWAVE ;display a wave
;*************************************************
; DIGITAL ANALOG CONVERTER
;*************************************************
DACINIT:
;-------------------------------------------------
; Initialize waveform pointers.
;INIT
mov R2,#0 ;table entries pntr
mov R3,#0 ;+half wave=0,neg=1
ret
;-------------------------------------------------
DACWAVE:
;-------------------------------------------------
; Produce a wave.
;PROCESS
mov DPTR,#TabWave1;table with wave
DACW1: mov A,R2 ;table pointer
movc A,A+DPTR ;get wave value
cjne R3,#1,DACW2 ;<>1,then pos
cpl A ;else,neg:comp table
DACW2: add A,#80h ;rang shift monolith

Chapter 39 Project 13 – D to A Converter 301
mov Pio,A ;out to DAC
inc R2 ;incr table pointer
cjne R2,#entry,DACW3 ;if<entry, next
sjmp DACW5
DACW3: lcall DELAY ;frequency delay
sjmp DACW1 ;next value
;COMPLEMENT CYCLE
DACW5: mov R2#0 ;reinit data pointer
mov A,R3 ;to neg half cycle
cpl A
mov R3,A
sjmp DACW1 ;repeat cycle
DACW9: ret
;-------------------------------------------------
TabWave1:
;-------------------------------------------------
; Make entries for half cycle.
db 01h ;first entry
;-------------------------------------------------
end
ž Ÿ

40
PROJECT 14 - PHOTOSENSOR
Thought
Principle 1:
Never criticize, ever.
Dale Carnegie
Project 14: Barcode reader ________
Purpose: To interface a uC with a common sensor system.
To use another interpretation of time pulse.
Preamble:
Sensors are the most important part of data acquisition systems used
in real-time computing. The photosensor is a widely used detector.
Interfacing a sensor system to the uC depends on the application. As
with any engineering problem, each method has its advantages.
Bar code readers are commonly used in department stores, on food
items, and on rail cars. Hence bar code readers are an important
system for microcomputer applications.
Plan:
Write a program that will utilize a software design for the reader.
The sensor uses a simple photo-detector device.

Chapter 40 Photosensor 303
Preparation:
The sensor is provided, but an interface circuit is required.
Procedure:
The sensor input is a bit from the port of your choice. The software
will count the width of dark and bright bars in the bar codes. The
width is determined by counting the number of samples while the
input bit is at a one-state. The program must calibrate the width of
the entire bar code. Next, group the number of dark & bright bars
into wide & narrow categories. There will be four groups. These are
dark wide, dark narrow, bright wide, and bright narrow. Use this
data with an encryption algorithm.
Presentation:
Print the bar code to the corresponding number on the display or
serial line.
ž Ÿ

41
PROJECT 15 – ANALOG CONTROL
Thought
Principle 2:
Give positive affirmation.
Dale Carnegie
Project 15: Pulse width modulation _
Purpose: To use a timer interrupt.
To interface with a pulse width modulation type device.
Preamble:
There are many analog devices that are controlled by a digital based
system. Variable speed motor control is one popular example. The
digital control of a physical variable such as speed can be
implemented in several ways.
In one approach, the digital signal is used with a D/A converter. The
converter is followed by a linear power amplifier to produce voltage
with a variable amplitude. The voltage is applied to the motor. The
cost of the D/A converter and linear amplifier are significant
disadvantages of this scheme.
An alternative approach is often used in digital control. The digital
signal is used to modulate the on-off application of constant

Chapter 41 Project 15 - Pulse Width Modulation 305
amplitude voltage to the motor. The modulation can take two forms:
(1) on-off pulses of constant duration but with a variable number of
pulses per second, and (2) on-off pulses at a constant repetition rate
but with a variable width for each pulses.
The microprocessor has a powerful feature that includes two built-in
timers. Each timer can be set to different modes to suit the
application. Programs should use Timer 0, since Timer1 is used for
serial communications.
Using an internal timer interrupt is strongly recommended for this
project. The timer interrupt operates much like the external
interrupt. Consult the reference information for further details.
The device used to modulate the currents is essentially a solid-state
switch. A transistor is used in the DC case and an optically coupled
silicon controlled rectifier (SCR) in the AC case. Transistor
operation should be familiar, but SCR operation may not be as well
understood.
A simplified model of an SCR is as follows.
1. With no gate current, the SCR will act like an open circuit
in a "blocking state".
2. With a small gate current, the SCR "snaps on" and conducts
like a high current diode in a "conducting state".
3. The SCR remains in the conducting state until the current
through the SCR goes to zero. Then the SCR returns to the
"blocking state".
Silicon controlled rectifiers are well suited for modulating AC
current. They can be triggered (turned on) at any given time in the
AC positive half-cycle. The device will turn off when the current
goes to zero at the beginning of the negative half-cycle. Thus a
variable width pulse derived from a half cycle may be produced.
In order to time the trigger pulse, the beginning of the cycle must be
known. An additional circuit called a zero crossing detector (ZCD)
takes care of this task. The ZCD output is a TTL level square wave
whose transitions indicate a zero crossing of the AC line. Note that

the trigger pulse to the SCR must be turned-off before the next cycle
begins or the SCR will turn-on again at the beginning of the cycle.
Plan:
This project uses both types of modulation for the control of light
intensity to 6 V DC and 120 V AC lamps.
Preparation:
Wire the circuit to contain a modulating circuit and lamp. Each
circuit has a TTL level control line, which is to be connected to an
I/O port of the microprocessor.
The control lines in the circuit may source up to 15 mA of current.
This is more than the I/O ports can sink so hardware buffering is
needed. A 7404 may be used as the buffer.
The keyboard system on the microprocessor is used to select the
desired light intensity for each of the control schemes. The keys
should be priority encoded as shown in the following table. The
keys settings should also be echoed to the display system.
Keypad light intensity
8 full on
7
6 .
. .
. .
. .
1 .
0 off
Procedure:
I. 6 VDC Lamp Control

Chapter 41 Project 15 - Pulse Width Modulation 307
Write a program to control the DC lamp. The program should
generate a constant pulse with a 1 ms "on" time. The intensity is
controlled by varying the number of pulses per second from 0 pulses
per 8.0 ms to 8 pulses per 8.0 ms as selected from the keyboard.
Remember that the gate of the PNP transistor must be at 0 V for the
transistor to turn-on. An internal timer interrupt will be handy for
the delay routine.
II. 120 VAC lamp control
Write a program to control an AC lamp. The program should poll
the ZCD for the beginning of the AC cycle, delay for the selected
amount of time, and then send a trigger pulse to the SCR. The half
cycle should be divided into 8 parts corresponding to the 8 intensity
levels. Note that 8 equally spaced turn-on times for the SCR will not
give 8 equally spaced intensity levels
Presentation:
Change the intensity of the light when a keypad key is selected for
each procedure.
ž Ÿ

42
PROJECT 16 - DIGITAL FEEDBACK
Thought
Principle 3:
Find out what the other person wants
and help them get it.
Then you will get what you want.
Dale Carnegie
Project 16: DC motor speed control _
Purpose: To utilize a pulse width modulation mechanism.
To implement feedback theory in a digital system.
Preamble:
A robot uses a motor as the engine. Hence, controlling the motor
will be the most crucial task in the robotic system. Most robots are
more concerned with position rather than with the speed, but a high
performance robot will deal with both speed and position.
A feedback control system is a critical component of speed motor
control. Several techniques may be applied to solve the feedback
problem. One solution is a phase-lock-loop (PLL) technique.
However, a simple feedback system may be used which can neglect
several factors necessary for a more sophisticated control scheme.

Chapter 42 Project 16 – Digital Feedback 309
The speed is detected by a photo-sensor. Since the output signal of
the sensor is very small, an amplifier is needed to obtain an
adequate output level. The system will maintain the speed of the
motor by comparing the detected speed to the desired (setting)
speed.
This mechanism is the fundamental feedback control system from
basic electrical circuit theory. From machine theory, it is apparent
that the DC motor speed will increase at a rate which is proportional
to the total current in a period of time.
A simple driver for the motor can establish the total current applied
in a certain period of time. This driver is easily turned on and off. A
driver that can be interfaced to a digital system will be able to detect
digital signals (hi-lo or on-off).
Use the microprocessor as a controller. Maintain the system stability
by comparing the output value of the system (motor speed) to the
input value (desired speed). The task should use only digital values.
Plan:
Drive the motor from the microcomputer using a pulse-width-
modulation technique. The pulse width will be the length of time
when the motor driver is turned-on.
Preparation:
To control the DC motor speed, build the circuits to convert the
speed to a digital input signal and the digital output signal to a
current. Using an optoisolator is strongly recommended. The
isolator protects the computer system from high current/voltage
transients.

Procedure:
Implement the simple feedback system for DC motor speed control.
The preset speed will be input from the keyboard. Operate the
system with at least 4 different speeds. Choose the appropriate
speeds for your circuit and motor. Use an internal timer interrupt
added convenience
Presentation:
The report should include a brief explanation about the following:
1. Why you chose the speeds used.
2. How you calculated the speed of the DC motor.
3. How the system performs at various speeds in term of
stability & response.
ž Ÿ

43
MATH FUNCTIONS
Thought
1+1 = 10
Binary thinking
Arithmetic ______________________
It has been argued that all arithmetic is simply addition. Subtraction
is addition with negative numbers. Multiplication is successive
addition. Division is a contorted combination of subtraction.
The processor actually has an elegant arithmetic set of operations.
The fundamental operations are add (add) and subtract with borrow
(subb). Many other machine use shift for multiply and divide by 2.
A very powerful hardware multiply (mul) and divide (div) are
available.
All the operations are structured for eight bits. In most calculations
that is too limited. The following routines were developed to expand
to a larger quantity of bits. These are based on routines in very early
versions of Intel applications notes. These routines were modified to
fit my applications, so the original copy is no longer available to
identify as a reference.

Extended precision _______________
;*************************************************
; MATH
;*************************************************
;
; Math routines are used in virtually every
; process. The range is addition to division and
; square root.
;
; I have used a variety of math structures over
; the years. My objective here is threefold.
; 1. minimize the number of special routines
; 2. minimize the number of lines of code
; 3. minimize the number of ram variables.
;
; Interestingly, all these criteria are
; synergistic. The one possible downside is the
; routines may be slightly longer in time.
;
; The key was to reduce the routines to the most
; basic concepts. This is alternative to doing
; every possible combination in detail.
;
; Indirect addressing is used extensively.
; To minimize ram space, this means variables
; are redefined in every procedure.
; Nevertheless, there is a common structure.
;
; Ram memory is precious. The following variables
; are used throughout. Locations are allocated
; for data. Some address use two different names,
; just for convenience.
;
;FraD equ 3BH ;4 bytes for fraction
;FraA equ 38H ;same space as upper Gap
;RemD equ 33H ;4 bytes for remainder
;RemA equ 30H ;same space as upper Tmp
;GapH equ 3BH ;8 bytes
;GapA equ 34H ;BCD digits for display & RESULTS
;TmpH equ 33H ;8 byte
;TmpA equ 2CH ;temporary or scratch
;HexD equ 2BH ;double word
;HexA equ 28H ;hexidecimal for all calculations

Chapter 43 Math Functions 313
; R6 = size, mul
; R5 = carry, mul
; R4 = mul
; R3 = loop control iteration
; R2 = loop control & size number of bytes
; R1 = @ source
; R0 = @ destination
;
; To enter the basic routines, place variables
; and size to count info in R0, R1, R2
; using something like the JHEX4 subroutine.
;-------------------------------------------------
JHEX4:
;-------------------------------------------------
; Standard arrangement for most math.
; Do this or a similar procedure before calling
; the math manipulation process.
mov R0,#HexA ;destination
mov R1,#TmpA ;source
mov R2,#4 ;4 bytes
ret
;-------------------------------------------------
JCLEAR:
;-------------------------------------------------
; Clear variables @ R0.
; R2 = # bytes
;ZERO GOES IN
mov @R0,#0 ;clear register
inc R0 ;next bit
djnz R2,JCLEAR ;continue loop
ret
;-------------------------------------------------
JCOPY:
;-------------------------------------------------
; Copy source to destination.
; Do not change source.
mov A, @R1 ;source

inc R1
mov @R0, A ;move to destination
inc R0 ;next
djnz R2, JCOPY ;loop
ret
;-------------------------------------------------
JCPL:
;-------------------------------------------------
; Complement the destination.
; Simply change all bits.
mov A, @R0 ;destination
cpl A ;complement
mov @R0, A ;save at same locate
inc R0 ;next
djnz R2, JCPL ;loop
ret
;-------------------------------------------------
JNEG:
;-------------------------------------------------
; Negate the destination
; It is complement and add 1
; That is called twos complement.
setb C ;set c
JNEG1: mov A, @R0 ;destination
cpl A ;complement
addc A, #0 ;add 1
mov @R0, A ;save
inc R0 ;next
djnz R2, JNEG1 ;loop
ret
;-------------------------------------------------
JINCSC:
;-------------------------------------------------
; Increment the destination.
; SC: set carry, signed
; WC: with carry that comes in.
setb C ;set
;---------

JINCWC:
;---------
addc A, #0 ;add C to byte
mov @R0, A ;save
inc R0 ;next
djnz R2,JINCWC ;loop
ret
;-------------------------------------------------
JDECSC:
;-------------------------------------------------
; Decrement the destination
; SC: set carry
; WC: with carry that comes in.
setb C
;--------- ;WITH CARRY
JDECWC:
;---------
subb A, #0 ;take away the carry
mov @R0, A ;save
inc R0 ;next
djnz R2,JDECWC ;loop
ret
;-------------------------------------------------
JINC10:
;-------------------------------------------------
; Increment the value by 10.
; This is a decimal increment
inc @R0 ;destination
mov A, @R0
cjne A,#10,JINC19 ;<>10, so exit
mov @R0, #0 ;=10,so set digit= 0
inc R0 ;next
djnz R2,JINC10 ;loop
JINC19: ret
;-------------------------------------------------
JSHIFTL:
;-------------------------------------------------

; Rotate left carry by 1 bit at time.
; SHIFT clears the C
; ROTATE leaves C bit as it comes in.
; R2 = # bits
;ROTATE C=0
clr C
;--------- ;ROTATE CARRY
JROTATEL:
;---------
mov A,@R0 ;point to HexS etc
rlc A
mov @R0,A ;update variable
inc R0 ;next bit
djnz R2,JROTATEL ;continue loop
ret
;-------------------------------------------------
JIFZERO:
;-------------------------------------------------
; Test if the value is zero.
; C=1, if zero
; C=0, in not zero
clr C
JZER1: mov A, @R0 ;get value
jnz JZER9 ;notZero
inc R0 ;next
djnz R2, JZER1 ;loop
setb C
JZER9: ret
;-------------------------------------------------
JIFLESS:
;-------------------------------------------------
; If @R0 < @R1, then C=1.
; Subtract but do not save results.
; R2 = # bytes
;subb BUT NOT SAVE
clr C
JLES1: mov A,@R0 ;point to HexS etc
subb A,@R1 ;subtract but not
save

inc R0 ;next bit
inc R1
djnz R2,JLES1 ;continue loop
ret
;-------------------------------------------------
JSUBSC:
;-------------------------------------------------
; Subtract @R0 = @R0 - @R1.
; SC: set carry, Signed
; CC: clr carry, Unsigned
; R2 = # bytes
;SIGNED
setb C ;set borrow
sjmp JSUB1
;--------- ;UNSIGNED
JSUBCC:
;---------
clr C
JSUB1: mov A,@R0 ;point to dest
subb A,@R1 ;@R0 = @R0 - @R1
mov @R0,A
inc R1
inc R0 ;next bit
djnz R2,JSUB1 ;loop
ret
;-------------------------------------------------
JADDCC:
;-------------------------------------------------
; @R0 = @R0 + @R1
; R2 = # bytes
;add & C=0
clr C
JADD1: mov A, @R0
addc A, @R1
mov @R0, A
inc R0
inc R1
djnz R2, JADD1 ;loop

ret
;-------------------------------------------------
JMULX16:
;-------------------------------------------------
mov R2,#16
mov R0,#HexA
ljmp JMULXR1
;-------------------------------------------------
JMULXR1:
;-------------------------------------------------
; Multiply @R0 by one byte in R1.
; R2 = number of bytes in R0
; R3:(@R0) = R1*(@R0)
;
mov R3, #0 ;clear carry byte
;LOOP
JMXR1: mov A, @R0 ;LSByte
mov B, R1 ;1 byte multiplier
mul AB ;1 byte multiply
add A, R3 ;add Carry byte
mov @R0, A ;save LSB of mult
inc R0 ;next higher byte
mov A, B ;get MSB of mult
addc A, #0 ;add Carry bit
mov R3, A ;save as Carry byte
djnz R2, JMXR1 ;loop on size
;TERMINATE
ret
;+++++++++++++++++++++++++++++++++++++++++++++++++
JDIV32:
;-------------------------------------------------
; Standard arrangement for multiply.

