R05 Mp Lab Journal

MICROPROCESSORS LAB MANUAL ECE Dept.

CONTENTS

Page no.

1. List of Experiments (as per JNTU) -1, 2

2. Microprocessor 8086 Programming

2.1 Introduction to MASM and TASM - 5

2.2 Arithmetic operationsa) Arithmetic Operations on 8 bit data - 8

b) Arithmetic Operations on 16 bit data -

14

c) Multibyte addition and subtraction -

20

d) Signed Operations on 8 bit data -

26

e) ASCII Arithmetic Operations -

31

2.3 Logic Operationsa) BCD to ASCII Conversion - 37

b) ASCII to BCD Conversion - 41

c) No. of Positive no’s & Negative no’s. - 45

d) No. of Odd no’s & Even no’s. - 51

e) Packed BCD to Unpacked BCD Conversion - 57

2.4 String Operationsa) Transfer Block of Data - 62

b) Reversal of given String - 66

c) Sorting of ‘N’ numbers - 72

d) Length of the given String - 78

e) Comparison of two Strings - 82

2.5 DOS/BIOS Programming

a) Reading Keyboard buffered with echo - 88

ADITYA ENGINEERING COLLEGE 1


b) Reading Keyboard buffered with out echo - 91

c) Display String by using DOS Commands - 94

Page no.

3. Interfacing with 8086

3.1. Programmable Peripheral Interfacing (8255) - 97

a) Digital to Analog Converters - 110

3.2. Keyboard Display Interfacing (8279) - 130

ADDITIONAL PROGRAMS

Dual Slope Analog to Digital Converter - 143

Stepper Motor Interfacing Using 8255 PPI - 150


MICROPROCESSORS LAB MANUAL ECE Dept.LIST OF EXPERIMENTS

III Year B.Tech, ECE –II semester.II Year B.Tech, CSE & IT – II Semester

I MICROPROCESSOR 8086:

1. Introduction to MASM and TASM

2. Arithmetic Operations – Multi byte addition and subtraction, Multiplication and division-signed and unsigned Arithmetic operation, ASCII-arithmetic operation

3. Logic operations-Shift and rotate-Converting packed BCD to unpacked BCD, BCD to ASCII conversion.

4. By using string operation and Instruction prefix: Move Block, Reverse string, Sorting, Inserting. Deleting, length of the string, String comparison.

5. DOS/BIOS programming: Reading keyboard (with Buffered and without echo)-Display characters, Strings.

II. INTERFACING:

1. 8259-Interrupt controller : Generate an interrupt using 8259 timer.

2. 8279-Keyboard Display : Write a small program to display a string of Characters.

3. 8255-PPI : Write an ALP to generate sinusoidal wave Using PPI

4. 8251-USART : Write a program in ALP to establishcommunication between two processors

III. MICROCONTROLLER 8051:

1. Reading and Writing on a parallel port.

2. Timer in different modes.

3. Serial communication implementation.


MICROPROCESSORS LAB MANUAL ECE Dept.LIST OF EXPERIMENTS

IV YEAR B.TECH EEE – I SEMESTERI MICROPROCESSOR 8086:

1. Introduction to MASM and TASM

2. Arithmetic Operations – Multi byte addition and subtraction, Multiplication and division-signed and unsigned Arithmetic operation, ASCII-arithmetic operation

3. Logic operations-Shift and rotate-Converting packed BCD to unpacked BCD, BCD to ASCII conversion.

4. By using string operation and Instruction prefix: Move Block, Reverse string, Sorting, Inserting. Deleting, length of the string, String comparison.

5. DOS/BIOS programming: Reading keyboard (with Buffered and without echo)-Display characters, Strings.

II. INTERFACING:

1. 8259-Interrupt controller : Generate an interrupt using 8259 timer.

2. 8279-Keyboard Display : Write a small program to display a string of Characters.

3. 8255-PPI : Write an ALP to generate sinusoidal wave Using PPI

4. 8251-USART : Write a program in ALP to establishcommunication between two processors

III. MICROCONTROLLER 8051:

1. Reading and Writing on a parallel port.

2. Timer in different modes.

3. Serial communication implementation.

4. Understanding three memory areas of 00-FF(Programs using above areas).

5. Using external Interrupts.

6. Programs using special instructions like swap, bit/byte, set/reset etc.

7. Programs based on short, page, absolute addressing.



INTRODUCITION TO MASM/TASM

ASSEMBLY LANGUAGE PROGRAMMING USING MASM SOFTWARE:

This software used to write a program (8086, Pentium processors etc.)The programs are written using assembly language in editor then compile it. The complier converts assembly language statements into machine language statements/checks for errors. Then execute the compiled program. Programs for different processor instructions (Pentium, 8086) programming manner differ for each model.

There are different soft wares developed by different companies for assembly language programming are:

MASM - Microsoft Company. TASM - Bore Land Company.

MERIT OF MASM:

1. produces binary code2. Referring data items by their names rather than by their address.

HOW TO ENTER INTO MASM EDITOR:

Click “Start” on the desktop.

Then select Run

Then it Shows inbox

Then type Command (CMD) which enters You into DOS prompt

Path setting

Suppose it display path as C:\ DOCUME-\ADMIN>

Then type CD\

i.e.; C:\DOCUME\\ADMIN>CD\

Then the path is C :\>

Then type CD MASM

Then the path is C: MASM>

Then type edit i.e.; C: MASM>edit


MICROPROCESSORS LAB MANUAL ECE Dept.Then you enter into MASM text editor.

Then enter to file and select New.

And name it and then write the ALP (Assembly Language Program) in this editor.After that save it as filename’sThen exit from the editor and go to prompt.Then type MASM filename.ASMI.e. C: MASM>MASM filename.ASM or C: MASM filename.ASM, , ;Then link this file using C: MASM>LINK filename.OBJ or C: MASM>LINK filename.OBJ , , ;i.e link the program in assembly with DOS then to debug to create exe fileC:MASM>debug filename. EXEThen it display “--” on the screenAfter that type ‘R’ displays the registers contents steps and starting step of the program.

‘T’ Tracing at contents of program step by step.Suppose you need to go for break point debugging. Then type that instruction no where you need to check your register. For example T10 it will display the contents of register after executing 10 instructions.

DEBUG:This command utility enables to write and modify simple assembly language programs in

an easy fashion. It provides away to run and test any program in a controlled environment.We can change any part of the program and immediately execute the program with an

having to resemble it. We can also run machine language(Object files) directly by using DEBUG

DEBUG COMMANDS:

ASSEMBLE A [address] ; Assembly the instructions at a particular address

COMPARE C range address ; Compare two memory ranges

DUMP D [range] ; Display contents of memory

ENTER E address [list] ; Enter new or modifies memory contents beginning at specific

Location

FILL F range list ; Fill in a range of memory

GO G [=address] [addresses] ; Execute a program in memory

HEX H value1 value2 ; Add and subtract two Hex values

INPUT I port

LOAD L [address] [drive] [first sector] [number]


MICROPROCESSORS LAB MANUAL ECE Dept.MOVE M range address

NAME N [pathname] [arg list]

OUTPUT O port byte

PROCEED P [=address] [number]

QUIT Q

REGISTER R [register]

SEARCH S range list

TRACE T [=address] [value]

UNASSEMBLE U [range]

WRITE W [address] [drive] [first sector] [number]

ALLOCATE expanded memory XA [#pages]

DEALLOCATE expanded memory XD [handle]

MAP expanded memory pages XM [Lpage] [Ppage] [handle]

DISPLAY expanded memory status XS



2.2 a) ARITHMETIC OPERATIONS ON 8BIT DATA

ABSTRACT: Assembly language program to perform all arithmetic operations on 8bit data

PORTS USED: None

REGISTERS USED: AX, BL

ALGORITHM:

Step1: Start

Step2: Initialize data segment

Step3: Copy the contents from memory location [0000] to AL

Step4: Copy the contents from memory location [0001] to BL

Step 5: Perform addition

Step6: Move the result to the memory location [0002]


Step8: Perform subtraction



Step11: Perform multiplication



Step14: Perform division


Step16: stop.



