Ec2308 microprocessor and_microcontroller__lab1

SMK FOMRA INSTITUTE OF TECHNOLOGY
Old Mahabalipuram Road, (I.T.High Way),
Kelambakkam,Chennai, Tamilnadu 603103

DEPARTMENT ELECTRONICS AND
COMMUNICATION ENGINEERING

EC 2308 – MICROPROCESSOR AND
MICROCONTROLLER LAB

V SEMSTER
2010-2011

NAME:___________________________________

YEAR/SEM: _______________BRANCH:_______

REG. NO.:________________________________

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB DEPT OF ECE

EXPT NO: 8086 PROGRAMMING DATE:

ADDITION & SUBTRACTION

AIM:
To write an Assembly Language Program (ALP) for performing the
addition and subtraction operation of two byte numbers.

APPARATUS REQUIRED:

SL.N ITEM SPECIFICATION QUANTITY
O
1. Microprocessor kit 8086 kit 1
2. Power Supply +5 V dc 1

PROBLEM STATEMENT:
Write an ALP in 8086 to add and subtract two byte numbers stored in the
memory location 1000H to 1003H and store the result in the memory location
1004H to 1005H.Also provide an instruction in the above program to consider
the carry also and store the carry in the memory location 1006H.
ALGORITHM:
(i) 16-bit addition
h)Initialize the MSBs of sum to 0
i)Get the first number.
j)Add the second number to the first number.
k)If there is any carry, increment MSBs of sum by 1.
l)Store LSBs of sum.
m)Store MSBs of sum.

(ii) 16-bit subtraction
f)Initialize the MSBs of difference to 0
g)Get the first number
h)Subtract the second number from the first number.
i)If there is any borrow, increment MSBs of difference by 1.
j)Store LSBs of difference
k)Store MSBs of difference.

2


FLOWCHART

ADDITION SUBTRACTION

START START

SET UP COUNTER (CY) SET UP COUNTER (CARRY)

GET FIRST OPERAND GET FIRST
OPERAND TO A

GET SECOND OPERAND SUBTRACT
TO A SECOND OPERAND
FROM MEMORY

YES
A=A+B
IS THERE
ANY CY
YES
NO COUNTER =
IS THERE
ANY COUNTER + 1
COUNTER =
COUNTER + 1
STORE THE
DIFFERENCE
NO

STORE THE SUM
STORE THE CARRY

STORE THE CARRY STOP

STOP

3


ADDITION
PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg

ADD AX,BX Add the contents of both the regs AX & BX

JNC L1 Check for carry

INC CX If carry exists, increment the CX

L1 : MOV [1206],CX Store the carry

MOV [1204], AX Store the sum

HLT Stop the program

SUBTRACTION
PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg

SUB AX,BX Subtract the contents of BX from AX

JNC L1 Check for borrow

INC CX If borrow exists, increment the CX

L1 : MOV [1206],CX Store the borrow

MOV [1204], AX Store the difference


RESULT:.

4


ADDITION

MEMORY

DATA

SUBTRACTION

MEMORY

DATA

MANUAL CALCULATION

Thus addition & subtraction of two byte numbers are performed and the result is stored.

5



MULTIPLICATION & DIVISION

AIM:
To write an Assembly Language Program (ALP) for performing the
multiplication and division operation of 16-bit numbers .

APPARATUS REQUIRED:

O
1. Microprocessor kit 8086 1

PROBLEM STATEMENT:

Write an ALP in 8086 MP to multiply two 16-bit binary numbers and
store the result in the memory location. Write instructions for dividing the data
and store the result.

ALGORITHM:
(i)Multiplication of 16-bit numbers:
a)Get the multiplier.
b)Get the multiplicand
c)Initialize the product to 0.
d)Product = product + multiplicand
e)Decrement the multiplier by 1
f)If multiplicand is not equal to 0,repeat from step (d) otherwise store the
product.

(ii)Division of 16-bit numbers.
a)Get the dividend
b)Get the divisor
c)Initialize the quotient to 0.
d)Dividend = dividend – divisor
e)If the divisor is greater, store the quotient. Go to step g.
f)If dividend is greater, quotient = quotient + 1. Repeat from step (d)
g)Store the dividend value as remainder.

6


FLOWCHART

MULTIPLICATION DIVISION

Start
Start

Load Divisor &
Get Multiplier & Multiplicand Dividend
MULTIPLICAND

QUOTIENT = 0
REGISTER=00

DIVIDEND =
DIVIDEND-DIVISOR
REGISTER =
REGISTER +
MULTIPLICAND

QUOTIENT = QUOTIENT+1

Multiplier=MU
LTIPLIER – 1

IS
NO DIVIDEN
D<
DIVISOR
NO IS
?
MULTIPLIER
=0?
YES

YES STORE QUOTIENT
STORE REMAINDER
STORE THE RESULT = DIVIDEND NOW

STOP STOP

7


MULTIPLICATION
PROGRAM COMMENTS

MOV AX,[1200] Get the first data

MOV BX, [1202] Get the second data

MUL BX Multiply both

MOV [1206],AX Store the lower order product

MOV AX,DX Copy the higher order product to AX

MOV [1208],AX Store the higher order product


DIVISION
PROGRAM COMMENTS

MOV AX,[1200] Get the first data

MOV DX, [1202] Get the second data

MOV BX, [1204] Divide the dividend by divisor

DIV BX Store the lower order product

MOV [1206],AX Copy the higher order product to AX

MOV AX,DX Store the higher order product

MOV [1208],AX Stop the program

HLT Get the first data

8


RESULT:.
MULTIPLICATION

MEMORY

DATA

DIVISON

MEMORY

DATA

MANUAL CALCULATION

Thus multiplication & division of two byte numbers are performed and the result is stored.

9



ASCENDING & DESCENDING

AIM:
To write an Assembly Language Program (ALP) to sort a given array in
ascending and descending order.

APPARATUS REQUIRED:

O

PROBLEM STATEMENT:

An array of length 10 is given from the location. Sort it into descending
and ascending order and store the result.

ALGORITHM:
(i)Sorting in ascending order:
a.Load the array count in two registers C1 and C2.
b.Get the first two numbers.
c.Compare the numbers and exchange if necessary so that the two numbers are
in ascending order.
d.Decrement C2.
e.Get the third number from the array and repeat the process until C2 is 0.
f.Decrement C1 and repeat the process until C1 is 0.

(ii)Sorting in descending order:
a.Load the array count in two registers C1 and C2.
c.Compare the numbers and exchange if necessary so that the two numbers are
in descending order.
d.Decrement C2.
e.Get the third number from the array and repeat the process until C2 is 0.
f.Decrement C1 and repeat the process until C1 is 0.