mov Size,#4 ;number of bytes
ljmp JDIVIDE
;-------------------------------------------------
JDIV16:
;-------------------------------------------------
mov Size,#2 ;number of bytes
ljmp JDIVIDE
;-------------------------------------------------
JDIVIDE:
;-------------------------------------------------
;--Entry: HexA, TmpA
;--Exit: GapA, FraA
; Divide routine for up to Size=4 bytes.
;
; Operator / Divisor = Quotient.Fraction + Remain
; HexA / TmpA = GapA . FraA + RemA
;
; R3 = ;counter for # units
; R2 = ;number of bytes
; R1 = ;source
; R0 = ;destination
;
; Divide and Fraction loop have the following.
; Loop times: R3= 8*SizeX; R3>0; R3 is decrement
; Results: Rem:Op<<=1; C=Rem/Divisor
; Rem -=Divisor
;ZERO VARIABLES
mov R0, #RemA ;remain
mov R2, #4 ;all
lcall JCLEAR ;Rem = 0
mov R0, #GapA ;quotient
mov R2, #4 ;all
lcall JCLEAR ;Quo = 0
;DIVIDE LOOP SIZE
mov A, Size ;Size

rl A
rl A
rl A
mov R3, A ;R3=8*Size =loop
;DIVIDE LOOP
JDIV1: lcall JDIV ;divide
mov R0, #GapA ;quotient
mov R2, Size
lcall JROTATEL ;Quo= Quo<<1 | C
djnz R3, JDIV1 ;loop
;CLEAR FRACTION
mov R0, #FraA ;fraction
mov R2, Size
lcall JCLEAR ;Fract = 0
;FRACTION LOOP SIZE
mov A, Size ;size
rl A
rl A
rl A
mov R3, A ;R3 = 8*SizeX
;FRACTION LOOP
JFRAC: lcall JDIV ;divide
Mov R0, #FraA ;fraction
Mov R2, Size ;size
lcall JROTATEL ;Frac= Fract<< 1 | C
djnz R3, JFRAC ;loop
ret
;-------------------------------------------------
JDIV:
;-------------------------------------------------
; Rem:Op<<=1; C=Rem/Divisor; Rem -= Divisor
;DIVIDE
clr C
mov R0, #HexA ;operator
mov R2, Size ;# bytes
lcall JROTATEL ;C:Op = Op*2

mov R0, #RemA ;remainder
lcall JROTATEL ;Rem= Rem*2 + C
mov R1, #TmpA ;divisor
lcall JIFLESS ;C=1 if @R0<@R1
jc JDIV9 ;if Rem>= Divisor
mov R1, #TmpA ;divisor
lcall JSUBCC ;Rem -=Divisor,unsgn
JDIV9: cpl C ;C = !C
ret
;-------------------------------------------------
JMUL32:
;-------------------------------------------------
mov ArgH,#HexA ;destination
mov ArgT,#TmpA ;source
mov ArgG,#GapA ;results
mov SizeT,#4 ;4 bytes in Tmp
mov Size,#4 ;4 bytes in Hex
ljmp JMULTIPLY
;-------------------------------------------------
JMUL16:
;-------------------------------------------------
mov ArgH,#HexA ;destination
mov ArgT,#TmpA ;source
mov ArgG,#GapA ;results
mov SizeT,#2 ;2 bytes in Tmp

mov Size,#2 ;2 bytes in Hex
ljmp JMULTIPLY
;-------------------------------------------------
JMULTIPLY:
;-------------------------------------------------
; Multiply multiple bytes. Variables are used
; as pointers to the values.
; @ArgG = @ArgH * @ArgT
; Gap = Hex * Tmp
;
;SizeT equ 33H ;same space as Tmp upper bytes
;ArgG equ 32H ;used in multiply
;ArgT equ 31H ;
;ArgH equ 30H ;same space as Tmp upper bytes
;GapH equ 3BH ;8 digit
;GapA equ 34H ;result
;TmpD equ 2FH ;4 byte
;TmpA equ 2CH ;math low byte
;HexD equ 2BH ;double word
;HexA equ 28H ;low byte of variable
;
; R6 = size, mul
; R5 = carry, mul
; R4 = mul
; R3 = loop control iteration
; R2 = loop control & size number of bytes
; R1 = @ source
; R0 = @ destination
;
; Move the location of the source and destination
; to the ArgS variables.
; Move the size of the Arguments to SizeS.
; The Arg and Size will be incremented as the
; process goes through each byte.
;
; Xcand= multiplicand, Xer= multiplier
mov R5, #0 ;clear carry byte
;LOOP
JMUX1: mov R0, ArgH ;Xcand addr
mov A, @R0 ;Xcand data

mov R2, A ;Xcand data
inc ArgH ;R2= *ArgX++
mov R0, ArgG ;Res addr
inc ArgG ;R0= ArgZ++
mov R1, ArgT ;Xer addr
mov R3, SizeT ;Xer loop
mov R4, #0
jz JMUX9 ;result=0, exit
;NIBBLE MULTIPLY
JMUX2: mov A, R2 ;Xcand data
mov B, @R1 ;Xer data
inc R1 ;next byte
mul AB ;B:A= *R1++ * R2
add A, R4 ;add carry byte
xch A, B
addc A, #0 ;add carry bit
xch A, B ;B:A += R4;
add A, @R0 ;add Xcand data
mov @R0, A ;save Xer
inc R0 ;next Xer
mov A, B
addc A, #0 ;add carry bit
mov R4, A ;R4:*R0++ =B:A+ *R0
djnz R3, JMUX2
mov R3, Size ;bytes in Xcand
mov A, R4 ;carry byte
;CARRY BY 1
JMUX3: add A, @R0 ;Xcand data
mov @R0, A ;save new Xcand
inc R0 ;C:*R0++ += *R0 + A
jnc JMUX9 ;C<>0, next loop
mov A, #1
djnz R3, JMUX3 ;next Xcand bit
;
inc R5 ;inc carry byte

JMUX9: djnz Size,JMUX1 ;next Xcand
;TERMINATE
ret
;-------------------------------------------------
JSQROOT:
;-------------------------------------------------
; QuoS = square root of HexS
; Enter with a value.
; Rotate the 2 MSB into a new variable.
; Compare comparison with new.
; If comparison < new, then C=1.
; Then new is greater than nearest multiple of 2.
; So shift another bit into answer.
; Calculate new = new-compare-C = new-(compare+1)
; This removes last compare from re consideratio.
; Multiply comparison by 2.
; If Comparison < new then orl #4
; This sets bit equiv to nest multiple.
; Continue.
;
; SHL 2 bits into D If X<D, C=1, C->Q, D=D-X-C
; If X>D, C=0. C->Q
; SHL 1 bits in X If X>D, X=X orl #4
;
; HexA = Input 4 bytes
; TmpA = comparison 4 bytes
; RemA = remaining 4 bytes
; GapA = result 2 bytes
;
; R0, R1, R2, R3
;INITIALIZE
mov Size,#4 ;#bytes of source
;CLEAR VARIABLES
mov R0,#GapA ;result
mov A,Size ;# bytes
rr A ;only need 1/2
mov R2,A ;2 bytes
lcall JCLEAR ;clear
mov R0,#RemA ;remaining
mov R2, Size ;4 bytes

lcall JCLEAR ;clear
mov R0,#TmpA ;comparison
lcall JCLEAR ;clear
mov A,Size ;# bytes
rl A ;mul by 2
rl A ;mul by 2
mov R3,A ;bits shift to answ
;SHIFT 2BITS=mul BY4
JSQR1: mov R0,#HexA ;input
lcall JSHIFTL ;msb to C
lcall JROTATEL ;c to lsb
mov R0,#HexA ;input
lcall JSHIFTL ;msb to C
lcall JROTATEL ;c to lsb
;COMP REMAIN& INTERM
mov R0,#TmpA ;interim comparison
lcall JIFLESS ;TmpA<TmpE, C=1
mov FgC,C ;c=resul,abit>2*bbit
mov R0,#GapA ;result
mov A,Size ;# bytes
rr A ;divide by 2
mov R2,A ;2 bytes
lcall JSHIFTL ;C to lsb
jnb FgC,JSQR2 ;c=0,no change reman
;UPD REMAIN& INTERIM

mov R1,#TmpA ;interim
lcall JSUBSC ;sign,rem=rem-intr-1
JSQR2: mov R0,#TmpA ;interim
lcall JSHIFTL ;MSB to C
jnb FgC,JSQR3 ;<>sq,no change intm
;IF 2**2 PUT 4 INTRM
mov A, TmpA ;first interim
orl A, Size ;sq,put 2**2 in intm
JSQR3: djnz R3,JSQR1 ;loop in 16 bit answ
mov HexA,GapA ;answer
mov HexB,GapB ;answer
ret
ž Ÿ

SECTION IV – HARDWARE
ž Ÿ

44
PARTS AND PIN-OUTS
Thought
The universal engineering unit = $
Dr. D
Watch your money _______________
Parts are separated into categories. The Proto then uC group is the
basic components to make a working computer. All these will be
later used on a development board. This also includes cable for the
SPI download of programs
The uC only list is the remaining components for the development
board. These will provide interface to most projects. This also
includes cables and connectors for RS 232 communications with a
PC.
The optional list includes liquid crystal display, analog to digital
converter, static ram, and expansion headers. Not all these will
always be needed for every project.
The project list is components that will be used with a proto-board
to design various projects. This list will change depending on the
particular application.

Chapter 44 Parts & Pin-outs 329
Proto then uC board ______________
Part Device Package Where Vendor Number Qty Cost
89S8252 uP Dip40 U1 1
11.059MHz Xtal 2 pin Y1 1
10 K : Res Axial 0.4 1
1.5 K : Res Axial 0.4 R3 1
330 : Res pak Sip8 Rp8 1
10 uF Cap RB.2/.4 2 2
39 pF Cap Axial 0.3 C1,2 2
7 Segment Display CC DS1 1
V Reg U12 1
Quad latch Dip14 U3 1
5 V Pwr Sup Wall wart JP16 1
RJ45 Jack ISP 2
RJ45 Adapter to DB25 ISP 1
RJ45 Adapter to DB9 RS232 1
5’ Cat5E cable RJ45 ISP 2
uC board only____________________
Develop Board 2 sided B1 TUEE 1
Socket uP Dip40 U1 1
74573 Octal lat Dip20 U4, 5, 6 4
233 Rs232 Dip16 U9 1
P22v10 Peel Dip24 U8 1
Socket Peel Dip24 U8 1
2.2 K : Res Axial 0.4 1
2.2 K : Res pak Sip10 BR1 1
R Y R LED Diode 0.2 DS2 1
Pushbutton S1, 2 2
Diode Diode 0.2 D1
0.1 uF Cap Axial 0.3 C4
4.7 uF Cap polar Axial 0.3 C6
mm Jack Power in 1

uC board optional ________________
6.8V Zener LCD D2 1 n/o
2N3906 Xistor LCD To-92A Q4,5 2 n/o
50K Thermistor Axial 0.4 TH1 1 n/o
300K Res LCD Axial 0.4 R4 1 n/o
3.3K Res LCD Axial 0.4 R9 1 n/o
2x40 LCD Sip14 JP2 1
Socket LCD Sip14 JP2 1
Cable LCD 3” ribbon JP2 1
32Kx8 Sram Dip28 U2 1 n/o
Socket Sram Dip28 U2 1 n/o
ADC Dip8 U10 1
FET Sip3 Q1,2,3 3
1K Resistor Axial 0.4 R4
485 Rs485 Dip8 U10
InfraRed Detector Sip3 U11
Projects ________________________
7 Segment Display CC 2
3x4 keypad 1
330 Ohm Res Axial 0.4 6 n/o
R Y G LED Diode 0.2 6 n/o
2 Switches Dip 4 1
7406 Hex Inverter 1
Breadboard 3 Column 1

uC board headers & jumpers_______
8 pin Expansion Sip 8 4
16 pin LCD Sip16 Jp2 1
2 pin Analog in Sip2 Jp3 1
11 pin 7-Segment Sip11 Jp4 1
4 pin mmio Sip4 Jp6 1
8 pin Key in Sip8 Jp12 1
2 pin Power in Sip2 Jp13 1
8 pin Address Sip8 Jp14 1
2 pin Power out Sip2 Jp17 1
2 EA’ Jumper J1 1
2 PLD bypass Jumper J15 1
2 Serial hand Jumper Jp5 1
2 7seg select Jumper Jp9 1
ž Ÿ
PLD / PEEL pin-out ______________
* V
IO/CK 1 24 VCC
I1 2 23 F9
I2 3 22 F8
I3 4 21 F7
I4 5 20 F6
I5 6 19 F5
I6 7 18 F4
I7 8 17 F3
I8 9 16 F2
I9 10 15 F1
I10 11 14 F0
Ground 12 13 I11
22V10 PLD

Microprocessor pin-out ___________
* V
P1.0 1 T2 40 VCC
P1.1 2 T2 EX AD0 39 P0.0
P1.2 3 AD1 38 P0.1
P1.3 4 AD2 37 P0.2
P1.4 5 SS/ AD3 36 P0.3
P1.5 6 MOSI AD4 35 P0.4
P1.6 7 MISO AD5 34 P0.5
P1.7 8 SCK AD6 33 P0.6
Reset 9 AD7 32 P0.7
P3.0 10 RXD VPP 31 /EA
P3.1 11 TXD PROG/ 30 ALE
P3.2 12 INT0/ 29 /PSEN
P3.3 13 INT1/ A15 28 P2.7
P3.4 14 T0 A14 27 P2.6
P3.5 15 T1 A13 26 P2.5
P3.6 16 /WR A12 25 P2.4
P3.7 17 /RD A11 24 P2.3
Xtal2 18 A10 23 P2.2
Xtal1 19 A9 22 P2.1
Ground 20 A8 21 P2.0
MCS 51 / 8031 / 8051 / AT89S8252

Buffer pin-out ___________________
* V
//OE 1 20 VCC
D0 2 19 Q0
D1 3 18 Q1
D2 4 17 Q2
D3 5 16 Q3
D4 6 15 Q4
D5 7 14 Q5
D6 8 13 Q6
D7 9 12 Q7
Ground 10 11 LE
Operating INP OUT
Modes OE LE Dn Qn
enable & read L H L L
register transp L H H H
latch & read L L L L
register L L H H
latch register& H L L Z
disable outputs H L H Z
The data for the latch is setup when the latch
enable (LE) pin is driven high. When the pin
is pulled low, the latch traps the data.
74573 Octal D latch
* V
A1 1 14 VCC
/Y1 2 13 A6
A2 3 12 /Y6
/Y2 4 11 A5
A3 5 10 /Y5
/Y3 6 9 A4
Ground 7 8 /Y4
Open-collector requires connection of 2.2K
pull-up resistor between each output and VCC.
7406 Hex inverter,open-collector buffer driver

RS232 & RS233 pin-out __________
* V
C1+ 1 16 VCC
V+ 2 15 Ground
C1- 3 14 T1OUT
C2+ 4 13 R1 IN
C2- 5 12 R1OUT
V- 6 11 T1IN
T2OUT 7 10 T2IN
R2IN 8 9 R2OUT
Max C1 – C5
232 1 uF
232A 0.1 uF
RS 232 Interface
* V
T2IN 1 20 R2OUTC
T1IN 2 19 R2IN
R1OUT 3 18 T2OUT
R1IN 4 17 V-
T1OUT 5 16 C2-
GND 6 15 C2+
Vcc 7 14 V+
C1+ 8 13 C1-
GND 9 12 V-
CS- 10 11 C2+
RS 233 Interface

7-Segment & LCD pin-out _________
* V
a 1 14 common
f 2 a 13 b
common 3 f / / b 12 nc
dp pre 4 g 11 g
nc 5 e / / c 10 c
nc 6 d 9 dp post
e 7 8 d
Pin Function
1 a
13 b
10 c
8 d
7 e
2 f
11 g
4 dp pre
9 dp post
3 common
14 common
7 Segment display
Pin Symbol Level Function
1 DB7 H / L data bus line
5 DB3 H / L data bus line / no-connection for 4-bit operation
9 E1 H, H->L enable signal (no pull-up resistor)
10 R-/W H / L read/write select signal, h : read l : write
11 RS H / L register select signal
12 VEE – power supply for LCD drive
13 VSS – power supply (0v, ground)
14 VCC – power supply for logic
15 E2 H, H->L enable signal (no pull-up resistor)
16 NC – no-connection
17 LED K – led cathode terminal
18 LED A – led anode terminal
Liquid crystal display

A/D Converter pin-out ____________
* V
CH0 1 16 VCC
CH1 2 15 CCLK
COM 3 14 SCLK
/DOR 4 13 DI
EOC 5 12 DO
VREF- 6 11 /CS
VREF+ 7 10 /CONV
DGnd 8 9 VA+
LTC 1098 analog / digital converter
* V
REF+ 1 8 VCC
AN-IN 2 7 SCLK
REF- 3 6 DOUT
GND 4 5 /CS
TLC549 analog / digital converter

Memory pin-out _________________
* V
A14 1 VPP ROM 28 VCC
A12 2 ROM A14 27 /WE
A7 3 26 A13
A6 4 25 A8
A5 5 24 A9
A4 6 23 A11
A3 7 22 /OE
A2 8 21 A10
A1 9 20 //CE
A0 10 19 DQ7
DQ0 11 18 DQ6
DQ1 12 17 DQ5
DQ2 13 16 DQ4
Ground 14 15 DQ3
Memory 256K (32K x 8)
* V
A15 1 28 VCC
A12 2 27 A14
A7 3 26 A13
A6 4 25 A8
A5 5 24 A9
A4 6 23 A11
A3 7 VPP 22 /OE
A2 8 21 A10
A1 9 20 /CE
A0 10 19 DQ7
DQ0 11 18 DQ6
DQ1 12 17 DQ5
DQ2 13 16 DQ4
Ground 14 15 DQ3
ROM 512K (64K x 8)
ž Ÿ

Cable pin-out, SPI & serial ________
Two connectors are used on the board. One is required for the
RS232 communications which will connect by cable to the PC serial
port. The other is required for in-system-programming (ISP) which
will connect by cable to the PC parallel port.
Note, connect a 1.5K resistor between the cable and the reset.
Funct Printer To Cable Cable
db 25 RJ45 TIA568B Telco
Pin Pin Color Alt
MOSI 7 1 W/Orange Black
MISO 10 2 Orange Yellow
3 W/Green White
4 Blue Red
Gnd 18 5 W/Blue Green
6 Green Blue
Sck 8 7 W/Brown Brown
Reset 6 8 Brown Orange
SPI - Parallel RJ45 Cable
Funct Serial To Board Cable Cable
db 9 RJ11 RJ45 TIA568B Telco
Pin Pin Pin Color Alt
Jump 1 1 W/Orange Black
Jump 6 2 Orange Yellow
TXD 3 1 3 W/Green White
RXD 2 2 4 Blue Red
Gnd 5 3 5 W/Blue Green
nc 4 4 6 Green Blue
Jump 7 7 W/Brown Brown
Jump 8 8 Brown Orange
RS232 – Serial RJ45 Cable

The RJ45 connectors have multiple functions. The following table
illustrates the relationship between Ethernet, telephone, and the
connections used with the processor schematic.o
Funct Printer To Cable Cable Board Funct To Serial Funct
db 25 RJ45 TIA568B Telco RJ45 RJ11 db 9
Pin Pin Color Alt Pin Pin Pin
MOSI 7 1 W/Orange Black 1 P1.5 1 Jump
MISO 10 2 Orange Yellow 2 P1.6 6 Jump
3 W/Green White 3 T1X 1 3 TXD
4 Blue Red 4 R1X 2 2 RXD
Gnd 18 5 W/Blue Green 5 Gnd 3 5 Gnd
6 Green Blue 6 R2x 4 4 nc
Sck 8 7 W/Brown Brown 7 P1.7 7 Jump
Reset 6 8 Brown Orange 8 Reset 8 Jump
RJ45 Cables
ž Ÿ

45
DEVELOPMENT BOARD
Thought
A design – build contract
means you do it all.
Design _________________________
The development board is a general-purpose microcontroller
system. It has a multipurpose design. First, it allows addition of
substantial hardware without the complexity of wires on a proto-
board. It can be used as a prototype for various devices. Finally, the
configuration may be installed in standard electrical enclosures, so it
can be used as a production controller. The schematic follows.
Options ________________________
Obviously, numerous options can be implemented by simply not
installing various devices. However, this does give substantial
flexibility in uses and applications.
Perhaps the most significant option is the memory. Either EPROM
or SRAM can be installed in the socket. The pin functions are
different on these two devices. There are three pins that control the
variations. Chip enable is listed simply for reference.