ADITYA ENGINEERING COLLEGE

Start

Copy the contents from memory location [0000] to AL

Perform addition

Move the result to the memory location [0002]


Perform subtraction



Perform multiplication

1

Initialize data segment

FLOW CHART

9



1



Perform division


Stop

10


PROGRAM:ASSUME CS: CODE, DS: DATA

DATA SEGMENTN1 EQU 04HN2 EQU 06HRESULT DB 04H DUP (00)DATA ENDSCODE SEGMENTSTART:

MOV AX , DATA

MOV DS , AX

MOV AL , N1

MOV BL, N2

ADD AL, BL

MOV [RESULT], AL

MOV AL, N1

SUB AL, BL

MOV [RESULT+1], AL

MOV AL, N1

MUL BL

MOV [RESULT+2], AL

MOV [RESULT+3], AH

MOV AL, N1

MOV AH, 00H

DIV BL

MOV [RESULT+4], AL

MOV [RESULT+5], AH

MOV AH, 4CHINT 21H

CODE ENDSEND START



code table:Physical address Label Hex code Mnemonics Comments Segment address

Effective address

Opcode Operand

OD64 0000

0003

0005

0008

000C

000E

0011

0014

0016

0019

001C

001E

0021

0025

0028

002A

002C

002F

0033

0035

B8630D

8ED8

A00000

8A1E0100

02C3

A20200

A00000

2AC3

A20300

A00000

F6E3

A20400

88260500

A00000

B400

F6F3

A20600

88260700

B44C

CD21

MOV AX,0D63

MOV DS,AX

MOV AL,[0000]

MOV BL,[0001]

ADD AL,BL

MOV [0002],AL

MOV AL,[0000]

SUB AL,BL

MOV [0003],AL

MOV AL,[0000]

MUL BL

MOV [0004],AL

MOV [0005],AH

MOV AL,[0000]

MOV AH,00

DIV BL

MOV [0006],AL

MOV [0007],AH

MOV AH,4C

INT 21

;Initialize data segment

;Move the contents to AL

;Move the contents to BL

;Add BL contents with AL

;Move the contents to memory;Move the contents to AL

;Subtract BL contents with AL;Move the contents to memory;Move the contents to AL

;Multiply the contents of BLwithAL;Move the contents from AH

;Move the contents to AH

;Move the contents to AL

;Move 00 into AH register

;Division the contents of BL with AL;Move the contents to AL

;Move the contents from AH

;Terminate the Program

;Stop

RESULT: D 0000 04 06 0A FE 18 00 00 04



VIVA-VOCE QUESTIONS

1. What is meant by microprocessor?

2. What was the first company to bring out the microprocessor?

3. What was the semiconductor technology used for fabrication of the 8086P?

4. What are the main blocks of the microprocessor?

5. What is the word size of the 8086 P?



2.2(b) ARITHMETIC OPERATIONS ON 16BIT DATA

ABSTRACT: Assembly language program to perform all arithmetic operations on 16bit data

PORTS USED: None

REGISTERS USED: AX, BX, SI

ALGORITHM:

Step1: Start


Step3: Load SI with memory location 5000

Step4: Move the contents from memory location [0000] to AX

Step5: Move the contents from memory location [0001] to BX

Step6: Perform addition

Step7: Move the result to the memory location specified

Step8: Copy the contents from the memory location [0000] to AX


Step10: Move the result to the memory location [SI+02]



Step13: Move the result to the memory location [SI+04] &[SI+06]



Step16: Move the result to the memory location [SI+08]&[SI+0A]

Step17: Stop.


MICROPROCESSORS LAB MANUAL ECE Dept. :


Start

Move the contents from memory location [0000] to AX

Move the contents from memory location [0001] to BX

Perform addition

Move the result to the memory location specified

Copy the contents from the memory location [0000] to AX]

Perform subtraction

Move the result to the memory location [SI+02]

1


Load SI with memory location 5000

FLOW CHART

15



1

Copy the contents from the memory location [0000] to AX


Move the result to the memory location [SI+04] &[SI+06]

Copy the contents from the memory location [0000] to AX

Stop

Perform division

Move the result to the memory location [SI+08] &[SI+0A]

16


PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENTN1 EQU 8888HN2 EQU 4444HDATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV SI, 5000H

MOV AX, N1

MOV BX, N2

ADD AX, BX

MOV [SI], AX

MOV AX, N1

SUB AX, BX

MOV [SI+2], AX

MOV AX, N1

MUL BX

MOV [SI+4], AX

MOV [SI+6], DX

MOV AX, N1

MOV DX, 0000

DIV BX

MOV [SI+8], AX

MOV [SI+0AH], DX

MOV AH, 4CH

INT 21H

CODE ENDS

END START

CODE TABLE:


MICROPROCESSORS LAB MANUAL ECE Dept.Physical address Label Hex code Mnemonic Operand CommentsSegment address

Effective address

0D64 0000

0003

0005

0008

000B

000F

0011

0013

0016

0018

001B

001E

0020

0023

0026

0029

002C

002E

0031

0034

0036

B8 63 0D

8E D8

BE 00 50

A1 00 00

8B 1E 02 00

03 C3

89 04

A1 00 00

2B C3

89 44 02

A1 00 00

F7 E3

89 44 04

89 54 06

A1 00 00

BA 00 00

F7 F3

89 44 08

89 54 0A

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV SI,5000

MOV AX,[0000]

MOV BX,[0002]

ADD AX,BX

MOV [SI],AX

MOV AX,[0000]

SUB AX,BX

MOV [SI+02],AX

MOV AX,[0000]

MUL BX

MOV [SI+04]AX

MOV [SI+06],BX

MOV AX,[0000]

MOV DX,0000

DIV BX

MOV [SI+08],AX

MOV [SI+0AH],AX

MOV AH,4C

INT 21h


;Initialize SI with 5000

;copy the contents of 0000;Copy the contents of 0002;Add the contents of AX with BX;Copy the result in [SI]

;copy the contents of 0000;Subtract the contents of AX with BX;Move the result into [SI+2];Copy the contents of 0000;Multiply AX with BX

;Move the lower word of result into [SI+4];Move the higher word of result into [SI+6];Copy the contents of 0000;Initialize DX with 0000

;Perform division with BX;Move the quotient into [SI+8];Move the remainder into [SI+0A];Terminate the program

RESULT: D 5000: CC CC 44 44 20 64 68 24 02 00 00 00


MICROPROCESSORS LAB MANUAL ECE Dept.VIVA-VOCE QUESTIONS

1. Which technique is used in Stacks?

2. Give the number of flags in 8086 P?

3. What are the different interrupts in 8086 P?

4. How many address lines are there in 8086 P?

5. What are the different addressing modes available in 8086 P?



2.2 (C) MULTIBYTE ADDITIONS AND SUBTRACTION

ABSTRACT: Assembly language program to perform multibyte addition and subtraction

PORT USED: None

REGISTERS USED: AL, BX, CX

ALGORITHM:

Step1: Start


Step3: Load the CX register with count

Step4: Load the BX register with no: of bytes

Step5: Copy the contents from the memory location n1 [bx] to AL

Step6: Perform addition with second number n2 [bx]

Step7: Move the result to the memory location sum [bx]

Step8: Decrement BX

Step9: Decrement CX, if CX not equal to Zero jump to step5

Step10: Load CX register with count

Step11: Load the BX register with no: of bytes

Step12: Move the contents from memory location n1 [bx] to AL

Step13: Perform subtraction with second number n2 [bx]

Step14: Move the result to the memory location sum [bx]

Step15: Decrement BX

Step16: Decrement CX, if CX not equal to Zero jump to step12

Step17: Stop




Start

Copy the contents from the memory location n1 [bx] to AL

Perform addition with second number n2 [bx]

Move the result to the memory location sum [bx]

Decrement BX

1


Get count into CX register

Load the BX register with no: of bytes

Decrement CX, &CX = 0

NO

YES

FLOW CHART

21



1

Move the contents from memory location n1 [bx] to AL

Perform subtraction with second number n2 [bx]

Move the result to the memory location sum [bx]

Stop

Decrement BX

Decrement CX, &If CX = 0

Get count into CX register &Load BX register with no of

bytes

NO

YES

22


PROGRAM:ASSUME CS: CODE, DS: DATADATA SEGMENTN1 DB 33H, 33H, 33HN2 DB 11H, 11H, 11HCOUNT EQU 0003HSUM DB 03H DUP (00)DIFF DB 03H DUP (00)DATA ENDSCODE SEGMENTORG 1000HSTART: MOV AX, DATA

MOV DS, AX

MOV CX, COUNT

MOV BX, 0002H

CLCBACK: MOV AL, N1 [BX]

ADC AL, N2 [BX]

MOV SUM [BX], AL

DEC BX

LOOP BACK

MOV CX, COUNT

MOV BX, 0002H

BACK1: MOV AL, N1 [BX]

SBB AL, N2 [BX]

MOV DIFF [BX], AL

DEC BX

LOOP BACK1

MOV AH, 4CH

INT 21H

CODE ENDSEND START



CODE TABLE:

Physical address Label

Hex code Mnemonic Operand commentsSegment address

Effective address

0D64 1000

1003

1005

1008

100B

100F

1013

1017

1018

101A

101D

1020

1024

1028

102C

102D

102F

1031

L1

L2

B8 63 0D

8E D8

B9 03 00

BB 02 00

8A 87 00 00

12 87 00 00

88 87 06 00

4B

E2 F1

B9 03 00

BB 02 00

8A 87 00 00

1A 87 03 00

88 87 09 00

4B

E2 F1

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV CX,0003

MOV BX,0002

MOV AL,[BX+0000]

ADC AL,[BX+0003]

MOV [BX+0006],AL

DEC BX

LOOP L1

MOV CX,0003

MOV BX,0002

MOV AL,[BX+0000]

SBB AL,[BX+0003]

MOV [BX+0009],AL

DEC BX

LOOP 1020

MOV AH,4C

INT 21

;Initialize the data segment

;Initialize the count CX with 0003;Move the value 02 in BX register;Copy the contents of BX+0000

;Perform ADC with[BX+0000]

;Move the result in BX+0006 location;Decrement the contents of BX

;Decrement the counter and go to L1 until CX is zero;Initialize the counter CX with 0003;Move the number 02 in BX register ;Copy the contents of BX+0000

;Perform SBB with [BX+0003]

;Move the result in BX+0009 location;Decrement the contents of BX

;Decrement the counter and go to L2 until CX is zero

;Terminate the program

RESULT:

D 0000: 33 33 33 11 11 11 44 44 44 22 22 22



6. What is the word length of 8086 P?

7. Which signal is used to separate address and data bus?

8. How many multiplexed lines are there in 8086 P?

9. What is the maximum memory addressing and I/O addressing capability of 8086 P?

10. From which the 8086 P starts execution after reset?



2.2 d) SIGNED OPERATIONS ON 8 BIT DATA

ABSTRACT: Assembly language program to perform signed operations

PORT USED: None

REGISTERS USED: AL, BL

ALGORITHM:

Step1: Start


Step3: Load AL with first number

Step4: Negate the contents of AL

Step5: Load BL with second number

Step6: Perform signed Multiplication



Step9: Negate the contents of AL

Step11: Perform signed division


Step13: Stop




Start

Negate the contents of AL

Perform signed Multiplication

1


Get a first number in AL register

Load BL with second number

Get a first number in AL register

Negate the contents of AL


FLOW CHART

27



1

Perform signed division


Stop

Decrement CX, if CX not equal to Zero jump to step12

28


PROGRAM:

ASSUME CS: CODE , DS: DATADATA SEGMENTN1 DB 08HN2 DB 04HRESULT DW 02 DUP (00)DATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV AL, N1

NEG AL

CBW

MOV BL, N2

IMUL BL

MOV [RESULT], AX

MOV AL, N1

NEG AL

CBW

IDIV BL

MOV [RESULT+2], AX

MOV AH, 4CH

INT 21H

CODE ENDSEND START


MICROPROCESSORS LAB MANUAL ECE Dept.CODE TABLE:

Physical address Label Hex code Mnemonic Operand comments

Segment address

Effective address

0D64 0000

0003

0005

0008

000A

000B

000F

0011

0014

0017

0019

001A

001C

001F

0021

B8 63 0D

8E D8

A0 00 00

F6D8

98

8A 1E 01 00

F6 EB

A3 02 00

A0 00 00

F6 D8

98

F6 FB

A3 04 00

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV AL,[0000]

NEG AL

CBW

MOV BL,[0001]

IMUL BL

MOV [0002],AX

MOV AL,[0000]

NEG AL

CBW

IDIV BL

MOV [0004],AX

MOV AH,4C

INT 21


;Copy the [0000] in AL

;Negate contents of AL

;Convert Byte to Word

;Get second number in BL

;Perform IMUL with BL

;Move the result into [0002]

;Get first number in AL

;Negate the contents of AL

;Convert Byte to Word

;Perform IDIV with BL



RESULT:D 0000: 08 04 E0 FF FE 00 00 00

VIVA-VOCE QUESTIONS

11. How does the 8086 P differentiate between an opcode and instruction data?

12. Which interrupt is having the highest priority?

13. what do you mean by addressing modes?

14. What is the addressing mode of an instruction communicated to the cpu?

15. What are the assembler directives?



2.2 e) ASCII ARITHMETIC OPERATIONS

ABSTRACT: Assembly language program to perform ASCII arithmetic operations

PORT USED: None

REGISTERS USED: AL, BL, SI

ALGORITHM:

Step1: Start


Step3: Load SI with Memory location


Step5: Load BL with Second number

Step6: Perform addition

Step7: Perform ASCII adjustment after addition

Step8: Copy the result to the memory location [SI]



Step11: Perform ASCII adjustment after subtraction

Step12: Copy the result to the memory location [SI+01]



Step15: Perform ASCII adjustment after multiplication




Step19: Perform ASCII adjustment before division


Step21: Stop




Start

Perform addition

Copy the result to the memory location [SI]

Perform subtraction

Perform ASCII adjustment after addition

1


Initialize SI with memory location

Get a first ASCII no in AL register

Get second ASCII no in BL register

Get first ASCII no in AL register & second ASCII no in BL register

FLOW CHART

32



1

Perform ASCII adjustment after subtraction

Copy the result to the memory location [SI+01]

Stop


Perform ASCII adjustment after multiplication


Perform division

Perform ASCII adjustment before division




Decrement CX, if CX not equal to Zero jump to step12

33

MICROPROCESSORS LAB MANUAL ECE Dept.PROGRAM:

ASSUME CS: CODE, DS: DATADATA SEGMENTN1 DB ‘8’N2 DB ‘4’DATA ENDS

CODE SEGMENTORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV SI, 1000H

XOR AX, AX

XOR BX, BX

MOV AL, N1

MOV BL, N2

ADD AX, BX

AAA

MOV [SI], AX

XOR AX, AX

MOV AL, N1

SUB AL, BL

AAS

MOV [SI+2], AX

MOV AL, 08H

MOV BL, 04H

MUL BL

AAM

MOV [SI+4], AX



AAD

DIV BL

MOV [SI+6], AX

MOV AH, 4CH

INT 21H

CODE ENDS

END START

CODE TABLE:


Segment address

Effective address

0D64 1000

1003

1005

1008

100A

100C

100F

1013

1015

1016

1018

101B

101D

101E

B8 63 0D

8E D8

BE 00 10

33 C0

33 DB

A0 00 00

8A 1E 01 00

02 C3

37

89 04

A0 00 00

2A C3

3F

89 44 02

MOV AX,0D63

MOV DS,AX

MOV SI,1000

XOR AX,AX

XOR BX,BX

MOV AL,[0000]

MOV BL,[0001]

ADD AL,BL

AAA

MOV [SI],AX

MOV AL,[0000]

SUB AL,BL

AAS

MOV [SI+02],AX



;Perform XOR operation on AX;Perform XOR operation on BX;Get first no. in AL

;Get second no. in BL

;Perform ADD AL , Bl

;Perform AAA

;Move the result in[SI]

;Get first no. in AL

;Perform SUB AL, BL

;Perform AAS

;Move the result in [SI+2]


MICROPROCESSORS LAB MANUAL ECE Dept.1021

1023

1025

1027

1029

102C

102E

1031

1033

1035

B0 08

B3 04

F6 E3

D4 0A

89 44 04

D5 0A

F6 F3

89 44 06

B4 4C

CD 21

MOV AL,08

MOV BL,04

MUL BL

AAM

MOV [SI+04],AX

AAD

DIV BL

MOV [SI+06],AX

MOV AH,4C

INT 21

;Load AL with 08h

;Load BL with 04h

;Perform MUL with BL

; Perform AAM


; Perform AAD

;Perform division with BL



RESULT:D 1000: 02 01 06 00 02 03 20 00

VIVA-VOCE QUESTIONS

16. What are the Programmable peripheral interfacing chips available?

17. What is the memory capacity of 8086 P?

18. How data can be transferred without using the microprocessor?

19. How does the CPU identify between 8-bit and 16-bit operations?

20. What are the different instruction types of 8086 P?



2.3 a) BCD TO ASCII CONVERSION

ABSTRACT: Assembly language program to convert BCD number to ASCII number

PORT USED: None

REGISTERS USED: AL, AH, CX

ALGORITHM:

Step1: Start


Step3: Load AL with BCD number

Step4: Load CX register with count 04

Step5: Copy the contents from AL to AH

Step6: Perform AND operation AL with 0F

Step7: Perform AND operation AL with F0

Step8: Rotate the AH contents

Step9: Perform OR operation AL with 30

Step10: Perform OR operation AH with 30

Step11: Move the result to the memory location

Step12: Stop


MICROPROCESSORS LAB MANUAL ECE Dept.FLOW CHART :


Start

Perform AND operation AL with 0F

Rotate the AH contents

Perform OR operation AH with 30

Copy the contents from AL to AH

Perform AND operation AL with F0

Move the result to the memory location

Stop

Perform OR operation AL with 30


Get BCD no in AL register

Load the count 04 in CX register

38


PROGRAM:


DATA SEGMENTBCD DB 17HASCII DW?DATA ENDS


START: MOV AX, DATA

MOV DS, AX

MOV AL, BCD

MOV CL, 04

MOV AH, AL

AND AL, 0FH

AND AH, 0F0H

ROR AH, CL

OR AL, 30H

OR AH, 30H

MOV ASCII, AX

MOV AH, 4CH

INT 21H CODE ENDS

END START




Segment address

Effective address

0D64 1000

1003

1005

1007

1009

100B

100D

1010

1012

1014

1017

101A

101C

B8630D

8ED8

B017

B104

8AE0

240F

80E4F0

D2CC

0C30

80CC30

A30100

B44C

CD21

MOV AX,0D63

MOV DS,AX

MOV AL,17

MOV CX,0004

MOV AH,AL

AND AL,0F

AND AH,F0

ROR AH,CL

OR AL,30

OR AH,30

MOV [0001],AX

MOV AH,4C

INT 21

;initialize data segment

;Get a number in AL

;Initialize CX with 0004

;Copy the contents of AL

;Perform AND ALwith 0F

;Perform AND AHwith F0

;Rotate right AH, CL

;Perform OR AL with 30

;Perform OR AH with 30

;Move the result in [0001]


RESULT:

D 0000: 17 37 31VIVA-VOCE QUESTIONS

21. What is a linker?

22. What are the DOS function calls?

23. How a CALL instruction will be executed?

24. What is the role of stack?

25. What is an interrupt vector table of 8086 P?



2.3 b) ASCII TO BCD CONVERSION

ABSTRACT: Assembly language program convert ASCII number to BCD number

PORT USED: None

REGISTERS USED: AL, BL, CX

ALGORITHM:

Step1: Start


Step3: Load AL with first ASCII number

Step4: Load CX register with count 04

Step5: Load BL with second ASCII number

Step6: Perform AND operation AL with 0F

Step7: Perform AND operation BL with 0F

Step8: Rotate the contents of AL with CL count

Step9: Perform OR operation AL with BL


Step11: Stop




Start

Perform AND operation AL with 0F

Rotate the contents of AL with CL count


Perform AND operation BL with 0F

Stop

Perform OR operation AL with BL


Get first ASCII no in AL register

Load CX register with a count 04

Get second ASCII no in BL register

42


PROGRAM:


DATA SEGMENTASCII1 DB “1”ASCII2 DB ‘7’BCD DB ?DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV CL, 04H

MOV AL, ASCII1

MOV BL, ASCII2

AND AL, 0FH

AND BL, 0FH

ROR AL, CL

OR AL, BL

MOV BCD, AL

MOV AH, 4CH

INT 21H

CODE ENDS

END START




Segment address

Effective address

0D64 1000

1003

1005

1007

100A

100E

1010

1013

1015

1017

101A

101C

B8 63 0D

8E D8

B1 04

A0 00 00

8A 1E 01 00

24 0F

80 E3 0F

D2 C8

0A C3

A2 02 00

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV CL,04

MOV AL,[0000]

MOV BL.[0001]

AND AL,0F

AND BL,0F

ROR AL,CL

OR AL,BL

MOV [0002],AL

MOV AH,4C

INT 21


;Initialize count CX with 04

;Get 1st no. in AL register

;Get 2nd no. in BL register

;Perform AND AL with 0Fh

; Perform AND BL with 0Fh

;Rotate right AL , CL

;Perform OR, AL with BL

;Move the result in [0002]

; Terminate the program

RESULT:D 0000: 31 37 17

VIVA-VOCE QUESTIONS

26. What is the difference between near and far procedure?

27. What is the difference between macro and sub-routine?

28. What is a nested macro?

29. What is the difference between software and hardware interrupt?

30. What are the various modes available in 8255?



2.3 c) NO OF POSITIVE AND NEGATIVE NUMBERS

ABSTRACT: Assembly language program to count number of positive and negative numbers