10


FLOWCHART
ASCENDING ORDER DESCENDING ORDER

START
START

INITIALIZE POINTER INITIALIZE POINTER

COUNT = COUNT – 1 COUNT = COUNT – 1

YES
YES IS
IS
POINTER POINTER

NO
NO

TEMP = POINTER
TEMP = POINTER POINTER = POINTER + 1
POINTER = POINTER + 1 POINTER + 1 = TEMP
POINTER + 1 = TEMP

POINTER = POINTER +1
POINTER = POINTER +1 COUNT = COUNT + 1
COUNT = COUNT + 1

NO NO
IS
IS
COUNT
COUNT
YES
YES

NO
IS IS

FLAG FLAG
YES
YES

STOP
STOP

11


ASCENDING

PROGRAM COMMENTS
MOV SI,1200H Initialize memory location for array size
MOV CL,[SI] Number of comparisons in CL
L4 : MOV SI,1200H Initialize memory location for array size
MOV DL,[SI] Get the count in DL
INC SI Go to next memory location
MOV AL,[SI] Get the first data in AL
L3 : INC SI Go to next memory location
MOV BL,[SI] Get the second data in BL
CMP AL,BL Compare two data’s
JNB L1 If AL < BL go to L1
DEC SI Else, Decrement the memory location
MOV [SI],AL Store the smallest data
MOV AL,BL Get the next data AL
JMP L2 Jump to L2
L1 : DEC SI Decrement the memory location
MOV [SI],BL Store the greatest data in memory location
DEC DL Decrement the count
JNZ L3 Jump to L3, if the count is not reached zero
MOV [SI],AL Store data in memory location
DEC CL Decrement the count
HLT Stop

DESCENDING

PROGRAM COMMENTS
MOV SI,1200H Initialize memory location for array size
MOV CL,[SI] Number of comparisons in CL
L4 : MOV SI,1200H Initialize memory location for array size
MOV DL,[SI] Get the count in DL
MOV AL,[SI] Get the first data in AL
MOV BL,[SI] Get the second data in BL
CMP AL,BL Compare two data’s

12


JB L1 If AL > BL go to L1
DEC SI Else, Decrement the memory location
MOV [SI],AL Store the largest data
MOV AL,BL Get the next data AL
JMP L2 Jump to L2
L1 : DEC SI Decrement the memory location
MOV [SI],BL Store the smallest data in memory location
DEC DL Decrement the count
MOV [SI],AL Store data in memory location
DEC CL Decrement the count
HLT Stop

RESULT:.
ASCENDING

MEMORY

DATA

DESCENDING

MEMORY

DATA

Thus given array of numbers are sorted in ascending & descending order.

13



LARGEST& SMALLEST

AIM:
To write an Assembly Language Program (ALP) to find the largest and
smallest number in a given array.

APPARATUS REQUIRED:

O

PROBLEM STATEMENT:

An array of length 10 is given from the location. Find the largest and
smallest number and store the result.

ALGORITHM:
(i)Finding largest number:

a.Load the array count in a register C1.
c.Compare the numbers and exchange if the number is small.
d.Get the third number from the array and repeat the process until C1 is 0.

(ii)Finding smallest number:

e.Load the array count in a register C1.
f.Get the first two numbers.
g.Compare the numbers and exchange if the number is large.
h.Get the third number from the array and repeat the process until C1 is 0.

14


FLOWCHART
LARGEST NUMBER IN AN ARRAY SMALLEST NUMBER IN AN ARRAY

START START

INITIALIZE
INITIALIZE
COUNT
COUNT
POINTER MAX =
POINTER MIN = 0

PONITER = PONITER =
POINTER + 1 POINTER + 1

YES
IS MAX IS MIN ≥
≥ POINTE
YES
NO NO

MAX = POINTER MIN = POINTER

COUNT = COUNT-1 COUNT = COUNT-1

NO NO

IS COUNT = IS COUNT =
0 0
YES
YES
STORE MAXIMUM STORE MINIIMUM

STOP STOP

15


LARGEST

PROGRAM COMMENTS

MOV SI,1200H Initialize array size
MOV CL,[SI] Initialize the count
MOV AL,[SI] Move the first data in AL
DEC CL Reduce the count
L2 : INC SI Move the SI pointer to next data
CMP AL,[SI] Compare two data’s
JNB L1 If AL > [SI] then go to L1 ( no swap)
MOV AL,[SI] Else move the large number to AL
L1 : DEC CL Decrement the count
JNZ L2 If count is not zero go to L2
MOV DI,1300H Initialize DI with 1300H
MOV [DI],AL Else store the biggest number in 1300 location
HLT Stop

SMALLEST

PROGRAM COMMENTS

MOV SI,1200H Initialize array size
MOV CL,[SI] Initialize the count
MOV AL,[SI] Move the first data in AL
DEC CL Reduce the count
L2 : INC SI Move the SI pointer to next data
CMP AL,[SI] Compare two data’s
JB L1 If AL < [SI] then go to L1 ( no swap)
MOV AL,[SI] Else move the large number to AL
L1 : DEC CL Decrement the count
JNZ L2 If count is not zero go to L2
MOV DI,1300H Initialize DI with 1300H
MOV [DI],AL Else store the biggest number in 1300 location
HLT Stop

16


RESULT:.
LARGEST

MEMORY

DATA

SMALLEST

MEMORY

DATA

Thus largest and smallest number is found in a given array.

17


COPYING A STRING
AIM:
To move a string of length FF from source to destination.

ALGORITHM:
a. Initialize the data segment .(DS)
b. Initialize the extra data segment .(ES)
c. Initialize the start of string in the DS. (SI)
d. Initialize the start of string in the ES. (DI)
e. Move the length of the string(FF) in CX register.
f. Move the byte from DS TO ES, till CX=0.

START

`
Initialize DS,ES,SI,DI

CX=length of string,
DF=0.

Move a byte from source string (DS)
to destination string (ES)

Decrement CX

`
NO Check
for ZF=1

STOP

18


COPYING A STRING
PROGRAM COMMENTS

MOV SI,1200H Initialize destination address

MOV DI,1300H Initialize starting address

MOV CX,0006H Initialize array size

CLD Clear direction flag

REP MOVSB Copy the contents of source into destination until
count reaches zero
HLT Stop

RESULT:

INPUT

MEMORY

DATA

OUTPUT

MEMORY

DATA

Thus a string of a particular length is moved from source segment to destination segment.

19



SEARCHING A STRING
AIM:
To scan for a given byte in the string and find the relative address of the byte from the
starting location of the string.

ALGORITHM:
a. Initialize the extra segment .(ES)
b. Initialize the start of string in the ES. (DI)
c. Move the number of elements in the string in CX register.
d. Move the byte to be searched in the AL register.
e. Scan for the byte in ES. If the byte is found ZF=0, move the address pointed by ES:DI
to BX.

START

Initialize DS,ES ,SI,DI

CX=length of the string,
DF=0.

Scan for a particular character
specified in AL Register.

NO
Check for
ZF=1

Move DI to BX

STOP

20


SEARCHING FOR A CHARACTER IN THE STRING
PROGRAM COMMENTS

MOV DI,1300H Initialize destination address

MOV SI, 1400H Initialize starting address

MOV CX, 0006H Initialize array size


MOV AL, 08H Store the string to be searched

REPNE SCASB Scan until the string is found

DEC DI Decrement the destination address

MOV BL,[DI] Store the contents into BL reg

MOV [SI],BL Store content of BL in source address

HLT Stop

RESULT:

INPUT

MEMORY

DATA

OUTPUT

MEMORY LOCATION

DATA

Thus a given byte or word in a string of a particular length in the extra segment(destination)
is found .

