UNIT-5.ppt

UNIT-5
INTERFACING
MICROCONTROLLER
1

UNIT 5 Syllabus
• Programming 8051 Timers
• Serial Port Programming
• Interrupts Programming
• LCD & Keyboard Interfacing
• ADC, DAC & Sensor Interfacing
• External Memory Interface
• Stepper Motor & Waveform generation
2

8051 Timer Modes
Timer 0
Mode 3
Mode 2
Mode 1
Mode 0
Mode 2
Mode 1
Mode 0
Timer 1
8051 TIMERS
4

TMOD Register
GATE:
When set, timer/counter x is enabled, if INTx pin is high and TRx
is set.
When cleared, timer/counter x is enabled, if TRx bit set.
C/T*:
When set(1), counter operation (input from Tx input pin).
When clear(0), timer operation (input from internal clock).
5

TMOD Register
The TMOD byte is not bit addressable.
6
00- MODE 0
01- MODE 1
10- MODE 2
11- MODE 3

TCON Register
7
TF 1-Timer 1 overflow flag
TF0-
TR1- Timer 1 Run control bit
TR0-
IE1- Interrupt 1
IE0-
IT1- Timer 1 interrupt
IT0-

8051 Timer/Counter
OSC ÷12
TLx
(8 Bit)
/ 0
C T 
/ 1
C T 
INT PIN
Gate
TR
T PIN
THx
(8 Bit)
TFx
(1 Bit)
INTERRUPT
8

OSC ÷12
TL0
/ 0
C T 
/ 1
C T 
0
INT PIN
Gate
0
TR
0
T PIN
TH0
INTERRUPT
TIMER 0
TF0
9

TL0
(5 Bit)
INTERRUPT
TIMER 0 – Mode 0
OSC ÷12
/ 0
C T 
/ 1
C T 
0
INT PIN
Gate
0
TR
0
T PIN
TH0
(8 Bit)
TF0
13 Bit Timer / Counter
Maximum Count = 1FFFh (0001111111111111)
10

TL0
(8 Bit)
INTERRUPT
TIMER 0 – Mode 1
OSC ÷12
/ 0
C T 
/ 1
C T 
0
INT PIN
Gate
0
TR
0
T PIN
TH0
(8 Bit)
TF0
Maximum Count = FFFFh (1111111111111111)
11

TH0
(8 Bit)
Reload
TIMER 0 – Mode 2
8 Bit Timer / Counter with AUTORELOAD
TL0
(8 Bit)
OSC ÷12
/ 0
C T 
/ 1
C T 
0
INT PIN
Gate
0
TR
0
T PIN
TH0
(8 Bit)
TF0 INTERRUPT
Maximum Count = FFh (11111111)
12

TL0
(8 Bit)
INTERRUPT
TIMER 0 – Mode 3
OSC ÷12
/ 0
C T 
/ 1
C T 
0
INT PIN
Gate
0
TR
0
T PIN
TF0
Two - 8 Bit Timer / Counter
OSC ÷12
1
TR
TH0
(8 Bit)
INTERRUPT
TF1
13

OSC ÷12
TL1
/ 0
C T 
/ 1
C T 
Gate
TH1
INTERRUPT
TIMER 1
TF1
1
INT PIN
1
TR
1
T PIN
14

TL1
(5 Bit)
INTERRUPT
TIMER 1 – Mode 0
OSC ÷12
/ 0
C T 
/ 1
C T 
Gate
TH1
(8 Bit)
TF1
Maximum Count = 1FFFh (0001111111111111)
1
INT PIN
1
TR
1
T PIN
15

TL1
(8 Bit)
INTERRUPT
TIMER 1 – Mode 1
OSC ÷12
/ 0
C T 
/ 1
C T 
Gate
TH1
(8 Bit)
TF1
Maximum Count = FFFFh (1111111111111111)
1
INT PIN
1
TR
1
T PIN
16

TH1
(8 Bit)
Reload
TIMER 1 – Mode 2
8 Bit Timer / Counter with AUTORELOAD
TL1
(8 Bit)
OSC ÷12
/ 0
C T 
/ 1
C T 
Gate
TH1
(8 Bit)
TF1 INTERRUPT
Maximum Count = FFh (11111111)
1
INT PIN
1
TR
1
T PIN
17