PORT USED: None

REGISTERS USED: SI, DX, CX, AL

ALGORITHM:

Step1: Start



Step4: Initialize DX with 0000

Step5: Load SI with offset list

Step6: Move the contents from memory location SI to AL

Step7: Rotate left the contents of AL

Step8: Jump to step13 if carry

Step9: Increment DL

Step10: Increment SI

Step11: Decrement CX and jump to step6 if no zero

Step12: Jump to step16

Step13: Increment DH




Step17: Stop




Start

Initialize data segment and Load SI with offset list &

Get a count with CX register &Initialize DX with 0000

Move the contents from memory location SI to AL

Rotate the contents of AL to left

Increment DL and Increment SI

1

If carry equal to

zero

NOOo

FLOW CHART

YES

Decrement CX&If CX=0

YES

NOOo

2 JMP

3 4

46



1

Increment DH & Increment SI

Stop

Increment SI

YES

4

NO

3

2

Decrement CX

If CX=0


47


PROGRAM:


DATA SEGMENTLIST DB 0FFH, 0DDH, 04H, 05H, 98HRESULT DW ?DATA ENDS


START: MOV AX, DATA

MOV DS, AX

LEA SI, LIST

MOV CX, 0005H

MOV DX, 0000H

BACK: MOV AL, [SI]

ROL AL, 01H

JC NEGATIVE

INC DL

INC SI

LOOP BACK

JMP EXIT

NEGATIVE: INC DH

INC SI

LOOP BACK

EXIT: MOV [RESULT], DX

MOV AH, 4CH

INT 21H

CODE ENDSEND START



CODE TABLE:

Physical address Label Hex code Mnemonic Operand Comments

Segment address

Effective address

0D64 1000

1003

1005

1009

100C

100E

1010

1012

1014

1015

1017

1019

101A

101C

101D

101F

1023

1025

Back

L1

L2

B8630D

8ED8

8B360000

B90500

8A04

D0C0

7208

FEC2

46

E2F5

EB06

90

FEC6

46

E2ED

89160500

B44C

CD21

MOV AX,0D63

MOV DS,AX

LEA SI,[0000]

MOV CX,0005

MOV AL,[SI]

ROL AL,01

JB L1

INC DL

INC SI

LOOP Back

JMP L2

NOP

INC DH

INC SI

LOOP Back

MOV [0005],DX

MOV AH,4C

INT 21


;

; Load Effective addr. SI With 0000;Initialize the count CX with 0005;Copy the number of [SI] in AL;Rotate Left AL , 01

;If Barrow jump L1

; Increment value of DL

;Increment value of SI

;Decrement count and go to Back until CX = 0;Go to L2 without condition

;No Operation

:Decrement value of DH


;Decrement counter and go to Back until CX=0;Move the result in [0005] location


RESULT:

D 0000: FF DD 04 05 98 02 03



VIVA-VOCE QUESTIONS

31. What are the different registers in 8255?




35. What are the various modes of operation available in 8255?



2.3 d) NO OF ODD AND EVEN NUMBERS

ABSTRACT: Assembly language program to count number of odd and even numbers

PORT USED: None

REGISTERS USED: AL, CX, DL, DH, SI

ALGORITHM:

Step1: Start






Step7: Rotate right the contents of AL


Step9: Increment DL




Step13: Increment DH




Step17: Stop




Start

Initialize data segment and Load SI with offset list &

Get a count with CX register &Initialize DX with 0000

Move the contents from memory location SI to AL

Rotate the contents of AL to right

Increment DL and Increment SI

1

If carry equal to

zero

NOOo

FLOW CHART

YES

Decrement CX&If CX=0

YES

NOOo

2 JMP

3 4

52



1

Increment DH & Increment SI

Start

Increment SI

YES

4

NO

3

2

Decrement CX

If CX=0


53


PROGRAM:


DATA SEGMENT

LIST DB 05H,01H,03H,04H,08H,02H

COUNT DW 0006H

RESULT DW?

DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV CX, COUNT

MOV DX, 0000H

MOV SI, OFFSET LIST

BACK: MOV AL, [SI]

ROR AL, 01H

JC ODD

INC DL

INC SI

LOOP BACK

JMP EXIT

ODD: INC DH

INC SI

LOOP BACK

EXIT: MOV [RESULT], DX

MOV AH, 4CH

INT 21H

CODE ENDS

END START



CODE TABLE:


Segment address

Effective address

0D64 1000

1003

1005

1009

100C

100F

1011

1013

1015

1017

1018

101A

101C

101D

101F

1020

1022

1026

1028

Back

L1

L2

B8 63 0D

8E D8

8B 0E 06 00

BA 00 00

BE 00 00

8A 04

D0 C8

F2 08

FE C2

46

E2 F5

EB 06

90

FE C6

46

E2 ED

89 16 08 00

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV CX,[0006]

MOV DX,0000

MOV SI,0000

MOV AL,[SI]

ROR AL,01

JB L1

INC DL

INC SI

LOOP Back

JMP L2

NOP

INC DH

INC SI

LOOP Back

MOV [0008],DX

MOV AH,4C

INT 21


;Initialize counter with [0006];Initialize DX register with 0000;Initialize SI with 0000

;Load the 1st no. in AL

;Rotate right AL,01

;If Barrow go to L1

;Increment the value of DL

;Increment the value of SI

;Decrement counter and go to Back until CX=0;Go to L2 without unconditional;No Operation

;Increment the value of DH

;Increment the Value of SI

;Decrement counter and go to Back until CX=0;Move the result in register DX




RESULT:

D 0000: 05 01 03 04 08 02 06 00 03 03

VIVA-VOCE QUESTIONS

36. In how many modes does 8086 microprocessor works?

37. What are the conditions required to make 8086 to work in maximum mode?

38. What is the difference between maskable and non-maskable interrupts?

39. What is the difference between jump and loop instruction?

40. What is the difference between the respective shift and rotate instructions?



2.3 e) PACKED BCD TO UNPACKED BCD CONVERSION

ABSTRACT: Write a program to convert packed BCD number into Unpacked BCD number.

REGISTERS USED: AL, BL

PORTS USED: None.

ALOGARITHM:

Step1: Start

Step2: Initialize the data segment

Step3: Move packed number into AL register

Step4: Move packed number into BL register

Step5: Initialize the count CX with 04h.

Step6: AND operation AL with 0Fh.

Step7: AND operation BL with 0F0h.

Step8: Rotate right without carry operation on BL by CL times.

Step9: Move the result into location 0000 and 0001.

Step10: Stop.




Start

Initialize the data segment

FLOW CHART

Move the number 29 into AL register &

Move the value 29h into BL register

1

Initialize the counter CX with 0004

Logical AND ed between AL & 0Fh and

Logical AND ed between BL & 0F0h

58



1

Stop

Rotate right BL with 04h times

Store the result in 0000 and 0001 location

59


PROGRAM:


DATA SEGMENT

N EQU 29H

RESULT DB 02H DUP (0)

DATA ENDS

CODE SEGMENT

ORG 2000h

START: MOV AX, DATA

MOV DS, AX

MOV AL, N

MOV BL, N

MOV CL, 04H

AND AL, 0Fh

AND BL, 0F0h

ROR BL, CL

MOV [RESULT], BL

MOV [RESULT+1], AL

MOV AH, 4Ch

INT 21h

CODE ENDS

END START



CODE TABLE:

Physical address Label Hex code Mnemonic

Op code operands

Comments

Segment address

Effective address

0B3C 2000 B8 3B 0B MOV AX,0B3B ;Initialize the data segment

2003 8E D8 MOV DS, AX

2005 B1 04 MOV CL,04h ; Initialize the count with 04h

2007 B8 29 MOV AL, N ;Packed number in AL

2009 B3 29 MOV BL, N ;Packed number in BL

200B 24 0F AND AL,0Fh ;AND Operation AL with 0F

200D 80 E3 F0 AND BL, 0F0h ;AND Operation BL with F0

2010 D2 CB ROR BL,CL ;Rotate right BL, Cl

2012 88 1E 00 00 MOV [0000],AL ; Move the result in [0000]

2016 A2 01 00 MOV [0001], BL ;Move the result in [0001]

2019 8A 07 MOV AH, 4Ch ;Terminate the program

201B CD 21 INT 21 ;Stop

RESULT:

D 0B42:0000 29 02 09VIVA-VOCE QUESTIONS

41. How will you enter the single step mode of 8086?

42. What is LOCK prefix? What is its use?

43. What is REP prefix? What is its use?

44. What is the difference between conditional and unconditional control transfer

instructions?

