microcontrollers lab manual

8/10/2019 MICROCONTROLLERS LAB MANUAL

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K.Sridhar Raju www.engglabs.blogspot.com

MICROPROCESSORS

MICROCONTROLLERS

LAB MANUAL

By

K.Sridhar Raju

www.engglabs.blogspot.com


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INDEX

Description Page No

1. List of experiments 3

2. 8086 kit details and operating procedure 4

3.

Program and data sheet 6

4.

Addition,Subtraction,Multiplication,Division Programs 8-12

5.

Addressing Modes 13

6.

Factorial 14

7.

Positive and negative numbers from given series numbers. 15

8.

Program to generate Fibonacci Series 16

9. program for block exchange 17

10.

Find square of a number using XLAT instruction 18

11.Find the Greatest Common Divisor (GCD) of two numbers. 19

12.Copy a string of 6 bytes from memory to memory 20

13.Compare two strings 21

14.Packed BCD to unpacked BCD 22

15.

Number of logic ones in a byte 23

16.

Addition of a series of bytes from a given array 24

17.

Find the largest number 25

18.

Code conversions I.Binary to BCD,II.Hexadecimal to ASCII 26

19.

Linear search 2720.

Two multi-byte numbers addition 28

21.Sorting 29

22. Square wave generation and Rate generator using 8253 30

23.Read voltage & display in digital using ADC with 8255 32

24.

Square,triangular,sin wave generation using DAC with 8255 33

25.

Interfacing 7-seg LED (CC,CA, BCD to 7-seg) with 8255 36

26.

Interface stepper motor with 8255 37

27.

Interface key board with 8255 39

28.

Single LED interfacing with 8255 (bit set mode) 42

29.

Digital Input and output card with 8255 43

30.

ProgramInterface eight LEDs with 8051 44

31.

Interface stepper motor with 8051 45

32.

Interfacing 7-seg LED with 8051 46

33.

MASM based programs 49


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MICROPROCESSORS AND MICROCONTROLLERS LAB

CODE: EC 333 B.E 3/4 ECE- I Semester

PART A

[Experiments on assembly programming for 8086 using Assembler]

1. Study and use of 8086 microprocessor trainer kit and execution of programs

2.

Programs using different addressing modes.

3. Multiplication and division.

4. Single byte, multi byte binary and BCD addition and Subtraction.

5.

Code conversions.6. String Searching and Sorting.

7.

Generation of waveforms and gating applications using 8253/8254 timers.

8. Generation of waveforms using DAC interface.

9. Monitor utilities of 8086 kit for keyboard/displaying results.

PART B

[Experiments on assembly programming for 8051 using Assembler]

10.Familiarity and use of 8051/8031 Microcontroller trainer kit and execution of

programs.

11.Programs using different addressing modes.

12.Timer and counter programming.

13.Interfacing for D/A applications.

14.Interfacing for A/D applications

15.Interfacing traffic signal control.

16.Program to control stepper motor.

17.

7-segment display/LCD display interfacing.

18.

Keyboard interfacing.


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STUDY OF 8086 KIT

1. Study 8086 microprocessor and identify various components in 8086 trainer kit.

2.

Crystal Oscillator generates 20MHz clock.

Clk 2(divide by 2) pin 14 , (8086 clk is 10MHz)

Clk 4(divide by 4 )pin 15, (8254 clk is 5MHz)

Clk 8(divide by 8)pin 13,(8255 clk is 2.5 MHz)

Operating Commands for SDA 8086 kit:

Step 1: Switch on the kit/ Reset the trainer.

ALS SDA 86-STA

Step 2: Key in A to go to Assembler mode

ALS 86 ASSEMBLER

Step 3: User has to key in Enter key. Line Assembler Prompt is now displayed.

>

The Assembler is waiting for an assembler commands which can be one of the

following:

Step 4: SG 0 to set the segment to 0

Step 5: Label Clear LC to clear the previous label for new session.

Step 6: DA 4000: Disassemble and Assemble. The user can key in assembler mnemonics.

The assembler expects inputs from the user to assemble from 0000:4000.

4000 B83412 MOV AX,#1234

Step 7: Assembler has now disassembled the instruction at 0000:4000 and has displayed

on LCD and is waiting for N key. NDisplay by Assembler @

User has three choices:

1.

User can key in a fresh instruction followed by Enter key.

2. User can key in Enter key if he does not want to alter the display code.

3. User can key in ! to come out of Assembler mode and then key in EX and

Enter key.

Step 8: MOV AL, [4100]

User has entered the above instruction.


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MOV A00041 MOV AL,4100

User Response N

DA prompt @

Next instruction is disassembled and displayed 4003 BB3412 MOV BX,1234

User response N

DA prompt @

User can change the instruction and press Enter or to without change the instruction

Next instruction is disassembled and displayed 4006 01D8 ADD AX,BX

User response N

DA prompt @

User response INT 3

New instruction is disassembled and displayed 4008 CC INT 3

User response to exit from DA mode option 3 !

Assembler prompt >

User response EX

Return to monitor *

GO COMMAND

The execute user program command allows the user to execute a program that user has

entered. The message GO appears in the first line of the display, when the user presses

the G key. The current user CS:IP and the instruction are displayed and the user can

change the address, if required, by either entering a Segment: Offset or just an offset.

The command is executed when the user presses enter key. The user can stop execution of

the program at a specified address by including an INT 3.

EXAM/MODIFY THE REGISTER CONTENTS

Press twice R EXAM REGISTER

BX

Press P for previous register and , for next register.

EXAM / MODIFY THE MEMORY CONTENTS

Press E twice EXAMBYTE

Modify the address as required and press , for check data 5678 C1

Press W EXAMWORD

Give the address as required and press , for check data 5678 41C1


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PROGRAM SHEET:

Address Instruction code Label Mnemonic Comment


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DATA SHEET:

MEMORY LOCATION / REGISTER DATA (HEX)


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1. AIM: TO PERFORM ADDITION OF TWO BYTES

Apparatus:8086 kit, power supply

Procedure/ Algorithm:

1. Load the first number in register AX.

2. Load the second number in register BX.

3.