TCON Register (1/2)
• Timer control register: TMOD
– Upper nibble for timer/counter, lower nibble for
interrupts
• TR (run control bit)
– TR0 for Timer/counter 0; TR1 for Timer/counter 1.
– TR is set by programmer to turn timer/counter on/off.
• TR=0: off (stop) TR=1: on (start)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt
(MSB) (LSB)

TCON Register (2/2)
• TF (timer flag, control flag)
– TF0 for timer/counter 0; TF1 for timer/counter 1.
– TF is like a carry. Originally, TF=0. When TH-TL roll
over to 0000 from FFFFH, the TF is set to 1.
• TF=0 : not reach
• TF=1: reach
• If we enable interrupt, TF=1 will trigger ISR.
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt
(MSB) (LSB)

Equivalent Instructions for the Timer Control
Register
For timer 0
SETB TR0 = SETB TCON.4
CLR TR0 = CLR TCON.4
SETB TF0 = SETB TCON.5
CLR TF0 = CLR TCON.5
For timer 1
SETB TR1 = SETB TCON.6
CLR TR1 = CLR TCON.6
SETB TF1 = SETB TCON.7
CLR TF1 = CLR TCON.7
TF1 IT0
IE0
IT1
IE1
TR0
TF0
TR1
TCON: Timer/Counter Control Register

Timer Mode 1
• In following, we all use timer 0 as an example.
• 16-bit timer (TH0 and TL0)
• TH0-TL0 is incremented continuously when TR0 is set to 1. And
the 8051 stops to increment TH0-TL0 when TR0 is cleared.
• The timer works with the internal system clock. In other words,
the timer counts up each machine cycle.
• When the timer (TH0-TL0) reaches its maximum of FFFFH, it
rolls over to 0000, and TF0 is raised.
• Programmer should check TF0 and stop the timer 0.

Steps of Mode 1 (1/3)
1. Choose mode 1 timer 0
– MOV TMOD,#01H
2. Set the original value to TH0 and TL0.
– MOV TH0,#FFH
– MOV TL0,#FCH
3. You had better to clear the flag to monitor: TF0=0.
– CLR TF0
4. Start the timer.
– SETB TR0

5.The 8051 starts to count up by incrementing the
TH0-TL0.
– TH0-TL0=
FFFCH,FFFDH,FFFEH,FFFFH,0000H
FFFC FFFD FFFE FFFF 0000
TF = 0 TF = 0 TF = 0 TF = 0 TF = 1
TH0 TL0
Start timer
Stop timer
Monitor TF until TF=1
TR0=1 TR0=0
TF

6. When TH0-TL0 rolls over from FFFFH to 0000,
the 8051 set TF0=1.
TH0-TL0= FFFEH, FFFFH, 0000H (Now TF0=1)
7. Keep monitoring the timer flag (TF) to see if it is
raised.
AGAIN: JNB TF0, AGAIN
8. Clear TR0 to stop the process.
CLR TR0
9. Clear the TF flag for the next round.
CLR TF0

Mode 1 Programming
XTAL
oscillator ÷ 12
TR
TH TL TF
Timer
overflow
flag
C/T = 0
TF goes high when FFFF 0

Timer Delay Calculation for XTAL = 11.0592 MHz
(a) in hex
• (FFFF – YYXX + 1) × 1.085 s
• where YYXX are TH, TL initial values respectively.
• Notice that values YYXX are in hex.
(b) in decimal
• Convert YYXX values of the TH, TL register to
decimal to get a NNNNN decimal number
• then (65536 – NNNNN) × 1.085 s

Example 1 (1/3)
• square wave of 50% duty on P1.5
• Timer 0 is used
;each loop is a half clock
MOV TMOD,#01 ;Timer 0,mode 1(16-bit)
HERE: MOV TL0,#0F2H ;Timer value = FFF2H
MOV TH0,#0FFH
CPL P1.5
ACALL DELAY
SJMP HERE
50% 50%
whole clock
P1.5

Example 1 (2/3)
;generate delay using timer 0
DELAY:
SETB TR0 ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR0 ;stop timer 0
CLR TF0 ;clear timer 0 flag
RET
FFF2 FFF3 FFF4 FFFF 0000
TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 1

