EC2308 Lab Manual

SREE SASTHA INSTITUTE OF ENGINEERING AND TECHNOLOGY CHEMBARAMBAKKAM - 602 103

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

LAB MANUAL

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB

V SEMSTER 2010-2011 Under the Management of A.M. Kanniappa Mudaliar & A.M.K. Jambulinga Mudaliar Educational Trust # 122, Gengu Reddy Road, Near Hotel Dasaprakash , Opp. Blue Diamond Hotel, Egmore, Chennai - 600 008. Phone Trust Off.

: 28364541 28364542

Phone College Off.

:

26810122 26810114


DEPT OF ECE

LIST OF EXPERIMENTS EC2308

MICROPROCESSOR AND MICROCONTROLLER LAB

1.

Programs for 16 bit Arithmetic operations (Using 8086).

2.

Programs for Sorting and Searching (Using 8086).

3.

Programs for String manipulation operations (Using 8086).

4.

Programs for Digital clock and Stop watch (Using 8086).

5.

Interfacing ADC and DAC.

6.

Parallel Communication between two MP Kits using Mode 1 and Mode 2 of 8255.

7.

Interfacing and Programming 8279, 8259, and 8253.

8.

Serial Communication between two MP Kits using 8251.

9.

Interfacing and Programming of Stepper Motor and DC Motor Speed control.

10.

Programming using Arithmetic, Logical and Bit Manipulation instructions of 8051 microcontroller.

11.

Programming and verifying Timer, Interrupts and UART operations in 8051 microcontroller.

12.

Communication between 8051 Microcontroller kit and PC.

2


DEPT OF ECE

INDEX S.No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

Name of the Experiment 8086 PROGRAMS ADDITION & SUBTRACTION MULTIPLICATION & DIVISION ASCENDING & DESCENDING LARGEST & SMALLEST COPYING A STRING SEARCHING A STRING FIND & REPLACE INTERFACING EXPERIMENTS INTERFACING WITH ADC INTERFACING WITH DAC STEPPER MOTOR INTERFACE DC MOTOR INTERFACE INTERFACING WITH KEYBOARD/DISPLAY CONTROLLER INTERFACING WITH PROGRAMMABLE TIMER INTERFACING WITH 8251 INTERFACING WITH PPI 8255 INTERFACING WITH INTERRUPT CONTROLLER 8051 PROGRAMS 8 BIT ADDITION 8 BIT SUBTRACTION 8 BIT MULTIPLICATION 8 BIT DIVISION LOGICAL AND BIT MANIPULATION TIMER IN 8051 SERIAL COMMUNICATION USING 8051 DATA COMMUNICATIONS BETWEEN PC AND 8051 DIGITAL CLOCK DISPLAY STOP WATCH

Page No. 2 7 12 16 19 22 24 26 28 32 34 37 40 44 47

3

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EXPT NO:

8086 PROGRAMMING

DEPT OF ECE

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 O 1. 2.

ITEM Microprocessor kit Power Supply

SPECIFICATION 8086 kit +5 V dc

QUANTITY 1 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.

4


DEPT OF ECE

FLOWCHART ADDITION

SUBTRACTION

START

START

SET UP COUNTER (CY)

SET UP COUNTER (CARRY)

GET FIRST OPERAND

GET FIRST OPERAND TO A

GET SECOND OPERAND TO A

SUBTRACT SECOND OPERAND FROM MEMORY

YES

A=A+B IS THERE ANY CY YES NO

IS THERE ANY CARRY COUNTER = COUNTER + 1 NO STORE THE SUM

STORE THE CARRY

COUNTER = COUNTER + 1

STORE THE DIFFERENCE

STORE THE CARRY

STOP

STOP

5


DEPT OF ECE

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

HLT

Stop the program

6


DEPT OF ECE

PROGRAM USING TASM (ADDITION): data_here

data_here code_here start:

go:

code_here

segment firstno dw 0202h secondno dw 0202h sum dw 2 dup(0) ends segment assume cs:code_here,ds:data_here mov ax,data_here mov ds,ax mov ax,firstno mov dx,0000h add ax,secondno jnc go inc dx mov sum,ax mov sum+2,dx int 3 ends end start

; first No. ; second No. ; store sum here

; Initialize data segment ; Get first No. ; Initialize dx for carry. ; Add second to it.

; store the sum & carry.

PROGRAM USING TASM (SUBTRACTION): data_here


ahead:

code_here

segment minuend dw 2222h subtrahend dw 1111h result dw 2 dup(0) ends segment assume cs:code_here,ds:data_here mov ax,data_here mov ds,ax mov ax,minuend mov dx,subtrahend mov cx,0000h cmp ax,dx jnc ahead mov bx,dx mov dx,ax mov ax,bx mov cx,0001h sub ax,dx mov result,ax mov result+2,cx int 3 ends end start

; Minuend ; Subtrahend ; Store result here.

; Initialize data segment. ; Get minuend & store in Acc. ; Get subtrahend & store in dx. ; Initialize cx for carry ; compare minuend & subtrahend if minuend smaller than subtrahend , interchange minuend & subtrahend.

; Increment carry by one. ; subtract dx from ax ; store the result & carry.

7


DEPT OF ECE

RESULT:. ADDITION

MEMORY DATA

SUBTRACTION

MEMORY

DATA

MANUAL CALCULATION

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

8


8086 PROGRAMMING

DEPT OF ECE

DATE:

MULTIPLICATION & DIVISION

AIM:

To write an Assembly Language Program (ALP) for performing the multiplication and division operation of 16-bit numbers . APPARATUS REQUIRED: SL.N O 1. 2.


SPECIFICATION 8086 +5 V dc

QUANTITY 1 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.

9


DEPT OF ECE

FLOWCHART MULTIPLICATION

DIVISION Start

Start

Load Divisor & Dividend

Get Multiplier & Multiplicand

QUOTIENT = 0

REGISTER=00

DIVIDEND = DIVIDEND - DIVISOR

REGISTER = REGISTER + MULTIPLICAND

QUOTIENT = QUOTIENT + 1

Multiplier=MULTIPLIER – 1

IS

IS NO

NO

MULTIPLIER

DIVIDEND < DIVISIOR?