21



FIND AND REPLACE
AIM:
To find a character in the string and replace it with another character.

ALGORITHM:
a. Initialize the extra segment .(E S)
b. Initialize the start of string in the ES. (DI)
c. Move the number of elements in the string in CX register.
d. Move the byte to be searched in the AL register.
e. Store the ASCII code of the character that has to replace the scanned byte in BL
register.
f. Scan for the byte in ES. If the byte is not found, ZF≠1 and repeat scanning.
g. If the byte is found, ZF=1.Move the content of BL register to ES:DI.

START

Initialize DS, ES, SI, DI

CX=length of the string in ES,
DF=0.
DF=0.

Scan for a particular character
specified in AL Register.

NO

Check for
ZF=1

YES

Move the content of BL
to ES:DI
¿

STOP

22


FIND AND REPLACE A CHARACTER IN THE STRING
PROGRAM COMMENTS

MOV DI,1300H Initialize destination address

MOV SI,1400H Initialize starting address



MOV AL, 08H Store the string to be searched

MOV BH,30H Store the string to be replaced

REPNE SCASB Scan until the string is found

DEC DI Decrement the destination address

MOV BL,[DI] Store the contents into BL reg

MOV [SI],BL Store content of BL in source address

MOV [DI],BH Replace the string

HLT Stop

RESULT:

INPUT

MEMORY

DATA

OUTPUT

MEMORY

DATA

Thus a given byte or word in a string of a particular length in the extra segment(destination)
is found and is replaced with another character.

23


EXPT NO: 8086 INTERFACING DATE:
INTERFACING ANALOG TO DIGITAL CONVERTER
AIM:
To write an assembly language program to convert an analog signal into a digital signal
using an ADC interfacing.
APPARATUS REQUIRED:
SL.NO ITEM SPECIFICATION QUANTITY
2. Power Supply +5 V dc,+12 V dc 1
3. ADC Interface board - 1
PROBLEM STATEMENT:
The program is executed for various values of analog voltage which are set with the
help of a potentiometer. The LED display is verified with the digital value that is stored in a
memory location.
THEORY:
An ADC usually has two additional control lines: the SOC input to tell the ADC when
to start the conversion and the EOC output to announce when the conversion is complete. The
following program initiates the conversion process, checks the EOC pin of ADC 0809 as to
whether the conversion is over and then inputs the data to the processor. It also instructs the
processor to store the converted digital data at RAM location.
ALGORITHM:
(i) Select the channel and latch the address.
(ii) Send the start conversion pulse.
(iii) Read EOC signal.
(iv) If EOC = 1 continue else go to step (iii)
(v) Read the digital output.
(vi) Store it in a memory location.
FLOW CHART:
STAR

SELECT THE CHANNEL AND LATCH
ADDRESS

SEND THE START CONVERSION PULSE

NO
IS EOC =
1?
YES

READ THE DIGITALOUTPUT

STORE THE DIGITAL VALUE IN THE
MEMORY LOCATION SPECIFIED

STOP

24


PROGRAM TABLE
PROGRAM COMMENTS

MOV AL,00 Load accumulator with value for ALE high
OUT 0C8H,AL Send through output port
MOV AL,08 Load accumulator with value for ALE low
OUT 0C8H,AL Send through output port
MOV AL,01 Store the value to make SOC high in the accumulator
OUT 0D0H,AL Send through output port
MOV AL,00
MOV AL,00 Introduce delay
MOV AL,00
MOV AL,00 Store the value to make SOC low the accumulator
OUT 0D0H,AL Send through output port
L1 : IN AL, 0D8H
AND AL,01 Read the EOC signal from port & check for end of
conversion
CMP AL,01
JNZ L1 If the conversion is not yet completed, read EOC signal
from port again
IN AL,0C0H Read data from port
MOV BX,1100 Initialize the memory location to store data
MOV [BX],AL Store the data
HLT Stop
RESULT:
ANALOG DIGITAL DATA ON LED HEX CODE IN MEMORY
VOLTAGE DISPLAY LOCATION

Thus the ADC was interfaced with 8086 and the given analog inputs were converted
into its digital equivalent.
EXPT NO: 8086 INTERFACING DATE:
INTERFACING DIGITAL – TO – ANALOG CONVERTER
AIM :

25


1. To write an assembly language program for digital to analog conversion
2. To convert digital inputs into analog outputs & To generate different waveforms
APPARATUS REQUIRED:
1. Microprocessor kit 8086 Vi Microsystems 1
2. Power Supply +5 V, dc,+12 V dc 1
3. DAC Interface board - 1
PROBLEM STATEMENT:
The program is executed for various digital values and equivalent analog voltages are
measured and also the waveforms are measured at the output ports using CRO.
THEORY:
Since DAC 0800 is an 8 bit DAC and the output voltage variation is between –5v
and +5v. The output voltage varies in steps of 10/256 = 0.04 (approximately). The digital
data input and the corresponding output voltages are presented in the table. The basic idea
behind the generation of waveforms is the continuous generation of analog output of DAC.
With 00 (Hex) as input to DAC2 the analog output is –5v. Similarly with FF H as input, the
output is +5v. Outputting digital data 00 and FF at regular intervals, to DAC2, results in a
square wave of amplitude 5v.Output digital data from 00 to FF in constant steps of 01 to
DAC2. Repeat this sequence again and again. As a result a saw-tooth wave will be generated
at DAC2 output. Output digital data from 00 to FF in constant steps of 01 to DAC2. Output
digital data from FF to 00 in constant steps of 01 to DAC2. Repeat this sequence again and
again. As a result a triangular wave will be generated at DAC2 output.
ALGORITHM:
Measurement of analog voltage:
(i) Send the digital value of DAC.
(ii) Read the corresponding analog value of its output.
Waveform generation:
Square Waveform:
(i) Send low value (00) to the DAC.
(ii) Introduce suitable delay.
(iii) Send high value to DAC.
(iv) Introduce delay.
(v) Repeat the above procedure.
Saw-tooth waveform:
(i) Load low value (00) to accumulator.
(ii) Send this value to DAC.
(iii) Increment the accumulator.
(iv) Repeat step (ii) and (iii) until accumulator value reaches FF.
(v) Repeat the above procedure from step 1.
Triangular waveform:
(i) Load the low value (00) in accumulator.
(ii) Send this accumulator content to DAC.
(iii) Increment the accumulator.
(iv) Repeat step 2 and 3 until the accumulator reaches FF, decrement the
accumulator and send the accumulator contents to DAC.
(v) Decrementing and sending the accumulator contents to DAC.
(vi) The above procedure is repeated from step (i)
FLOWCHART:
MEASUREMENT OF ANALOG VOLTAGE SQUARE WAVE FORM
START
START