Example 1 (3/3)
Solution:
In the above program notice the following steps.
1. TMOD = 0000 0001 is loaded.
2. FFF2H is loaded into TH0 – TL0.
3. P1.5 is toggled for the high and low portions of the pulse.
4. The DELAY subroutine using the timer is called.
5. In the DELAY subroutine, timer 0 is started by the “SETB TR0”
instruction.
6. Timer 0 counts up with the passing of each clock, which is provided by the
crystal oscillator.
As the timer counts up, it goes through the states of FFF3, FFF4, FFF5, FFF6,
FFF7, FFF8, FFF9, FFFA, FFFB, FFFC, FFFFD, FFFE, FFFFH. One more
clock rolls it to 0, raising the timer flag (TF0 = 1). At that point, the JNB
instruction falls through.
7. Timer 0 is stopped by the instruction “CLR TR0”. The DELAY subroutine
ends, and the process is repeated.
Notice that to repeat the process, we must reload the TL and TH
registers, and start the timer again (in the main program).

Example 2 (1/2)
• This program generates a square wave on pin P1.5 Using timer 1
• Find the frequency.(dont include the overhead of instruction delay)
• XTAL = 11.0592 MHz
MOV TMOD,#10H ;timer 1, mode 1
AGAIN:MOV TL1,#34H ;timer value=3476H
MOV TH1,#76H
SETB TR1 ;start
BACK: JNB TF1,BACK
CLR TR1 ;stop
CPL P1.5 ;next half clock
CLR TF1 ;clear timer flag 1
SJMP AGAIN ;reload timer1

Example 2 (2/2)
Solution:
FFFFH – 7634H + 1 = 89CCH = 35276 clock
count
Half period = 35276 × 1.085 s = 38.274 ms
Whole period = 2 × 38.274 ms = 76.548 ms
Frequency = 1/ 76.548 ms = 13.064 Hz.
Note
Mode 1 is not auto reload then the program must reload the
TH1, TL1 register every timer overflow if we want to have a
continuous wave.

Find Timer Values
• Assume that XTAL = 11.0592 MHz .
• And we know desired delay
• how to find the values for the TH,TL ?
1. Divide the delay by 1.085 s and get n.
2. Perform 65536 –n
3. Convert the result of Step 2 to hex (yyxx )
4. Set TH = yy and TL = xx.

Example 3 (1/2)
• Assuming XTAL = 11.0592 MHz,
• write a program to generate a square wave of 50 Hz
frequency on pin P2.3.
Solution:
1. The period of the square wave = 1 / 50 Hz = 20 ms.
2. The high or low portion of the square wave = 10 ms.
3. 10 ms / 1.085 s = 9216
4. 65536 – 9216 = 56320 in decimal = DC00H in hex.
5. TL1 = 00H and TH1 = DCH.

Example 3 (2/2)
MOV TMOD,#10H ;timer 1, mode 1
AGAIN: MOV TL1,#00 ;Timer value = DC00H
MOV TH1,#0DCH
SETB TR1 ;start
BACK: JNB TF1,BACK
CLR TR1 ;stop
CPL P2.3
CLR TF1 ;clear timer flag 1
SJMP AGAIN ;reload timer since
;mode 1 is not
;auto-reload

Basics of Serial Communication
• Serial data communication uses two methods
– Synchronous method transfers a block of data at a time
– Asynchronous method transfers a single byte at a time
• There are special IC’s made by many manufacturers
for serial communications.
– UART (universal asynchronous Receiver transmitter)
– USART (universal synchronous-asynchronous Receiver-
transmitter)
39

Asynchronous – Start & Stop Bit
• Asynchronous serial data communication is
widely used for character-oriented transmissions
• The start bit is always a 0 (low) and the stop
bit(s) is 1 (high)
42

Asynchronous – Start & Stop Bit
43

Data Transfer Rate
• The rate of data transfer in serial data
communication is stated in bps (bits per second).
• Another widely used terminology for bps is baud
rate.
– It is modem terminology and is defined as the
number of signal changes per second
44

8051 Serial Port
• Synchronous and Asynchronous
• SCON Register is used to Control
• Data Transfer through TXd & RXd pins
• Some time - Clock through TXd Pin
• Four Modes of Operation:
Mode 0 :Synchronous Serial Communication
Mode 1 :8-Bit UART with Timer Data Rate
Mode 2 :9-Bit UART with Set Data Rate
Mode 3 :9-Bit UART with Timer Data Rate
45