=0? YES STORE THE RESULT

YES STORE QUOTIENT STORE REMAINDER = DIVIDEND NOW

STOP

STOP

10

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB MULTIPLICATION PROGRAM

DEPT OF ECE

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 [1208],DX

Store the higher order product

HLT

Stop the program

DIVISION PROGRAM

COMMENTS

MOV AX,[1200]

Get the first data (Dividend)

MOV DX, [1202]

Initialize DX register with 0000h

MOV BX, [1204]

Get the second data(Divisor)

DIV BX

Divide the dividend by divisor

MOV [1206],AX

Store the Quotient

MOV AX,DX

Move the remainder to AX

MOV [1208],AX

Store the content of AX to memory location. Stop the program

HLT

11


DEPT OF ECE

PROGRAM USING TASM (MULTIPLICATION): data_here


code_here

segment multiplicand dw 0202h multiplier dw 0202h product dw 2 dup(0) ends segment assume cs:code_here,ds:data_here mov ax,data_here mov ds,ax mov ax,multiplicand mul multiplier mov product,ax mov product+2,dx int 3 ends end start

; Multiplicand ; Multiplier ; store product here.

; Initialize data segment. ; Get multiplicand ; multiply multiplier with it. ; Store the result.

PROGRAM USING TASM (DIVISION): data_here


code_here

segment dividend divisor result ends

dw 2222h dw 1111h dw 2 dup(0)

segment assume cs:code_here,ds:data_here mov ax,data_here mov ds,ax mov ax,dividend div divisor mov result,ax mov result+2,dx int 3 ends end start

; Dividend ; Divisor ; Store result here.

; Initialize data segment. ; Get dividend ; Divide it by divisor. ; Store result.

12


DEPT OF ECE

RESULT:. MULTIPLICATION

MEMORY DATA

DIVISON

MEMORY

DATA

MANUAL CALCULATION

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

13


8086 PROGRAMMING

DEPT OF ECE

DATE:

ASCENDING & DESCENDING

AIM:

To write an Assembly Language Program (ALP) to sort a given array in ascending and descending order. APPARATUS REQUIRED: SL.N O 1. 2.



QUANTITY 1 1

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. b. Get the first two numbers. 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.

14

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB FLOWCHART ASCENDING ORDER

DEPT OF ECE

DESCENDING ORDER START

START

INITIALIZE POINTER

INITIALIZE POINTER

COUNT = COUNT – 1

COUNT = COUNT – 1

YES IS POINTER  POINTER +1

YES

IS POINTER  POINTER NO

NO TEMP = POINTER POINTER = POINTER + 1 POINTER + 1 = TEMP

POINTER = POINTER +1 COUNT = COUNT-1

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

POINTER = POINTER +1 COUNT = COUNT + 1

NO NO

IS COUNT =0

IS COUNT = 0

YES YES IS FLAG NO

IS FLAG

= 0

=0 YES YES STOP STOP

15


DEPT OF ECE

ASCENDING PROGRAM MOV CL,COUNT L3: MOV DL,COUNT

COMMENTS Get the count in CL Get the count in DL

DEC DL

Number of comparisons in DL

MOV SI,1200

Initialize memory location

L2: MOV AL,[SI]

Get the first data in AL

INC SI

Go to next memory location

MOV BL,[SI]

Get the second data in BL

CMP AL,BL

Compare two data’s

JC /JNC

If AL < BL go to L1 (Ascending/Descending)

L1

DEC SI

Else, Decrement the memory location

MOV [SI],BL

Store the smallest data

INC SI

Get the next data AL

MOV [SI],AL

Get the next data AL

L1: DEC DL

Decrement the no of comparisons

JNZ L2

Jump to loop2

DEC CL

Decrement the count

JNZ L3

Jump to L3, if the count is not reached zero

HLT

Stop

PROGRAM USING TASM(ASCENDING/DESCENDING) Dataseg

segment series db

12h,23h,14h,25h,18h

; datas to be sorted

datasegends codeseg start:

loop2: loop1:

segment assume cs:codeseg,ds:dataseg mov ax,dataseg mov ds,ax mov cl,04h mov bl,04h lea si,series mov al,[si] inc si cmp al,[si] jc loop mov dl,[si] mov [si],al dec si mov [si],dl

; initialize data segment.

; move the datas to source index

; Compare first data with second data. If second data smaller than first one, interchange the datas

16


loop:

inc si dec bl jnz loop1

DEPT OF ECE

; Repeat this comparison with

other

codeseg

dec cl jnz loop2 int 3 ends end start

other datas in series.

RESULT:. ASCENDING

MEMORY DATA

DESCENDING

MEMORY

DATA

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

17


8086 PROGRAMMING

DEPT OF ECE

DATE:

LARGEST& SMALLEST

AIM:

To write an Assembly Language Program (ALP) to find the largest and smallest number in a given array. APPARATUS REQUIRED: SL.NO ITEM 1. Microprocessor kit 2. Power Supply


QUANTITY 1 1

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. b. c. d. (ii) e. f. g. h.

Load the array count in a register C1. Get the first two numbers. Compare the numbers and exchange if the number is small. Get the third number from the array and repeat the process until C1 is 0. Finding smallest number: Load the array count in a register C1. Get the first two numbers. Compare the numbers and exchange if the number is large. Get the third number from the array and repeat the process until C1 is 0.

18

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB FLOWCHART LARGEST NUMBER IN AN ARRAY

DEPT OF ECE

SMALLEST NUMBER IN AN ARRAY

START

START

INITIALIZE COUNT POINTER MAX = 0

INITIALIZE COUNT POINTER MIN = 0

PONITER = POINTER + 1

PONITER = POINTER + 1

YES IS MAX  POINTER?