Chapter 45 Development Board 341
Memory EPROM SRAM
/CE Ground Ground
/OE /PSEN jumper /RD or Gnd jumper
Pin 1 VPP A14
Pin 27 A14 /WE
To change between memory chips only two things are required.
First place the correct chip in the socket. Next, change the jumper
for output enable, /OE. The pin conversions are made inside the
PEEL, and the chip is permanently enabled.
HyperTerminal __________________
Several programs can be loaded to assist in using the
microprocessor development system. HyperTerminal is a Windows
accessory communications program. It is setup to use the serial port
for communications with the microcontroller. Baud rate is 9600.
Programs can be dropped onto the serial port. The program will also
display any information arriving on the serial port.
Test Program ____________________
The program is run from the development board. Use the sample
test program MODTest shown as a simple software illustration. Run
the program through an assembler. Take the *.hex file results and
download to the board memory. This program is very simple, but
illustrative.
Schematic _______________________
This board is custom designed for system projects. It incorporates
numerous features and characteristics, including selecting various
processors and devices. The design is protected by international
copyright.
ž Ÿ

Board specifications ______________
The board design is based on numerous performance specifications.
These specs are identified by the device. Therefore, most criteria
will be listed twice, once as an output from one device, then as a
input to another.
PROCESSOR
1. 89S8252 (8051 core) w/ in-system programming, internal
program & data memory
2. Crystal 11.059 MHz
3. Reset can be from auto circuit, pushbutton, and ISP.
4. Package is 40 pin, board to fit electrical gang-box.
EXTERNAL MEMORY
5. Expansion memory socket in a common socket
a. 32K eeprom
b. 32K sram w/ socket for DIP or wide
c. Memory pin 1, 27, and OE from PLD
POWER
6. Wall-wart mini-jack power connector pin
7. LP2954 voltage regulator enlarge hole size
8. Add pins for 5V & Ground
9. Place filter capacitor on line.
ISP
10. Quad buffer on in-system-programming lines for port 1.
11. Enable buffer with uc reset complemented from PLD.
12. Connect in and out lines as required.
RESET
13. Place capacitor and resistor to auto reset on pin 9.
14. Place pushbutton for manual reset.
PUSHBUTTON / LED ON P35
15. Pushbutton as input on P35 (T1).
16. Add LED to P35 (T1) as output.
17. Connect pull-up to led so can work together with switch.

PORT 1
18. Connect IIC to pins 6 & 7, share with SPI
19. Connect 7-segment select lines to pins 2-4.
20. Connect SPI on pins 5-7.
PORT 3
21. Connect RS232 on pins 0&1
22. Connect handshake on pin 2 (INT0).
23. Connect infrared receive/ transmit module to P33 (Int1).
24. If Int1 is used separately, the infrared must be covered with
black tape.
25. Connect RS485 select to P34 (T0).
26. Connect LED & pushbutton to P35 (T1).
I/O LATCHES
27. All external I/O uses memory-mapped connections.
28. Display-Out Latch for 7-segment data.
a. Latch Enable from PLD.
b. Output Enable is connected to ground.
c. Take data to on-board 7-segment display through current
limiting resistors.
d. Take data out after resistor to expansion header.
29. Key-Out Latch for keypad column output.
a. Latch Enable from PLD. Enable when write to column.
b. Output Enable is connected to ground.
c. Use upper nibble for columns out to keypad.
d. Use other bits for digital out.
30. Key-In Latch for keypad row input.
a. Output Enable from PLD. Enable when read from row.
b. Latch Enable connected to 5 volts.
c. Use lower nibble for rows in from keypad..
d. Connect pull-up resistors on keypad inputs.
e. Other bits are for digital input.
31. Address Latch to separate address from data on Port 0.
a. LE is connected to ALE.
b. OE is connected to ground.
DISPLAY CONTROL

32. Assign three bits to select three different 7-segment displays.
a. Add drivers for select lines, 7406 open-collector with pull-
ups or 30 A MosFets.
b. On-board 7-segment has header jumper to select ground or
a bit to control.
c. Connect control lines to P12, P13, P14.
LCD
33. Bypass auto-contrast circuit with a jumper to ground.
34. Add backlight connector adjacent to LCD header
35. LCD has 3 control lines, RW’, RS, En. Shared with 7-segment.
a. Take controls to LCD header.
b. Connect Enable to P1.2.
c. Connect Write enable to P1.1.
d. Connect data/instruction register select to P1.0.
A/D CONVERT
36. Use with serial peripheral interface.
a. Connect serial I/O to port 1 SPI bus using MISO, SCK
b. Connect /CS to P1.4, /SS.
37. Make reference voltage 0-5 volts for rail to rail operation.
38. Make header for A/D input with analog ground.
39. Connect analog ground to digital ground at 1 point via a jumper
that can be isolated.
INFRARED
40. Add infrared receiver / transmitter
41. Connect to P33 (Int1) pin
SERIAL
42. Use Max 233 since it does not require external caps.
a. Because of limited port space, only 1 handshake is used..
b. Connect Int0 as handshake control.
c. Header selects in or out line for connection to RJ45
43. Add Max 485 chip as option.
a. Connect control lines RE and DE together to P34 (T0).
b. Connect serial differential lines to header.
c. These will parallel the serial lines for the Max233.

IIC
44. Connect IIC to Port 1.
a. SDA connect to P17.
b. SCL connect to P16
RJ45 CONNECTORS
45. Connect 3 in-system-programming lines plus common to ISP
connector.
46. Connect serial TX, RX, and handshake to serial connector.
a. Connect handshake to a jumper between out and in
handshake
b. Configurable for computer or modem
PLD (22V10 PEEL)
47. Connect input lines
a. PSEN
b. WR’
c. RD’
d. Reset
48. Connect input address lines (08001h)
a. A15, A14, A2, A1, A0.
49. Connect input data line, D0
50. Leave pin 1 open for clock
51. One spare input & output line.
52. Connect output lines
a. Key OE
b. Key LE
c. Display LE
d. ISP chip select
53. Connect memory control output
a. Memory OEn
b. Memory pin MP1
c. Memory pin MP27
d. MmWr
e. MmRd
OFF-BOARD HEADERS
54. In-system-program port through 4-bit isolation latch
a. Connect in and out lines as required.

55. Seven-segment
a. Data from Display-Out latch
b. Seven-segment control lines from Port 1 through 7406.
56. LCD
a. Data from P0
b. Control from microprocessor.
c. Add backlight connector. It can have separate power source.
57. Address latch output connect for expansion MMIO
58. Keypad
a. Column from latch upper nibble.
b. Row to latch lower nibble with pull-up resistors.
59. Address latch
a. Output address for MMIO or expansion
b. Have pins including ALE, PSEN, MMIO RD, MMIO WR.
60. Analog input
a. Connect to ADC
b. Analog ground
61. RS485 needs only 2 differential data lines
62. IIC needs 2 lines.
63. Power pins for 5V & Ground
SELECT JUMPERS
64. External memory EA’ select 5V or Ground
65. Seven-segment onboard select ground or 7406 / MosFet
66. Memory Output Enable select ground or PLD
67. LCD contrast jumper to Ground
68. Handshake select for RS232
GENERAL
69. Be cautious that fan out is not a problem on port 0.
PORT CONNECTIONS
70. Port 1
a. 10 – SegA / RS
b. 11 – SegB / RW’
c. 12 – SegC / En
d. 13 –
e. 14 – /SS
f. 15 – SPI MOSI

g. 16 – SPI MISO
h. 17 – SPI SCK
71. Port 3
a. 30 – serial RXD
b. 31 – serial TXD
c. 32 – Int1 serial handshake
d. 33 – Int0 infrared input
e. 34 – T0 RS485 select
f. 35 – pushbutton and LED
g. 36 – write’
h. 37 – read’
ž Ÿ

46
IN SYSTEM PROGRAMMING
Thought
What is the tradeoff between
hardware and software?
Serial downloading _______________
Atmel’s AT89S8252 flash microcontroller offers 8K bytes of in-
system re-programmable flash code memory and 2K bytes of
EEPROM data memory. This information and much more is
available from the Atmel data sheet.
Both the Code and Data memory arrays can be programmed using
the serial SPI bus while RST is pulled to VCC. The serial interface
consists of pins SCK, MOSI (input) and MISO (output).
P1.7 P1.6 P1.5 P1.4
SCK MISO MOSI /SS
After RST is set high, the Programming Enable instruction must be
executed first before program/erase operations can be executed.
An auto-erase cycle is built into the self-timed programming
operation (in the serial mode ONLY). It is unnecessary to first
execute the Chip Erase instruction, unless the lock bits have been
programmed. The Chip Erase operation turns the contents of every
memory location in both the Code and Data arrays into 0FFH.

The Code and Data memory arrays have separate address spaces:
0000H to 1FFFH for Code memory and 000H to 7FFH for Data
memory.
Either an external system clock is supplied at pin XTAL1 or a
crystal needs to be connected across pins XTAL1 and XTAL2. The
maximum serial clock (SCK) frequency should be less than 1/40 of
the crystal frequency. With a 24 MHz oscillator clock, the
maximum SCK frequency is 600 kHz.
Programming algorithm __________
To program and verify the flash controller is in the serial
programming mode, the following sequence is recommended:
1. Power-up sequence: Apply power between VCC and GROUND
pins. Set RST pin to Hi. If a crystal is not connected across pins
XTAL1 and XTAL2, apply a 3 MHz to 24 MHz clock to
XTAL1 pin and wait for at least 10 milliseconds.
2. Enable serial programming by sending the Programming Enable
serial instruction to pin MOSI/P1.5. The frequency of the shift
clock supplied at pin SCK/P1.7 needs to be less than the CPU
clock at XTAL1 divided by 40.
3. The Code or Data array is programmed one byte at a time by
supplying the address and data together with the appropriate
Write instruction. The selected memory location is first
automatically erased before new data is written. The write cycle
is self-timed and typically takes less than 2.5 ms at 5V.
4. Any memory location can be verified by using the Read
instruction, which returns the content at the selected address at
serial output MISO/P1.6.
5. At the end of a programming session, RST can be set low to
commence normal operation.

Chapter 46 In-System Programming 351
If a power-off sequence is required, use only three tasks. Set
XTAL1 to Low, if a crystal is not used. Set RST to Low. Turn off
VCC power.
Data polling is used to indicate the end of a write cycle, which
typically takes less than 2.5 ms at 5V.
Programming instruction __________
The Instruction Set for serial programming follows a 3-byte
protocol and is shown in the following table:
Operation Byte 1 Byte 2 Byte 3
Programming
Enable
1010 1100 0101 0011 xxxx xxxx Enable serial programming
interface after RST goes high.
Chip Erase 1010 1100 xxxx x100 xxxx xxxx Chip erase both 8K & 2K
memory arrays.
Read Code
Memory
aaaa a001 low addr xxxx xxxx Read data from Code memory
array at the selected address. The
5 MSBs of the first byte are the
high order address bits. The low
order address bits are in the
second byte. Data are available at
pin MISO during the third byte.
Write Code
Memory
aaaa a010 low addr data in Write data to Code memory
location at selected address. The
address bits are the 5 MSBs of
the first byte together with the
second byte.
Read Data
Memory
00aa a101 low addr xxxx xxxx Read data from Data memory
array at selected address. Data
are available at pin MISO during
the third byte.
Write Data
Memory
00aa a110 low addr data in Write data to Data memory
location at selected address.
Write Lock
Bits
1010 1100 x x111 xxxx xxxx Write lock bits. Set LB1, LB2 or
LB3 = “0” to program lock bits.

Programming schematic __________
The connections to the microcontroller are very simple as shown in
the schematic. The standard connections are power, ground, crystal,
and reset. Only three additional lines are required for loading a
program into the memory.
Peripheral timing ________________
Serial peripheral interface (SPI) is a protocol to connect several
devices on a common data input and a common data output lines.
Both input and output happen simultaneously, as illustrated in the
diagram.
The SCLK line shifts data into the device (MOSI) on a rising edge.
So SCLK must be made LO then HI to input to the device from the
controller.
The SCLK line shifts data out from the device (MISO) on a falling
edge. So SCLK must be made HI then LO to output from the device
to the controller.
VCC
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset
Xtal2
Xtal1
Ground
P1.5 MOSI
P1.6 MISO
P1.7 SCK
Reset

Programming and printer _________
There are several software packages for performing the program
code download from the personal computer to the internal PROM
on the microcontroller. One of the easier to use programs is a small
executable file called AEC_ISP.EXE. This file can be downloaded.
Connections are needed on the personal computer for the three data
lines and a reset line. The only access with that many control wires
is the parallel port for the printer.
Note place a 1.5K resistor between the connector and the reset line.
The AEC_ISP program uses the pattern of connection shown in the
schematic below to make the tie between the printer port and the
corresponding SPI lines on port 1 of the microcontroller.
Once the program code has been downloaded, the microcontroller
must be reset. The SPI connection to the computer printer port can
be removed. It is important that the first few lines of executable
code in the program must not change port 1. This can cause a
conflict and lock-up. If that happens, the buffer must be cleaned and
the program downloaded again.

P1.0 1
P1.1 2
P1.2 3
P1.3 4
P1.4 5
MOSI P1.5 6
MISO P1.6 7
SCK P1.7 8
RESET 9
Ground 20
Microprocessor
1
2
7
8
5
RJ45
1 Strobe Ground25
2 D0 Ground24
3 D1 Ground23
4 D2 Ground22
5 D3 Ground21
6 D4 Ground20
7 D5 Ground19
8 D6 Ground18
9 D7
10ACK
Printer port
Connectors _____________________
Making a viable, custom termination between a printer connector
and a circuit board or proto board can be a tedious process. The best
connection technique uses available technology.
One very dependable technology is RJ45 Ethernet connectors. First,
obtain a printer port (DB25P) to Ethernet (RJ45S) adapter. Wire the
adapter to the appropriate pins according to the following table.
Use the required length of Ethernet cable with RJ45P connectors on
both ends. Typically, 5 feet is adequate. The standard pin
arrangement and color coding is shown in the chapter on
Networking.
Obtain a RJ45S connector that can be configured with pigtails to
mount to the proto board.
These three components provide a dependable, flexible, and
relatively inexpensive interconnection between the PC printer and
the microcontroller.

Funct Printer To Cable Cable Board Funct
db 25 RJ45 TIA568B Telco RJ45
Pin Pin Color Alt Pin
MOSI 7 1 W/Orange Black 1 P1.5
MISO 10 2 Orange Yellow 2 P1.6
3 W/Green White 3
4 Blue Red 4
Ground 18 5 W/Blue Green 5 Ground
6 Green Blue 6
Sck 8 7 W/Brown Brown 7 P1.7
Reset 6 8 Brown Orange 8 Reset
ž Ÿ

SECTION V – ARCHITECTURE
ž Ÿ

47
INSTRUCTION SET
Thought
Triad Principle:
Any item that can be uniquely identified,
can be further explained by three components
MOD
Microcontroller instruction set______
The instruction set is very powerful for a microprocessor. As a
result, this core is one of the most common used in microcontroller
applications.
Instructions can be divided into categories consisting of data
transfer, arithmetic, logic, program branch, and bit manipulation.
Instructions can be demonstrated many ways. The first group of
tables represents the categories of instructions. Then, two tables are
provided that are a matrix of the numeric instruction codes.
There are many ways data can be obtained. These are called
addressing modes. The machine has seven unique modes -
immediate, direct, register, indirect, relative, absolute, and bit.