26


INTIALISE THE ACCUMULATOR
SEND ACC CONTENT TO DAC
SEND THE
DIGITALVALUE TO
ACCUMULATOR
DELAY

TRANSFER THE
ACCUMULATOR CONTENTS LOAD THE ACC WITH MAX
VALUE SEND ACC CONTENT
TO DAC

READ THE CORRESPONDING DELAY
ANALOG VALUE

STOP TRIANGULAR WAVEFORM
STAR
SAWTOOTH WAVEFORM
STAR INITIALIZE
ACCUMULATOR

INITIALIZE SEND ACCUMULATOR
ACCUMULATOR CONTENT TO DAC

SEND ACCUMULATOR INCREMENT
ACCUMULATOR
CONTENT TO DAC
YES
INCREMENT
IS ACC ≤
ACCUMULATOR
FF
CONTENT NO
NO YES
DECREMENT
IS ACCUMULATOR CONTENT
ACC ≤

SEND
ACCUMULATOR

MEASUREMENT OF ANALOG VOLTAGE: IS ACC ≤
YES 00 NO
PROGRAM COMMENTS

27


MOV AL,7FH Load digital value 00 in accumulator
OUT C0,AL Send through output port
HLT Stop

DIGITAL DATA ANALOG VOLTAGE

PROGRAM TABLE: Square Wave
PROGRAM COMMENTS

L2 : MOV AL,00H Load 00 in accumulator
CALL L1 Give a delay
MOV AL,FFH Load FF in accumulator
CALL L1 Give a delay
JMP L2 Go to starting location
L1 : MOV CX,05FFH Load count value in CX register
L3 : LOOP L3 Decrement until it reaches zero
RET Return to main program

PROGRAM TABLE: Saw tooth Wave
PROGRAM COMMENTS

L1 : OUT C0,AL Send through output port
INC AL Increment contents of accumulator
JNZ L1 Send through output port until it reaches FF

28


PROGRAM TABLE: Triangular Wave
PROGRAM COMMENTS

INC AL Increment contents of accumulator
JNZ L1 Send through output port until it reaches FF
MOV AL,0FFH Load FF in accumulator
DEC AL Decrement contents of accumulator
JNZ L2 Send through output port until it reaches 00

RESULT:WAVEFORM GENERATION:

WAVEFORMS AMPLITUDE TIMEPERIOD

Square Waveform
Saw-tooth waveform

Triangular waveform

MODEL GRAPH:

Square Waveform Saw-tooth waveform

Triangular waveform

Thus the DAC was interfaced with 8085 and different waveforms have been generated.

29


EXP.NO: STEPPER MOTOR INTERFACING DATE:

AIM:
To write an assembly language program in 8086 to rotate the motor at different speeds.

APPARATUS REQUIRED:

2. Power Supply +5 V, dc,+12 V dc 1
3. Stepper Motor Interface board - 1
4. Stepper Motor - 1

PROBLEM STATEMENT:

Write a code for achieving a specific angle of rotation in a given time and particular
number of rotations in a specific time.

THEORY:

A motor in which the rotor is able to assume only discrete stationary angular
position is a stepper motor. The rotary motion occurs in a stepwise manner from one
equilibrium position to the next.Two-phase scheme: Any two adjacent stator windings are
energized. There are two magnetic fields active in quadrature and none of the rotor pole faces
can be in direct alignment with the stator poles. A partial but symmetric alignment of the
rotor poles is of course possible.

ALGORITHM:

For running stepper motor clockwise and anticlockwise directions
(i) Get the first data from the lookup table.
(ii) Initialize the counter and move data into accumulator.
(iii) Drive the stepper motor circuitry and introduce delay
(iv) Decrement the counter is not zero repeat from step(iii)
(v) Repeat the above procedure both for backward and forward directions.

SWITCHING SEQUENCE OF STEPPER MOTOR:

MEMORY A1 A2 B1 B2 HEX
LOCATION CODE
4500 1 0 0 0 09 H
4501 0 1 0 1 05 H
4502 0 1 1 0 06 H
4503 1 0 1 0 0A H

30


FLOWCHART:
STAR
T
INTIALIZE COUNTER FOR LOOK UP TABLE

GET THE FIRST DATA FROM THE ACCUMULATOR

MOVE DATA INTO THE ACCUMULATOR

DRIVE THE MOTOR
CIRCUITARY

DELAY

DECREMENT COUNTER

YES
IS B = 0 ?
NO

GET THE DATA FROM LOOK UP
TABLE

PROGRAM TABLE
PROGRAM COMMENTS
START : MOV DI, 1200H Initialize memory location to store the array of
number
LOOP 1 : MOV AL,[DI] Copy the first data in AL
OUT 0C0,AL Send it through port address
MOV DX, 1010H
L1 : DEC DX Introduce delay
JNZ L1
INC DI Go to next memory location
LOOP LOOP1 Loop until all the data’s have been sent
JMP START Go to start location for continuous rotation
1200 : 09,05,06,0A Array of data’s

RESULT: Thus the assembly language program for rotating stepper motor in both clockwise
and anticlockwise directions is written and verified.

31


EXP.NO: INTERFACING PRGRAMMABLE KEYBOARD AND DISPLAY
CONTROLLER- 8279
DATE:

AIM :
To display the rolling message “ HELP US “ in the display.

APPARATUS REQUIRED:
8086 Microprocessor kit, Power supply, Interfacing board.

ALGORITHM :
Display of rolling message “ HELP US “
1. Initialize the counter
2. Set 8279 for 8 digit character display, right entry
3. Set 8279 for clearing the display
4. Write the command to display
5. Load the character into accumulator and display it
6. Introduce the delay
7. Repeat from step 1.

1. Display Mode Setup: Control word-10 H

0 0 0 1 0 0 0 0
0 0 0 D D K K K

DD
00- 8Bit character display left entry
01- 16Bit character display left entry
10- 8Bit character display right entry
11- 16Bit character display right entry
KKK- Key Board Mode
000-2Key lockout.
2.Clear Display: Control word-DC H

1 1 0 1 1 1 0 0
1 1 0 CD CD CD CF CA

11 A0-3; B0-3 =FF

1-Enables Clear display
0-Contents of RAM will be displayed
1-FIFO Status is cleared

1-Clear all bits
(Combined effect of CD)

32


3. Write Display: Control word-90H

1 0 0 1 0 0 0 0
1 0 0
AI A A A A

Selects one of the 16 rows of display.

Auto increment = 1, the row address selected will be incremented after each of read and
write operation of the display RAM.