IS MIN  POINTER

YES

NO

NO

MIN = POINTER

MAX = POINTER

COUNT = COUNT-1

COUNT = COUNT-1 NO

NO IS COUNT = 0? YES

IS COUNT = 0? YES

STORE MAXIMUM

STOP

STORE MINIIMUM

STOP

19


DEPT OF ECE

LARGEST PROGRAM

COMMENTS

MOV SI,1200H

Initialize array size

MOV CL,[SI]

Initialize the count

INC SI


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

JNC L1/JC L1

If AL > [SI] then go to L1 ( Largest/Smallest)

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

store the biggest number in 1300 location

HLT

Stop

RESULT:. LARGEST

MEMORY DATA

SMALLEST

MEMORY

DATA

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

20


8086 PROGRAMMING COPYING A STRING

DEPT OF ECE

DATE:

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. FLOWCHART: 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

Check for ZF=1

NO

STOP

21

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB COPYING A STRING PROGRAM

DEPT OF ECE

COMMENTS

MOV SI,1200H

Initialize destination address

MOV DI,1300H

Initialize starting address

MOV CX,0006H


CLD

Clear direction flag

REP MOVSB

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

HLT

PROGRAM USING TASM data

data

segment test_mess count new_loc ends

db 23h,45h,67h,87h db 100 dup(?) equ 04h db 4 dup(0)

code segment assume cs:code,ds:data,es:data start: mov ax,data mov ds,ax mov es,ax lea si,test_mess lea di,new_loc cld

; data to move ; stationary block of data ; number of data to move ; data destination

; initialize data segment register ; initialize extra segment register ; Point si at source data ; Point di at destination data ; clear direction flag so pointers auto increment after each data element is moved

rep movsb code ends end start

22


DEPT OF ECE

RESULT: INPUT MEMORY DATA

OUTPUT MEMORY

DATA

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

23


8086 PROGRAMMING

DEPT OF ECE

DATE:

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


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

24


DEPT OF ECE

SEARCHING FOR A CHARACTER IN THE STRING PROGRAM COMMENTS MOV DI,1300H

Initialize destination address

MOV SI, 1400H


MOV CX, 0006H


MOV BL,00H

Initialize the relative address

CLD


MOV AL, 08H

Store the string to be searched

LOOP2: NOP

Delay

SCASB

Scan until the string is found

JNZ LOOP1

Jump if the string is found

MOV [SI],BL

Move the relative address to SI.

INC SI

Increment the memory pointer

LOOP1: INC BL

Increment the relative address

LOOP LOOP2

Repeat until the count reaches zero

HLT

Stop


OUTPUT MEMORY LOCATION DATA

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

25


EXPT NO:

8086 PROGRAMMING

DEPT OF ECE

DATE:

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


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

Scan for a particular character specified in AL

Check NO for ZF=1 YES Move the content of ¿ BL to ES: DI

STOP 26


DEPT OF ECE

FIND AND REPLACE A CHARACTER IN THE STRING PROGRAM COMMENTS MOV DI,1300H


MOV CX, 0006H


CLD


MOV AL, 08H

Store the string to be searched

MOV BH,30H

Store the string to be replaced

BACK:SCASB

Scan until the string is found

JNZ LOOP1

Is the string found

DEC DI

Decrement the destination address

MOV [DI],BL

Replace the string

LOOP1:LOOP BACK

Continue until count zero.

HLT

Stop


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.

27


DEPT OF ECE

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: ITEM SL.NO SPECIFICATION QUANTITY 1. 8086 1 Microprocessor kit 2. +5 V dc,+12 V dc 1 Power Supply 3. 1 ADC Interface board 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: START

SELECT THE CHANNEL AND LATCH SEND THE START CONVERSION PULSE NO IS EOC = 1? YES READ THE DIGITALOUTPUT STORE THE DIGITAL VALUE IN THE MEMORY LOCATION SPECIFIED

STOP

28


DEPT OF ECE

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


MOV AL,01

Store the value to make SOC high in the accumulator

OUT 0D0H,AL


MOV AL,00 Introduce delay

MOV AL,00 MOV AL,00 MOV AL,00

Store the value to make SOC low the accumulator

OUT 0D0H,AL


L1 : IN AL, 0D8H Read the EOC signal from port & check for end of conversion

AND AL,01 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 RESULT:

Stop

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. 29


DEPT OF ECE

8086 INTERFACING DATE: INTERFACING DIGITAL – TO – ANALOG CONVERTER

AIM : 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: SL.NO SPECIFICATION QUANTITY ITEM 1. 8086 Vi Microsystems 1 Microprocessor kit 2. +5 V, dc,+12 V dc 1 Power Supply 3. 1 DAC Interface board 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)

30

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB FLOWCHART: MEASUREMENT OF ANALOG VOLTAGE

DEPT OF ECE

SQUARE WAVE FORM START

START

SEND THE DIGITALVALUE TO ACCUMULATOR

INTIALISE THE ACCUMULATOR SEND ACC CONTENT TO DAC

DELAY TRANSFER THE ACCUMULATOR CONTENTS TO DAC

LOAD THE ACC WITH MAX VALUE SEND ACC CONTENT

READ THE CORRESPONDING ANALOG VALUE

DELAY

STOP TRIANGULAR WAVEFORM START SAWTOOTH WAVEFORM START

INITIALIZE ACCUMULATOR

INITIALIZE ACCUMULATOR

SEND ACCUMULATOR CONTENT TO DAC

SEND ACCUMULATOR CONTENT TO DAC

INCREMENT ACCUMULATOR CONTENT

INCREMENT ACCUMULATOR CONTENT

IS ACC  FF

YES

NO NO

IS ACC  FF

YES

DECREMENT ACCUMULATOR CONTENT SEND ACCUMULATOR CONTENT TO DAC YES

IS ACC  00

NO

31


DEPT OF ECE

MEASUREMENT OF ANALOG VOLTAGE: PROGRAM

COMMENTS

MOV AL,7FH

Load digital value 00 in accumulator

OUT C0,AL


HLT

Stop

DIGITAL DATA

ANALOG VOLTAGE

PROGRAM TABLE: Square Wave PROGRAM

COMMENTS

L2 : MOV AL,00H

Load 00 in accumulator

OUT C0,AL


CALL L1

Give a delay

MOV AL,FFH

Load FF in accumulator

OUT C0,AL


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

L2 : MOV AL,00H


L1 : OUT C0,AL


INC AL

Increment contents of accumulator

JNZ L1

Send through output port until it reaches FF

JMP L2


32


DEPT OF ECE

PROGRAM TABLE: Triangular Wave PROGRAM

COMMENTS

L3 : MOV AL,00H


L1 : OUT C0,AL


INC AL

Increment contents of accumulator

JNZ L1

Send through output port until it reaches FF

MOV AL,0FFH

Load FF in accumulator

L2 : OUT C0,AL


DEC AL

Decrement contents of accumulator

JNZ L2

Send through output port until it reaches 00

JMP L3


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. 33

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EXP.NO:

DEPT OF ECE

STEPPER MOTOR INTERFACING

DATE:

AIM: To write an assembly language program in 8086 to rotate the motor at different speeds. APPARATUS REQUIRED: SL.NO 1. 2. 3. 4.

ITEM Microprocessor kit Power Supply Stepper Motor Interface board Stepper Motor

SPECIFICATION 8086 +5 V, dc,+12 V dc -

QUANTITY 1 1 1 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 LOCATION 4500 4501 4502 4503

A1

A2

B1

B2

1 0 0 1

0 1 1 0

0 0 1 1

0 1 0 0

HEX CODE 09 H 05 H 06 H 0A H

34

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB FLOWCHART:

DEPT OF ECE

START 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

MOV CX, 0004H


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


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.

35


DC MOTOR INTERFACING

DEPT OF ECE DATE:

AIM: To write an assembly language program in 8086 to control the speed of the motor. APPARATUS REQUIRED: SL.NO 1. 2. 3. 4.

ITEM Microprocessor kit Power Supply DC Motor Interface board DC Motor

SPECIFICATION 8086 +5 V, dc,+12 V dc -

QUANTITY 1 1 1 1

PROBLEM STATEMENT: Write a code for controlling the speed of the motor and measure the speed. 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: FLOWCHART:

36


DEPT OF ECE

37


DEPT OF ECE

38


DEPT OF ECE

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 0

0 0

1 D

0 D

0 K

0 K

0 K

1 CD

1 CD

0 CF

0 CA

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

1 1

0 0

1 CD

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) 39


DEPT OF ECE

3. Write Display: Control word-90H 1

0 1

0 0

1

0

0

0

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 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

40


DEPT OF ECE

PROGRAM TABLE

PROGRAM

COMMENTS

START : MOV SI,1200H

Initialize array

MOV CX,000FH


MOV AL,10

Store the control word for display mode

OUT C2,AL


MOV AL,CC

Store the control word to clear display

OUT C2,AL


MOV AL,90

Store the control word to write display

OUT C2,AL


L1 : MOV AL,[SI]

Get the first data

OUT C0,AL


CALL DELAY

Give delay

INC SI

Go & get next data

LOOP L1

Loop until all the data’s have been taken

JMP START


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 1204

98 FF

68 1C

7C 29

C8 FF

RESULT: MEMORY LOCATION 1200H 1201H 1202H 1203H 1204H 1205H 1206H 1207H

d 1 0 0 1 1 0 0 1

7-SEGMENT LED FORMAT c b a dp e g 0 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 1 1 1

HEX DATA f 0 0 0 0 1 0 1 1

98 68 7C C8 FF 1C 29 FF

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

41

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EXP. NO:

DEPT OF ECE

INTERFACING PROGRAMMABLE TIMER-8253

DATE:

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

SL.NO 1. 2. 3. 4.

ITEM Microprocessor kit Power Supply 8253 interfacing kit CRO

SPECIFICATION 8086 Vi Microsystems +5V dc -

QUANTITY 1 1 1 1

THEORY:

i. ii. iii. iv.

The main features of the timer are, Three independent 16-bit counters Input clock from DC to 2 MHz Programmable counter modes 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 21. 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.

42


DEPT OF ECE

Mode 31. 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 3. COUNTER OF CHANNEL 1 4. COUNTER OF CHANNEL 2 5. O/P PORT OF CHANNEL 0 6. O/P PORT OF CHANNEL 1 7. O/P PORT OF CHANNEL 2

– -

CONTROL WORD FORMAT: D7 SC1

D6 SC0

D5 RL1

D4 RL0

D3 M2

D2 M1

D1 M0

D0 BCD

0

0

1

1

0

1

0

0 Mode 2 = 34 H

0

0

SC1

SC0

0 0 1 1 BCD

1

1

0

CHANNEL SELECT

0 CHANNEL 0 1 CHANNEL 1 0 CHANNEL 2 1 ------0 –BINARY COUNTER

M2

M1

M0

MODE

0 0 0 0 1 1

0 0 1 1 0 0

0 1 0 1 0 1

MODE 0 MODE 1 MODE 2 MODE 3 MODE 4 MODE 5

1

1

RL1

0

RL0

Mode 3 = 36 H READ/LOAD

0 0 LATCH 0 1 LSB 1 0 MSB 1 1 LSB FIRST, MSB NEXT 1 --BCD COUNTER

43

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB PORT PIN ARRANGEMENT

DEPT OF ECE

DEBOUNCE CIRCUIT CONNECTION

1 CLK 0

*

* *

2 GATE 0 3 OUT 0 4 CLK 1 5 GATE1 6 OUT 1 7 CLK 2 8 GATE 2 9 OUT 2 10 GND

MODE 2 – RATE GENERATOR: PROGRAM

COMMENTS

MOV AL, 34H

Store the control word in accumulator

OUT 0BH


MOV AL, 0AH

Copy lower order count value in accumulator

OUT 08H


MOV AL, 00H

Copy higher order count value in accumulator

OUT 08H


HLT

Stop

MODE 3 – SQUARE WAVE GENERATOR: PROGRAM

COMMENTS

MOV AL, 36H

Store the control word in accumulator

OUT 0BH


MOV AL, 0AH

Copy lower order count value in accumulator

OUT 08H


MOV AL, 00H

Copy higher order count value in accumulator

OUT 08H


HLT

Stop

44


DEPT OF ECE

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.