Perform addition operation on the register contents.

Program:

MOV AX, 1234 h

MOV BX, 2753h

ADD AX, BX

MOV CL, 00

ADC CL, 00

INT 3

Result:

SUM AX: 3987 H

CARRY CL: 0

Check with different data: ABCD and 9DEF


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2. AIM: TO PERFORM SUBTRACTION OF TWO BYTES

Apparatus:8086 kit, power supply


1. Load the minuend in a register AX.

2.

Load the subtrahend in the second register BX.

3. Perform subtraction operation on the register contents.

Program:

MOV AX, 5678H

MOV BX, 1234H

SUB AX, BX

INT 3

Result:

AX: 4444H

Check with different data as:

Sign Flag Zero Flag

AX > BX 0 0

AX = BX 0 1

AX < BX 1 0


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3. AIM: TO PERFORM MULTIPLICATION OF TWO BYTES

Apparatus: 8086 kit, power supply


1.

Load the multiplicand in a register AX.

2. Load the multiplier in the second register BX.

3. Perform multiplication operation on the register contents.

Data:

Program:

MOV AX, 2080

MOV BX, 0070

MUL BX

INT 3

Result:

AX: 38C0 H lower sixteen bit

DX: 02 H higher byte

i.e. 0238C0 H

Check with different data as:


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4. AIM: TO PERFORM DIVISION OF TWO BYTES



1.

Load the dividend in a register.

2. Load the divisor in the second register.

3. Perform division operation on the register contents.

Program:

MOV AX, 2056

MOV BX, 0004

DIV BX

INT 3

Result:

Quotient in AX: 0815 h

Reminder in DX: 02h

Check with different data:

Dividend is 16 bit and divisor is 8 bit

Divide error/ over flow

MOV AX, 2056

MOV BL, 04/45

DIV BL

INT 3

Quotient in AL: 77 h

Reminder in AH: 43h


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5. AIM: WRITE AN ALP FOR 8086 USING ALL ADDRESSING MODES.



1. Initialize the contents of segment registers or general purpose registers using

immediate addressing mode.

2. The contents of memory locations can also be initialized using direct and indirect

addressing modes.

Program:

MOV AX, 0000H

MOV DS, AX

MOV AX, 1234

MOV SI, 5000

MOV [SI], AX

MOV [5000], AX

MOV 2[SI], AX

MOV BX, 1000

MOV [BX][SI], AX

MOV 2[BX][SI], AX

INT 3

Result:

DS: 0000 AX: 1234H, BX: 1000 5000: 34 6000: 34

SI: 5000 5001: 12 6001: 34

5002: 34 6002: 34

5003: 12 6003: 12


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6. Aim: Write an ALP for finding the FACTORIAL of a given number.



1.

Initialize the data segment.

2. Set product to one.

3. Inialise the pointer and fetch the data from the memory location using a pointer.

4. Multiply the data with product.

5. Repeat this until data becomes zero.

6. Increment pointer to store the final product obtained as the factorial of the given

number.

Program

MOV AX, 0000H

MOV DS, AX

MOV AX,1

MOV SI, 4100h hint: 4101 data should be zero.

MOV CX, [SI]

next: MUL CX

LOOP next

INC SI

MOV [SI], AX

INT 3

Input Address Data:

4100 : 04 (Input should be 8)

Result: 4101 : 18h

4102 : 00


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INT 03

Result: Negative No BH:03 Positive No BL :03

Questions: Modify the above program for counting odd and even numbers in a given

series of numbers and compare instruction in 8086.


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8. AIM: WRITE A PROGRAM TO GENERATE FIBONACCI SERIES.


Algorithm/ Procedure:

1. Initialize a=0

2.

Initialize b=1

3. while i


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9. AIM: WRITE A PROGRAM FOR BLOCK EXCHANGE


Algorithms/ Procedure;

1. Set source and destination pointers.

2. Set counter.

3. Get content of source pointer into one of the register (AL).

4. Get the content of destination pointer into another register (BL).

5. Place BL content into the source block and place AL content into the source

block.

6. Increment both the pointers.

7. Decrement counter.

8. If count not equal to zero go to step 3.

9.

Stop the program.

Program:

MOV AX, 0000H; Data: Result:

MOV DS, AX; To initialize data

segment

5000:11 6000:AA 5000:AA 6000:11

MOV ES, AX; To initialize extra

segment

5001:22 6001: BB 5001: BB 6001:22

MOV SI, 5000H; Set source pointer 5002:33 6002: CC 5002: CC 6002:33

MOV DI, 6000H; Set destination pointer 5003:44 6003: DD 5003: DD 6003:44

MOV CX, 000AH; Set counter 5004:55 6004: EE 5004: EE 6004:55

L1: MOV AL, [SI]; 5005:66 6005: FF 5005: FF 6005:66

MOV BL, [DI];

MOV [DI], AL;

MOV [SI], BL;

INC SI;

INC DI; Increment pointers

LOOP L1;

INT 3; Break point

Questions:

1.

Write an ALP to copy data from one memory area to another.

2. Modify the above program to exchange reverse order.

3.

Explain LOOP, LOOPE, LOOPNE instructions.


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10. AIM: TO WRITE AN ALP TO FIND SQUARE OF A NUMBER USING

LOOK-UP (XLAT) TABLE.


Algorithm/ Procedure:

1. Store the values of squares from 0 to 9 in continuous memory locations.

2. Read the number for which square is required.

3. Form the address in BX register.

4. Store the result in another register.

Questions:

1. Can we find factorials using Look-up Tables?

2. Can we find sum of numbers using Look-up Tables?

3.

What will you store in memory for questions 1 & 2?