FLOWCHART:

SET UP
POINTER

INITIALIZE THE COUNTER

SET 8279 FOR 8-DIGIT CHARACTER
DISPLAY

SET 8279 FOR CLEARING THE
DISPLAY

WRITE THE COMMAND TO DISPLAY

LOAD THE CHARACTER INTO
ACCUMULATOR AND DISPLAY

DELAY

33


PROGRAM TABLE

PROGRAM COMMENTS
START : MOV SI,1200H Initialize array
MOV CX,000FH Initialize array size
MOV AL,10 Store the control word for display mode
MOV AL,CC Store the control word to clear display
MOV AL,90 Store the control word to write display
L1 : MOV AL,[SI] Get the first data
CALL DELAY Give delay
INC SI Go & get next data
LOOP L1 Loop until all the data’s have been taken
JMP START Go to starting location
DELAY : MOV DX,0A0FFH Store 16bit count value
LOOP1 : DEC DX Decrement count value
JNZ LOOP1 Loop until count values becomes zero
RET Return to main program

LOOK-UP TABLE:

1200 98 68 7C C8
1204 FF 1C 29 FF

RESULT:

MEMORY 7-SEGMENT LED FORMAT HEX DATA
LOCATION d c b a dp e g f
1200H 1 0 0 1 1 0 0 0 98
1201H 0 1 1 0 1 0 0 0 68
1202H 0 1 1 1 1 1 0 0 7C
1203H 1 1 0 0 1 0 0 0 C8
1204H 1 1 1 1 1 1 1 1 FF
1205H 0 0 0 0 1 1 0 0 1C
1206H 0 0 1 0 1 0 0 1 29
1207H 1 1 1 1 1 1 1 1 FF

Thus the rolling message “HELP US” is displayed using 8279 interface kit.

34


EXP. NO: INTERFACING PROGRAMMABLE TIMER-8253 DATE:

AIM :
To study different modes of operation of programmable timer 8253

APPARATUS REQUIRED:

1. Microprocessor kit 8086 Vi Microsystems 1
2. Power Supply +5V dc 1
3. 8253 interfacing kit - 1
4. CRO - 1

THEORY:

The main features of the timer are,
i. Three independent 16-bit counters
ii. Input clock from DC to 2 MHz
iii. Programmable counter modes
iv. Count binary or BCD
The control signals with which the 8253 interfaces with the CPU are CS, RD,
WR, A1, A2.The basic operations performed by 8253 are determined by these
control signals. It has six different modes of operation, viz, mode 0 to mode 5.

MODE 2 – RATE GENERATOR
It is a simple divide - by – N counter. The output will be low for one input clock
period. The period from one output pulse to the next equals the number of input counts in the
count register. If the count register is reloaded between output pulses, the present period will
not be affected, but the subsequent period will reflect the new value.

MODE 3 – SQUARE WAVE GENERATOR
It is similar to mode 2, except that the output will remain high until one half for even
number count, If the count is odd, the output will be high for (count+1)/2 counts and low for
(count-1)/2 counts

ALGORITHM:

Mode 2-
1. Initialize channel 0 in mode 2
2. Initialize the LSB of the count.
3. Initialize the MSB of the count.
4. Trigger the count
5. Read the corresponding output in CRO.

35


Mode 3-
1. Initialize channel 0 in mode 3
2. Initialize the LSB of the count.
3. Initialize the MSB of the count.
4. Trigger the count
5. Read the corresponding output in CRO.

PORT ADDRESS :
1. CONTROL REGISTER –
2. COUNTER OF CHANNEL 0 -
5. O/P PORT OF CHANNEL 0 -

CONTROL WORD FORMAT:

D7 D6 D5 D4 D3 D2 D1 D0
SC1 SC0 RL1 RL0 M2 M1 M0 BCD

0 0 1 1 0 1 0 0
Mode 2 = 34 H
0 0 1 1 0 1 1 0
Mode 3 = 36 H

SC1 SC0 CHANNEL SELECT RL1 RL0 READ/LOAD

0 0 CHANNEL 0 0 0 LATCH
0 1 CHANNEL 1 0 1 LSB
1 0 CHANNEL 2 1 0 MSB
1 1 ----- 1 1 LSB FIRST, MSB NEXT
BCD --0 –BINARY COUNTER 1 --BCD COUNTER

M2 M1 M0 MODE
1 CLK 0
0 0 0 MODE 0
0 0 2 CLK 1
1 MODE 1
0 1 0 MODE 2
0 1 3 CLK 2
1 MODE 3
1 0 0 MODE 4
1 0 4 OUT 0
1 MODE 5

5 OUT 1 PORT PIN ARRANGEMENT DEBOUNCE CIRCUIT
CONNECTION
6 OUT 2
* * *
7 GATE 0

8 GATE 1

9 GATE 2 36
10 GND


MODE 2 – RATE GENERATOR:

PROGRAM COMMENTS
MOV AL, 34H Store the control word in accumulator
OUT 0BH Send through output port
MOV AL, 0AH Copy lower order count value in accumulator
OUT 08H Send through output port
MOV AL, 00H Copy higher order count value in accumulator
HLT Stop

MODE 3 – SQUARE WAVE GENERATOR:

PROGRAM COMMENTS
MOV AL, 36H Store the control word in accumulator
OUT 0BH Send through output port
MOV AL, 0AH Copy lower order count value in accumulator
MOV AL, 00H Copy higher order count value in accumulator
HLT Stop

37


MODEL GRAPH:

RATE GENERATOR SQUARE WAVE GENERATOR

FLOW CHART

START

INITIALIZE ACCUMULATOR
WITH MODE SET WORD

INITIALIZE LSB OF COUNT

INITIALIZE MSB OF COUNT

TRIGGER THE COUNT

STOP

RESULT: Thus an ALP for rate generator and square wave generator are written and
executed.

38


EXP. NO: INTERFACING USART 8251 DATE:

AIM:
To study interfacing technique of 8251 (USART) with microprocessor 8086 and
write an 8086 ALP to transmit and receive data between two serial ports with RS232 cable.

APPARATUS REQUIRED:
8086 kit (2 Nos), RS232 cable.
THEORY:
The 8251 is used as a peripheral device for serial communication and is
programmed by the CPU to operate using virtually any serial data transmission technique.
The USART accepts data characters from the CPU in parallel format and then converts them
into a continuous serial data stream for transmission. Simultaneously, it can receive serial
data streams and convert them into parallel data characters for the CPU. The CPU can read
the status of the USART at any time. These include data transmission errors and control
signals. The control signals define the complete functional definition of the 8251. Control
words should be written into the control register of 8251.These control words are split into
two formats: 1) Mode instruction word & 2) Command instruction word. Status word format
is used to examine the error during functional operation.