Registers related to Serial Communication
1. SBUF Register
2. SCON Register
3. PCON Register
46

SBUF Register
• SBUF is an 8-bit register used solely for serial communication.
• For a byte data to be transferred via the TxD line, it must be
placed in the SBUF register.
• SBUF holds the byte of data when it is received by 8051 RxD
line.
47

SBUF Register
• Sample Program:
48

SCON Register
SM0 SM1 SM2 REN TB8 RB8 TI RI
Enable Multiprocessor
Communication Mode
Set to Enable
Serial Data
reception
9th Data Bit
Transmitted
in Mode 2,3
9th Data Bit
Received in Mode 2,3
Set when Stop bit Txed
Set when a Cha-
ractor received
49

8051 Serial Port – Mode 0
The Serial Port in Mode-0 has the following
features:
1. Serial data enters and exits through RXD
2. TXD outputs the clock
3. 8 bits are transmitted / received
4. The baud rate is fixed at (1/12) of the oscillator frequency
50

features:
1. Serial data enters through RXD
2. Serial data exits through TXD
3. On receive, the stop bit goes into RB8 in SCON
1. Start bit (0)
2. Data bits (8)
3. Stop Bit (1)
5. Baud rate is determined by the Timer 1 over flow rate.
51

features:
3. 9th data bit (TB8) can be assign value 0 or 1
4. On receive, the 9th data bit goes into RB8 in SCON
1.Start bit (0)
2.Data bits (9)
3.Stop Bit (1)
6. Baud rate is programmable
52

features:
3. 9th data bit (TB8) can be assign value 0 or 1
4. On receive, the 9th data bit goes into RB8 in SCON
1.Start bit (0)
2.Data bits (9)
3.Stop Bit (1)
6. Baud rate is determined by Timer 1 overflow rate.
53

Programs in 8051 serial port
TIMER 1 MODE 2 {AUTO RELOAD}

MOV TMOD, #20H ; Timer 1, mode 2
MOV TH1,#-06 TH1 is loaded to set the baud rate.
MOV SCON, #50H
SETB TR1 ; Run Timer 1
L2 : MOV SBUF, # ’A’
Loop: JNB TI, Loop ; Monitor RI
MOV A, SBUF
CLR TI
SJMP L2
Write a program for the 8051 to transfer letter ‘A’
serially at 4800 baud rate, continuously.

MOV TMOD, #20H ; Timer 1, mode 2
MOV TH1,#-06 TH1 is loaded to set the baud rate.
MOV SCON, #50H
SETB TR1 ; Run Timer 1
Loop: JNB RI, Loop ; Monitor RI
MOV A, SBUF
CLR RI
SJMP Loop
Program the 8051 to receive bytes of data serially, and
put them in P1. Set the baud rate at 4800.

INTERRUPTS
• An interrupt is an external or internal event
that interrupts the microcontroller to inform it
that a device needs its service
• A single microcontroller can serve several
devices by two ways:
1. Interrupt
2. Polling
61

Interrupt
–Upon receiving an interrupt signal, the
microcontroller interrupts whatever it is doing
and serves the device.
–The program which is associated with the
interrupt is called the interrupt service routine
(ISR) .
62

Steps in Executing an Interrupt
1. It finishes the instruction it is executing and saves the address of
the next instruction (PC) on the stack.
2. It also saves the current status of all the interrupts internally (i.e:
not on the stack).
3. It jumps to a fixed location in memory, called the interrupt vector
table, that holds the address of the ISR.
4. The microcontroller gets the address of the ISR from the
interrupt vector table and jumps to it.
5. It starts to execute the interrupt service subroutine until it
reaches the last instruction of the subroutine which is RETI
(return from interrupt).
6. Upon executing the RETI instruction, the microcontroller returns
to the place where it was interrupted.
63

Steps in executing an interrupt
• Finish current instruction and saves the PC on stack.
• Jumps to a fixed location in memory depend on type
of interrupt
• Starts to execute the interrupt service routine until
RETI (return from interrupt)
• Upon executing the RETI the microcontroller returns
to the place where it was interrupted. Get pop PC
from stack

Interrupt Sources
• Original 8051 has 6 sources of interrupts
– Reset (RST)
– Timer 0 overflow (TF0)
– Timer 1 overflow (TF1)
– External Interrupt 0 (INT0)
– External Interrupt 1 (INT1)
– Serial Port events (RI+TI)
{Reception/Transmission of Serial Character}