45


INTERFACING USART 8251

DEPT OF ECE 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)

46


DEPT OF ECE

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: START

Check TX/RX Ready No

Is it High Yes

Write Data into data register

STOP 47


DEPT OF ECE

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 OUT C8,AL Send through port address MOV AL,4E Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit) OUT C2,AL Send through port address MOV AL,37 Set command instruction(enables transmit enable & receive enable bits) OUT C2,AL Send through port address 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 OUT C0,AL Send through port address 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) OUT C8,AL Send through port address MOV AL,00 Initialize AL with higher value of count OUT C8,AL Send through port address MOV AL,4E Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit) OUT C2,AL Send through port address MOV AL,37 Set command instruction(enables transmit enable & receive enable bits) OUT C2,AL Send through port address 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.

48


DEPT OF ECE

INTERFACING WITH 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 START

Store control word in control register

Mode 1 & 2 START 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

49


DEPT OF ECE

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

OUT C6,AL


MOV AL,09H

Control for BSR mode

OUT C6,AL


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 MOV AL,0C0H

COMMENTS Set the control word for mode 2

OUT C6,AL


MOV AL,09H

Control for BSR mode

OUT C6,AL


MOV AL,13H

Interrupt generation

OUT 30,AL MOV AL,0AH

Through 8259

OUT 32,AL

50

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB MOV AL,0FH

DEPT OF ECE

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 BSR mode

Stop

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 Bit select: (Taking Don't care's as 0)

B2

B1

B0

S/R (1=S,0=R)

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: Input

Mode 0 Output

Mode 1 Input

Output

Mode 2 Input

Output

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

51


DEPT OF ECE

Interfacing 8259 Programmable Interrupt Controller to 8085 p Aim: To initialize 8259 with the following specifications: 1) ICW4 needed 2) single 8259 3) Interval of 4 4) Edge triggered mode 5) A7 A6 A5 = 0 0 0 6) Interrupt Service Routine address for IR0 = 5000 H 7) 8085 mode 8) Normal EOI 9) Non buffered mode 10) not special fully nested mode 11) mask all Interrupts except IR0 Apparatus Required: 8085p kit, 8259 Interface Board, Regulated Power supply, vxt parallel bus Theory: The 8259 is a programmable Interrupt Controller that can resolve eight levels of interrupt priorities in a variety of modes. It can vector an interrupt anywhere in the memory map. It is also capable of masking each interrupt request individually and read the status of pending interrupts, in_service interrupts and masked interrupts. The 8259 can be expanded to 64 priority levels by cascading network. Initialisation command words (ICW s): Before the normal operation can begin , each 8259 in the system must be initialised using 2 to 4 bytes of ICW s. Operation command words (OCW s): These are the command words which command the 8259 to operate in various interrupt modes. Program: MVI OUT MVI OUT MVI OUT MVI

A,17 C0H A,50 C2H A,00 C2H A,FE OUT C2H

; Initialize ICW 1 for ICW 4 needed , ; interval of 4 , single , edge triggered mode ; Initialize ICW 2 ; ; Initialize ICW 4 ; Initialize OCW 1 , set mask for all except ; IR0

HLT ISR : MVI A,20 MVI MVI ADD STA HLT

A,08 B,08 B 4500

; Initialize OCW 2 for non specific EOI OUT C0H ; addition program

Procedure: The program is entered in user RAM locations. On pressing IR0 , the CPU jumps to location 5000H. 8259 will not accept any more interrupt at IR0 since AEOI is not set. After it is set , the CPU goes back to main program after performing service addition subroutine. The result addition can be viewed at 4500H location.

52


DEPT OF ECE

Exercise: Program the 8259 in special mask mode. (In the special mask mode ,when the mask bit is set in 0CW1, it inhibits further interrupts at that level and enables interrupts from all other levels(lower as well as higher) that are not masked. Thus any interrupts may be selectively enabled by loading the mask register. Result: 1) Thus 8259 was initialized for the required specifications 2) All the LED s except one corresponding to IRO is seen glowing. 3) The result of addition subroutine was verified at memory location 4500H.

53


8051 PROGRAMMING

DEPT OF ECE DATE:

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

54

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB 8 Bit Addition (Immediate Addressing) ADDRESS LABEL MNEMONIC OPERAND 4100

CLR

C

HEX CODE C3

4101

MOV

A, data1

74,data1

4103

ADDC

A, # data 2

24,data2

4105

MOV

90,45,00

4108

MOVX

DPTR, # 4500H @ DPTR, A

SJMP

L1

80,FE

4109

L1

F0

DEPT OF ECE

COMMENTS

Get the data1 in Accumulator Add the data1 with data2 Initialize the memory location Store the result in memory location Stop the program

Clear CY Flag

RESULT:

OUTPUT MEMORY LOCATION

DATA

4500

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

55


DEPT OF ECE

EXPT NO:

DATE:

8 BIT SUBTRACTION

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

56

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB 8 Bit Subtraction (Immediate Addressing) ADDRESS LABEL MNEMONIC OPERAND 4100

CLR

C

HEX CODE C3

4101

MOV

A, # data1

74, data1

4103

SUBB

A, # data2

94,data2

4105

MOV

DPTR, # 4500

90,45,00

4108

MOVX

@ DPTR, A

F0

SJMP

L1

80,FE

4109

L1

DEPT OF ECE

COMMENTS

Store data1 in accumulator Subtract data2 from data1 Initialize memory location Store the difference in memory location Stop

Clear CY flag

RESULT:


DATA

4500

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

57


8051 PROGRAMMING

DEPT OF ECE DATE:

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

58


DEPT OF ECE

8 Bit Multiplication ADDRESS LABEL

MNEMONIC

OPERAND

4100

MOV

A ,#data1

HEX CODE 74, data1

4102

MOV

B, #data2

75,data2

Store data1 in accumulator Store data2 in B reg

4104

MUL

A,B

F5,F0

Multiply both

4106

MOV

90,45,00

4109

MOVX

DPTR, # 4500H @ DPTR, A

401A

INC

DPTR

A3

Initialize memory location Store lower order result Go to next memory location

410B

MOV

A,B

E5,F0

410D

MOV

@ DPTR, A

F0

Store higher order result

SJMP

STOP

80,FE

Stop

410E

STOP

F0

COMMENTS

RESULT: INPUT MEMORY LOCATION

DATA


4500

4502

4501

4503

DATA

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

59


8051 PROGRAMMING

DEPT OF ECE DATE:

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

60


DEPT OF ECE

8 Bit Division ADDRESS LABEL MNEMONIC

OPERAND

4100

MOV

A, # data1

HEX CODE 74,data1

4102

MOV

B, # data2

75,data2

Store data1 in accumulator Store data2 in B reg

4104

DIV

A,B

84

Divide

4015

MOV

DPTR, # 4500H

90,45,00

4018

MOVX

@ DPTR, A

F0

Initialize memory location Store remainder

4109

INC

DPTR

A3

410A

MOV

A,B

E5,F0

410C

MOV

@ DPTR, A

F0

SJMP

STOP

80,FE

COMMENTS

Go to next memory location Store quotient

410D

STOP

Stop

RESULT: INPUT MEMORY LOCATION 4500 (dividend) 4501

(divisor)

DATA

OUTPUT MEMORY LOCATION 4502 (remainder) 4503

DATA

(quotient)

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

61


LOGICAL AND BIT MANIPULATION

DEPT OF ECE DATE:

AIM: To write an ALP to perform logical and bit manipulation operations using 8051 microcontroller. APPARATUS REQUIRED: 8051 microcontroller kit ALGORITHM: 1. 2. 3. 4. 5. 6. 7. 8.

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

62


DEPT OF ECE

PROGRAM TABLE ADDRESS HEX CODE 4100 90,45,00

LABEL

MNEMONICS

OPERAND

COMMENT

MOV

DPTR,#4500

Initialize memory location Get the data in accumulator Set CY bit

4103

74,FF

MOV

A,#FF

4105

D3

SETB

C

4016

82,EF

ANL

C, ACC.7

4018 410A

E5,D0 F0

MOV MOVX

A,D0H @DPTR,A

Store the result

410B

A3

INC

DPTR

Go to next location

410C

72,AA

ORL

C, IE.2

410E

C2,D6

CLR

PSW.6

OR CY bit with 2 nd bit if IE reg Clear 6 th bit of PSW

4110 4112

E5,D0 F0

MOV MOVX

A,DOH @DPTR,A

Store the result

4113

A3

INC

DPTR

Go to next location

4114 4116

D2,90 C2,99

SETB CLR

SCON.5 SCON.1

Set 5th of SCON reg Clear 1st bit of SCON reg

4118 411A

E5,98 F0

MOV MOVX

A,98H @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

4120

F0

MOVX

@DPTR,A

4122

80,FE

SJMP

L2

Perform AND with 7th bit of accumulator

Store the result

L2

Stop

63


DEPT OF ECE

RESULT:

MEMORY

SPECIAL FUNCTION REGISTER FORMAT

LOCATION

BEFORE

AFTER

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

09H

AC

FO

RS1

RS0 OV

-

P

FFH

Thus the bit manipulation operation is done in 8051 microcontroller.

64

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EX.NO

DEPT OF ECE

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 square wave to be generated be Fs KHz. And the time period of the square wave 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 L1:

; To select timer1 & mode1 operation

MOV TL1,#LOWERORDER BYTE OF THE COUNT MOV TH1,#HIGHER ORDER BYTE OF THE COUNT

BACK:

SETB TR1

; to start the timer (TCON.6)

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 65

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB MOV SBUF,A

DEPT OF ECE

; 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

66


DEPT OF ECE

SETTING BITS IN AN 8-BIT NUMBER PROBLEM STATEMENT:

Write a program to set specific bits of an 8 bit number.

ALGORITHM:

1. Get an 8-bit number in accumulator. 2. The bit which is to be set is ‘OR’ed by 1. 3. Store the answer at another memory location.

FLOW CHART:

START

INPUT THE DATA TO (A)

OR (A) WITH AN 8-BIT IMMEDIATE DATA

STORE THE RESULT IN MEMORY

67


DEPT OF ECE

STOP

PROGRAM: ADDRESS 4100 4101 4102 4103

OPCODE LABEL 74 START: 2F 44 45

4104 4105 4106 4107

90 45 00 F0

4108 4109

80 FE

HERE:

MNEMONICS MOV

OPERAND A, #DATA1

ORL

A, #DATA2

MOV

DPTR, #4500

MOVX

@DPTR, A

SJMP

HERE

COMMENT Copy to A Register first number OR the A register with an immediate data 45h Store in Data Pointer the memory address Store the result in memory Halt here

RESULT:

INPUT DATA1 0010 1111 (2Fh) DATA2 0100 0000 (40h)

4500

OUTPUT 0110 1111 (6Fh) -- 6th bit is getting set

CONCLUSION:

Thus the setting of bits in an 8-bit number has been done using 8051 C.

68


DEPT OF ECE

69


DEPT OF ECE

MASKING BITS IN AN 8-BIT NUMBER PROBLEM STATEMENT:

Write a program to mask specific bits of an 8 bit number.

ALGORITHM: 1.Get an 8-bit number in accumulator. 2.The bit, which is to be masked, is ANDed by 1. 3.Store the result in memory.