1...transmit enable
1...data terminal ready
1... receive enable
1... send break character
1.... reset error flags (pe,oe,fe)
1..... request to send (rts)
1...... internal reset
1....... enter hunt mode (enable search for sync characters)

1 ransmitter ready
1. receiver ready
1.. transmitter empty
1... parity error (pe)
1.... overrun error (oe)
1..... framing error (fe), async only
1...... sync detect, sync only
1....... data set ready (dsr)

39


ALGORITHM:
1.Initialize 8253 and 8251 to check the transmission and reception of a character
2.Initialize8253 to give an output of 150Khz at channel 0 which will give a 9600 baud rate of
8251.
3.The command word and mode word is written to the 8251 to set up for subsequent
operations
4.The status word is read from the 8251 on completion of a serial I/O operation, or when the
host CPU is checking the status of the device before starting the next I/O operation
FLOW CHART:
STAR

Check TX/RX Ready

No

Is it High
Yes

Write Data into data register

STOP

40


PROGRAM: TRANSMITTER END
PROGRAM COMMENTS
MOV AL,36 Initialize 8253 in mode 3 square wave generator
OUT CE,AL Send through port address
MOV AL,10 Initialize AL with lower value of count (clock frequency 150KHz)
OUT C8,AL Send through port address
MOV AL,00 Initialize AL with higher value of count
MOV AL,4E Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit)
MOV AL,37 Set command instruction(enables transmit enable & receive enable bits)
L1:IN AL,C2 Read status word
AND AL,04 Check whether transmitter ready
JZ L1 If not wait until transmitter becomes ready
MOV AL,41 Set the data as 41
INT 2 Restart the system

RECEIVER END
PROGRAM COMMENTS
MOV AL,36 Initialize 8253 in mode 3 square wave generator
OUT CE,AL Send through port address
MOV AL,10 Initialize AL with lower value of count (clock frequency 150KHz)
MOV AL,00 Initialize AL with higher value of count
MOV AL,4E Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit)
MOV AL,37 Set command instruction(enables transmit enable & receive enable bits)
L1:IN AL,C2 Read status word
AND AL,02 Check whether receiver ready
JZ L1 If not wait until receiver becomes ready
IN AL,C0 If it is ready, get the data
MOV BX,1500 Initialize BX register with memory location to store the data
MOV [BX],AL Store the data in the memory location
INT 2 Restart the system

RESULT:

Thus ALP for serial data communication using USART 8251 is written and the equivalent
ASCII 41 for character ‘A’ is been transmitted & received.

41


EXP. NO: INTERFACING PPI 8255 DATE:

AIM:
To write ALP by interfacing 8255 with 8086 in mode 0, mode 1 and mode 2

APPARATUS REQUIRED:
8086 kit, 8255 interface kit.

ALGORITHM:
Mode 0
1. Initialize accumulator to hold control word
2. store control word in control word register
3. Read data port A.
4. Store data from port A in memory
5. Place contents in port B
Mode 1 & Mode 2
1. Initialize accumulator to hold control word (for port A)
2. Store control word in control word register
3. Initialize accumulator to hold control word (for port B)
4. Place contents in control word register.
5. Disable all maskable interrupts, enable RST 5.5
6. send interrupt mask for RST 6.5 & 7.5
7. Enable interrupt flag
8. Read data from port A, place contents in port B

FLOWCHART
Mode 0 Mode 1 & 2
STAR
STAR
Store control word in control register

Store control word in
control register
Input to be read from port A

Input to be read from port A
Disable all interrupts except RST
6.5
Store into accumulator
Store output to port B
Output written on port B

STOP
STOP

42


MODE 0
PROGRAM COMMENTS
MOV AL,90H Set the control word
OUT C6,AL Send it to control port
IN AL,C0 Get the contents of port A in AL
OUT C2,AL Send the contents of port B to port address
HLT Stop

MODE 1
PROGRAM COMMENTS
MOV AL,0B0H Set the control word for mode 1
MOV AL,09H Control for BSR mode
MOV AL,13H Interrupt generation
OUT 30,AL
MOV AL,0AH Through 8259
OUT 32,AL
MOV AL,0FH Using IR2 interrupt(lower order count)
OUT 32,AL
MOV AL,00H Higher order count
OUT 32,AL
STI Set trap flag
HLT Stop
ISR: Subroutine
IN AL,C0 Read from Port A
OUT C2,AL Send it to Port B
HLT Stop
MODE 2
PROGRAM COMMENTS
MOV AL,0C0H Set the control word for mode 2
MOV AL,09H Control for BSR mode
MOV AL,13H Interrupt generation
OUT 30,AL
MOV AL,0AH Through 8259
OUT 32,AL

43


MOV AL,0FH Using IR2 interrupt(lower order count)
OUT 32,AL
MOV AL,00H Higher order count
OUT 32,AL
STI Set trap flag
HLT Stop
ISR: Subroutine
IN AL,C0 Read from Port A
OUT C2,AL Send it to Port B
HLT Stop
BSR mode
Bit set/reset, applicable to PC only. One bit is S/R at a time. Control word:
D7 D6 D5 D4 D3 D2 D1 D0
0 (0=BSR) X X X B2 B1 B0 S/R (1=S,0=R)
Bit select: (Taking Don't care's as 0)
B2 B1 B0 PC bit Control word (Set) Control word (reset)
0 0 0 0 0000 0001 = 01h 0000 0000 = 00h
0 0 1 1 0000 0011 = 03h 0000 0010 = 02h
0 1 0 2 0000 0101 = 05h 0000 0100 = 04h
0 1 1 3 0000 0111 = 07h 0000 0110 = 06h
1 0 0 4 0000 1001 = 09h 0000 1000 = 08h
1 0 1 5 0000 1011 = 0Bh 0000 1010 = 0Ah
1 1 0 6 0000 1101 = 0Dh 0000 1100 = 0Ch
1 1 1 7 0000 1111 = 0Fh 0000 1110 = 0Eh

I/O mode
D7 D6 D5 D4 D3 D2 D1 D0
1 (1=I/O) GA mode select PA PCU GB mode select PB PCL

• D6, D5: GA mode select:
o 00 = mode0
o 01 = mode1
o 1X = mode2
• D4(PA), D3(PCU): 1=input 0=output
• D2: GB mode select: 0=mode0, 1=mode1
• D1(PB), D0(PCL): 1=input 0=output

Result:
Mode 0 Mode 1 Mode 2
Input Output Input Output Input Output

The programs for interfacing 8255 with 8085 are executed & the output is obtained for modes
0,1 & 2

44



8 BIT ADDITION

AIM:
To write a program to add two 8-bit numbers using 8051 microcontroller.

ALGORITHM:

1. Clear Program Status Word.
2. Select Register bank by giving proper values to RS1 & RS0 of PSW.
3. Load accumulator A with any desired 8-bit data.
4. Load the register R 0 with the second 8- bit data.
5. Add these two 8-bit numbers.
6. Store the result.
7. Stop the program.

FLOW CHART:

START

Clear PSW

Select Register
Bank

Load A and R 0
with 8- bit datas

Add A & R 0

Store the sum

STOP

45


8 Bit Addition (Immediate Addressing)
ADDRESS LABEL MNEMONIC OPERAND HEX COMMENTS
CODE
4100 CLR C C3 Clear CY Flag

4101 MOV A,# data1 74,data1 Get the data1 in
Accumulator
4103 ADDC A, # data 2 24,data2 Add the data1 with
data2
4105 MOV DPTR, # 90,45,00 Initialize the memory
4500H location
4108 MOVX @ DPTR, A F0 Store the result in
memory location
4109 L1 SJMP L1 80,FE Stop the program

RESULT:

OUTPUT
MEMORY LOCATION DATA

4500

Thus the 8051 ALP for addition of two 8 bit numbers is executed.

46


EXPT NO: 8 BIT SUBTRACTION DATE:

AIM:
To perform subtraction of two 8 bit data and store the result in memory.

ALGORITHM:
a. Clear the carry flag.
b. Initialize the register for borrow.
c. Get the first operand into the accumulator.
d. Subtract the second operand from the accumulator.
e. If a borrow results increment the carry register.
f. Store the result in memory.