Addressing modes _______________
R Register R7-R0 of the currently selected Register Bank.
Direct eight bit internal data location’s address. This could be an
Internal Data RAM location (0-127) or a SFR [i.e., I/O port,
control register, status register, etc. (128-255)].
@R eight bit internal data RAM location (0-255) addressed
indirectly through register R1or R0.
#data eight bit constant included in instruction.
#data 16 sixteen bit constant included in instruction.
addr 16 sixteen bit destination address. Used by lcall and ljmp. A
branch can be anywhere within the 64K byte Program
Memory address space.
addr 11 11-bit destination address. Used by acall and ajmp. The branch
will be within the same 2K byte page of program memory as
the first byte of the following instruction.
Rel Signed (two’s complement) eight bit offset byte. Used by sjmp
and all conditional jumps. Range is -128 to +127 bytes relative
to first byte of the following instruction.
Bit Direct Addressed bit in Internal Data RAM or Special
Function Register

Chapter 47 Instruction Set 359
Data transfer ____________________
Op Mnem Register Description Byt Cyc
74 mov A,#data Move immediate data to Accumulator 2 1
75 mov direct,#data Move immediate data to direct byte 3 2
76 mov @R0,#data Move immediate data to indirect RAM 2 1
77 mov @R1,#data Move immediate data to indirect RAM 2 1
7* mov Rn,#data Move immediate data to register 2 1
85 mov direct,direct Move direct byte to direct 3 2
86 mov direct,@R0 Move indirect RAM to direct byte 2 2
87 mov direct,@R1 Move indirect RAM to direct byte 2 2
8* mov direct,Rn Move register to direct byte 2 2
A6 mov @R0,direct Move direct byte to indirect RAM 2 2
A7 mov @R1,direct Move direct byte to indirect RAM 2 2
A* mov Rn,direct Move direct byte to register 2 2
E5 mov A,direct Move direct byte to Accumulator 2 1
E6 mov A,@R0 Move indirect RAM to Accumulator 1 1
E7 mov A,@R1 Move indirect RAM to Accumulator 1 1
E* mov A,Rn Move register to Accumulator 1 1
F2 mov @R0,A Move Accumulator to indirect RAM 1 1
F3 mov @R1,A Move Accumulator to indirect RAM 1 1
F5 mov direct,A Move Accumulator to direct byte 2 1
F* mov Rn,A Move Accumulator to register 1 1
90 mov DPTR,#datal6 Load Data Pointer w/ a 16-bit constant 3 2
83 movc A,@A+PC Move Code byte relative to PC to Acc 1 2
93 movc A,@A+DPTR Move Code byte relat to DPTR to Acc 1 2
E0 movx A,@DPTR Move Exter RAM (16-bit addr) to Acc 1 2
E2 movx A,@R0 Move Exter RAM (8- bit addr) to Acc 1 2
E3 movx A,@R1 Move Exter RAM (8- bit addr) to Acc 1 2
F0 movx @DPTR,A Move Acc to Exter RAM (1 6-bit addr) 1 2
F2 movx @R0,A Move Acc to Exter RAM (eight bit addr) 1 2
F3 movx @R1,A Move Acc to Exter RAM (eight bit addr) 1 2
C0 push direct Push direct byte onto stack 2 2
D0 pop direct Pop direct byte from stack 2 2
C5 xch A,direct Exchange direct byte w/ Accumulator 2 1
C6 xch A,@R0 Exchange indirect RAM w/ Accum 1 1
C7 xch A,@R1 Exchange indirect RAM w/ Accum 1 1
C* xch A,Rn Exchange register with Accumulator 1 1
D6 xchd A,@R0 Exc low-order digit indirect RAM w/ A 1 1
D7 xchd A,@R1 Exc low-order digit indirect RAM w/ A 1 1

Arithmetic operations ____________
04 inc A Increment Accumulator 1 1
05 inc Direct Increment direct byte 2 1
06 inc @R0 Increment direct RAM 1 1
07 inc @R1 Increment direct RAM 1 1
0* inc Rn Increment register 1 1
14 dec A Decrement Accumulator 1 1
15 dec direct Decrement direct byte 2 1
16 dec @R0 Decrement indirect RAM 1 1
17 dec @R1 Decrement indirect RAM 1 1
1* dec Rn Decrement Register 1 1
24 add A,#data Add immediate data to Accumulator 2 1
25 add A,direct Add direct byte to Accumulator 2 1
26 add A,@R0 Add indirect RAM to Accumulator 1 1
27 add A,@R1 Add indirect RAM to Accumulator 1 1
2* add A,Rn Add register to Accumulator 1 1
34 addc A,#data Add immediate data to Acc with Carry 2 1
35 addc A,direct Add direct byte to Acc with Carry 2 1
36 addc A,@R0 Add indirect RAM to Acc with Carry 1 1
37 addc A,@R1 Add indirect RAM to Acc with Carry 1 1
3* addc A,Rn Add register to Accumulator w/ Carry 1 1
84 div AB Divide A by B 1 4
94 subb A,#data Subt immediate data from A w/borrow 2 1
95 subb A,direct Subt direct byte from Acc w/ borrow 2 1
96 subb A,@R0 Subt indirect RAM from A w/ borrow 1 1
97 subb A,@R1 Subt indirect RAM from A w/ borrow 1 1
9* subb A,Rn Subtract Register from Acc w/ borrow 1 1
A3 inc DPTR Increment Data Pointer 1 2
A4 mul AB Multiply A & B 1 4
D4 da A Decimal Adjust Accumulator 1 1

Program branching ______________
00 nop No Operation 1 1
02 ljmp addr16 Long Jump 3 2
*1 ajmp addr11 Absolute Jump 2 2
12 lcall addr16 Long Subroutine Call 3 2
*1 acall addrl 1 Absolute Subroutine Call 2 2
22 ret Return from Subroutine 1 2
32 reti Return from interrupt 1 2
60 jz rel Jump if Accumulator is Zero 2 2
70 jnz rel Jump if Accumulator is Not Zero 2 2
73 jmp @A+DPTR Jump indirect relative to the DPTR 1 2
80 sjmp rel Short Jump (relative addr) 2 2
B4 cjne A,#data,rel Compare immediate to Acc & Jump if
Not Equal
3 2
B5 cjne A,direct,rel Compare direct byte to Acc & Jump if
Not Equal
3 2
B6 cjne @R0,#data,rel Compare immediate to indirect & Jump if
Not Equal
3 2
B7 cjne @R1,#data,rel Compare immediate to indirect & Jump if
Not Equal
3 2
B* cjne Rn,#data,rel Compare immediate to register & Jump if
Not Equal
3 2
D5 djnz direct,rel Decrement direct byte & Jump if Not
Zero
3 2
D* djnz Rn,rel Decrement register & Jump if Not Zero 2 2

Logical operations _______________
03 rr A Rotate Accumulator Right 1 1
13 rrc A Rotate Accumulator Right through C 1 1
23 rl A Rotate Accumulator Left 1 1
33 rlc A Rotate Accumulator Left through C 1 1
42 orl direct,A OR Accumulator to direct byte 2 1
43 orl direct,#data OR immediate data to direct byte 3 2
44 orl A,#data OR immediate data to Accumulator 2 1
45 orl A,direct OR direct byte to Accumulator 2 1
46 orl A,@R0 OR indirect RAM to Accumulator 1 1
47 orl A,@R1 OR indirect RAM to Accumulator 1 1
4* orl A,Rn OR register to Accumulator 1 1
52 anl direct,A AND Accumulator to direct byte 2 1
53 anl direct,#data AND immediate data to direct byte 3 2
54 anl A,#data AND immediate data to Accumulator 2 1
55 anl A,direct AND direct byte to Accumulator 2 1
56 anl A,@R0 AND indirect RAM to Accumulator 1 1
57 anl A,@Rl AND indirect RAM to Accumulator 1 1
5* anl A,Rn AND Register to Accumulator 1 1
62 xrl direct,A Exclusive-OR Accum to direct byte 2 1
63 xrl direct,#data Exclusive-OR immed data to direct by 3 2
64 xrl A,#data Exclusive-OR immed data to Accum 2 1
65 xrl A,direct Exclusive-OR direct byte to Accum 2 1
66 xrl A,@R0 Exclusive-OR indirect RAM to Accum 1 1
67 xrl A,@R1 Exclusive-OR indirect RAM to Accum 1 1
6* xrl A,Rn Exclusive-OR register to Accumulator 1 1
C4 swap A Swap nibbles within the Accumulator 1 1
E4 clr A Clear Accumulator 1 1
F4 cpl A Complement Accumulator 1 1

Bit manipulation _________________
10 jbc bit,rel Jump if direct Bit is set & clear bit 3 2
20 jb bit,rel Jump if direct Bit is set 3 2
30 jnb bit,rel Jump if direct Bit is Not set 3 2
40 jc rel Jump if Carry is set 2 2
50 jnc rel Jump if Carry not set 2 2
72 orl C,bit OR direct bit to Carry 2 2
A0 orl C,/bit OR complement of direct bit to Carry 2 2
82 anl C, bit AND direct bit to Carry 2 2
B0 anl C./bit AND complement of direct bit to Carry 2 2
92 mov bit,C Move Carry to direct bit 2 2
A2 mov C,bit Move direct bit to Carry 2 1
B2 cpl bit Complement direct bit 2 1
B3 cpl C Complement Carry 1 1
C2 clr bit Clear direct bit 2 1
C3 clr C Clear Carry 1 1
D2 setb bit Set direct bit 2 1
D3 setb C Set Carry 1 1
Instructions that affect flags ________
In addition to impacting data, the instructions can influence control
bits called flags. These flags represent the results of a completed
operation. The flags are carry (C), overflow (OV), and auxiliary
carry (AC), which can be used for binary coded decimal arithmetic.
Instruction Flag Bit Instruction Flag
C OV AC C
add X X X clr C O
addc X X X cpl C X
subb X X X anl C,bit X
mul O X anl C,/bit X
div O X orl C,bit X
da X orl C,/bit X
rrc X mov C,bit X
rlc X setb C 1
cjne X

Instruction set ___________________
The op code is obtained by using the number for each column
across the top as the first character. The number down the side for
each row is the second character.
0 1 2 3 4 5 6 7
0 nop jbc jb jnb jc jnc jz jnz
bit,rel bit, rel bit, rel rel rel rel rel
I ajmp acall ajmp acall ajmp acall ajmp acall
(P0) (P0) (P 1) (P 1) (P2) (P2) (P3) (P3)
2 ljmp lcall ret reti orl anl xrl orl
addr16 addr16 [2C] [2C] dir, A dir, A dir, a C, bit
3 rr rrc rl rlc orl anl xrl jmp
A A A A dir, #data dir, #data dir, #data @A+DPTR
4 inc dec add addc orl anl xrl mov
A A A, #data A, #data A, #data A, #data A, #data A, #data
dir dir A, dir A, dir A, dir A, dir A, dir dir, #data
@R0 @R0 A, @R0 A, @R0 A, @R0 A, @R0 A, @R0 @R0,@dat
@R1 @R1 A, @R1 A, @R1 A, @R1 A, @R1 A, @R1 @R1,#dat
R0 R0 A,R0 A.R0 A,R0 A,R0 A,R0 R0, #data
R1 R1 A, R1 A, R1 A, R1 A, R1 A, R1 R1,#data
A inc dec add addc orl anl xrl mov
R2 R2 A,R2 A,R2 A, R2 A,R2 A,R2 R2,#data
R3 R3 A,R3 A,R3 A,R3 A,R3 A,R3 R3,#data
C inc dec add addc orl anl xrl mov
R4 R4 A,R4 A,R4 A, R4 A,R4 A,R4 R4,#data
D inc dec add addc orl anl xrl mov
E inc dec add addc orl anl xrl mov
F inc dec add addc orl anl xrl mov

8 9 A B C D E F
0 sjmp mov orl anl push pop movx movx
REL DPTR,#d16 C, /bit C, /bit dir dir A,@DPTR @DPTR, A
1 ajmp acall ajmp acall ajmp acall ajmp acall
(P4) (P4) (P5) (P5) (P6) (P6) (P7) (P7)
2 anl mov mov cpl clr setb movx movx
C, bit bit, C C, bit bit bit bit A, @R0 @R0, A
3 movc A, movc A, inc cpl clr setb MOW movx
@A+PC @A+DPTR DPTR C C C A, @RI @RI, A
4 div subb mul cjne A, swap da clr cpl
AB A, #data AB #data, rel A A A A
5 mov subb cjne xch djnz mov mov
dir, dir A, dir A, dir, rel A, dir dir, rel A, dir dir, A
6 mov subb mov cjne xch xchd mov mov
dir,@R0 A, @R0 @R0, dir @R0,
#data, rel
A, @R0 A, @R0 A, @R0 @R0, A
7 mov subb mov cjne xch xchd mov mov
dir,@R1 A, @R1 @R1, dir @R1,
#data, rel
A, @R1 A, @R1 A, @R1 @R1, A
8 mov subb mov cjne xch djnz mov mov
dir, R0 A, R0 R0, dir R0, #data,
rel
A, R0 R0, rel A, R0 R0, A
9 mov subb mov cjne xch djnz mov mov
rel
A, R1 RI, rel A, R1 R1, A
A mov subb mov cjne xch D jnz mov mov
dir, R2 A. R2 R2 dir R2, #data,
rel
B mov subb mov cjne xch djnz mov mov
rel
A, R3 R3, rel A. R3 R3, A
C mov subb mov cjne xch djnz mov mov
dir, R4 A. R4 R4, dir R4, #data,
rel
D mov subb mov cjne xch djnz mov mov
rel
E mov subb mov cjne xch djnz mov mov
rel
A. R6 R6, rel A, R6 R6, A
F mov subb mov cjne xch djnz mov mov
dir, R7 A, R7 R7, dir R7 #data,
rel
ž Ÿ

48
MEMORY ORGANIZATION
Thought
Experience is great to have.
I just do not like getting it.
MOD
Harvard vs. Princeton ____________
Computer memory organization is typically divided into two types
of architecture. The Princeton architecture, also called Von
Neumann, has a common memory for data and program code. This
was the most common technology when magnetic cores were used
for storage.
Harvard architecture has separate memory devices. This became
more popular with solid-state memory. Programmable read-only
memory (PROM) was one technology that could be used for code.
Random access memory (RAM) was a different technology used for
data.
The Motorola scheme uses the Princeton approach, while the Intel
philosophy developed around the Harvard technique. This applies
both to the personal computer systems as well as the microcontroller
devices.
Code memory can exist internally as well as on an external chip. In
contrast, all versions have a limited internal data memory. External

Chapter 48 Memory Organization 367
data memory can also be used. In some models, additional internal
memory is available in the form of RAM or EEPROM.
Code addresses __________________
The processor memory is arranged in a variety of structures. The
lower code memory has reserved space for interrupt vectors.
Symbol Hex Meaning
RESET 00 Power on (reset)
EXTI0 03 External interrupt 0
TIMER0 0B Timer 0 interrupt
EXTI1 13 External interrupt 1
TIMER1 1B Timer 1 interrupt
SINT 23 Serial port interrupt
2B Expansion interrupts
Code memory is accessed based on the chip line /EA. If the line is
low, control goes to external memory. If the line is high, initial
control goes to internal memory. The next line of code after the top
of internal memory is external memory. This transition is regardless
of /EA setting. The next address will be in sequence. In other words,
the low external memory is inaccessible.
FFFF
56K Bytes
External
2000
1FFF 8K Bytes
Internal
0000 /EA pin high
FFFF
64K Bytes
External
0000 /EA pin low
Program Memory

Access to code memory causes the program storage enable (/PSEN)
line to be asserted low. This provides the Chip Select for the storage
chip.
Instructions that access code memory use the movc mnemonic. This
also activates the /PSEN line.
External memory is accessed by sixteen-bit addressing lines. The
low lines are located on port 0 and the upper lines are on port 2.
This provides a 64k byte page.
Since both data and code are multiplexed on these lines, it is
necessary to capture the address. An address latch enable (ALE) line
is triggered when addressing is placed on the line by the processor.
The ALE is connected to the chip select line of a latch.
External data addresses ___________
External data memory is typically static ram. It can occupy up to
64K bytes in one page. This is accomplished by sharing the sixteen-
bit addressing lines with the code memory. Similarly, the ALE
selects the address latch.
The instruction mnemonic for accessing external data memory is
movx. The instruction will activate either the read (/RD) or the write
(/WR) lines. The write line is connected to the memory-chip write-
enable pin. Similarly, the read line is connected to the output enable.
Data memory expansion __________
Data memory can be segregated into three types - internal ram,
external ram, and internal EEPROM.
Internal and external ram capability is standard across the
microcontroller line. The internal ram contains the lower 128 bytes
with only 21 special function registers in the upper 128 area.

Internal
FF SFR
Direct
80 Addressing
7F Internal RAM
Direct & Indirect
00 Addressing
External
FFFF
64K Bytes
0000
Data Memory
Enhanced microcontrollers permit access to the remaining locations
of the upper 128 bytes of internal ram.
In addition, some models have EEPROM or other memory that is
internal on the chip. However, addressing this memory is done
exactly as if it were external. Therefore, movx instructions are
required.
Internal
Expansion
Memory
FF
FF SFR Indirect only
Direct 80
80 Addressing
7F Internal RAM
Direct & Indirect EEPROM 7FF
00 Addressing movx
Addressing 000
External
FFFF
64K Bytes
0000
Data Memory Expansion
Internal data addresses ___________
The basic processor has 128 bytes of internal data that is available.
The entire area may be used as general-purpose data. In addition,

the lower part can be used as registers. Another group of bytes is bit
addressable.
The basic instruction that accesses this area is mov. It does not
influence any external pins or lines. Numerous other instructions
can also be used, depending on the addressing mode.
The next 128 bytes of internal memory depend on the manufacturer
design. For the basic microcontroller, 21 special function registers
(SFR) are implemented from this memory. That leaves a substantial
portion that is unused. In terms of memory space, a table identifies
which locations are reserved for the registers. Direct addressing
accesses the data in the SFR.
The special function registers permit control of many features and
capabilities. The next chapter is dedicated to the functions and
implementation.
Some versions of the microcontroller allow access to the remaining
bytes of the internal memory. The internal memory will be stated as
256 bytes. Indirect addressing accesses data in the upper 128 bytes.
Other internal memory such as EEPROM is available in certain
models. Discussion on this topic is specific to the device and will be
broached in a different section.

Internal RAM low ________________
<= 8 Bytes =>
78 7F
70 77
68 6F
60 67
58 5F Scratch
50 57 Pad
48 4F Area
40 47
38 3F
30 37
28 …7F 2F Bit
20 0… 27 Segment
18 Bank 3 1F
10 Bank 2 17 Register
08 Bank 1 0F Banks
00 Bank 0 07

Internal RAM high ______________
Locations that are divisible by eight are bit addressable. Therefore,
any location that has a hex address that ends with 0 or 8 has bit
access. The first bit corresponds to the byte address.
These locations are also called special function registers. Registers
that are identified with a plus (+) symbol are added features that
may be in various designs.
Symbol Name Address
ACC Accumulator 0E0H
B B or multiplication register 0F0H
PSW Program Status Word 0D0H
TH2+ Timer/Counter 2 High Byte 0CDH
TL2+ Timer/Counter 2 Low Byte 0CCH
RCAP2H+ T/C 2 Capture Reg. High Byte 0CBH
RCAP2L+ T/C 2 Capture Reg. Low Byte 0CAH
T2MOD+ Timer/Counter 2 Mode Control 0C9H
T2CON+ Timer/Counter 2 Control 0C8H
IP Interrupt Priority Control 0B8H
P3 Port 3 0B0H
IE Interrupt Enable Control 0A8H
P2 Port 2 0A0H
SBUF Serial Data Buffer 99H
SCON Serial Control 98H
P1 Port 1 90H
TH1 Timer/Counter 1 High Byte 8DH
TH0 Timer/Counter 0 High Byte 8CH
TL1 Timer/Counter 1 Low Byte 8BH
TL0 Timer/Counter 0 Low Byte 8AH
TMOD Timer/Counter Mode Control 89H
TCON Timer/Counter Control 88H
PCON Power Control 87H
DPTR Data Pointer 2 Bytes
DPL Low Byte 82H
DPH High Byte 83H
SP Stack Pointer 81H
P0 Port 0 80H

Predefined bit addresses ___________
Bit addresses are the components of bit definable special function
registers (SFR). These can be accessed either by the byte or by the
individual bits.
Sym. Position Hex Meaning
CY PSW.7 D7 Carry flag
AC PSW.6 D6 Auxiliary carry flag
FO PSW.5 D5 Flag 0
RS1 PSW.4 D4 Register bank select bit 1
RSO PSW.3 D3 Register bank select bit 0
OV PSW.2 D2 Overflow flag
P PSW.O D0 Parity flag
TF1 TCON.7 8F Timer 1 overflow flag
TR1 TCON.6 8E Timer 1 run control bit
TFO TCON 5 8D Timer 0 overflow flag
TRO TCON.4 8C Timer 0 run control bit
IE1 TCON 3 8B Interrupt 1 edge flag
IT1 TCON.2 8A Interrupt 1 type control bit
IE0 TCON.1 89 Interrupt 0 edge flag
IT0 TCON.0 88 Interrupt 0 type control bit
SMO SCON.7 9F Serial mode control bit 0
SM1 SCON.6 9E Serial mode control bit 1
SM2 SCON.5 9D Serial mode control bit 2
REN SCON.4 9C Receive enable
TB8 SCON.3 9B Transmit bit 8
RB8 SCON.2 9A Receive bit 8
TI SCON.1 99 Transmit interrupt flag
RI SCON.0 98 Receive interrupt flag
EA IE.7 AF Enable all interrupts
ES IE.4 AC Enable serial port interrupt
ET1 IE.3 AB Enable timer 1 interrupt
EX1 IE.2 AA Enable external interrupt 1
ETO IE.1 A9 Enable timer 0 interrupt
EXO IE.O A8 Enable external interrupt 0
PS IPA BC Priority of serial port interrupt
PT1 IP.3 BB Priority of timer i interrupt
PX1 IP.2 BA Priority of external interrupt 1
PTO IP.1 B9 Priority of timer 0
PXO IP.O B8 Priority of external interrupt 0

Predefined bits port 3 ____________
Port 3 is an enhanced function location. It can be accessed as an
input / output port. In addition, each bit is associated with control of
external devices.
RD P3.7 87 Read data for external memory
WR P3.6 B6 Write data for external memory
T1 P3.5 B5 Timer/counter 1 external flag
TO P3.4 B4 Timer/counter 0 external flag
INT1 P3.3 B3 Interrupt 1 input pin
INTO P3.2 B2 Interrupt 0 input pin
TXD P3.1 B1 Serial port transmit pin
RXD P3.0 BO Serial port receive pin
ž Ÿ

49
SPECIAL FUNCTION REGISTERS
Thought
What is so special
about function registers?
First time user
Reserved memory ________________
The microprocessor has many unique capabilities that give it
features of very powerful machines. One of those discussed has
been some of the characteristics of a reduced instruction set (RISC)
machine. Another was sequential processing as well as powerful
stack capability. Another is serial or parallel data transfer. The next
is internal memory is register accessible.
There are 21 special function registers (SFR) in the standard
microcontroller. These are located in the upper 128 bytes of the
internal static RAM. Hence, the bytes are both memory and
registers. These registers can be accessed using instructions for
memory. In addition, each register has instructions that are special
just for that register.
The registers that will be addressed in this section are the ports as
well as the control and mode for status flags, power control, timers,
and serial.