FLOW CHART:

START

INPUT THE DATA TO (A)

AND (A) WITH AN 8-BIT IMMEDIATE DATA

STORE THE RESULT IN MEMORY

STOP

70


DEPT OF ECE

PROGRAM: ADDRESS 4100 4102 4102 4103 4104 4105 4106 4107

OPCODE LABEL 74 START: 87 54 7E 90 45 00 F0

4108 4109

80 FE

HERE:

MNEMONICS MOV

OPERAND A, #DATA1

ANL

A, #DATA2

MOV

DPTR, #4500

MOVX

@DPTR, A

SJMP

HERE

COMMENT Copy 1st Number to A register AND A register with 2nd number Copy the memory address to Data Pointer Store the result in memory Halt here

RESULT:

INPUT 1000 0111 (87h) DATA2 0111 1110 7Eh (to mask bits 0 and 7) DATA1

4500

OUTPUT 0000 0110 (06h) -- Bits 0 and 7 are getting masked

CONCLUSION:

Thus the masking of bits in an 8-bit number has been done using 8051 C.

71

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EX.NO.

DEPT OF ECE

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 6. Exit 72


DEPT OF ECE

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

73


DEPT OF ECE

8051>E EDIT (R,B,M,P,D)…B - BITS Enter STA address = 00 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

74


DEPT OF ECE

8051>E EDIT (R,B,M,P,D)…-M - INTERNAL RAM 0000 = 00 : 12 0001 = 00 : 34 0002 = 00 : 00

DISPLAY COMMAND

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

Enter STA address = 0400

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 75


DEPT OF ECE

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

76

EC 2308 – MICROPROCESSOR AND MICROCONTROLLER LAB EX. NO.

DEPT OF ECE

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

77


DEPT OF ECE

CLOCK displaying Program Problem Statement: To write an 8086 assembly language program to display the Computer’s real-time clock on the screen as in digital clock. Theory: Every PC is equipped with a real-time clock operated by a small battery. Even when the computer is switched off, this clock continues to function to maintain the clock in its CMOS registers. When the PC is on, the Operating System software reads this clock and stores the value in memory. User programs can access this data. The OS software provides facilities for the programmer to read this clock value using Software Interrupts. The IBM PC system BIOS also provides interrupt services to access this real time clock. The BIOS services provide the hours, minutes and seconds in CPU registers as BCD data. Hence they can be converted easily into ASCII value for displaying on screen.

Algorithm: Main Program: 1. Initialize data segment to hold the clock data as hours, minutes and seconds 2. Display a message on screen for the user 3. Call the procedure to get the time from PC’s real-time clock using BIOS interrupt 4. Call the procedure to convert the clock data into ASCII and display it on the screen 5. Check for user key press using BIOS interrupt service 6. If no key is pressed continue updating the clock on screen. 7. If a key is pressed terminate the program Procedure GetTime: 1. Use Function 02 in Interrupt 1Ah service to read the PC’s real-time clock. This function returns hours in CH, minutes in CL and seconds in DH registers. 2. Divide each of these registers by 10h to separate the BCD digits 3. Convert each separated digit into ASCII by adding 30h with it. 4. Store each digit in the appropriate location in the memory. 5. Return from the procedure Procedure ShowTime: 1. Use BIOS interrupt 10h function 02h to set video cursor at desired location on the screen 2. Use DOS interrupt 21h function 09h to display the stored clock data from Data segment memory as a string terminated by ‘$’ 3. Call a procedure to delay for few milliseconds doing nothing 4. Return from the procedure

78


DEPT OF ECE

Procedure Delay: 1. Load a value in CX register 2. Do nothing 3. Decrement CX register by one. 4. loop until CX register is zero 5. Return from the procedure Flow chart: START

Initialize Data Segment

Display a friendly Message

Call Procedure GetTime

Call Procedure ShowTime

Is there a key pressed?

No

Yes STOP

79


DEPT OF ECE

Procedure GetTime: Start

Read System Clock

Separate the digits from hours, minutes and seconds

Convert each of these digits into its ASCII equivalent

Return

Procedure ShowTime: Start

Position Video Cursor at desired location

Display the clock data stored in memory as a string

Execute a delay loop before going for next update of time

Return

80


DEPT OF ECE

Assembly Listing: .model small .stack .data message db 0ah, 0dh, "This program displays the time " message1 db 0ah, 0dh, "Press any key to quit...", '$', 0ah, 0dh prompt db "Time :: " hours1 db 0 hours2 db 0 separator1 db ':' minutes1 db 0 minutes2 db 0 separator2 db ':' seconds1 db 0 seconds2 db 0 last db '$' divisor db 10h .code mov ax, @data mov ds, ax ; initialize data segment mov ah, 09h ; display a friendly message lea dx, message int 21h again: call GetTime ; call the procedure to get the time call ShowTime ; show the time mov ah, 01h ; check for user key press int 16h jz again ; if no key pressed do again mov ah, 04ch int 21h

; terminate the program

GetTime proc mov ah, 02h ; BIOS function call to read real time clock int 1Ah mov al, ch ; resulting hours value in BCD is to be split div divisor ; into digits add al, 30h ; convert the digits into ASCII values add ah, 30h mov hours1, al ; store them in memory mov hours2, ah xor ax,ax ; clear AX register mov al, cl ; Now convert minutes in the same way and store div divisor add al, 30h add ah, 30h mov minutes1, al mov minutes2, ah

81


DEPT OF ECE

xor ax,ax ; clear AX register mov al, dh ; convert seconds in the same way and store div divisor add al, 30h add ah, 30h mov seconds1, al mov seconds2, ah ret GetTime endp

ShowTime proc mov ah, 02h ; set video cursor mov bh, 0 mov dh, 20d ; row and col position on screen mov dl, 10d int 10h mov ah, 09h ; display the time already stored in memory lea dx, prompt int 21h call delay ret ShowTime endp delay proc push cx mov cx, 0ffffh idleout: nop nop nop nop loop idleout pop cx ret delay endp end

Result: The program should function as expected after assembling and linking into an executable .EXE program. It will display a clock on the screen. It will stop when the user presses a key.

82


DEPT OF ECE

STOP WATCH Display

Aim: To display a digital stopwatch on the PC screen allowing the user to start and stop the watch.