45. What are different string instructions?



2.4 a) TRANSFER BLOCK OF DATA

ABSTRACT: Assembly language program to transfer a block of data.

PORT USED: None.

REGISTERS USED: AX, BL.

ALGORITHM:

Step1: Start

Step2: Initialize data segment & extra segment


Step4: Initialize DI with memory location


Step6: Repeat the process of moving string byte from SI to DI until count equals to zero

Step7: Stop




Start

Initialize data segment and Initialize extra segment

FLOW CHART

Repeat the process of moving string byte from SI to DI until count equals to zero

Load CX register with count

Initialize DI with memory location & load SI with offset

list

Stop

63


PROGRAM:

ASSUME CS: CODE, DS: DATA, ES: DATA

DATA SEGMENT

LIST DB ‘ADITYA’

COUNT EQU 06H

DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV ES, AX

MOV CX, COUNT

MOV DI, 5000H

LEA SI, LIST

REP MOVSB

MOV AH, 4CH

INT 21H

CODE ENDS

END START



CODE TABLE:

Physical address Label Hex code Mnemonic Operand Comments

Segment address

Effective address

0D64 1000

1003

1005

1007

100A

100D

1011

1013

1015

B8 63 0D

8E D8

BE C0

B9 06 00

BF 00 50

8D 36 00 00

F3 A4

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV ES,AX

MOV CX,0006

MOV DI,5000

LEA SI,[0000]

REPZ MOVSB

MOV AH,4C

INT 21


Initialize extra segment

;Initialize counter with 06

;Initialize DI with 5000

;Load SI with effective adr

;Move the contents of SI into DI;Terminate the program

;Stop

RESULT:

D 0000: 61 64 69 74 79 61

D 5000: 61 64 69 74 79 61

A D I T Y A

VIVA-VOCE QUESTIONS

46. What is the difference between inter-segment call and intra-segment call?

47. What is meant by maskable interrupts?

48. What is meant by non-maskable interrupts?

49. What is meant by Program Status Word (PSW)?

50. Draw the flag register format?



2.4 b) REVERSAL OF GIVEN STRING

ABSTRACT: Assembly language program to reverse a given string

PORT USED: None


ALGORITHM:

Step1: Start



Step4: Copy the contents from CX to AX


Step6: Initialize DI with (count-1)

Step7: Initialize BX with 02

Step8: Perform division with BX

Step9: Copy the contents from AX to CX


Step11: Exchange the contents of AL with [DI]

Step12: Move the contents from memory location AL to SI


Step14: Decrement DI


Step16: Stop




Start


FLOW CHART

Copy the contents from CX to AX and perform division with BX

load CX with count and SI with offset list

Initialize DI with count-1 and Initialize BX with 02h

1

67



Exchange the contents of AL to [DI] and move the contents from AL to SI

Increment SI and Decrement DI

Decrement CX and if CX=0

YES

Stop

NO

Copy the contents from AX to CX and move the contents from SI to AL

1

68



DATA SEGMENT

LIST DB ‘MICRO PROCESSOR’

COUNT EQU ($-LIST)

DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV ES, AX

MOV CX, COUNT

MOV AX, CX

MOV SI, OFFSET LIST

MOV DI, (COUNT-1)

MOV BX, 02

DIV BXS

MOV CX, AX

BACK: MOV AL,[SI]

XCHG AL,[DI]

MOV [SI], AL

INC SI

DEC DI

LOOP BACK

MOV AH, 4CH

INT 21H

CODE ENDS

END START



CODE TABLE:

Physical address Label Hex code Mnemonic Operand CommentsSegment address

Effective address

0D64 1000

1003

1005

1007

100A

100C

100F

1012

1015

1017

101A

101B

101D

101F

1020

1021

1023

1025

Back

B8 63 0D

8E D8

8E C0

B9 0F 00

8B C1

BE 00 00

BF 0E 00

BB 02 00

F7 F3

8B C8

8A 04

86 05

88 04

46

4F

E2 F6

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV ES,AX

MOV CX,000F

MOV AX,CX

MOV SI,0000

MOV DI,000E

MOV BX,0002

DIV BX

MOV CX,AX

MOV AL,[SI]

XCHG AL,[DI]

MOV [SI],AL

INC SI

DEC DI

LOOP Back

MOV AH,4C

INT 21


;Initialize the extra segment

;Initialize the counter with 000F;Load AX with CX value


;Initialize DI with 000e

;Move the value 02 in BX

;Division operation with BX

;Move AX value in CX

;Load the 1st no. in AL

;Exchange AL,[DI]

;Move the no in AL into [SI]


;Decrement the value of DI

;Decrement the counter and go to Back until CX=0;Terminate the program

;Stop

RESULT:D 0000: 72 6F 73 73 65 63 6F 72 70 20 6F 72 63 69 6D

R O S S E C O R P O R C I M



VIVA-VOCE QUESTIONS

51. What are the general purpose registers in 8086?

52. What is meant by stack pointer?

53. Which instructions are used for stack operations?

54. What is the length of physical address?

55. Distinguish between segment registers and index registers?



2.4 C) SORTING OF ‘N’ NUMBERS

ABSTRACT: Assembly language program to do sorting of numbers in a given series

PORT USED: None

REGISTERS USED: CX, DX, AL, SI

ALGORITHM:

Step1: Start



Step4: Copy the contents from CX to DX


Step6: Copy the contents from DX to CX


Step8: Increment SI

Step9: Compare AL contents with [SI]


Step11: Exchange the contents of AL with [SI]

Step12: Decrement SI

Step13: Move the contents from AL to memory location SI



Step16: Decrement DX and jump to step5 if no zero

Step17: Stop




Start

Initialize data segment and Load SI with offset list

&Get a count with CX register

Copy the contents from CX to DX

Copy to contents from SI to AL register

Compare the contents of AL and SI

1

Increment SI

If carry equal to

zero

NOOo

Exchange the contents of AL & SI

FLOW CHART

2

YES

3

4

73



1

Decrement SI & move the contents of AL into SI

Start

Increment SI

Decrement CX

If CX =0

Decrement DX

If DX=0

YES

3

NO

2

NO

YES

4

74


PROGRAM:


DATA SEGMENTLIST DB 56H, 12H, 72,32HCOUNT EQU 0003HDATA ENDS


START: MOV AX, DATA

MOV DS, AX

MOV CX, COUNT

MOV DX, CX

AGAIN: MOV SI, OFFSET LIST

MOV CX, DX

BACK: MOV AL, [SI]

INC SI

CMP AL, [SI]