FLOWCHART:
START

CLEAR CARRY
FLAG

GET I’ST
OPERAND IN
ACCR

SUBTRACT THE
2’ND OPERAND
FROM ACCR

N
IS CF=1

Y
INCREMENT
THE BORROW
REGISTER

STORE
RESULT IN
MEMORY

STOP

47


8 Bit Subtraction (Immediate Addressing)
CODE
4100 CLR C C3 Clear CY flag

4101 MOV A, # data1 74, data1 Store data1 in
accumulator
4103 SUBB A, # data2 94,data2 Subtract data2 from
data1
4105 MOV DPTR, # 4500 90,45,00 Initialize memory
location
4108 MOVX @ DPTR, A F0 Store the difference
in memory location
4109 L1 SJMP L1 80,FE Stop

RESULT:

OUTPUT
MEMORY LOCATION DATA

4500

Thus the 8051 ALP for subtraction of two 8 bit numbers is executed.

48



8 BIT MULTIPLICATION

AIM:
To perform multiplication of two 8 bit data and store the result in memory.

ALGORITHM:
a. Get the multiplier in the accumulator.
b. Get the multiplicand in the B register.
c. Multiply A with B.
d. Store the product in memory.

FLOWCHART:

START

GET
MULTIPLIER
IN ACCR

GET
MULTIPLICAND
IN B REG

MULTIPLY A
WITH B

STORE
RESULT IN
MEMORY

STOP

49


8 Bit Multiplication
CODE
4100 MOV A ,#data1 74, data1 Store data1 in
accumulator
4102 MOV B, #data2 75,data2 Store data2 in B reg

4104 MUL A,B F5,F0 Multiply both

4106 MOV DPTR, # 90,45,00 Initialize memory
4500H location
4109 MOVX @ DPTR, A F0 Store lower order
result
401A INC DPTR A3 Go to next memory
location
410B MOV A,B E5,F0
Store higher order
410D MOV @ DPTR, A F0 result

410E STOP SJMP STOP 80,FE Stop

RESULT:

INPUT OUTPUT
MEMORY LOCATION DATA MEMORY LOCATION DATA

4500 4502

4501 4503

Thus the 8051 ALP for multiplication of two 8 bit numbers is executed.

50



8 BIT DIVISION

AIM:
To perform division of two 8 bit data and store the result in memory.

ALGORITHM:
1. Get the Dividend in the accumulator.
2. Get the Divisor in the B register.
3. Divide A by B.
4. Store the Quotient and Remainder in memory.

FLOWCHART:

START

GET DIVIDEND
IN ACCR

GET DIVISOR IN
B REG

DIVIDE A BY B

STORE
QUOTIENT &
REMAINDER
IN MEMORY

STOP

51


8 Bit Division
CODE
4100 MOV A, # data1 74,data1 Store data1 in
accumulator
4102 MOV B, # data2 75,data2 Store data2 in B reg

4104 DIV A,B 84 Divide

4015 MOV DPTR, # 4500H 90,45,00 Initialize memory
location
4018 MOVX @ DPTR, A F0 Store remainder

4109 INC DPTR A3 Go to next memory
location
410A MOV A,B E5,F0
Store quotient
410C MOV @ DPTR, A F0

410D STOP SJMP STOP 80,FE Stop

RESULT:

INPUT OUTPUT
MEMORY LOCATION DATA MEMORY LOCATION DATA
4500 (dividend) 4502 (remainder)

4501 (divisor) 4503 (quotient)

Thus the 8051 ALP for division of two 8 bit numbers is executed.

52


EXP. NO: LOGICAL AND BIT MANIPULATION DATE:

AIM:

To write an ALP to perform logical and bit manipulation operations using 8051
microcontroller.

APPARATUS REQUIRED:
8051 microcontroller kit

ALGORITHM:

1. Initialize content of accumulator as FFH
2. Set carry flag (cy = 1).
3. AND bit 7 of accumulator with cy and store PSW format.
4. OR bit 6 of PSW and store the PSW format.
5. Set bit 5 of SCON.
6. Clear bit 1 of SCON.
7. Move SCON.1 to carry register.
8. Stop the execution of program.

FLOWCHART:

START

Set CY flag, AND CY with MSB of
ACC
Store the PSW format, OR CY with bit 2 IE reg

Clear bit 6 of PSW, Store PSW

Set bit 5 of SCON , clear bit 1 and store
SCON

Move bit 1 of SCON to CY and store PSW

STOP

53


PROGRAM TABLE

ADDRESS HEX LABEL MNEMONICS OPERAND COMMENT
CODE
4100 90,45,00 MOV DPTR,#4500 Initialize memory
location
4103 74,FF MOV A,#FF Get the data in
accumulator
4105 D3 SETB C Set CY bit

4016 82,EF ANL C, ACC.7 Perform AND with 7th
bit of accumulator
4018 E5,D0 MOV A,DOH
410A F0 MOVX @DPTR,A Store the result

410B A3 INC DPTR Go to next location

410C 72,AA ORL C, IE.2 OR CY bit with 2nd bit
if IE reg
410E C2,D6 CLR PSW.6 Clear 6th bit of PSW

4110 E5,D0 MOV A,DOH
4112 F0 MOVX @DPTR,A Store the result

4113 A3 INC DPTR Go to next location

4114 D2,90 SETB SCON.5 Set 5th of SCON reg
4116 C2,99 CLR SCON.1 Clear 1st bit of SCON
reg
4118 E5,98 MOV A,98H
411A F0 MOVX @DPTR,A Store the result

411B A3 INC DPTR Go to next location

411C A2,99 MOV C,SCON.1 Copy 1st bit of SCON
reg to CY flag
411E E5,D0 MOV A,DOH
Store the result
4120 F0 MOVX @DPTR,A

4122 80,FE L2 SJMP L2 Stop

54


RESULT:

MEMORY SPECIAL FUNCTION REGISTER FORMAT BEFORE AFTER
LOCATION EXECUTION EXECUTION
4500H (PSW) CY AC FO RS1 RS0 OV - P 00H 88H

4501H (PSW) CY AC FO RS1 RS0 OV - P 40H 88H

4502H (SCON) SM0 SM1 SM2 REN TB8 RB8 TI RI 0FH 20H

4503H (PSW) CY AC FO RS1 RS0 OV - P FFH 09H

Thus the bit manipulation operation is done in 8051 microcontroller.