Ports __________________________
Four ports are connected to external pins. As such, they are the
input and output connections to the processor. All ports can be used
for general-purpose input and output. In addition, ports 0, 2, and 3
have special applications. On some machines, even port 1 has
additional functions. The four ports are similar, but the hardware for
each is different because of the functions.
In all cases, data stored in the special function register or data
memory address will be displayed on the ports. This information
may be identified as a byte or individual bits.
To make the port pins perform as an input, first place a 1 on each bit
that is to be read. This will pull the line high so that external
switches may be connected. Then the data that is read from the port
address is the status of the external switches.
Port 0 __________________________
Port 0 is at internal data address 80h. When external memory is
accessed, the port becomes multiplexed. First, the lower eight bits of
the address are displayed on the port. Then the data for that address
is on the port. Depending on the instruction, the data is either
written to the address or read from the address.
AD7 AD6 AD5 AD4 AD3 AD2 AD1 AD0
The port has very weak internal pull-up resistors. When used as an
output, the port functions like the other ports. However, it is limited
in the driving current. When the port is used as an input, it must
have external pull-up or pull-down resistors attached. Connect a 2.2
K: resistorto 5 volts or to ground as required.

Chapter 49 Special Function Registers 377
Port 1 __________________________
Port 1 is at internal data address 90h. It is the general-purpose input
output register. Most external connections are made to this port. It
has strong internal pull-up resistors; therefore, it does not float.
SCK MISO MOSI P1.4 P1.3 P1.2 P1.1 P1.0
On machines that have serial peripheral interface (SPI) capability,
the upper bits of the port are used. Devices that have in-system
programming, use the SPI connections.
After the SPI exchange, the port pins can revert to standard input /
output. However, the port pins should not be used for several
instruction cycles after the in system programming, or the machine
may lock up.
Port 2 __________________________
Port 2 is at internal data address 0A0h. When external memory is
accessed, the upper eight bits of the address are displayed on the
port. The port is not readily shared as an input / output if external
memory is used.
A15 A14 A13 A12 A11 A10 A9 A8
Port 3 __________________________
Port 3 is at internal data address 0B0h. It is used as a function
management location. Each bit is associated with control of external
devices. This includes external memory, serial communications,
interrupts, and timers.
RD WR T1 T0 INT1 INT0 TXD RXD

RD P3.7 87 Read data for external memory
WR P3.6 B6 Write data for external memory
INT1 P3.3 B3 Interrupt 1 input pin
INTO P3.2 B2 Interrupt 0 input pin
TXD P3.1 B1 Serial port transmit pin
RXD P3.0 B0 Serial port receive pin
PSW: Program status word _______
The program status word is a bit addressable location. The register
bank selected is determined by the combination of bits RS1-RS0.
The result is bank 0 to 3.
CY AC F0 RS1 RS0 OV — P
CY PSW.7 Carry flag.
AC PSW.6 Auxiliary carry flag.
F0 PSW.5 Flag 0 available to the user for general purpose.
RS1 PSW.4 Register Bank selector bit 1.
RS0 PSW.3 Register Bank selector bit 0.
OV PSW.2 Overflow flag.
— PSW.1 User definable flag.
P PSW.0 Parity flag. Set/cleared by hardware each instruction cycle to
indicate an odd/even number of 1 bits in the accumulator.
RS1 RS0 Bank Address
0 0 0 00H-07H
0 1 1 08H-0FH
1 0 2 10H-17H
1 1 3 18H-1FH

PCON: Power control register _____
The PCON register is not bit addressable. Therefore values must be
moved into the location.
SMOD — — — GF1 GF0 PD IDL
SMOD Double baud rate bit. If Timer 1 is used to generate baud rate and SMOD =
1, the baud rate is doubled when the Serial port is used in modes 1, 2, or 3.
Serial port is used in modes 1, 2, or 3.
— Not implemented, reserved for future use.
GF1 General purpose flag bit.
GF0 General purpose flag bit.
PD Power Down bit. Setting this bit activates Power Down operation in select
processors.
IDL Idle Mode bit. Setting this bit activates Idle Mode operation in select
processors. If 1s are written to PD and IDL at the same time, PD takes
precedence.

Interrupts ______________________
To use any of the interrupts in the microcontroller, take the
following three steps.
1. Set the EA (enable all) bit in the IE register to 1.
2. Set the corresponding individual interrupt enable bit in the IE
register to 1.
3. Begin the interrupt service routine at the corresponding vector
address of that interrupt. See the following table.
If the bit is 0, the corresponding interrupt is disabled. If the bit is 1,
the corresponding interrupt is enabled.
In addition, for external interrupts, pins INT0 and INT1 (P3.2 and
P3.3) must be set to 1, and depending on whether the interrupt is to
be level or transition activated, bits IT0 or IT1 in the TCON register
may need to be set to 1. ITx = 0 is used for level activated, while
ITx = 1 makes the interrupt transition activated.
Both the interrupt enable and the interrupt priority registers are bit
addressable. These can be changed with setb, clr, anl, orl, and xrl
instructions.
Function. Interrupt
Source
Vector
Address Hex
External 0 IE0 0003H
Timer 0 TF0 000BH
External 1 IE1 0013H
Timer 1 TF1 001BH
Serial R1 & T1 0023H
Timer 2 TF2 & EXF2 002BH
To assign higher priority to an interrupt, the corresponding bit in the
IP register must be set to 1. While an interrupt service is in progress,
it cannot be interrupted by an interrupt of the same or lower priority.
The only purpose of priority within a level is to resolve
simultaneous requests of the same priority level. If the bit is 0, the

corresponding interrupt has a lower priority. If the bit is 1, the
corresponding interrupt has a higher priority. The interrupt sources
are listed below in order from highest to lowest priority.
IE0 TF0 IE1 TF1 RI / TI TF2 / EXF2
IE: Interrupt enable register _______
EA — ET2 ES ET1 EX1 ET0 EX0
EA IE.7 Disables all interrupts. If EA = 0, no interrupt is acknowledged.
If EA = 1, each interrupt source is individually enabled or
disabled by setting or clearing its enable bit.
— IE.6 Not implemented, reserved for future use.(1)
ET2 IE.5 Enables or disables the Timer 2 overflow or capture interrupt
on select models.
ES IE.4 Enables or disables the serial port interrupt.
ET1 IE.3 Enables or disables the Timer 1 overflow interrupt.
EX1 IE.2 Enables or disables External Interrupt 1.
ET0 IE. 1 Enables or disables the Timer 0 overflow interrupt.
EX0 IE.0 Enables or disables External Interrupt 0.
IP: Interrupt priority register ______
— — PT2 PS PT1 PX1 PT0 PX0
— IP. 7 Not implemented, reserved for future use.
— IP. 6 Not implemented, reserved for future use.
PT2 IP. 5 Defines the Timer 2 interrupt priority level.+
PS IP. 4 Defines the Serial port interrupt priority level.
PT1 IP. 3 Defines the Timer 1 interrupt priority level.
PX1 IP. 2 Defines External Interrupt 1 priority level.
PT0 IP. 1 Defines the Timer 0 interrupt priority level.
PX0 IP. 0 Defines the External Interrupt 0 priority level.

Timer / counters _________________
The microcontroller has two timers. These can be used as timers or
counters. A timer simply counts the number of clock cycles. Two
registers operate the timer/counters.
The Control register determines if the timer is running, has
completed its count, and what type signal provides the trigger. The
Control register is bit addressable. Therefore, discrete values can be
changed.
The Mode register operates the timers as 8, 13, or sixteen bit
devices. The Mode register is not bit addressable, so only mov
instructions are appropriate.
Two tables are given for selecting the mode. The last table has
TMOD values that can be used to set up Timer 0. It is assumed that
only one timer is used at a time. If Timers 0 and 1 must run
simultaneously in any mode, the value in TMOD for Timer 0 must
be ORed with the value required for Timer 1.
For example, if Timer 0 is in mode 1 with Gate (external control),
and Timer 1 is run in mode 2 as a counter, then the value that must
be loaded into TMOD is 69H (09H for Timer 0 ORed with 60H for
Timer 1.
Moreover, it is assumed that the timer is not turned on at this point.
It can be started at another point in the program by setting bit TRx
(in TCON) to 1.
1. Timer x is turned ON/OFF by setting/clearing bit TRx in the
software.
2. Timer 0 is turned ON/OFF by the 1 to 0 transition on INT0
(P3.2) when TR0 = 1 (hardware control).
3. Timer 1 is turned ON/OFF by the 1 to 0 transition on INT1
(P3.3) when TR1 = 1 (hardware control).

TCON: Timer/counter control register
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
TF1 TCON. 7 Timer 1 overflow flag. Set by hardware when the
Timer/Counter 1 overflows. Cleared by hardware as the
processor vectors to the interrupt service routine.
TR1 TCON. 6 Timer 1 run control bit. Set/cleared by software to turn
Timer/Counter 1 ON/OFF.
TF0 TCON. 5 Timer 0 overflow flag. Set by hardware when the
Timer/Counter 0 overflows. Cleared by hardware as the
processor vectors to the service routine.
TR0 TCON. 4 Timer 0 run control bit. Set/cleared by software to turn
Timer/Counter 0 ON/OFF.
IE1 TCON. 3 External Interrupt 1 edge flag. Set by hardware when the
External Interrupt edge is detected.
Cleared by hardware when the interrupt is processed.
IT1 TCON. 2 Interrupt 1 type control bit. Set/cleared by software to specify
falling edge/low level triggered External Interrupt.
IE0 TCON. 1 External Interrupt 0 edge flag. Set by hardware when
External Interrupt edge detected. Cleared by hardware when
interrupt is processed.
IT0 TCON. 0 Interrupt 0 type control bit. Set/cleared by software to specify
falling edge/low level triggered External Interrupt.
TMOD: Timer/counter mode register
GATE-1 C/T -1 M1 –1 M0 -1 GATE-0 C/T -0 M1 -0 M0 -0
GATE When TRx (in TCON) is set and GATE = 1, TIMER/COUNTERx runs
only while the INTx pin is high (hardware control). When GATE = 0,
TIMER/COUNTERx will run only while TRx = 1 (software control).
C/T Timer or Counter selector. Cleared for Timer operation (input from
internal system clock). Set for Counter operation(input from Tx input
pin).
M1 Mode selector bit.
M0 Mode selector bit.

M1 M0 Mode Operation
0 0 0 13-bit Timer
0 1 1 sixteen bit Timer/Counter
1 0 2 eight bit Auto-Reload Timer/Counter
1 1 3 Split Timer Mode: (Timer 0) TL0 is an eight bit
Timer/Counter controlled by the standard Timer 0
control bits, TH0 is an eight bit Timer and is
controlled by Timer 1 control bits.
1 1 3 (Timer 1) Timer/Counter 1 stopped.
Mode Function TMOD Count TMOD Time
Internal External Internal External
0 13-bit Timer 00H 08H 04H 0CH
1 sixteen bit Timer 01H 09H 05H 0DH
2 eight bit Auto-Reload 02H 0AH 06H 0EH
3 two eight bit Timers 03H 0BH 07H 0FH

Serial __________________________
Serial communications is directed by a universal asynchronous
receive transmit (UART) circuit. Timer 1, or Timer 2 if it is
available, determines the baud rate. The serial control register is bit
addressable.
Operation mode determines the use as a shift, variable baud, or
fixed baud register.
SCON: Serial control register ______
SM0 SM1 SM2 REN TB8 RB8 TI RI
SM0 SCON. 7 Serial port mode selection.
SM1 SCON. 6 Serial port mode selection.
SM2 SCON. 5 Enables the multiprocessor communication feature in modes
2 and 3. In mode 2 or 3, if SM2 is set to 1, then RI is not
activated if the received 9th data bit (RB8) is 0. In mode 1, if
SM2 = 1, then RI is not activated if a valid stop bit was not
received. In mode 0, SM2 should be 0.
REN SCON. 4 Set/Cleared by software to Enable/Disable reception.
TB8 SCON. 3 The 9th bit that is transmitted in modes 2 and 3. Set/Cleared
by software.
RB8 SCON. 2 In modes 2 and 3, is the 9th data bit that was received. In
mode 1, if SM2 = 0, RB8 is the stop bit that was received. In
mode 0, RB8 is not used.
TI SCON. 1 Transmit interrupt flag. Set by hardware at the end of the 8th
bit time in mode 0 or at the beginning of the stop bit in the
other modes. Must be cleared by software.
RI SCON. 0 Receive interrupt flag. Set by hardware at the end of the 8th
bit time in mode 0 or halfway through the stop bit time in the
other modes (except see SM2). Must be cleared by software.

SM0 SM1 Mode Description Baud Rate
0 0 0 Shift register Fosc./12
0 1 1 eight bit UART Variable
1 0 2 9-Bit UART Fosc./64 OR
Fosc./32
1 1 3 9-Bit UART Variable
MODE SCON SM2 VARIATION
0 10H Single Processor
1 50H Environment
2 90H (SM2 = 0)
3 D0H
0 NA Multiprocessor
1 70H Environment
2 B0H (SM2 = 1)
3 F0H
ž Ÿ

50
SFR EXTENDED
Thought
Be strong, be courageous,
Be not afraid.
General Joshua, ~1500 BC
Enhanced registers _______________
The previous registers are standard to the family of
microprocessors. However, there are numerous other registers that
various enhanced microprocessor versions offer. Some of the more
common registers handle additional timers, on-board memory,
additional data pointer, and interfaces such as SPI. Illustrations of
some of these extended group or registers are included.
Timer/counter 2 _________________
Some versions of the microcontroller have a third timer/counter
register. It is operated very similar to the standard Timer 0 and 1.
T2CON is located at address 0C8h and T2MOD is address 0C9h.
The control register is bit addressable, but the mode register is not.
The reset value for the mode register is = xxxx xx00b.
Four data registers are associated, RCAP2L, RCAP2H, TL2, and
TH2.

A table provides typical initialization values. Except for the baud-
rate generator mode, the values given for T2CON do not include the
setting of the TR2 bit. Therefore, bit TR2 must be set separately to
turn the Timer on.
1. Capture/Reload occurs only on Timer/Counter overflow.
2. Capture/Reload occurs on Timer/Counter overflow and a 1 to 0
transition on T2EX (P1.1) pin except when Timer 2 is used in
the baud-rate generating mode.

Chapter 50 SFR Extended 389
T2CON: Timer/counter 2 control register
Address = 0C8h Reset Value = 0000 0000b
TF2 EXF2 RCLK TCLK EXEN2 TR2 C/T2 CP/RL2
TF2 T2CON. 7 Timer 2 overflow flag set by hardware and cleared by
software. TF2 cannot be set when either RCLK = 1 or CLK =
1
EXF2 T2CON. 6 Timer 2 external flag set when either a capture or reload is
caused by a negative transition on T2EX, and EXEN2 = 1.
When Timer 2 interrupt is enabled, EXF2 = 1 causes the CPU
to vector to the Timer 2 interrupt routine. EXF2 must be
cleared by software.
RCLK T2CON. 5 Receive clock flag. When set, causes the Serial port to use
Timer 2 overflow pulses for its receive clock in modes 1 and
3. RCLK = 0 causes Timer 1 overflow to be used for the
receive clock.
TLCK T2CON. 4 Transmit clock flag. When set, causes the Serial port to use
Timer 2 overflow pulses for its transmit clock in modes 1 and
3. TCLK = 0 causes Timer 1 overflows to be used for the
transmit clock.
EXEN2 T2CON. 3 Timer 2 external enable flag. When set, allows a capture or
reload to occur as a result of negative transition on T2EX if
Timer 2 is not being used to clock the Serial port. EXEN2 = 0
causes Timer 2 to ignore events at T2EX.
TR2 T2CON. 2 Software START/STOP control for Timer 2. A logic 1 starts
the Timer.
C/T2 T2CON. 1 Timer or Counter select. 0 = Internal Timer. 1 = External
Event Counter (triggered by falling edge).
CP/RL2 T2CON. 0 Capture/Reload flag. When set, captures occur on negative
transitions at T2EX if EXEN2 = 1. When
cleared, auto-reloads occur either with Timer 2 overflows or
negative transitions at T2EX when
EXEN2 = 1. When either RCLK = 1 or TCLK = 1, this bit is
ignored and the Timer is forced to auto- reload on Timer 2
overflow.