4. Where are the changes required? Program or data or both?

5. Write the programs for above logic.

Program: Data:

MOV AX, 0000 H 6000: 00

MOV DS, AX ; To initialize Data segment register

MOV BX, 6000H ; To set Offset address 6001: 01MOV AL, 04 ; Load code into AL 6002: 04

XLAT ; Translate one code to another code 6003: 09

MOV CL, AL ; Store result in CL 6004: 16

INT 3 ; Breakpoint 6005: 25

6006: 36

Result: AL = 16 6007:49

6008:64

XLAT AL = [BX+AL]6009:81


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11. AIM: WRITE A PROGRAM FOR GIVEN TWO POSITIVE INTEGERS N

AND M, TO FIND THE GREATEST COMMON DIVISOR (GCD) OF TWO

NUMBERS.


PROGRAM: DATA

MOV SI,4050H Input Result

MOV DI,4060H 4050: 08 4060: 02

MOV AL,[SI] 4051: 02

MOV BL,1[SI]

CMP AL,BL

JNC GO

XCHG AL,BL

GO:MOV AH,00

DIV BL

CMP AH,00

JZ AGAIN

MOV AL,BL

MOV BL,AH

JMP GO

AGAIN:MOV [DI],BL

INT 3

Questions:

1. Write an ALP to reverse the given digits.

2. Explain IMUL and IDIV Instructions with help of examples.


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12. AIM: WRITE AN ALP TO COPY A STRING OF 6 BYTES FROM MEMORY

STARTING FROM 2000:0004H INTO MEMORY STARTING FROM

3000:0014H.

Apparatus: 8086 kit, power supply.

Algorithm/Procedure:

1. Clear Direction Flag.(DF=0 i.e auto increment mode)

2. Set data segment and source index pointer.

3. Set extra segment and destination index.

4. Set counter.

5.

Copy the string bytes from source to destination until count =0.

Program:

CLD ; clear Direction Flag for auto increment Data:

MOV AX, 2000H Input Result:

MOV DS, AX ; Initialize DS 2000:0004: 11 3000:0014:AA

MOV AX, 3000H 2000:0005: 22 3000:0015:BB

MOV ES, AX ; Initialize ES 2000:0006: 33 3000:0016:CC

MOV SI, 0004H ; Offset of source location in DS 2000:0007: 44 3000:0017:DD

MOV DI, 0014H; Offset of destination location in ES 2000:0008: 55 3000:0018:EE

MOV CX, 0006H ; length of the string 2000:0009: 66 3000:0019:FF

REP ; repeat the same up to CX is zero

MOVSB ;copy a byte from source to destination ,if DF=0,increment SI & DI

INT 3


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13. AIM: WRITE AN ALP TO COMPARE TWO STRINGS.


Algorithm / Procedure:

1. Clear Direction Flag.( DF = 0 i.e auto increment mode)

2.

Set Data Segment, extra segment, source and destination index.

3. Set counter.

4. Compare the source string bytes with destination string bytes until count =0.

Program:

CLD ; clear the Direction Flag

MOV AX, 0000H;

MOV DS, AX ; Initialize DS

MOV ES, AX ; Initialize ES

MOV SI, 5000H ; Set source pointer

MOV DI, 6000H ; Set destination pointer

MOV CX, 0003H ; Set length of the string

REPE ; cx=cx-1, if ZF=1 repeat the same maximum CX time

CMPSB ; Compare with source to destination byte set flags according

the result, if DF=0 increment the SI and DI

INT 3

Result: After the execution of the program, if both the strings are completely equal, i.e

CX becomes zero, the ZF is SET, and otherwise ZF is RESET.

D15 D0

Over flow Zero

Direction Sign Parity

Interrupt

Trap Auxiliary carry Carry

Questions:

1.

Write an ALP to search a byte in a given string using SCANSB instruction.2. What are REP, REPE, REPNE prefix instructions? What is its use?


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14.AIM: WRITE AN ASSEMBLY LANGUAGE PROGRAM FOR PACKED BCD

TO UNPACKED BCD


Procedure / Algorithm:

1. Set data segment and source pointer

2. Get number into two registers

3. Mask off upper nibble.

4. Increment pointer.

5. Place register value in the memory.

6. Get lower nibble into memory.

7. Mask off lower nibble

8. Rotate four times right.

9. Increment pointer.10. Store the result in memory.

Program:

MOV AX, 0000H Data:

MOV DS, AX; To initialize data segment 0000:5000: 43

MOV SI, 5000; Set pointer Result:

MOV AL, [SI];Get number into AL 0000:5001: 03

MOV BL, AL; as well as in to BL 0000:5002: 04

AND AL, 0FH; Mask upper nibble

INC SI; Increment pointer

MOV [SI], AL; Place least significant digit in memory

AND BL, 0F0H;Mask lower nibble

MOV CL, 04;Set count for number of rotations

ROR BL, CL; Rotate contents of BL

INC SI; Increment pointer

MOV [SI], BL; Place most significant digit in memory location

INT 3; Break Point

Questions:

1. What is the difference between SHR & SAR?

2. Write an ALP for combining least significant bits of memory locations 6000H &

6100H. Store the result in 6002H.

3.

Write an ALP to clear the memory locations 6010H & 601FH.


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15.AIM: WRITE AN ALP TO FIND THE NUMBER OF LOGIC 1S IN A

BYTE.

Apparatus: 8086 Kit, Power Supply

Procedure:

1. Set count value to 08.

2.

Set one of the registers to 00 for storing number of 1s.

3. Load number into one of the registers.

4. Rotate left the contents of the register.

5. Check the carry flag.

6. If the carry flag is set then increment register otherwise go to step 3.

7. Decrement count register.

8.

If count = 0 then stop the program, otherwise go to step 4.