55


EX.NO PROGRAMS TO VERIFY TIMER, INTERRUPTS & UART
OPERATIONS IN 8031 MICROCONTROLLER
DATE :
a) Program to generate a square wave of frequency --------.
Steps to determine the count:
Let the frequency of sqaurewave to be generated be Fs KHz.
And the time period of the squarewave be Ts Sec.
Oscillator Frequency = 11.0592MHz.
One machine cycle = 12 clock periods
Time taken to complete one machine cycle=12*(1/11.0592MHz)= 1.085microsec.
Y(dec) = (Ts/2)/(1.085microsec)
Count(dec) = 65536(dec) – Y(dec)
= Count(hexa)

MOV TMOD,#10h ; To select timer1 & mode1 operation
L1: MOV TL1,#LOWERORDER BYTE OF THE COUNT
MOV TH1,#HIGHER ORDER BYTE OF THE COUNT
SETB TR1 ; to start the timer (TCON.6)
BACK: JNB TF1,BACK ; checking the status of timerflag1(TCON.7) for
overflow
CPL Px.x ; get the square wave through any of the portpins
; eg. P1.2 (second bit of Port 1)
CLR TR1 ; stop timer
CLR TF1 ; clear timer flag for the next cycle
SJMP L1
b) Program to transfer a data serially from one kit to another.
Transmitter:
MOV TMOD,#20H ; Mode word to select timer1 & mode 2
MOV TL1,#FDH ; Initialize timer1 with the count
MOV TH1,#FFH
MOV SCON,#50H ; Control word for serial communication to
to select serial mode1
SETB TR1 ; Start timer1
MOV A,#06h

56


MOV SBUF,A ; Transfer the byte to be transmitted to serial
Buffer register.
LOOP: JNB TI, LOOP ; checking the status of Transmit interrupt
flag
CLR TI
HERE: SJMP HERE

Receiver:
MOV TMOD,#20H
MOV TL1,#FDH
MOV TH1,#FFH
MOV SCON,#50H
SETB TR1
LOOP: JNB RI,LOOP
MOV A,SBUF
MOV DPTR,#4500H
MOVX @DPTR,A
CLR RI
HERE: SJMP HERE

57


EX.NO. COMMUNICATION BETWEEN 8051 MICROCONTROLLER
KIT & PC
DATE :

SERIAL COMMUNICATION

8051>H
HELP MENU

D Display data, program, internal, bit memory or registers
E Edit data, program, internal, bit memory or registers
S Single step from specified address, press SP to terminate
G Execute the program till user break
B Set address till where the program is to be executed
C Clear break points
F10 Key followed by 4 key at the PC to upload data to a file (DOS)
T Test the onboard peripherals
: Download a file from PC mem to the SDA-SI-MEL kit (DOS)
A Assembler
Z Disassembler

TEST FOR ONBOARD PERIPHERALS

For SDA SI-MEL kit, following menu is displayed on pressing the option "T"

8051>T
ALS-SDA SI-MEL Kit Test monitor

1. Test internal Data RAM
2. Test external Data Memory (U6)
3. Test external Data memory (U7)
4. 8255 loop test
5. Test 8253

58


6. Exit

Select (1-6):

Suppose the user presses the key '1', following message is displayed if the internal data RAM
is OK.

Testing internal data RAM: Pass

After displaying the message, the menu is displayed once again waits for user to enter a key

EDITING MEMORY COMMAND:

8051>E
EDIT (R,B,M,P,D)…D - EXTERNAL DATA RAM
Enter STA address = 0400
0400 = 7F:55 Press 'N' key to go to the next address
0401 = D5:66
0402 = D3:77
0403 = 73:88
0404 = 6F:12
0405 = CB:01
0406 = A7:02 Press 'P' key to go to the previous address
0407 = 6F:03
0408 = 7B:04
0409 = 29:05
040A = 6F:06
040B = 73:07
040C = FF:08
040D = 7D:09 Press 'CR' key to have the same address
040E = 09:90 Press 'ESC' Key to abort the command

8051>E
EDIT (R,B,M,P,D)…B - BITS

59


00 = 0:1
01= 0:1
02 = 0:0
03 = 0:1
03 = 1:
03 = 1:
02 = 0:

8051>E
EDIT (R,B,M,P,D)…R- REGISTERS
ACC = 00:33
PSW = 00:44
DPH = 00:55
DPL = 00:00
DPL = 00:00

8051>E
EDIT (R,B,M,P,D)…-P = PROGRAM CODE
8000 = FF:78
8001 = FF:10
8002 = FF:79
8003 = FF:20
8004 = FF:7A
8005 = FF: 12
8007 = FF : 00
8008 = FF : 03
8009 = FF : 0F

8051>E
EDIT (R,B,M,P,D)…-M - INTERNAL RAM
0000 = 00 : 12
0001 = 00 : 34

60


0002 = 00 : 00

DISPLAY COMMAND

8051>D
EDIT (R,B,M,P,D)…-EXTERNAL DATA RAM


Enter END address = 040F

0500 55 66 77 88 12 01 02 03 04 05 06 07 08 09 04 D7

SETTING BREAK COMMAND :

8051>B
BR _ NO: R
BR_ADD 0000
ERROR! ONLY A BREAKS ALLOWED

8051>B
BR _ NO: 0
ERROR! BREAK NUMBERS MUST BE BETWEEN 1 & 8

CLEAR BREAK COMMAND:

8051>C
BR_N0:A Clears all the break point set by the user

8051>C
BR_N0:1 Clears the break point number 1

61


PROGRAMME EXECUTION COMMAND:
8051>G
PROGRAM EXECUTION

ENTER START ADDRESS = 8000

ACC PSW DPH DPL PCH PCL SP B R0 R1 R2 R3 R3 R4 R5 R6 R7
33 44 55 00 10 34 00 00 00 00 00 00

ASSEMBLE MEMORY COMMAND

8051>A
ENTER START ADDRESS = 8000

DISASSEMBLE MEMORY COMMAND
8051>Z

62


EX. NO. PROGRAMS FOR DIGITAL CLOCK AND STOP WATCH (USING
8086)
DATE:

Program that places a message on the screen every 10 seconds, using int 1ah;

CODE SEGMENT
TIMEDELAY:
MOV SP,1000H
MOV DI,10XD
TIME OUT:
MOV AH,00H
INT 1AH
MOV BX,DX
TIMER:
MOV AH, 00H
INT 1AH
SUB DX, BX
CMP DX, 182XD
JC TIMER
MOV AH, 09H
CS MOV DX,MSG
INT 21H
DEC DI
JNZ TIMEOUT
MOV AX,4C00H
INT 21H
MSG:
DB 'TEN MORE SECONDS HAVE PASSED $'
CODE ENDS

63


EX.NO. Interfacing 8259 with 8085 microprocessor DATE:

Initialize 8259 with the following specifications:
ICW4 needed, single 8259, Interval of ------(8085), Edge triggered mode, Vector
address(8085)/ Type number(8086) of IR0 ------, ------- mode, Normal EOI, Non-buffered
mode, Not special fully nested mode, Mask all interrupts except ---------.

A0 – Address with A0 = 0
A1 – Address with A0 = 1
Using 8086:

MOV AL,ICW1
OUT A0,AL
MOV AL,ICW2
OUT A1,AL
MOV AL,ICW4
OUT A1,AL
MOV AL,OCW1
OUT A1,AL
STI
HLT

Vector Table: offset address - Interrupt type * 4
0000: offsetaddress - Segment Address : Offset address of the ISR

ISR: MOV AL,OCW2
OUT A0,AL
HLT

64

Ec2308 microprocessor and_microcontroller__lab1

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