T2MOD: Timer 2 mode register ____
Address = 0C9hh Reset Value = xxxx xx00b
- - - - - - T2OE DCEN
- Not implemented, reserved for future use
T2OE Timer 2 Output Enable bit
DCEN When set, this bit allows Timer 2 to be configured as an up/down
counter.
Function TMOD2 Time TMOD2 Count
Internal External Internal External
Sixteen bit Auto-Reload 00H 08H 02H 0AH
Sixteen bit Capture 01H 09H 03H 0BH
Receive only 24H 26H
Transmit only 14H 16H
Baud rate generator receive and transmit
same baud rate
34H 36H
Timer 2 data registers ____________
RECAP2L: Address = 0CAh Reset Value = 0000 0000b
RECAP2H Address = 0CBh Reset Value = 0000 0000b
TL2 Address = 0CCh Reset Value = 0000 0000b
TH2 Address = 0CDh Reset Value = 0000 0000b
D7 D6 D4 D3 D2 D1 D0

Serial peripheral interface _________
The serial peripheral interface (SPI) allows high-speed synchronous
data transfer between a master and peripheral chips. The
characteristics are listed.
1. Full-Duplex, 3-Wire Synchronous Data Transfer
2. Master or Slave Operation
3. 1.5 MHz Bit Frequency (max.)
4. LSB First or MSB First Data Transfer
5. Four Programmable Bit Rates
6. End of Transmission Interrupt Flag
7. Write Collision Flag Protection
8. Wakeup from Idle Mode (Slave Mode Only)
A unique select line is connected to each device. In addition, there
are three common lines.
MOSI – master out, slave in.
MISO – master in, slave out.
SCK – clock generated by the master.
The SCK pin is the clock output in the master mode but is the clock
input in the slave mode. Writing to the SPI data register of the
master CPU starts the SPI clock generator, and the data written
shifts out of the MOSI pin and into the MOSI pin of the slave CPU.
After shifting one byte, the SPI clock generator stops, setting the
end of transmission flag (SPIF). If both the SPI interrupt enable bit
(SPIE) and the serial port interrupt enable bit (ES) are set, an
interrupt is requested.
The Slave Select input, SS/P1.4, is set low to select an individual
SPI device as a slave. When SS/P1.4 is set high, the SPI port is
deactivated and the MOSI/P1.5 pin can be used as an input.
There are four combinations of SCK phase and polarity with respect
to serial data, which are determined by control bits CPHA and
CPOL.
The data register (SPDR) is double buffered, so data can be written
both directions simultaneously.

The clock frequency is determined by the control register bits 0 and
1. The frequency is the microprocessors clock divided by the
selected divisor. The divisor is generated in the table.
SPR1 SPR0 DIVISOR
0 0 4
0 1 16
1 0 64
1 1 128
A generic technique for SPI is shown in the SPI projects chapter. It
will work with any processor, including those without SPI registers.
Another procedure for implementing on-board SPI for read and
write is also included.
SPCR: SPI control register ________
Address = 0D5h Reset Value = 0000 01xxb
SPIE SPE DORD MSTR CPOL CPHA SPR1 SPR0

SPIE SPCR. 7 SPI Interrupt Enable. This bit, in conjunction with the ES bit
in the IE register, = 1 enable SPI interrupts. SPIE = 0 disables
SPI interrupts.
SPE SPCR. 6 SPI Enable. SPI = 1 enables the SPI channel and connects
SS, MOSI, P1.7. SPI = 0 disables the SPI channel.
DORD SPCR. 5 Data Order. DORD = 1 selects LSB first data transmission.
DORD = 0 selects MSB first data transmission
MSTR SPCR. 4 Master/Slave Select. MSTR = 1 selects Master SPI mode.
MSTR = 0 selects slave SPI mode
CPOL SPCR. 3 Clock Polarity. When CPOL = 1, SCK is high when idle.
When CPOL =0, SCK of the master device is low when not
transmitting.
CPHA SPCR. 2 Clock phase. The CPHA bit together with the CPOL bit
controls the clock and data relatinship between master and
slave.
SPR0 SPCR. 1
SPR1 SPCR. 0
SPI clock rate select. These 2 bits control the SCK rate of the
device configured as amster. They have no effect on the
slave. The SCK frequency is the oscillator frequency divided
by the values.
SPR1 SPR0 Divisor
0 0 4
0 1 16
1 0 64
1 1 128
SPSR: SPI status register __________
Address = 0AAh Reset Value = 00xx xxxxb
SPIF WCOL
SPIF SPSR. 7 SPI Interrupt Flag. When a serial transfer is complete, the
SPIF bit is set and an interrupt is generated if SPIE = 1 and
ES = 1. The SPIF bit is cleared by reading the SPI status
register with SPIF and WCOL bits set, and then accessing the
SPI data register
WCOL SPSR. 6 Write collision flag. The bit is set if the SPI data register is
written during a data transfer. During transfer, the result of
reading the SPDR register may be incorrect, and writing to it
has no effect. The WCOL bit and the SPIF bit are cleared by
reading the SPI status register with SPIF and WCOL set, and
then accessing the SPI data register.

SPDR: SPI data register __________
Address = 86h Reset Value = unchanged
SPD7 SPD6 SPD5 SPD4 SPD3 SPD2 SPD1 SPD0
WMCON: Watchdog _____________
The watchdog and memory control register (WMCON) contains
control bits for several expansion functions. One is the Watchdog
Timer. The EEMEN and EEMWE bits are used to select the 2K
bytes on-chip EEPROM, and to enable byte-write. The DPS bit
selects which of the two DPTR registers are active.
Address = 96h Reset Value = 0000 0010b
PS2 PS1 PS0 EEMWE EEMEN DPS WDTRST WDTEN
PS2 WMCON.
7
Prescaler Bits for the Watchdog Timer..
PS1 WMCON.
6
When all three bits are set to “0”, the watchdog timer has a
nominal period of16 ms.
PS0 WMCON.
5
When all three bits are set to “1”, the nominal period is
2048 ms
EEMWE WMCON.
4
EEPROM Data Memory Write Enable Bit. Set this bit to
“1” before initiating byte write to on-chip EEPROM with
the
EEMEN WMCON.
3
movx instruction. User software should set this bit to “0”
after EEPROM write is completed.
DPS WMCON.
2
Internal EEPROM Access Enable. When EEMEN = 1, the
movx instruction with DPTR will access on-chip
EEPROM
WDTRST WMCON.
1
Watchdog Timer Reset and EEPROM Ready/Busy Flag.
Each time this bit is set to “1” by user software, a pulse is
RDY/BSY WMCON.
0
generated to reset the watchdog timer. The WDTRST bit is
then automatically reset to “0” in the next instruction
cycle.

Using onboard EEPROM __________
The onboard eeprom requires the WMCON register. This is very
similar to an external IIC or SPI function.
;-------------------------------------------------
;Program: EEPROM.ASM
;Initial: July 28, 2004
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c) 2004. All rights reserved
; Original adapted from Atmel.
;
;Purpose:
; A set of routines are provided to write and
; read from the on board EEPROM.
;
;#################################################
; ASSIGNMENTS
;#################################################
;CONSTANTS
;-------------------------------------------------
; ;WMCON REGISTER
Spcr data 0d5h ;SPI control register
Wmcon data 96h ;watchdog & memory register
Eemen equ 00001000b ;EEPROM access enabl bit
Eemwe equ 00010000b ;EEPROM write enable bit
Wdtrst equ 00000010b ;EEPROM RDY/BSY bit
;#################################################
; PROGRAM
;#################################################
org 00h

START: ljmp INITIAL
;-------------------------------------------------
INITIAL:
MAIN:
;-------------------------------------------------
lcall WMEEPRD ;read eeprom
MAN9: sjmp MAIN
;-------------------------------------------------
WMEEPRD:
;-------------------------------------------------
; WMCON is used to access internal EEPROM.
; WMCON is not bit addressable, so Boolean
; functions are necessary to control bits.
; EEPROM read
orl Wmcon,#Eemen ;enable EEPROM access
mov DPTR,#Address;address to read
movx A,@DPTR ;read EEPROM
xrl Wmcon,#Eemen ;disable EEP access
ret ;home again
;-------------------------------------------------
WMEEPWRD:
;-------------------------------------------------
; EEPROM write example, utilizing fixed delay
; for write cycle. Delay is worst case (10 ms).
; Write is followed by verify (read & compare)
; Code for delay is not shown.
; Code to handle verification failure not shown.
orl Wmcon,#Eemwe ;enable EEPROM writes
mov Dptr,#Address;address to write
mov A,#Data ;data to write
movx @Dptr,A ;write EEPROM

CALL DELAY_10_MS ;wait 10 ms
movx A,@Dptr ;read EEPROM
cjne A,#Data,Error;data compare fails
xrl Wmcon,#Eemwe ;disable EEPROM write
xrl Wmcon,#Eemen ;disabl EEPROM access
ret ;home again
;-------------------------------------------------
WMEEPWRB:
;-------------------------------------------------
; EEPROM write example, utilizing RDY/BSY
; to determine the end of the write cycle.
; Write is followed by verify (read and compare)
Loop: mov A,Wmcon ;EEPROM write status
anl A,#Wdtrst ;check RDY/BSY
jz Loop ;Jump if busy
movx A,@Dptr ;read EEPROM
cjne A,#Data,Error;data compare fails
ret ;home again
;-------------------------------------------------
WMEEPWRP:
;-------------------------------------------------
; EEPROM write example, utilizing data polling
; to determine the end of; the write cycle.
; After data is loaded, the code loops on read
; until data is returned true.
; Write verification is implicit in this method.

Loop: movx A,@Dptr ;read EEPROM
cjne A,#Data,loop ;data compare (busy)
ret ;home again
ž Ÿ

SECTION VI – COMMUNICATION
ž Ÿ

51
ASCII
Thought
Seldom does one problem produce a failure.
It takes at least two problems to cause a catastrophe.
MOD
What is it _______________________
Traditionally, communications between teletypes was done using
the American Standard Code for Information Interchange (ASCII).
As computer equipment developed and replaced the early Teletype
systems, communications between computer devices used the same
ASCII format.
It is interesting that the Teletype technology operated with a current
loop. The instrumentation system of 4-20 mA is a direct descendent.
Therefore both computer communications and instrumentation
signaling can trace their lineage to the Teletype.
Computers internally are considered to operate with a hexadecimal
(hex) format. This is simply a selection or combination of four bits.
Therefore, software within the computer must convert between hex
and ASCII. This table will provide the normal standards. An
additional table provides the next 127 characters. However, the
additional special characters are seldom used.

Chapter 51 ASCII 401
Because of the history, most of the first characters are somewhat
arcane. These were used as control messages for the early machines.
Only a few of them have application to current transmissions.
The remaining tables provide the characters, numbers, and
punctuation used by most messages.
ASCII-hex table _________________
ASCII Hex Symbol
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
NUL
SOH
STX
ETX
EOT
ENQ
ACK
BEL
BS
TAB
LF
VT
FF
CR
SO
SI
null, empty
start of heading
start of text
end of test
end of xmission
enquire
acknowledge
bell ring
back space
horizontal tab
line feed
vertical tab
form feed
carriage return
shift out
shift in
ASCII Hex Symbol
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
DLE
DC1
DC2
DC3
DC4
NAK
SYN
ETB
CAN
EM
SUB
ESC
FS
GS
RS
US
data link escape
device control 1
device control 2
device control 3
device control 4
not acknowledge
synchronous idle
EOT block
cancel
end of medium
substitute
escape
file separator
group separator
record separator
unit separator

ASCII Hex Symbol
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
20
21
22
23
24
25
26
27
28
29
2A
2B
2C
2D
2E
2F
(space)
!
"
#
$
%
&
'
(
)
*
+
,
-
.
/
ASCII Hex Symbol
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
30
31
32
33
34
35
36
37
38
39
3A
3B
3C
3D
3E
3F
0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?
ASCII Hex Symbol
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
40
41
42
43
44
45
46
47
48
49
4A
4B
4C
4D
4E
4F
@
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
ASCII Hex Symbol
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
50
51
52
53
54
55
56
57
58
59
5A
5B
5C
5D
5E
5F
P
Q
R
S
T
U
V
W
X
Y
Z
[

]
^
_
ASCII Hex Symbol
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
60
61
62
63
64
65
66
67
68
69
6A
6B
6C
6D
6E
6F
`
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
ASCII Hex Symbol
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
70
71
72
73
74
75
76
77
78
79
7A
7B
7C
7D
7E
7F
p
q
r
s
t
u
v
w
x
y
z
{
|
}
~
ž Ÿ

52
RS 232
Thought
God does not throw dice.
Dr. Albert Einstein
Connections _____________________
The Electrical Industry Apparatus (EIA) has established numerous
standards for connections and communications. One of the originals
was RS232. This is for serial exchange of data between two
machines. Other standards, such as RS 422/485 are used for
communications with multiple devices.
These standards were developed for connections in environments
that are uncontrolled. Any device can be connected and it can be
remote. Therefore, the signal levels are basically 9 - 25 volts. Any
device connected using the standard must be able to operate under
these voltage constraints.
Numerous connection possibilities exist with serial cables. The most
common designations are shown in the tables. The function is
compared to the standard db 25 designations. A ‘P’ indicates plug
while ‘S’ is a socket.
When connecting to other termination types such as a
microcontroller board, treat the cable as a null modem. Make the
terminations according to the function. With a null cable, the

functions are crossed. For example TX connects to RX. Only lines
2, 3, & 7 are required for bi-directional communications.
If using an RJ45 Ethernet adapter, consider the pins to correspond
with the 9-pin db-9. Since RI is typically jumpered to DSR, leave it
as the pin that is unconnected. For a jumper used as a loop back,
only pins 2, 3, and 5 would be connected.
RS 232 pin outs __________________
25 to 9 straight 25 to 25 straight 25 to 25 null
25-pin 9-pin 25-pin 25-pin 25-pin 25-pin
Funct DB25 DB25P DB9S DB25P DB25S DB25P DB25S
CGround 1 1 1 1 1
TX 2 2 3 2 2 2 3
RX 3 3 2 3 3 3 2
RTS 4 4 7 4 4 4 5
CTS 5 5 8 5 5 5 4
DSR 6 6 6 6 6 6 20
GROUND 7 7 5 7 7 7 7
DCD 8 8 1 8 8
DTR 20 20 4 20 20 20 6
RI 22 22 9 22 22
25 to 3 jumper ]
25-pin 3-pin 3-pin
Funct DB25P DB25S DB9S
TX 2 2 3
RX 3 3 2
4 ]
5 ]
6 ]
8 ]
Gnd 7 7 5

Chapter 52 RS 232 405
Schematic _______________________
M O DURHAM, PhD
Dwg no.
Drn: 08/07/2002
Rev:
NTSScale:
WIRING DIAGRAM
STD-H2D1T H E W A Y C O R P.
RS 232 SERIAL COMMUNICATION
www.ThewayCorp.com
mod@superb.org
Box 33124
Tulsa, OK 74153
918-496-8709
TRANSMIT DATA
RECEIVE DATA
GROUND1
2
3
4
5
8
6
22
20
7
1
2
3
4
5
8
6
22
20
7
REQUEST TO SEND
CLEAR TO SEND
CARRIER DETECT
DATA SET READY
RING INDICATOR
DATA TERMINAL READY
SIGNAL GROUND
GROUND
TRANSMIT DATA
RECEIVE DATA
REQUEST TO SEND
CLEAR TO SEND
CARRIER DETECT
DATA SET READY
RING INDICATOR
DATA TERMINAL READY
SIGNAL GROUND
COMPUTER TERMINAL
TRANSMIT DATA
RECEIVE DATA
1
2
3
4
5
8
6
22
20
7
REQUEST TO SEND
DATA SET READY
SIGNAL GROUND
COMPUTER MODEM
1
2
3
4
5
8
6
22
20
7
TRANSMIT DATA
RECEIVE DATA
REQUEST TO SEND
DATA SET READY
SIGNAL GROUND
TERMINALMODEM

Development board pin outs _______
The RJ45 connector on the board supplies serial RS232
communications. The pins are separated by function.
Cable Cable Board Funct To Serial Funct
TIA568B Telco RJ45 RJ11 db 9
Color Alt Pin Pin Pin
W/Orange Black 1 1 Jump
Orange Yellow 2 6 Jump
W/Green White 3 T1X 1 3 TXD
Blue Red 4 R1X 2 2 RXD
W/Blue Green 5 Ground 3 5 Ground
Green Blue 6 R2x 4 4 nc
W/Brown Brown 7 7 Jump
Brown Orange 8 8 Jump
ž Ÿ

53
NETWORK CONNECTION
Thought
To err is human.
To forgive is not a computer function.
Popular quip
Network ________________________
Two fundamental networks are used with computer systems. These
are the analog telephone or telecom system and digital Ethernet
networking connections. Since termination to these networks are
very similar, but different, it is prudent to have the pin-out
diagrams.
The Telecommunications Industry Association (TIA) develops the
standards for interconnection. The standards for this type wiring is
TIA568. The telecom system uses the Universal System Ordering
Codes (USOC). USOC was developed by the Bell system to provide
standard access. It has been adopted by most current users of
telecommunications equipment.
Wire for these networks should be enhanced category 5, commonly
called Cat 5E or better. This cable contains four twisted pairs. To
prevent cross talk, these have different twist rates. When properly
installed and terminated, it permits communications up to 1000
megabits per second.

The diagrams show the color-coding and pin arrangements for these
designs. A design designation can be used with any connector that
has that number of pins or more. The common connector
designation is based on the maximum number of pins.
Designation Standard Pairs Pins
RJ48 USOC 4
RJ45 8
RJ25 USOC 3
RJ14 USOC 2
RJ12 6
RJ11 USOC 1 4
568 TIA
Plug designation
The pins are numbered when looking at a plug with the contacts
facing up. Pin number 1 is on the left. The corresponding pin 1 is
observed in the jack, when the contacts are on the top. The strip
designation is the standard termination strip for networks.
| | | | | | | |
RJ45 1 2 3 4 5 6 7 8
RJ12 1 2 3 4 5 6
RJ11 1 2 3 4
Plug Pin out
The USOC standards start with Pair 1 on the middle two pins. Pair 2
is on the next two pins outward. Pair 3 is on the next two pins
outward. Finally, Pair 4 is on the outermost two pins.
The TIA standards also start with Pair 1 on the middle two pins,
4&5. For TIA568A, Pair 2 is on the next two pins outward, 3 & 6,
and Pair 3 is on the first two pins, 1 & 2. For TIA568B, Pair 3 is on
the next two pins outward and Pair 2 is on the first two pins. Finally,
Pair 4 is on the last two pins, 7 & 8.