9. Stop the program.

Program: Data:

MOV AX,0000 Input

MOV DS,AX 0000:4200 : 57

MOV SI,4200H

MOV CL,08 Result

MOV BL,00 0000:4201 : 05

MOV AL,[SI]

L2:ROL AL,01

JNC L1

INC BL

L1:LOOP L2

INC SI

MOV [SI],BL

INT 3

Questions:

1. Modify the above program to find the number of zeros in a given byte.

2. Write the above program using loop instruction.

3.

Explain NOT & NEG instructions in 8086.


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16.AIM: TO WRITE AN ALP FOR THE ADDITION OF A SERIES OF

BYTES FROM A GIVEN ARRAY. THE SERIES CONTAINS 10 BYTES.

Apparatus: 8086 Kit, Power supply


1.

Set the data segment and source pointer.

2. Set sum to zero.

3. Set counter

4.

Add the contents specified by the pointer to the variable sum and store the sum in

the same variable.

5.

Increment pointer and decrement counter.

6.

If count is zero then total = sum, otherwise go to step 4.7. Stop the program.

PROGRAM: DATA

MOV AX, 0000H; Initialize data segment 0000:6000 01

MOV DS, AX; 0000:6001. 01

MOV CX, 0AH; set counter 0000:6002. 01

XOR AX, AX; set sum = 0 0000:6003. 01

MOV SI, 6000H; set pointer 0000:6004.

01AGAIN: MOV BL, [SI]; Take the first number 0000:6005. 01

ADD AL, BL 0000:6006. 01

ADC AH, 00H; 0000:6007. 01

INC SI; Increment pointer 0000:6008. FF

DEC CX; Decrement count 0000:6009. FF

JNZ AGAIN; count is not zero do the same RESULT

MOV [SI], AL; Store the sum 0000:600A 06INC SI 0000:600B 02

MOV [SI], AH; Store the carry

INT 03H; Break point

Questions:

1. Explain function of DAA instruction.

2. Modify above program for decimal addition.

3. Write an ALP for decimal addition & decimal subtraction.


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17.AIM: WRITE A PROGRAM TO FIND THE LARGEST NUMBER FROM A

GIVEN ARRAY OF NUMBERS STORED IN THE LOCATIONS STARTING

FROM A KNOWN ADDRESS.

Apparatus: 8086 Kit, Power supply


1. Set data segment and pointer

2. set max=0 and counter = desired count value

3. is max > [pointer], if yes decrement count. If no increment pointer

4. if count = 0result = max, if no go to step 3

5.

result = max

6. Stop the program.

Program: Data:

MOV AX, 0000H 0000:5000 09

MOV DS, AX 0000:5001 03

MOV SI, 5000H 0000:5002 54

MOV CX, 0005H 0000:5003 65

L2: CMP AL, [SI] 0000:5004 02

JNC L1 Result

MOV AL, [SI] 0000:5005 65

L1: INC SI

LOOP L2

MOV [SI], AL

INT 03H


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18.. AIM: WRITE A PROGRAM FOR CODE CONVERSION

(a)BCD to Binary

(b)Hexadecimal to ASCII


Algorithm/ Procedure: BCD to Binary Program:

1.

Set data segment and pointer MOV AX, 0000H

2.

Take number to AL. MOV DS, AX

3. Keep a copy in AH Register. MOV SI, 5000H

4. Mask off upper nibble. MOV AL, [SI];

5. Keep least significant nibble in CH Register. MOV AH,AL

6.

Get byte from CL to AL. AND AL, 0FH; Mask off upper nibble

7.

Mask off lower nibble. MOV BL, AL;

8. Rotate AL content 4 times right. MOV AL, AH;

9. Multiply AL content with 0Ah. AND AL, 0F0H; Mask off lower nibble

10.Add CH content to AL. MOV CL, 04H;

11.Store the result. ROR AL, CL; Rotate AL content 4 times

12.

Stop the program. MOV CH, 0AH;MUL CH;

ADD AL,BL;

INC SI

MOV [SI],AL; Store the result

INT 3; Break point


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This program converts hexadecimal no: 00 to FF only

Hexadecimal to ASCII conversion MOV AX,0000

1.

Set data segment and pointer MOV DS,AX

2.

Take number to AL. MOV SI,4100H3. Keep a copy in AH Register. MOV AL,[4100H]

4. Mask off upper nibble. MOV AH,AL

5. Compare number with 09 AND AL,0FH

6. If AL >09 ; AL = AL+07 or AL=AL CALL CONVT

7.

AL= AL+30 = ASCII value of lower nibble XCHG AL,AH

8. Take number and mask lower nibble MOV CL,04

9. Similarly convert the as above AND AL,0F0H

10.Store the value in memory ROR AL,CL

Data CALL CONVT

0000:4100 FF h MOV [4101H],AX

0000:4101 46 INT 30000:4102 46 CONVT: CMP AL,09

JNC AGAIN

Data ADD AL,30H

0000:4100 01 JMP GO

0000:4101 30 AGAIN: ADD AL,07

0000:4102 31 ADD AL,30H

GO:RET

Questions: Modify the above programs for converting ASCII to Hexadecimal code.


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19. AIM: WRITE AN ALP TO SEARCH A GIVEN ITEM FROM THE LIST

USING LINEAR SEARCH TECHNIQUE.

Apparatus Requires: 8086 kit, Power Supply

Program: Data:

MOV AX, 0000H; 0000:5000 86

MOV DS, AX; Initialize Data Segment Register 0000:5001 93

MOV CX, 0007H; Count of number of elements 0000:5002 34

MOV SI, 5000H; Set pointer 0000:5003 45

MOV DL, 01; loc=1 0000:5004 XX

MOV AL,XX ; Number to be searched 0000:5005 87

BACK: CMP AL, [SI]; 0000:5006 64

JZ AHEAD; If number found, stop

INC SI;

INC DL;

LOOP BACK; Result:

AHEAD: MOV [5020], DL; Result location 0000:5020 05 Location

INT 03h;

Questions:

1. Write an ALP to search a given item from the list using Binary Search technique.

2.