JC NEXT

XCHG [SI], AL

DEC SI

MOV [SI], AL

INC SINEXT: LOOP BACK

DEC DX

JNZ AGAIN

MOV AH, 4CH

INT 21HCODE ENDSEND START



CODE TABLE:

Physical address Label Hex code Mnemonic Operands Comments

Segment address

Effective address

0D64 1000

1003

1005

1008

100A

100D

100F

1011

1012

1014

1016

1018

1019

101B

101C

101E

101F

1021

1023

L2

Back

L1

B8 63 0D

8E D8

B9 03 00

8B D1

BE 00 00

8B CA

8A 04

46

3A 04

72 06

86 04

4E

88 04

46

E2 F1

4A

75 E9

B4 4C

CD 21

MOV AX,0D63

MOV DS,AX

MOV CX,0003

MOV DX,CX

MOV SI,0000

MOV CX,DX

MOV AL,[SI]

INC SI

CMP AL,[SI]

JB L1

XCHG AL,[SI]

DEC SI

MOV [SI],AL

INC SI

LOOP Back

DEC DX

JNZ L2

MOV AH,4C

INT 21


;Initialize counter with 0003

;Load DX with Value of CX


;Get 1st no. in AL


;Compare AL,[SI]

;If barrow go to L1

;Exchange the values of AL , SI;Decrement the value of SI

;Copy the AL reg into [SI]


;Decrement the counter and go to Back;Decrement the DX

;Jump no zero go to L2


RESULT: D 0000 12 32 56 72



VIVA-VOCE QUESTIONS

56. How the physical address is calculated?

57. What is the difference between overlapping and non-overlapping segmentation?

58. What is the use of memory segmentation?

59. What is the clock frequency of 8086 microprocessor?

60. What is the function of overflow flag?


MICROPROCESSORS LAB MANUAL ECE Dept. 2.4 d) LENGTH OF THE GIVEN STRING

ABSTRACT: Assembly language program to find the Length of a string

PORT USED: None


ALGORITHM:

Step1: Start


Step3: Load AL with ‘$’



Step6: Scan string byte from DI memory location until AL =ES: DI

Step7: Jump to step10 if equal

Step8: Increment DX



Step11: Stop




Start


FLOW CHART

Scan string byte from DI memory location until AL =ES: DI

Load AL with “$” and SI with offset list

Initialize DX with 0000h

If equal

Increment DX and jump to back


YES

NO

Stop

79


PROGRAM:


DATA SEGMENT

LIST DB ‘ADITYA$’

LEN DB ?

DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV ES, AX

MOV AL,’$’

LEA SI, LIST

MOV DX, 0000H

BACK: SCASB

JE EXIT

INC DX

JMP BACK

EXIT: MOV LEN, DX

MOV AH, 4CH

INT 21H

CODE ENDS

END START



Physical address Label Hex code Mnemonic Operands Comments

Segment address

Effective address

0D64 1000

1003

1005

1007

1009

100D

1010

1011

1013

1014

1016

101A

101C

L2

L1

B8 63 0D

8E D8

8E C0

B0 24

8D3600 00

BA0000

AE

74 03

42

EBFA

8916 0700

B44C

CD21

MOV AX,0D63

MOV DS,AX

MOV ES,AX

MOV AL,24

LEA SI,[0000]

MOV DX,0000

SCASB

JZ L1

INC DX

JMP L2

MOV [0007],DX

MOV AH,4C

INT 21


;Initialize the extra segment

;Move 24 into AL register

;Load SI Effective address00

;Initialize the DX with 0000

;Scan string byte with SI

;If equal to zero go to L1

;Increment the value of DX

;Go to L2 with out condition



RESULT:

D 0000: 61 64 69 74 79 61 24 06

A D I T Y A $

VIVA-VOCE QUESTIONS

61. What are the advantages of 8086 microprocessor over 8085 microprocessor?

62. What is the default register used in string operations?

63. What is the feature size of 8086 microprocessor?

64. What is meant by volatile memory?

65. What are the different non-volatile RAMs?



2.4 e) COMPARISON OF TWO STRINGS

ABSTRACT: Assembly language program to compare two strings.

PORT USED: None


ALGORITHM:

Step1: Start


Step3: Load AX with length1

Step4: Load BX with length2

Step5: Compare AX with BX

Step6: Jump step14 if not equal

Step7: Copy the contents from AX to CX

Step8: Load SI with first string

Step9: Load DI with second string

Step10: Repeat comparing string byte until count equals to zero

Step11: jump to step 14 if not equal


Step13: Jump to step 15

Step14: Move another result to the memory location

Step15: Stop




Start


FLOW CHART

Compare AX with BX

Load AX with length 1 and Load BX with length 2

1

If equal

YES

NO

Copy the contents from AX to CX

Load SI with first string &Load DI with second string

Repeat comparing string byte until count equals to zero

2

83



1

If equal NO

YES


Move another result to the memory location

Stop

2

84


PROGRAMS:


LIST1 DB ‘ADITYA’

LEN1 EQU ($-LIST1)

LIST2 DB ‘ADITYA’

LEN2 EQU ($-LIST2)

RESULT DW ?

DATA ENDS

CODE SEGMENT

ORG 1000H

START: MOV AX, DATA

MOV DS, AX

MOV ES, AX

MOV AX.LEN1

MOV BX, LEN2

CMP AX, BX

JNE EXIT

MOV CX, AX

MOV SI, OFFSET LIST1

MOV DI, OFFSET LIST2

REP CMPSB

JNE EXIT

MOV RESULT, 5555H

JMP NEXT

EXIT: MOV RESULT, 0FFFFH

NEXT: MOV AH, 4CH

INT 21H

CODE ENDS

END START



CODE TABLE:Physical address Label Hex code Mnemonics Operands Comments

Segment address

Effective address

0D64 1000

1003

1005

1007

100A

100D

100F

1011

1013

1016

1019

101B

101D

1023

1025

1026

102C

102E

L1

L2

B8 63 0D

8E D8

8E C0

B8 06 00

BB 06 00

3B C3

75 15

8B C8

BE 00 00

BF 06 00

F3 A6

75 09

C7060C005555

EB07

90

C7060C00FFFF

B44C

CD21

MOV AX,0D63

MOV DS,AX

MOV ES,AX

MOV AX,0006

MOV BX,0006

CMP AX,BX

JNZ L1

MOV CX,AX

MOV SI,0000

MOV DI,0006

REPZ CMPSB

JNZ L1

MOV PTR[000C],5555

JMP L2

NOP

MOV PTR[000C],FFFF

MOV AH,4C

INT 21

;Initialize the data

segment

;Initialize extra segment

;Load AX with 0006

;Load BX with 0006

;Compare AX with BX

;If no 0 go to L1

;Copy the contents of AX into CX;Initialize SI with 0000

;Load DI with 0006

;Compare with SI

;If no 0 go to L1

;Get 5555 in 000C location;Go to unconditional L2

;No Operation

;Get FFFF into 000C location; Terminate the program

RESULT:



66. What is meant by SRAM?

67. What does instruction-pipelining mean?

68. What is the difference between SRAM and DRAM?

69. What do ROM, EEPROM, and UVPROM mean?

70. What does masked ROM mean?

71. How data can be transferred in interrupt I/O?



2.5 a) READING KEYBOARD (WITH Echo)

ABSTRACT: To Reading the Keyboard (Buffered with Echo).

REGISTERS USED: AH, AL, SI.

PORTS USED: None.

ALGORITHM:

Step1: Start.

Step2: Load the number 13h into AL register.