Chapter 53 Network Connection 409
The expansion connection from the microcontroller design board
also uses an RJ45 connector. This is the same as the connection
used for computer networking. As a result, the color-coding and
pin-outs will be the same.
Diagram – digital network _________
Networks use an 8-pin connector called an RJ45. T568B is the most
common connection pattern and is used by AT&T. T568A was
developed to more closely match telephone pin-outs.
Only two pairs are used for computer networking. Therefore, the
other two pairs are available for other digital service, such as digital
telephone.
PAIR NETWORK T568B T568A
StandardŸ ATT Tele
ConnectorŸ RJ45 RJ45
Function 8-Pin 8 Pin
White/blue 5 5
Blue/white 4 4
White/orange TX+ 1 3
Orange/white TX- 2 6
White/green RX+ 3 1
Green/white RX- 6 2
White/brown 7 7
Brown/white 8 8
Diagram – analog telephone _______
Standard telephones are analog with a ringing voltage exceeding 90
volts. Since computer networking is digital, these signals should not
be mixed in a cable. Because of these considerations, it is standard
practice to run one cable for digital networking and one cable for
analog service.

PAIR PHONE USOC
StandardŸ ATT
ConnectorŸ RJ45 RJ12 RJ11 Term
Function 8-Pin 6-Pin 4-Pin Strip Telco
White/blue Line 1 tip + 5 4 3 1 Green
Blue/white Line 1 ring - 4 3 2 2 Red
White/orange Line 2 tip + 3 2 1 3 Black
Orange/white Line 2 ring - 6 5 4 4 Yellow
White/green Line 3 tip + 2 1 5 White
Green/white Line 3 ring - 7 6 6 Blue
White/brown Line 4 tip + 8 7 Brown
Brown/white Line 4 ring - 1 8 Orange
Diagram – analog audio ___________
Since networking cable is often used for audio systems, the
following format is one that has been used. At the time of writing,
there was not broad standard for this use. Notice that the selection of
the stereo pairs permits them to be placed on the same cable as the
analog telephone.
PAIR AUDIO
Function Strip Jack Speaker Plug
White/blue Surround R + 1 Red low Red Green
Blue/white Surround R - 2 Red low Black
White/orange Surround L + 5 White low Red White
Orange/white Surround L - 6 White low Black
White/green Stereo R + 3 Red top Red Red
Green/white Stereo R - 4 Red top Black
White/brown Stereo L + 7 White top Red Black
Brown/white Stereo L - 8 White top Black

54
PROGRAMMABLE LOGIC DEVICE
Thought
What you believe
is what happens.
MOD
It is just logic ____________________
A programmable logic device (PLD) is used to provide
combinational logic and register based sequential circuits. The
simple PLDs are rather inexpensive and powerful devices with
significant flexibility. They can be programmed as firmware.
The earlier versions are referred to as programmable array logic
(PAL). An enhanced version by one vendor is called generic array
logic (GAL). The more powerful version is a programmable
electrically erasable logic (PEEL) device which is a superset of the
basic devices.
A common superset peel variation is a 22cv10a. The 22 indicates
there are 22 I/O pins on the device. The 10 indicates that 10 of those
can be used as input/output while the remaining 12 are input only.
The package is a 24 pin skinny DIP.
These devices are programmed using software for the particular
chip. This is often manufacturer specific. Many variations of the

devices are reprogrammable either using ultraviolet or electrically
erasable technology.
The speed is quiet fast. Typical propagation delays are less that 15
ns and as low as 5 ns - faster than standard TTL 7400 series logic.
Combinational logic ______________
A peel is structured with input pins and input/output cell pins.
Boolean logic is used to describe the relationships with the pins.
The process of developing a PLD program is very simple. It is a
direct substitution for combinational logic. A very unpretentious
circuit will effectively illustrate the process.
A Enable
W
The Boolean equation is not complicated.
Enable = A and not W
When written in the software syntax, the equation is equally
obvious.
Enable = A & !W
This is actually the equation for a latch enable when A is the address
and W is the write not line.
From observing the programmable chip pin-out, inputs can be on
pins 2 through 11 and on 13. Similarly, the output can be placed on
pin 14 through 23. Pin one is a clock for register logic and an input
for combinational logic.

Chapter 54 Programmable Logic Device 413
First time user ___________________
Create a text file using a word processor or editor. Save it as
unformatted text. Save with the extension ‘*.PSF’ for the
programming assembly software used in these examples.
The file begins with header information. The only required field is
the filename. The device type must be specified. Then the file lists
assignments for the input pins. Next comes the specification for the
input/output cell pins. Finally, the equations are listed. Comments or
descriptions can be placed anywhere within the program.
Title: ‘File: ModPeel.psf’
PEEL22CV10A
A pin 2
W pin 3
IOC (14 ‘Enable’ POS COM FEED_PIN)
EQUATIONS
Enable.com = A & !W
That is it. That is all and there is no more. An explanation of the
IOC statements is given in the example file.
These programs are processed using ICT WinPlace software. It can
be downloaded from their website.
The PLD software prefers a *.PSF extension. The comments are
removed in a *.RED file. The output of the PLD software is
assembled to a *.MAP file. This is translated to a format that is used
to burn the PEEL. The industry standard is a JEDEC file with the
extension *.JED.
The JED file is loaded into a programmer / burner to change the
links in the PLD.

Gated latch _____________________
The combinational logic circuit implementation is very
straightforward. Much more complex circuits can be implemented
with the PLD. These include internal flip/flops and registers. In
addition the feedback type can be described as combinational,
register, or latch. Nevertheless, the definitions will be only
marginally more complex than the simple example.
Numerous projects use latches for external data expansion. One of
the problems with latches is hazards. The inputs will not change at
exactly the same time. Therefore, there will be a glitch in the output.
This may not be critical in some circuits, but it can be catastrophic
in timing systems.
Therefore, it is appropriate to address the design of a gated latch
circuit.
Data
Q Gated Latch Enable
MmWr
The center AND gate is a redundant term to prevent a hazard or
race. Therefore, if the software reduces the logic to minimum, the
minimization feature will need to be off for this section of code.
@R-
GatedLatch = (MmWr & D)#(!MmWr & GL)#(D & GL)
@R+
OLMC and pin assigments ________
Each of the output pins is controlled by an output logic macro cell
(OLMC). The OLMC allows the pin to act as input, combinational

output, register output, and input/output functions. There are 12
combinations of these variations. The decision is simply controlled
by the equations used to program the pin.
Pin 1 is the clock for registered logic or an input for combinational.
Pins 2 through 11 can be used for any input. For earlier devices, pin
13 is a common output enable for registers or an input for
combinational. For 22cv10 and similar peel devices, each output has
its own enable, so this pin is simply another input. Pins 14 through
23 are controlled by the OLMC.
With this much flexibility, it is necessary to define the input output
cell (IOC) logic with a specification statement for each pin. The
format follows.
IOC (pin-number ‘name’ polarity logic feedback)
Name is the label for that pin. Polarity can be pos(itive) or
neg(ative). Logic can be com(binational) or reg(ister) based.
Feedback can be from pin feed_pin (output pin) or feed_reg
(register out).
The peel devices have a common synchronous preset (SP) and an
asynchronous clear (AC) for the registers. These terms can be
defined by a software equation.
SP := logic equation
AC := logic equation
The superset peel devices also have one buried combinational
feedback, one buried combinational feedback with register, and one
buried register feedback.
Registers ________________________
If the OLMC is used as a register, it can be implemented as a D
flip/flop. By use of the registers, sequential logic such as counters
and shift registers can be constructed.

If any of the outputs are configured as a register, then pin 1 is the
clock input to the register.
Combinational logic uses a .COM extension on the label to set up
the OLMC. All the previous examples have been this type sum of
products.
Enable.COM = A & !W
Registered logic uses a .D extension, as a D flipflop, to set up the
OLMC.
Q.D = D
Q will take the value of D following the rising edge of the clock on
pin 1. Multiple registers can be configured to form counters. Each
intermediate state will be an output pin.
For earlier versions, the registered outputs have a common output
enable on pin 13. When the output enable is low, all the registered
outputs are enabled. When the pin is high, the registered outputs are
tri-stated (high impedance). Nevertheless, the registers can still be
used internally as feedback to another equation.
Combination output enable ________
Combinational logic can define an output enable. It is limited to a
single product term.
Result.OE = B & C
In actuality, it is simply an additional term to the descriptive
equation.

Limitations ______________________
One of the limitations of the simple PLD such as the 22V10 is the
limit on internal logic. Every term that is used must be either an
input or an output. There are no internal states. As a result,
intermediate states use an output pin and preclude its use for other
connections. This limitation simply results from the device being a
simple array that has connections ‘burned’.
The enhanced versions, such as the peel, have limited internal
feedback states that overcome some of the problems.
Nevertheless, it is excellent for direct implementation of
combinational logic and simple gates.
Program: combinational logic (*.psf)
TITLE 'FILE: PeelII.psf'
DESIGNER 'Dr. Marcus O. Durham'
DATE '28 July 2004'
DESCRIPTION
The description is optional. The title block
information must be enclosed within single
quotation marks.
The device must be specified.
end_DESC;
PEEL22CV10

" ___ ____
" | / |
" CLK { 1 24 } Vcc
" I1 { 2 23 } F9
" I2 { 3 22 } F8
" I3 { 4 21 } F7
" I4 { 5 20 } F6
" I5 { 6 19 } F5
" I6 { 7 18 } F4
" I7 { 8 17 } F3
" I8 { 9 16 } F2
" I9 { 10 15 } F1
" I10 { 11 14 } F0
" Gnd { 12 13 } I11
" |_________|
"Pins
"Registered clock or input: 1
"Input: 2,3,4,5,6,7,8,9,10,11, 13
"Input or output logic macro cell (OLMC):
" 14,15,16,17,18,19,20,21,22,23
"
"-------
"Symbols are used for programming functions
" Quote = comments not printed
" ' = comments for header to be printed'
" ! = not
" & = AND
" # = OR
" $ = exclusive-or
" = is equation for combination logic
" ; = last character in an equation
"-------
"Application
"Memory can be changed from EPROM to SRAM
"by changing the chip & /OE jumper.
"PLD / PEEL program selects MemP1 and MemP27.
"EPROM: MemP1=Vpp, MemP27=A14 requires /OE=/PSEN
jumper
"SRAM: MemP1=A14, MemP27=/WE, requires /OE=/RD or
Ground jumper

"MMIO: uses A15
"SRAM: !A15
"CHECK MemP27: IT CAN GO LOW IF !PSENn, IE RDn
ACTIVATES A14
"-------
"Address assignments.
"Because of the Don't Cares, the effective
"memory-mapped address can be 8xxn.
"Memory-mapped read or write is address A15 only.
"8000h
"Keypad latches are read row & write column
"8001h
"DisLE latch is selected for 7Segment & LCD.
"8002h
"
"Isp OE is select low on invert reset input.
"-------
"Memory switching latch was for old BIOS
"Kept for record purposes, original file is lost.
"CE = WRFF & ucA7 & ucD0 # !WRFF&CE
"OE = PSEN&RD # RD&!CE
"---------------
"PIN ASSIGNMENTS
"INPUTS
A00 pin 1
A01 pin 2
A02 pin 3
A14 pin 4
A15 pin 5
AD00 pin 6
Reset pin 7
" pin 8
PSEN pin 9
RDn pin 10
WRn pin 11
" pin 13

"INPUT / OUTPUT CELLS
"The input/output cells require specification.
"The format is
"IOC (pin-number 'name' polarity logic feedback)
"Polarity can be pos(itive) or neg(ative).
"Logic can be com(binational) or reg(ister)
"Feedback can be from the feed_pin or feed_reg.
IOC (14 'MmRd' POS COM FEED_PIN)
IOC (15 'MmWr' POS COM FEED_PIN)
"IOC (16 'Spare' POS COM FEED_PIN)
IOC (17 'DisLE' POS COM FEED_PIN)
IOC (18 'KeyLE' POS COM FEED_PIN)
IOC (19 'KeyOEn' POS COM FEED_PIN)
IOC (20 'ISPOEn' POS COM FEED_PIN)
IOC (21 'MemP27' POS COM FEED_PIN)
IOC (22 'Oen' POS COM FEED_PIN)
IOC (23 'MemP1' POS COM FEED_PIN)
DEFINE
"Define variables do not have feedback.
MmAd = A15
EQUATIONS
"Equations define logic functions
"Register variables have .D extension for D FF.
"Combinational variables have .COM extension.
IspOEn.com = !Reset;
MemP1.com = A14 & (!A15 & (!WRn #!RDn));
MemP27.com = (!PSEN & A14) # (WRn & !A15);
Oen.com = PSEN # !(!RDn & !A15);
MmWr.com = A15 & !WRn;
MmRd.com = A15 & !RDn;
KeyOEn.com = !(MmRd& !A02& !A01& A00);
KeyLE.com = (MmWr& !A02& !A01& A00);
DisLE.com = (MMWr & !A02 & A01 & A00);

TEST_VECTORS
"Test vectors are used to verify the logic works.
"This is a partial truth table.
"The first line specifies input output variables.
"Next line are input output values.
"C is clock transition.
"0 is an input low, 1 is an input high
"L is output low, H is output high.
"X is don't care
(A00 A15 A00 Wrn MmWr DisLE -> KeyOEn)
C 0 X 0 L x X
C 0 X 1 L x X
C 1 X 0 H x X
C 1 X 1 L x X
Created Files ____________________
The redefine file (*.red) is an output of the program. It eliminates all
the comments and extraneous information. It only contains the
necessary information to execute the program. This is the title,
device, pin assignments for input, input output cell description, and
equations.
The assembled file (*.map) contains the pin node connections and
all the product terms.
The JEDEC file (*.jed) is the information that is sent to burn the
programmable logic device. This is a common language among
many vendors.
ž Ÿ

55
CIRCUIT TIME & PHASE SHIFT
Thought
Educating process:
Encourage – instruction – example.
Dr. Jerry Falwell
Background ____________________
Every electrical device and circuit has a difference in time between
the input and the output. This difference is called by a variety of
names, depending on the specialty within the electrical field.
Various names are time delay, propagation delay, time constant,
phase shift, and wait state. A trigonometric interpretation is referred
to as power factor or lead/lag angle.
The property characteristics of every material causes some
resistance. In addition, bending the circuit path creates an
inductance. In contrast, placing two conductors in proximity causes
a capacitance.
The combination of resistance with inductance or with capacitance
defines the delay in terms of a time constant. The time delay is
measured in seconds, while the resistance is Ohms, inductance is
Henries, and capacitance is Farads.
tD = L / R

Chapter 55 Circuit Time & Phase Shift 423
tD = R * C
The combination of inductance and capacitance causes a frequency
that may be called oscillation, ringing, vibration, revolutions, or
cycling. The frequency is in cycles per second called Hertz.
f = 1 / 2S—(L * C)
Integrated circuit devices have very complex internal connections.
Therefore, it is virtually impossible for the user to determine the
delay. The manufacturer will describe the lag between the input and
the output as propagation delay.
Digital circuits have either an on or off state. The state will change
based on the input. If the input branches have different time delays,
then the digital state will not change at precisely the time expected.
This different response due to a delay can result in a hazard
condition called a race. That is a situation when the response is not
absolutely defined due to differences in time for the state changes
on the input.
Delay ___________________________
Occasionally it is desired to create a delay. For example if an enable
pulse is just a little too short, it can be stretched with a AND gate.
The input to one side is delayed.
Wr’
Wr’ (stretched)
Suppose a total delay of 37 ns is required. The 74AC08 has an eight
ns propagation delay. The values for the resistor and capacitor are
chosen to make the difference of 37 – 8 = 29 ns.
The resistor value must be large enough to keep Wr’ from sinking
too much current. A value in excess of 2 k Ohms is generally

acceptable, as was found in the fundamentals chapter. A capacitor of
10 pF will yield a time value in the nanosecond region.
The resistor value should correspond to the exact timing needed. For
29 ns, the value would be 2.9 k Ohms.
tD = R * C = 2900 * 10 e-12
= 29 ns
The capacitor should have a matched negative temperature
coefficient so the RC time constant will not drift with temperature.
Voltage and temperature variations and noise can cause the above
circuit to malfunction. A safety pad of 10 ns should be subtracted
from the maximum stretched pulse width allowed to assure adequate
delay.
Clock signals ____________________
Clock signals are critical to sequential circuits. Likewise, they
establish the timing relationship between all the components
associated with a computer.
Clocks are simply signals that cycle between low and high values.
State variables change state only at the clock edge. State changes are
level independent. The edge may be a rising edge on the transition
from low to high. Alternately it can be a trailing or falling edge on
the transition from high to low.
tL tH
tPERIOD
Interaction _____________________
Signals must be asserted in the correct sequence and for a long
enough duration that the next circuit can respond. In computer

circuits, the clock timing is determined by an oscillator connected to
the chip. The chip in turn, must perform multiple chores within its
purview.
Similarly, the processor must interface to other devices such as
memory and latches. All these perform at a certain rate. Since the
computer is the system controller, it will generate the timing signals
for the other devices.
A timing diagram is developed by the manufacturer of each
integrated circuit chip to illustrate the duration of each signal. The
computer primarily interfaces to external chips in one of three ways.
Read code is used to access program memory. Read data is used to
access input devices. Write data is used to access output devices.
The timing diagrams of the microprocessor for these three
operations are shown on the next page.
Consider the timing for addressing. The processor asserts the ALE
(address latch enable) line high. This is usually connected to a
74573 latch enable. Then the processors asserts the address on port
0. After adequate set-up time, the processor asserts the ALE low.
This causes the latch to trap the information that was on port 0.
Next the input control line is asserted low. This is the PSEN’ or
RD’. Then the data is brought in on port 0. For writing, the data is
first placed on port 0, then the WR’ is asserted low.
Any chips controlled by these lines, must respond within the time
frame that the line is asserted. Otherwise, the data will not transfer
between the chip and the processor.
ž Ÿ

Ext program memory read cycle____
Ext data memory read cycle________

Ext data memory write cycle _______

SECTION VII – DOCUMENTATION
ž Ÿ

56
EXTREME PROGRAMMING (XP)
HARMONIZATION
Thought
I change my life,
when I change my mind.
Jim Stovall
General guidelines _______________
Extreme programming is a team approach to project development.
An early chapter addressed the requirements of the practice.
Each team creates code that looks exactly like all the other. In
addition, variables are commonly defined so each team can access
them.
Since a large number of items are used to interface with a
microprocessor, many variables are commonly used. This segment
of program code illustrates the common values that are often
required. Therefore, it is prudent to avoid overlaying or changing
these values.
User variables can be placed in the upper locations of the internal
RAM between the defined assignments and where the stack has
been moved.