Write an ALP for searching using string instructions.


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20.WRITE AN ALP TO ADD TWO MULTI-BYTE NUMBERS AND STORE

THE RESULT AS A THIRD NUMBER. (The numbers are stored in the form

of byte list as the lowest byte is fi rst.)

Apparatus: 8086 trainer kit and power supply.

Algorithm:

Initially set segment and two index register to two given multi-byte number, base

pointer to result, accumulator, carry flag and count register. Copy the first byte from

source index to source register, add with carry of source register and first byte from

destination index. Copy the result from source register to base pointer. Every time

increment the base pointer for storing the result, loop continue till count is zero.

Finally check carry flag and store in base pointer location.

Program:

MOV AX, 0000 Data:

MOV DS, AX; set data segment to 0000 4100 A9 87 65 43 21 Ist No.

MOV CX, 05; CX is counter 4105- FE DC BA 98 76 IIndNo.

MOV SI, 4100H; first multi byte number 410A-01 A8 64 1F DB 97 Result

MOV DI, 4105H; second multi byte number

MOV BX, 410Ah; result multi byte number

XOR AX, AX; clear sum and carry flag

CLD index register auto increment

NEXT: LODSB; load AL from SI

ADC AL, [DI]; add with carry = AL + [DI]

MOV BYTE PTR [BX], AL; sum copy to [BX]

INC BX

INC DI

LOOP NEXT; loop continues up to CX is zero

JNC AHEAD; if carry flag is set after adding

MOV BYTE PTR [BX], 01; the end bytes, copied in

[BX]

AHEAD: INT 03h; halt.


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21 SORTING GIVEN HEXADECIMAL NUMBERS

Apparatus: 8086 trainer kit and power supply.

INPUT 0D 0C 0B 0A

Sorting STEP-1 0C 0B 0A 0D

STEP-2 0B 0A 0C 0D

STEP-3 0A 0B 0C 0D

Programe

MOV AX,0000 INPUT

MOV DS,AX 4200 04

MOV SI,4200H Count n 4201 0D

MOV BL,[SI] 4202 0A

DEC BL Outer n = n-1 4203 0C

OUTL00P: MOV CL,BL Inner n = n-1 4204 0B

MOV SI,4201H

INL00P: MOV AL,[SI]

INC SI

CMP AL,[SI] sort

JCPROCEED OUTPUT

XCHG [SI],AL 4201 0A

MOV -1[SI],AL 4202 0B

PROCEED: LOOP INL00P 4203 0C

DEC BL 4204 0D

JNZ OUTL00P

INT 3


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22.Aim: To interface a PROGRAMMABLE TIMER 8253to 8086 and generate a

SQUARE WAVEFORMof period 1ms.

(Mode 3) f = 1KHz

0000:4000 BA CF FF MOV DX,0FFCF Control word

0000:4003 B0 36 MOV AL,36h 0 0 1 1 0 1 1 0 = 36

0000:4005 EE OUT DX,AL

0000:4006 BA C9 FF MOV DX,0FFC9 Counter 0

0000:4009 B0 88 MOV AL,88h 16 bit count

0000:400B EE OUT DX,AL

0000:400C B0 13 MOV AL,13h Mode 3

0000:400E EE OUT DX,AL Binary count0000:400F CC INT 3

8086 Clock f = 5 M Hz O/P

Desired square wave f = 1K Hz

Count = 5 M Hz / 1K Hz 1 ms

= 5000 D = 1388 H

RATE GENERATOR(Mode 2)

0000:4000 BA CF FF MOV DX,0FFCF

0000:4003 B0 14 MOV AL,14


0000:4006 BA C9 FF MOV DX,0FFC9

0000:4009 B0 05 MOV AL,05


0000:400C CC INT 3

Control word 0 0 0 1 0 1 0 0

Counter 0

byte count

Mode 2 1- clock period

Binary count


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SDA -86 - MEL

8253ADDRESSES

Timer /Counter 0 : FFC9

Timer /Counter 1 : FFCB

Timer /Counter 2 : FFCD

Control Word Reg. : FFCF

CLK

8086A2

A1

I/O RD

I/O WR

D0 to D7 Vcc

CLK0

OUT0A1A2

RD GATE0WR

CS

To CRO

+5 V

8253


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23.Aim: Read the analog voltage and display in digital using A/D CONVERTOR

Apparatus: 8086 kit, power supply , ADC module, 26-pin FRC Connector.

MOV AL,90H set 8255 Port A I/P,

MOV DX,0FFC6 port B&C is O/P mode

OUT DX,AL

MOV AL,07 channel Selection

MOV DX,0FFC4H Channel 07

OUT DX,AL

MOV DX,0FFC6 BSR Mode

MOV AL,0EH Start of Conversion(SOC) = 0

OUT DX,AL

MOV AL,0CH Output Enable(0E)= 0

OUT DX,AL

MOV AL,0FH SOC =1

OUT DX,AL

MOV CX,3FFFH delay

L1: LOOP L1

MOV AL,0EH SOC = 0

OUT DX,AL

MOV DX,0FFFC0

WAT:IN AL,DX to Check for End of Conversion

AND AL,80 (EOC)i.e. conversion is

ROL AL,1 completed or not

JNC WAT

0 X X X 1 1 0 0

0 X X X 1 1 1 1

0

X

X

X

1

1

1

1

1 0 0 1 0 0 0 0


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A2A1

8086

I/ORD

I/OWR

D0 to D7

PC2

PC1

PC0

A1 PA7A0

PC7RST

RD PC6

WR

8255

Vref + Vref -

EOCALE

0809SOC

PA0-PA7 to O0O7CH 00

CH 07

00 to 5V DC

8086 Interfacing with ADC0809 through PPI8255

Questions: Kit No: NIFC 07A, Channel No.081.

Explain the different modes of operation of 8255.