Apparatus Required: Any Personal Computer running DOS or above OS. MASM software with debugger

Theory: Every PC is equipped with BIOS that contains software to initialize the built in Timer IC to produce ticks with a periodicity of 18.2 ticks per every second. This does not rely on the real-time clock usually incorporated in the PC. The BIOS maintains the number of ticks elapsed since midnight or power-on of the machine. BIOS Interrupt 1Ah service provides the number of ticks elapsed at any time. One can use this service repeatedly to check for completion of 18 ticks, which would mean an approximate 1-second time gap. This can be used to update a memory variable to count seconds. After 60 seconds counting, the minute variable can be updated. Thus a stopwatch can be made to operate at the press of a key.

Algorithm: Main Program: 1. Initialize Data Segment Register 2. Display a friendly message using DOS Int 21h function 09h 3. Call the procedure to show the watch with zero values 4. Check for user key press using BIOS Int 16h function 01h 5. If no key is pressed go to step 4 6. On key press, call the procedure to start the watch. 7. On return from the procedure terminate the program Procedure Show_watch: 1. Position the video cursor at the desired location on secreen using BIOS Int 10h function 02h 2. Display the watch data in memory using DOS Int 21h function 09h 3. Return from the procedure Procedure Start_Watch: 1. Get the current tick value using BIOS Int 1Ah function 00h 2. Store the last two digits of the returned value in DL into BL. This becomes the old value 3. Get the current tick value using BIOS Int 1Ah function 00h 4. Subtract the old value from the new value 5. Check whether it is 18 (which means 1 second) 6. If it is less than 18, goto step 3 7. If it is equal to 18, then increment LSD of seconds. 8. If LSD is greater than 9, proceed to increment MSD of seconds and make LSD of seconds equal to 0 9. Upon incrementing MSD of seconds if it is equal to 6, then make MSD equal to 0 and proceed to update minutes. 10. Increment LSD of minutes. 83


DEPT OF ECE

11. If LSD of minutes is greater than 9, increment MSD of minutes and make LSD equal to 0 12. If MSD of minutes is equal to 6, then make minutes and seconds equal to 0. 13. Call the procedure to Show the Watch. 14. Check for User key press using BIOS Int 16h function 01h. 15. If there is no key press proceed to step 1 16. If there is a key pressed, check whether that key is ‘q’. 17. If the key pressed is ‘q’, then return from the procedure. 18. Otherwise, go to step 1

Flow Chart: Main Program:

Start

Initialize Data Segment Register

Display a Friendly Message

Call Show_Watch

Is a key Pressed

No

Yes Call Start_watch

Stop

84


DEPT OF ECE

Show_watch Procedure: Start

Move Video Cursor to desired location

Display Timer data

Return

Start_Watch Procedure: Start B1 Get BIOS Timer Tick count and save it in another register Get BIOS Timer Tick count and compare it with previous value No Is the difference =18? Yes Update seconds LSD

A1

85


DEPT OF ECE

A1

Is Seconds LSD > 9?

N

Y Increment Seconds MSD; Seconds LSD = 0

Is Seconds MSD = 6?

N

Y Increment Minutes LSD; Seconds MSD = 0

Is Minutes LSD > 9?

N

Y Increment Minutes MSD; Minutes LSD = 0

Is Minutes MSD = 6?

N

Y Minutes MSD = 0

A2

86


DEPT OF ECE

A2

Update respective memory locations

Show Timer

Is Key Pressed

N

Y

Is the key = ‘q’

N B1

Y Return

Assembly Listing: .model small .stack .data message db 0ah, 0dh, "This programs displays a stop watch" message1 db 0ah, 0dh, "Press a key to start and q to stop", '$', 0ah, 0dh prompt db "Timer :: " minutes1 db '0' minutes2 db '0' sep1 db ':' seconds1 db '0' seconds2 db '0' sep2 db '.' last db '$' divisor db 10h

87


DEPT OF ECE

.code mov ax, @data mov ds, ax ; initialize data segment mov ah, 09h ; display a friendly message lea dx, message int 21h call show_watch ; show the watch again: mov ah, 01h ; check for user key press int 16h jz again ; if no key pressed then wait for user key call start_watch ; then start the watch mov ah, 04ch int 21h

; terminate the program

start_watch proc doitagain: mov ah, 00h ; BIOS function call to read timer count int 1Ah ; BIOS Timer interrupt service mov bl, dl ; save current tick value in BL checkagain: mov ah, 00h ; Function to read system clock tick counter int 1Ah ; under BIOS interrupt service sub dl, bl ; find how many ticks have elapsed cmp dl, 18 ; is it equal to 1 second? jb checkagain ; if not, read the counter again mov al, seconds2 ; increment the last digit inc al cmp al, 39h ; check whether it exceeds 9 jbe update_seconds2 ; if not update Seconds2 mov al, 30h ; if so, change seconds2 to 0 mov seconds2, al mov al, seconds1 ; and then increment seconds1 inc al cmp al, 36h ; upon incrementing, is it 6? jb update_seconds1 ; if not update Seconds1 and go ahead mov al, 30h ; if so, change seconds1 to 0 first mov seconds1, al mov al, minutes2 ; then increment minutes2 inc al cmp al, 39h ; is it above 9 jbe update_minutes2 ; if not update this digit and go ahead mov al, 30h ; if so, change this digit to 0 mov minutes2, al ; mov al, minutes1 ; and affect the previous digit inc al ; cmp al, 36h jb update_minutes1 mov al, 0 update_minutes1: mov minutes1, al jmp done update_minutes2: mov minutes2, al

88


DEPT OF ECE

jmp done update_seconds1: mov seconds1, al jmp done update_seconds2: mov seconds2, al done: call show_watch mov ah, 01h int 16h jnz check_key jmp doitagain check_key: mov ah, 00h int 16h cmp al, 'q' jnz doitagain ret start_watch endp

show_watch proc mov dh, 10d ; desired row, col position mov dl, 10d mov ah, 02h ; set video cursor mov bh, 0 ; video page int 10h mov ah, 09h ; display the watch lea dx, prompt int 21h ret show_watch endp end

Result: After assembling and linking into executable code, this program will display a timer as 00:00. When any key is pressed, it will start the watch and show the number of seconds and minutes. When the user presses the key ‘q’, the watch will stop and exit.

89

EC2308 Lab Manual

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