8051 Interrupt related Registers
• The various registers associated with the use of
interrupts are:
– TCON - Edge and Type bits for External Interrupts 0/1
– SCON - RI and TI interrupt flags for RS232 {SERIAL
COMMUNICATION}
– IE - interrupt Enable
– IP - Interrupts priority
67

Enabling and Disabling an Interrupt
• The register called IE (interrupt enable) that is
responsible for enabling (unmasking) and
disabling (masking) the interrupts.
68

Interrupt Enable (IE) Register
• EA : Global enable/disable.
• --- : Reserved for additional interrupt hardware.
• ES : Enable Serial port interrupt.
• ET1 : Enable Timer 1 control bit.
• EX1 : Enable External 1 interrupt.
• ET0 : Enable Timer 0 control bit.
• EX0 : Enable External 0 interrupt.
MOV IE,#08h
or
SETB ET1
--
69

Interrupt Priority (IP) Register
PS PT1 PX1 PT0 PX0
Reserved
Serial Port
Timer 1 Pin
INT 1 Pin Timer 0 Pin
INT 0 Pin
Priority bit=1 assigns high priority
Priority bit=0 assigns low priority
71

KEYBOARD INTERFACING
• Keyboards are organized in a matrix of rows
and columns
The CPU accesses both rows and columns
through ports .
• ƒ
Therefore, with two 8-bit ports, an 8 x 8
matrix of keys can be connected to a
microprocessor
When a key is pressed, a row and a
column make a contact
74

• Otherwise, there is no connection
between rows and columns
• ‰
A 4x4 matrix connected to two ports
The rows are connected to an
output port and the columns are
connected to an input port
75

Connection with keyboard matrix

Stepper Motor Interfacing
• Stepper motor is used in applications such as;
dot matrix printer, robotics etc
• It has a permanent magnet rotor called the shaft
which is surrounded by a stator. Commonly used
stepper motors have 4 stator windings
• Such motors are called as four-phase or unipolar
stepper motor.
81

Step angle:
• Step angle is defined as the minimum degree of
rotation with a single step.
• No of steps per revolution = 360° / step angle
• Steps per second = (rpm x steps per revolution) /
60
• Example: step angle = 2°
• No of steps per revolution = 180
85

A switch is connected to pin P2.7. Write an ALP to
monitor the status of the SW.
If SW = 0, motor moves clockwise and
If SW = 1, motor moves anticlockwise
SETB P2.7
MOV A, #66H
MOV P1,A
TURN: JNB P2.7, CW
RL A
ACALL DELAY
MOV P1,A
SJMP TURN
CW: RR A
ACALL DELAY
MOV P1,A
SJMP TURN 86
DELAY: MOV R1,#20
L2: MOV R2,#50
L1:DJNZ R2,L2
DJNZ R2,L1
RET

LCD Interfacing
{before discussed in Unit 3 LCD
interfacing using 8086}
87

A/D Interfacing
{before discussed in Unit 3 A/D
91

Interfacing ADC to 8051
ADC0804 is an 8 bit successive approximation analogue to digital
converter from National semiconductors. The features of ADC0804
are differential analogue voltage inputs, 0-5V input voltage range, no zero
adjustment, built in clock generator, reference voltage can be externally
adjusted to convert smaller analogue voltage span to 8 bit resolution etc.
92

D/A Interfacing
{before discussed in Unit 3 D/A
94

program to send data to the DAC to
generate a stair-step ramp
96

SENSOR
INTERFACING
take temperature sensor for example
97

99
Shunt voltage diod
potentiometer

EXTERNAL
MEMORY
INTERFACING
100

Access to External Memory
• Port 0 acts as a multiplexed address/data bus. Sending the
low byte of the program counter (PCL) as an address.
• Port 2 sends the program counter high byte (PCH) directly to
the external memory.
• The signal ALE operates as in the 8051 to allow an external
latch to store the PCL byte while the multiplexed bus is made
ready to receive the code byte from the external memory.
• Port 0 then switches function and becomes the data bus
receiving the byte from memory.
101

Presented by C.GOKUL,AP/EEE , Velalar College of Engg & Tech, Erode
102

UNIT-5.ppt

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