2. Explain the Control Word format of 8255 in I/O and BSR mode.

3. What is the function of ADC 804 IC?

MOV DX,0FFC6H to enable the output to

MOV AL,0D be read by theprocessor

OUT DX,AL 0E = 1

MOV DX,0FFC0

IN AL,DX Read the digital

INT 3 data in AL register

8255 PORT ADDRESSES 5V = FF h (255)

PORT A FFC0H 1V = ?PORT B FFC2H 1V = 255/5V =(51)d=(33)h

PORT C FFC4H 2V = 66h, 3V = 99h

CWR FFC6H 4V = CCh, 5V = FFh

0 X X X 1 1 0 1


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FF

00 f = 1/T

T

24.Aim: Write an ALP to generate SQUARE WAVE USING DACmodule.

Apparatus: 8086 kit, power supply unit, DAC module, 26-pin FRC Connector, CRO.

Description:

1. Use Port A of 8255 in output mode and write control word of 8255 into control

word register in simple I/O mode for data transfer.

2. Send high value to Accumulator.

3.

Send Accumulator value to port A.

4. Call delay sub-routine.

5.

Send low value to Accumulator.

6. Send accumulator value to the port.

7. Call delay sub-routine.

8. Jump to Step 2, for continues square wave generation.

( 8-Bit Digital to Analog Converter) through 8255 PPI

0000:4005 B0 50 MOV AL,800000:4007 BA C6 FF MOV DX,0FFC6

0000:400A EE OUT DX

0000:400B BA C0 FF MOV DX,0FFC0

0000:400E B0 00 Rpt MOV AL,00

0000:4010 EE OUT DX

0000:4011 E8 CALL DELAY

0000:4014 B0 FF MOV AL,FF

0000:4016 EE OUT DX


0000:4020 EB JMP Rpt

Delay program0000:4022 B9 FF 00 MOV CX,00FF

0000:4025 E2 here LOOP here

0000:4027 C3 RET

DELAY CALCULATION:

CLOCK CYCLES

MOV CX,COUNT 4

HERE: LOOP HERE 17/5RET 8

DELAY: (N0. of clock cycles) X one clock period

1 K Hz = (4 + (count X 17)12 + 8) X

Count = 1 K Hz/(17 X 0.2s )

1 0 0 0 0 0 0 0

I/O Port A Port A Port B,C

Mode Mode 0 O/P Output Ports


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Interfacing Diagram:

Triangular wave generation0000:4005 B0 80 MOV AL,80

0000:4007 BA C6 FF MOV DX,FFC60000:400A EE OUT DX

0000:400B BA C0 FF MOV DX,FFC0

0000:400E B0 00 Repeat: MOV AL,00

0000:4010 EE Rise: OUT DX

0000:4011 FE C0 INC AL

0000:4013 3C FF CMP AL,FF

0000:4015 75 JNZ Rise

0000:4017 FE C8 Decay: DEC AL

0000:4019 EE OUT DX

0000:401A 3C 00 CMP AL,000000:401C 75 JNZ Decay

0000:401E EB JMP Repeat

IORD RDIOWR WR

A2 A1A1 A0

D0 to D7

8086

8255

cs

PORT A

B0 to B7 2.5K + 5 V

5K

V0741 O/P to CRO

+5 Vor 12 V

DAC0

800

13

1

4

FF

00

T = Time period f = 1/T


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Sin wave Generation

0000: 4000 MOV AL,80

0000: 4002 BA C6 FF MOV DX,0FFC6

0000: 4004 EE OUT DX,AL


0000:4008 FC CLD0000:4009 B1 18 cont MOV CL,18

0000:400B BE 50 41 MOV SI,4150

0000:400E 8A 04 next LODSB


0000:4011 E2 F8 LOOP next

0000:4013 EB F0 JMP cont

Lookup table

S.No Address Decimal Hex

01. 4150 128 8002. 4151 160 AD

03. 4152 192 C0

04. 4153 217 D9

05. 4154 238 EE

06. 4155 251 FB

07. 4156 255 FF

08. 4157 251 FB

09. 4158 238 EE

0A. 4159 217 D9

0B. 415A 192 C0

0C. 415B 160 A0

0D. 415C 128 80

0E. 415D 96 60

0F. 415E 64 40

10. 415F 37 25

11. 4160 17 11

12. 4161 4 04

13. 4162 0 00

14. 4163 4 04

15. 4164 17 11

16. 4165 37 2517. 4166 64 40

18. 4167 96 60

19. 4168 128 80


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38

25. Aim: To INTERFACE 7-SEGMNET LED DISPLAYto 8086.


Interfacing Diagram:

Address CA CC BCD Display

4100 C0 3F 00 0

4101 F9 06 01 1

4102 A4 5B 02 2

4103 B0 4F 03 3

4104 99 66 04 4

4105 92 6D 05 5

4106 82 7D 06 6

MOV AL, 80H

MOV DX, FFC6

OUT DX, AL

MOV DX, FFC0

CLD

REPEAT: MOV SI, 4100

MOV BL, 07H

NEXT: LODSB

OUT DX, AL

CALL DELAY

DEC BL

JNZ NEXT

JMP REPEAT

DELAY: MOV CX,FFFF

HERE: LOOP HERE

RET

FRC cable

7-seg display

CC= Common Cathode

PA 0=aPA 5=f PA 1=b

PA 6=g

PA 4=e PA 2=c

PA 3=d PA 7=h

FRC cable 7447 Pin Diagram

CA = Common Anode

CC= Common Cathod

BCD to 7 -Segment

82558086 PA7- PA0D - D

8086 8255

PA0 -PA3

D - D

7447BCD to 7-segment


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26.Aim: To INTERFACE STEPPER MOTORto 8086.

Apparatus: 8086 kit, power supply, Stepper motor, stepper motor interface board.