Program specifics ________________
;-----------------------------
;Program: MOD-SYSTEM.asm
;Update: 15 August 2002
;
; Tulsa, OK, USA
; mod@superb.org
;Copyright (c)1991 - 2002. All rights reserved
;
;Purpose:
; basic functions of a computer.
;
; These include ports, timers, serial RS232,
; mmio, keypad, seven-segment display,
; liquid crystal display, analog/digital convert,
; and math.
;
;Baud: 9600
;Handshake: not used at this speed
;
; There are three types of processing techniques
; that are used.
; MAIN - is the normal loop processing
; INTERRUPTS- occur based on events, such as
; regular time intervals.
; BUSY - is called for random tasks that
; can be processed during the delay
; in other tasks. Keypad scan
; is a prime example.
;
;-----------------------------
;ASSEMBLERS
;-----------------------------
;
; Several assemblers are available for the 8051.
; The major differences are in the syntax of

Chapter 56 Extreme Programming Harmonization 431
; the directives. The three that I have used are
; compared.
;
;
; TASM.EXE is a table lookup.
; tasm -51 -p -l Bios11v2.asm
; .org 0050h ;Next inst @ 0050H
;Value .equ 12h ;predefined value
;Table .byte 34h ;define code byte
; .text "123" ;define ASCII
; .end ;last thing
;
;
; A51.EXE is student version of PseudoCodes.
; A51 -s Bios11v2.asm
; .org 0050h ;Next inst @ 0050H
; .equ label,12h
;Table: .db 34h ;define code byte
; .db "123" ;define ASCII
;
;
; ASM51.EXE is Intel assembler with library.
; ASM51 Bios11v2.asm
; org 0050h ;Next inst @ 0050H
;Value equ 12h ;predefined value
;Table: db 34h, '123' ;define code storage
; end ;last thing
;
;
;-----------------------------
;FORMAT
;-----------------------------
;
; The format of the code lines is shown. This
; gives space for opcodes in the *.lst files
; to fit on one line of display & print.
;
;LABEL: MNEMO REGISTERS ;COMMENTS
;Columns 1 1 2 3 4|
; 1 7 4 1 9
;
;-----------------------------
;EXTERNAL -VS- INTERNAL PROM
;-----------------------------

;
; 8051 has external memory on prom
; 8951 has 4K (07ffH) internal flash eprom
; 898252 has 8K flash & 2K eeprom
;
; To program 8051, burn an external prom/eeprom.
; To program low memory of 8951, burn flash
; To program 898252, program using SPI to port 1.
; P15 = MOSI
; P16 = MISO
; P17 = SCLK
; Ground = common
; Use program AEC_ISP.exe to download from PC
;
; When the uP is reset, it looks at EA'.
; If EA'=0, then external eprom is read.
;
; IF EA'=1, then internal eprom is read.
; Internal eprom is executed until it hits end.
; Then external eprom is used, even if EA'=1.
;
; EA' is pin 31.
; To begin program from prom, strap EA' to 0.
; To use the internal flash, strap EA' to 1.
;
;-----------------------------
;
; The 8951 has 4K (0FFFh) internal flash prom.
; Make the last internal executable line be 0FFFh
; The next instruction will be 1000h on external
; prom/sram.
; So, org the external program to 1000H.
; Initially make program organize to hi memory.
; org 1000H ;external
;
; If it is moved to low memory (onboard flash),
; the address must be known so the hi memory can
; call the routine.
; For example in low memory
; org 0250H ;Next inst @ 0250H
;SCRNDATA:mov ;low mem routine
;
; The high memory PROM/SRAM overlays the same
; space. It is just not available to the uC.

; However, this little feature lets us place
; an org directive to the assembler at the same
; location. It will never be executed. Program
; Control will go to the low memory
; org 0250H ;pseudo location
;SCRNDATA:nop ;just for assembler
;#################################################
;
; ASSIGNMENTS
;
;#################################################
;
; Assignments include
; 1. constants for symbols and sizes
; 2. constants for ext mmio addresses
; 3. variables assigned to internal RAM locations
;+++++++++++++++++++++++++++++++++++++++++++++++++
;CONSTANTS
;-----------------------------
;SYMBOLS
Scrl equ 42d ;scroll key,"*",2AH
Entr equ 35d ;enter key,"#",23H
None equ 0FFH ;blank key
Flash equ 254 ;black cursor for flash
Colon equ 3AH ;colon
Zero equ 30H ;zero character
Slash equ 2FH ;slash
DecPt equ 2EH ;decimal point
Minus equ 2DH ;minus sign
CRet equ 13d ;carriage return
DigX equ 8 ;number of decimal digits
MathX equ 4 ;number of hexadecimal byte
;-----------------------------
;EXTERNAL ADDRESS
FilAdH equ 00 ;address file base for DPH

FilAdL equ 50H ;for DPL, base addr= 0050H
AdKey equ 8001H ;MMIO addr for Keypad latch
AdSeven equ 8002H ;MMIO addr for 7-segment
AdInst equ 8003H ;MMIO addr for LCD instruct
AdData equ 8004H ;MMIO addr for LCD data
AdRead equ 8005H ;MMIO addr for LCD read
;+++++++++++++++++++++++++++++++++++++++++++++++++
;STACK
;-----------------------------
;
; The stack is moved up high to prevent it from
; overwriting data. Stack counts up from the next
; location past the stack pointer base.
;
; The default stack location is 07H. It will over
; write the registers.
;
; Two bytes are used by each interrupt and each
; subroutine call. So enough stack space must be
; reserved for nested loops.
;
; ;STACK RESERVED
; equ 5FH ;Move stack to 60H & above
;
;+++++++++++++++++++++++++++++++++++++++++++++++++
;USER VARIABLES
;-----------------------------
;
; All the RAM locations I used are shown below.
; This is so the user will not overwrite them.
;
; Place any user variables within this range.
;
;-----------------------------
;USER DEFINED VARIABLES
;place user variable here
; equ 5Fh ;
;and everything in between
SbcTeam2 equ 40h ;count team 1

SbcTeam1 equ 3Fh ;count team 1
TimMin equ 3EH ;time minutes counter
TimSec equ 3DH ;time seconds counter
TimPer equ 3CH ;time periods counter
;+++++++++++++++++++++++++++++++++++++++++++++++++
;DEFINED VARIABLES
;-----------------------------
;
; Defined variables are used by many routines.
; These are fixed locations that should not be
; changed since the order is critical to SqROOT.
;
; Note that the math variables ArgX, RemX, FraX
; locations are shared with TmpX and GapX
; Math procedures are seldom done. When they are,
; the process is completed before other functions
; are attempted. Therefore, there is no conflict.
;Gap is scratch, display
;Tmp is temporary register
;Hex =32 bit working reg
;ORDER CRITICAL TO SQ ROOT
FraD equ 3BH ;4 bytes used in divide
FraC equ 3AH ;same space as Gap upper
FraB equ 39H
FraA equ 38H
RemD equ 33H ;4 digit remain
RemC equ 32H ;same space as Tmp upper by
RemB equ 31H
RemA equ 30H
SizeT equ 33H ;same space as Tmp upper by
ArgG equ 32H ;used in multiply
ArgT equ 31H
ArgH equ 30H
GapH equ 3BH ;8 digit
GapG equ 3AH ;BCD digits high byte
GapF equ 39H ;& intermediate value hold
GapE equ 38H

GapD equ 37H
GapC equ 36H
GapB equ 35H
GapA equ 34H
TmpH equ 33H ;8 byte
TmpG equ 32H
TmpF equ 31H
TmpE equ 30H ;Delay routine
TmpD equ 2FH
TmpC equ 2EH ;RAM Init Valu, Init ea pas
TmpB equ 2DH ;RAM Test Valu, cpl ea addr
TmpA equ 2CH ;math low byte
HexD equ 2BH ;double word
HexC equ 2AH
HexB equ 29H ;sixteen bit register
HexA equ 28H ;low byte of variable
;-----------------------------
; ;BITS LOCATION
; equ 27H ;last byte reserved for bit
; equ 20H ;first byte for bit usage
;
;-----------------------------
;SCREENS
ScnFil equ 1FH ;File where data is stored
ScnMem equ 1EH ;bytes of memory required
ScnGet equ 1DH ;cursor loc present
ScnCur equ 1CH ;cursor loc desired & temp
ScnDis equ 1BH ;bytes of display space req
ScnTyp equ 1AH ;type data, if #, it is dp
;-----------------------------
;BANK3
QikB equ 19H ;Interrupt HEX value, msb
QikA equ 18H ;Interrupt HEX value, lsb
;-----------------------------
;KEYS
KeyCol equ 17H ;present column number
KeyBit equ 16H ;byte moves a bit w/ column
KeyRow equ 15H ;row number pushed
KeyMul equ 14H ;multiple key count

CharK equ 13H ;key input character
CharD equ 12H ;debounced
;-----------------------------
;BANK2
;R1 equ 11H ;
;R0 equ 10H ;
;-----------------------------
CharP equ 0FH ;undebounced previous input
MenuL equ 0DH ;loop @MENUS
LatCr equ 0CH ;control latch bits status
ScnDph equ 0BH ;hold DPH
ScnDpl equ 0AH ;hold DPL
;-----------------------------
;BANK1
;R1 equ 09H ;SRAM address, interrupts
;R0 equ 08H ;SRAM address, background
;-----------------------------
;BANK 0, USE W/ MATH & LOOP
Size equ 06H ;multiply, divide, sqrt, GP
;R5 equ 05H ;carry in multiply, GP
;R4 equ 04H ;carry in mul,GP
;R3 equ 03H ;loop 2, math
;R2 equ 02H ;loop, destination size,mat
;R1 equ 01H ;source indirect addr,math
;R0 equ 00H ;destination indirect addr
;+++++++++++++++++++++++++++++++++++++++++++++++++
;BITS ASSIGNMENTS
;-----------------------------
;
;AT RAM byte 20H
FgC bit 01H ;temporary for C, etx
FgKeyH bit 00H ;flag key held down

;+++++++++++++++++++++++++++++++++++++++++++++++++
;SPECIAL FUNCTION REGISTERS
;-----------------------------
;
; SFR are listed for information. With limited
; assemblers, remove the ; since the address must
; be defined.
;
; ;SFR DEFINED
;B equ 0F0H ;second math register
;Acc equ 0E0H ;accumulator
;PSW equ 0D0H ;processor status word
;IP equ 0B8H ;interrupt priority
;P3 equ 0B0H ;port 3
;IE equ 0A8H ;interrupt enable
;P2 equ 0A0H ;port 2
;SBUF equ 99H ;serial buffer
;SCON equ 98H ;serial control
;P1 equ 90H ;port 1
;TH1 equ 8DH ;timer hi byte 1
;TH0 equ 8CH ;timer hi byte 0
;TL1 equ 8BH ;timer lo byte 1
;TL0 equ 8AH ;timer lo byte 0
;TMOD equ 89H ;timer mode
;TCON equ 88H ;timer control
PCON equ 87H ;power control reg
;DPH equ 83H ;data pointer hi
;DPL equ 82H ;data pointer low
;SP equ 81H ;stack pointer
;P0 equ 80H ;port 0
;-----------------------------
;bit IN SFR
Acc7 equ 0E7H ;Accumulator bit 7
Acc6 equ 0E6H ;accum bit 6
BankH equ 0D4H ;PSW register bank high bit
BankL equ 0D3H ;PSW register bank low bit
Over equ 0D2H ;PSW bit 2, overflow on add
;P equ 0D0H ;parity, PSW
;TI equ 99H ;transmit is complete
;RI equ 98H ;receive is complete
;TR1 equ 8EH ;timer 1 start
;TF0 equ 8DH ;timer 0 overflow

;TR0 equ 8CH ;timer 0 start
;-----------------------------
;PORT USE
P33 equ 0B3H ;Port33 Hi=PC Busy,uP no xm
;Not always used in prog
;to use, setb P33 for input
P35 equ 0B5H ;Port35 Hi=uP BUSY,stop ser
;xmit to uP in download mod
;SPI ADC LINES
;move when know real locate
SpMosi equ 0B2H ;SPI data out, INT0, P32
SpMiso equ 0B4H ;SPI data in, T0, P34
SpCLk equ 0B5H ;SPI serial clock, T1, P35
SpCS equ 4 ;SPI ADC CS Con latch D4
SpConv equ 5 ;SPI ADC Cnv Con latch D5
;#################################################
;
; PROGRAM
;
;#################################################
org 00H
START:
;-----------------------------
; When processor is reset, program control comes
; here. Jump to the first executable address
; after all interrupts reserved locations.
ljmp INITIAL
;*************************************************
;
; INTERRUPTS
;
;*************************************************
;
; Interrupt set-up is discussed in a later
; section. It is associated with timer/counters.
;
;-----------------------------

;INTERRUPT-External 0
;-----------------------------
org 03H
reti
;-----------------------------
;INTERRUPT-Timer 0
;-----------------------------
; The procedure provides direction when timer
; completes count.
org 0BH
ljmp TIMECAL ;interrupt processor
;-----------------------------
;INTERRUPT-External 1
;-----------------------------
org 13H
reti
;-----------------------------
;INTERRUPT-Timer 1
;-----------------------------
; Timer 1 is used for serial. So this procedure
; will never be executed.
org 1BH
;-----------------------------
;INTERRUPT-Serial
;-----------------------------
; This is not used. The bits are polled in
; various other routines.
org 23H
;-----------------------------
;COPYRIGHT
;-----------------------------
; Ownership of program is stored in memory.
org 0033H ;Address past vectors

db 'Copyright (C) 2004'
db 'Tulsa, OK USA'
;*************************************************
;
; INITIAL & MAIN
;
;*************************************************
org 0080h ;Addres past reserve
; org 2000h ;Address if external
;-----------------------------
INITIAL:
;-----------------------------
; Setup the initial conditions.
;-----------------------------
end
ž Ÿ

57
DOCUMENTATION
Thought
The job is not finished,
until the paperwork is done.
Popular quip
Report _________________________
A report should be prepared that is a complete documentation of the
project. The report will be a technical manuscript that you provide
to your boss or customer. The intent is to provide all the information
so the project can be duplicated or advanced. Your next contract
depends on the quality of your presentation.
All the information should be typed and professionally printed on a
laser-quality printer.
1. A Cover Sheet should include project name, course number,
your name, date the project is due, and date the project is
submitted. When the project is completed, have the page
initialed and dated by a supervisor.
2. A Table of Contents should list all the sections. This can be on
the same page as the cover sheet.
3. An Abstract should give the project definition. It should be
adequate that an unfamiliar person can read it and determine

Chapter 57 Documentation 443
what the project is about. The length should be less than 250
words.
4. An Executive Summary should give the conclusions and
solutions statement. There should be a brief statement of
problems. One sentence should discuss future applications and
improvements on what you did. The length should be less than
400 words. The abstract and executive summary should tell the
complete story. Keep It Simple, Sam!
5. A Hardware Block Diagram will illustrate the major building
blocks of the equipment.
6. A Software Block Diagram will be a flowchart of the major
components of the program.
7. A Schematic will show the interconnection of all the hardware
components. The file should be in a format that is importable for
display by a web browser.
8. A Software Listing of all programs with comments is a
necessary component. This should be an editable executable
version of the software.
9. A Spreadsheet contains the cost items. This is a list of parts,
which contains the following information. A total should be
given as a summation of all the costs.
Part Device Package Where Used Vendor Number Qty Cost
10. A Time sheet should breakout the time invested in each phase of
the project. As a minimum this will be planning, software,
hardware, trouble-shooting, and report preparation.
11. A Support Equipment list will include all extraneous equipment.
The list will include a personal computer, applications software
packages, EPROM burner, power supply, cables, and any other
items required.

It is advisable to keep a copy of the report for your records. Much of
the information will be used on succeeding projects. This is called
file-drawer engineering.
Computer aided design ___________
In today’s marketplace, all circuit diagrams should be made with a
computer aided design tool. This results in professional schematics
and lends itself to changes. If a printed circuit board design program
is used, all the following comments will be included.
Occasionally, it is desirable to make a sketch in a word processing
or other program. These may be beneficial for enhancing a
particular part of the circuit or for illustration purposes. Then the
following suggestions are very appropriate.
There are many ways to draw a circuit diagram. While there is no
universal standard for drawing schematics, a standard approach for
diagrams is a requirement for consistent results.
The first and most obvious practice is to draw a block for each chip.
Label the chip number and its function (e.g. uP, ROM, RAM, etc.)
inside the block. Labeling the chip's function is especially important
when using unusual chips. However, there is one exception to the
block labeling; discrete gate diagrams are discussed below.
Every pin of each chip should be labeled with its function on the
inside of the block and its pin number on the outside of the block.
Some groups of pins have a similar function such as address-bus
pins. These can be labeled as a group instead of drawing a pin-for-
pin representation. Make sure that the pin numbers listed on the
outside of the block correspond appropriately to the pin descriptions
on the inside of the block. For example, A0 corresponds to pin 39.
Discrete gates (drivers, flip/flops, etc.) should not be drawn in block
form. The number of interconnections to various parts of the circuit
would be too cumbersome. Instead, draw the gate in the circuit
diagram where it is needed. Draw the part number inside the gate.

Chapter 57 Documentation 445
Include the pin function if appropriate. Draw pin numbers outside
the gate.
Power supply connections for the gate need not be specified unless
they are different from the standard. In most cases, power is applied
to the upper right corner pin and ground is applied to the lower left
corner pin. Consider a 14-pin chip. The upper right pin is number 14
and has +5 V while the lower left pin is number 7 and has ground.
Interconnections between pins on different chips can be drawn in
whatever way works best. Avoid making the lines too convoluted.
Lines that cross are not considered to connect unless there is a dot at
the intersection. This procedure does not require humps every time
one line crosses another.
Lines of similar function, such as address lines and data lines, need
not be drawn separately. It is common practice to use a single bus
line as long as the line is wider than normal. Draw a slash through
the bus line with the number of lines on the bus. If several, but not
all, lines branch off from the bus, label the branches and make a
new bus line if necessary.
Unused pins on any chips, except discrete gates, must still be
included in the circuit diagram. Draw a short 3-4 mm line from the
pin, but do not connect the wire to anything.
If the pin is tied to ground or to the power supply, draw in a short
connection and label it ground or +5 V. Do not use bus lines for
ground and power supply connections. This would result in too
many lines on the circuit diagram.
ž Ÿ

System design and 8051

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