Theory:

A stepper motor is a device used to obtain an accurate position control of rotating

shafts. A stepper motor employs rotation of its shaft in terms of steps rather than

continuous rotation as in case of AC or Dc motors. To rotate the shaft of the stepper

motor, a sequence of pulses is needed to be applied to the windings of the stepper

motor, in proper sequence. A popular scheme for rotation of a stepper motor shaft

applies pulses to two successive windings at a time but these are shifted only by one

position at a time. This scheme for rotation of stepper motor is shown in the following

figure and table.

ROTATING STEPPER MOTOR SHAFT

MOTION STEP Code Winding 1 Winding 2 Winding 3 Winding 4

Clock wise

step angle 1.80

(full step)

1 03 0 0 1 1

2 06 0 1 1 0

3 0C 1 1 0 0

4 09 1 0 0 1

Anti-clock

wise

step angle 1.80

(full step)

1 03 0 0 1 1

2 09 1 0 0 1

3 0C 1 1 0 0

4 06 0 1 1 0

Step angle

0.90

(half step)

1 0C 1 1 0 0

2 09 1 0 0 1

3 03 0 0 1 1

4 06 0 1 1 0

5 08 1 0 0 0

6 01 0 0 0 1

7 02 0 0 1 0

8 04 0 1 0 0


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The stepper motor connections for the four windings can be shown as follows:

Program:

MOV DX, 0FFC6H

MOV AL, 80H

OUT DX, ALMOV DX, 0FFC0H

CONT: MOV AL, 33H; 33, 99,CC,66

NEXT: OUT DX, AL

CALL DELAY

ROL AL, 1 (for anti-clock rotation ROR AL, 1)

LOOP NEXT

JMP CONT

DELAY MOV CX, 0FFH (for decrease the speed 0FF to 0FFFH)

HERE: LOOP HERE

RET

KIT MOTORPA0 =13 ORANGE GREEN

PA1 =10 BLUE RED

PA2 =07 YELLOW YELLOW

PA3 =04 BROWN ORANGE

PA0 PA1 PA2 PA3


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27. Aim: To interface a 3*8 KEYBOARDwith 8086 using 8255 and write an ALP

for detecting a key closure and return the key code in AL. The debouncing period of

a key is 10ms. Use software key debouncing technique. DEBOUNCE is an available

10ms delay routine.

Apparatus: 8086 kit, power supply, key-board module.

Solution: Port C is used as output port for selecting row of keys while port A is used as

input port for sensing a closed key. Thus the key boards lines are selected one by one

through Port C and Port A lines are hold continuously till a key closure is sensed. Then

routine DEBOUNCE is called for key debouncing. The Key code is decided dependingupon the selected row and a low sensed column. The hardware circuit diagram is shown

in the figure.

Figure: Interfacing 3*8 keyboard


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The port A and port C lower are used. The addresses of port A and C will be

respectively FFC0H and FFC4h while the address of CWR is FFC6h. The flowchart

of the entire program is given in the following figure.


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Program:

MOV AL,90H Port AI/P, Port B&CO/P

MOV DX,0FFC6H Control Word Register

OUT DX,AL

MOV BL,00 initialize key code counter

MOV AL,0FFH making all rows highMOV DX,0FFC4H Port C

OUT DX,AL

MOV DX,0FFC0H Port A

WAIT: IN AL,DX Reading Columns Port A

CMP AL,00

JZ WAIT

CALL DEBOUNCE

MOV BH,03 To high no of rows

MOV AL,80H

NXROW: ROL AL,1H Row 1=0MOV CH,AL

MOV DX,0FFC4H Checking row 0

OUT DX,AL

MOV DX,0FFC0H

IN AL,DX Reading columns

MOV CL,08H

NEXT: ROR AL,1H Checking for key press

JC KEYCOD

INC BL

DEC CLChecking for key press in the

columnJNZ NEXT

MOV AL,CH

DEC BH Next row

JNZ NXROW

JMP WAT

KEYCOD: MOV AL,BL

INT 3

DEBOUNCE: MOV CX,094C H 10ms 0F 5Mhz frequence

L1: NOP

DEC CXJNZ L1

RET

Result:In AL register


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28. Aim: Interface single LED blinking with 8255(BSR Mode)

Apparatus: 8086 kit, power supply, key-board module, 26 pin connector with FRC

& LED.

Single LED interfacing

0000: 4000 B0 80 MOV AL,80 All ports are O/P

CWR0000: 4002 BA C6 FF MOV DX,0FFC6

0000: 4004 EE OUT DX,AL

0000: 4005 B0 0F cont MOV AL,0F Port C

7thbit is high0000:4008 EE OUT DX


0000:400C B0 0F MOV AL,0E Port C

7thbit is low0000:400E EE OUT DX

0000:400F E8 CALL DELAY

JMP cont

DELAY

0000: 4010 B9 FF FF MOV CX,0FFFF

0000: 4013 F2 FE here LOOP here

0000: 4015 C3 RET

AD0- AD7

8086

IORD

IOWR

A2

A1

D0- D7 PC7

8255

RD

WR

A1

A0 CS


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29.Aim: Digital I/O Interfacing.


& Digital I/O module.

Circuit diagram:

Program

0000: 4000 B0 82 MOV AL,82 PA - O/P &PB-I/P


CWR address0000: 4004 EE OUT DX,AL

0000: 4005 BA C2 FF cont MOV DX,0FFC2 PB address

0000:4008 EC IN AL,DX

0000:4009 BA C0 FF MOV DX,0FFC0 PA address


0000:400E EB F0 JMP cont

AD0- AD7

8086

IORD

IOWR

A2

A1

D0- D7 PB0-7

8255RD

WR

A1 PA0-7

A0 CS

Digital I/O module

+12V


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30.

Interfacing 8 LED with 8051.


& LED module.

Circuit diagram:

Program:8000 74 55 MOV A,#55 Load the into register (0101 0101)

8002 F5 90 CONT MOV 90, A Load data from Register to P1

8004 F4 CPL A Compliment the register A contents

8005 11 09 ACALL DELAY Call delay subroutine

8007 01 02 AJMP CONT Jump unconditionally

8009 7F FF Delay MOV R7, #0FF Load the data to register R7

800B 7E FF L2 MOV R6, #FF Load the data to register R6

800D DE FE L1 DJNZ R6, L1 Decrement Register by one & jump

on no zero condition to address800F DF FA DJNZ R7, L28010 22 RET Resume to called program

P 1.0-7

8051

RS

RS

LED

LED


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31.

Interface stepper motor with 8051.

Apparatus: 8051 kit, power supply, key-board module, 10 pin connector with

FRC, ULN2003 & stepper motor.

Circuitdiagram:

Program:

8000 74 55 MOV A,#33 Load the into register (0011 0011)

8002 F5 90 CONT MOV 90, A Load data from Register to P1

8004 F4 RR A Rotate register contents 66,99,CC,33

8005 11 09 ACALL DELAY Call delay subroutine

8007 01 02 AJMP CONT Jump unconditionally

8009 7F FF Delay MOV R7, #0FF Load the data to register R7

800B 7E FF L2 MOV R6, #0FF Load the data to register R6

800D DE FE L1 DJNZ R6, L1 Decrement Register by one & jump

on no zero condition to address800F DF FA DJNZ R7, L2

8010 22 RET Resume to called program

P 1.0-3

8051+12 V

ULN2003

Stepper

Motor


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32.Interfacing seven segment LED with 8051

Apparatus: 8051 kit, power supply, key-board module, 10 pin connector with

FRC, & 7-seg LED.

P0-7

8051


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STUDY OF MASM

1. To study the various directives of Macro Assembler.

2.

Switch on the computer, and go to DOS window displaying the C prompti.e. cd\ enter C :\>

3.

Open MASM directory C:\CD MASM enter

4. C:\>MASM>editfile_name.asm

5. Then the program is to be loaded in to the file. After entering, the program is to be

saved using same method.

6. Alt + f are to be pressed. The monitor lists the commands. Enter Exit command

and press Enter key.

7. Monitor displays C prompt C:\MASM.

8. Enter

C:\>MASM>masmfile_name.asm and then press enter key. Press enter 4 times, it

shows 0 warning errors

0 severe errors

C:\masm>

9.

Enter the following command

C:\>MASM>linkfile_name.obj and then press enter key 4 times, it shows

C:\masm>

10.Then

C :\> MASM> debugfile_name.exe and then press enter key.

11. Monitor then displays a hyphen -

Debug the program line by link t (Trap) and press enter key.

..Program executes line by line.

12.

R is to be pressed to get the register contents.13.G is to be pressed for entire program execution.

14.D is to be pressed to get the data in memory locations.

Ex: d segment address : offset address (d 1039:0000 enter)

15.u is to be pressed to get the entire program with addresses, op -codes &

mnemonics

16.Q is to be pressed to return to DOS prompt.


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MASM (Macro Assembler) Programs

25. Aim: To display a message The study of microprocessors is interesting. on the CRT

screen.

Apparatus: PC, MASM (Macro Assembler) Software Tool

Program:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

MESSAGE DB 0DH, 0AH, STUDY OF MICROPROCESSORS IS INTERESTING.,

0DH, 0AH, $

DATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DA, AX

MOV AH, 09H

MOV DX, OFFSET MESSAGE

MOV AH, 4CH

INT 3

CODE ENDS

END START


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1. Write instructions to find the product of two BCD numbers.

ASSUME CS:CODE, DS:DATA

DATA SEGMENT

NUM1 DB 45H;

NUM2 DB 22H;

PRO DW ??

DATA ENDS

CODE SEGMENT

START:

MOV AX, DATA;

MOV DS, AX; Initialize Data Segment

MOV AL, NUM1; First BCD number to AL

CALL BCDTOBIN; Call the subroutine to convert BCD to Binary

MOV CL, AL; save in CL

MOV AL, NUM2; second BCD number to AL

CALL BCDTOBIN; call the subroutine to convert BCD to binary

MUL CL; compute the product

MOV PRO, AX; Store the product

BCDTOBIN:

PUSHF; Save the flags

PUSH BX; save BX

PUSH CX; save CX

MOV BL, AL; copy the BCD number to BlAND BL, 0F; separate the BCD number into two nibbles

AND AL, 0F0H

MOV CL, 04

ROR AL, CL

MOV CH, 0A; convert the MSB digit of BCD number to binary

MUL CH

ADD AL, BL; add the LSB digit and hold binary value in AL

POP CX; restore CX

POP BX; restore BXPOPF; restore flags

RET; return to main program

CODE ENDS

END START


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2. Program to compare two strings.

MYDATA SEGMENT

STRING1 DB PASSWORD; first string

STRING2 DB PASSWORD; second string

LENGTH DW 0008H; the length of the string

RESULT DB 1 DUP(0); space to indicate result

MYDATA ENDS

MYCODE SEGMENT

ASSUME CS:MYCODE, DS:MYDATA

START: MOV AX, MYDATA; initialize DS and Es

MOV DA, AX

MOV ES, AX

MOV CX, LENGTH ; Load length of the string in CX

MOV SI, OFFSET STRING1; point SI to first string

MOV DI, OFFSET STRING2; point DI to second string

CLD; clear direction flag to auto increment SI and DI

REPE CMPSB; Repeat comparison until the end of string or until compared bytes are not

equal

JNZ L1;

MOV RESULT, 0FFH;Store FF if the two strings are equal

INT 03H;Terminate the program

L1: MOV RESULT, 00H; store 00 if the two strings are not equal

INT 03H; Terminate the program

MYCODE ENDS

END START


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Address InputsCA

DISPLAY

DATA4100 CC BCD

4101 C0 3F 01 0

4102 F9 06 02 1

4103 A4 5B 03 2

4104 99 66 04 3

4105 92 6D 05 4

4106 82 7D 06 5F8 07 07 6

microcontrollers lab manual

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