Step3: Initialize the AH register with 00

Step4: Key board Interrupt

Step5: Initialize the AL register with 00

Step6: Key board Interrupt

Step7: Compare the data in AL register and character ‘q’.

Step8: If equal to zero go to step 12.

Step9: Move the number 0Fh into BL register.

Step10: Move the number 14h into AH register.

Step11: Keyboard Interrupt.

Step12: Load the number 4C in AH register.

Step13: Stop.




CODE SEGMENT

ORG 1000h

START: MOV AL, 13h

MOV AH, 00

INT 10h

BACK: MOV AH, 00h

INT 16h

CMP AL, ‘q’

JE EXIT

MOV BL, 0Fh

MOV AH, 14h

INT 10h

JMP BACK

EXIT: MOV AH, 4Ch

INT 21h

CODE ENDS

END START



CODE TABLE:

Segment base address

Effective address

Label Hex code Mnemonics Operands Comment

0B3D 1000 B0 13 MOV AL, 13 ;Move the 13 into AL register

1002 B4 00 MOV AH, 00 ;Initialize the AH register with 00

1004 CD 10 INT 10 ;Keyboard Interrupt

1006 B4 00 MOV AH, 00 ; Initialize the AH register with 00

1008 CD 16 INT 21 ; Keyboard Interrupt

100A 3C 71 CMP AL, 71 ;Compare the data 71 with AL register

100C 74 08 JZ L1 ;If equal to zero go to L1

100E B3 0F MOV BL, 0F ;Move 0F into the BL register

1010 B4 0F MOV AH, 0F ;Move 0F into the AH register

1012 CD 10 INT 10 ;Keyboard Interrupt

1014 EB F0 JMP L2 ;Jump to L2

1016 B4 4C MOV AH, 4C ;Move 4C intoAH register

1018 CD 21 INT 21 ;Go to DOS commands

RESULT:



2.5 b) READING KEYBOARD (without echo)

ABSTRACT: To Display the string or character (without ECHO) by using DOS Command

REGISTERS USED: AH, AL, SI

PORTS USED: None.

ALGORITHM:

Step1: Start

Step2: Initialize SI with Offset Result.

Step3: Initialize AH with 00h

Step4: Set the wait for KEY press..

Step5: Compare AL with q.

Step6: Copy the contents AL into SI register.

Step7: If equal to zero go to step 10

Step8: Increment the value of SI.

Step9: Go to step 3 without condition.

Step10: Terminate the program.

Step11: Stop.




DATA SEGMENT

ORG 3000h

RESULT DB 50h DUP (0)

DATA ENDS

CODE SEGMENT

ORG 1000h

START: MOV SI, OFFSET RESULT

BACK: MOV AH, 00h

INT 16h

CMP AL, ‘q’

MOV [SI], AL

JE EXIT

INC SI

JMP BACK

EXIT: MOV AH, 4Ch

INT 21h

CODE ENDS

END START



CODE TABLE:

Segment base address

Effective address


0E42 1000 BE 00 30 MOV SI, 3000 ;Initialize the SI with 3000h

1003 L2 B4 00 MOV AH, 00 ;Initialize the AH register with 00

1005 CD 16 INT 16 ;Bios keyboard Interrupt

1007 3C 71 CMP AL, 71 ;Compare 71 with AL register

1009 88 04 MOV [SI] , AL ;Move the data in AL register into [SI]

100B 74 03 JZ L1 ;If zero go to L1

100D 46 INC SI ;Increment the value of SI

100E EB F3 JMP L2 ;Go to L2 with out condition

1010 L1 B4 4C MOV AH, 4Ch ;Terminate the program

1012 CD 21 INT 21 ;Stop

RESULT:

- D 3000

- 6D 69 63 72 6F 70 72 6F 63 65 73 73 6F 72 73 20 6C 61 62

M I C R O P R O C E S S O R S L A B



2.5 c) DISPLAY STRING BY USING DOS COMMANDS

ABSTRACT: To display the string character by using DOS commands.

REGISTER USED: AL, AH, SI.

PORTS USED: None

ALGORITHM:

Step1: Start

Step2: Set the screen in Graphic mode

Step3: Initialize AH with 00h

Step4: Set the keyboard display mode.

Step5: Initialize SI with 0000h.

Step6: Copy the contents SI into AL register.

Step7: Compare AL register with null character ‘!’

Step8: If equal go to step 11.

Step9: Move the number 14h into AH register.

Step10: Move the number 05h into BL register.

Step11: Set keyboard display mode.

Step12: Go to step 6.

Step 13: Terminate the program.

Step14: Stop.



PROGRAM:


DATA SEGMENT

TEXT DB ‘ADITYA MICROPROCESSORS LAB!'

DATA ENDS

CODE SEGMENT

ORG 1000H

START: AX, DATA

MOV DS, AX

MOV AL, 13H; SET the screen in graphics mode

MOV AH, 00

INT 10H

MOV SI, 00H

BACK: MOV AL, TEXT [SI]

CMP AL,'!'

JE EXIT

MOV AH, 14

MOV BL, 05H

INT 10H

INC SI

JMP BACK

EXIT: MOV AH, 4CH

INT 21H

CODE ENDSEND START



CODE TABLE:

Segment base

address

Effective address


0B41 1000 B8 3F 0B MOV AX, 0B3F ;Initialize data segment

1003 8E D8 MOV DS, AX1005 B0 13 MOV AL,13 ;Move the13 into

AL register.1007 B4 00 MOV AH, 00 ;Initialize the AH

register with 00001009 CD 10 INT 10 ;Set the keyboard

display mode100B BE 00 00 MOV SI, 0000 ;Initialize SI with

0000 location100E L2 8A 80 40

00MOV AL,

[SI+00];Copy the contents of SI into AL

1012 3C 21 CMP AL, 21 ;Compare the value 21 with AL register

1014 74 09 JZ L1 ;If equal to zero go to L1

1016 B4 0E MOV AH, 0E ;Load AH register with 0Eh

1018 B3 05 MOV BL, 05 ;Load BL register with 05h

101A CD 10 INT 10 ;Set the keyboard display mode

101C 46 INC SI ;Increment value of SI

101D EB FF JMP L2 ;Jump go to L2 with out condition

101F L1 B4 4C MOV AH, 4Ch ;Terminate the program

1021 CD 21 INT 21 ;Dos command interrupt

RESULT:



3.1. PROGRAMMABLE PERIPHERAL INTERFACING(8255)

AIM: To study the functioning of programmable peripheral interfacing 8255. With port A & port B as out put port With port A as input and port B as output.

REGISTERS USED: AX, CX, DX.

PORTS USED: port A and port B.

CONNECTIONS: Study card; J1 of the study card adapter to I5 of ESA 8086/88.

DESCRIPTION:The 8255 is a general purpose programmable I/O device with 24 I/O

lines. These I/OLines are grouped as shown below:

Group A …. Port A (8 bits – PA0 to PA7) and Port C (4 bits – PC7 toPC4).

16 Group B …. Port B (8 bits – PB0 to PB7) and Port C (4 bits – PC3 toPC0).These groups can be operated in 3 different modes: mode 0, mode 1, and mode 2. InMode 0 the three ports, A, B, and C, may be programmed as input or output. In mode 1Ports A and B may be programmed as input or output, but port C is used to generateHandshake and interrupt signals. In mode 2, port A becomes a bi-directional port and 5I/O lines of port C are used for handshaking and interrupt signals. The pin-out of the8255 is given on the picture shown below.



Following is the description of the inputs and outputs used for interfacing to the 8255:CS’ – A low on this input pin enables communication between the 8255 and the deviceControlling it.47RD’ – A low on this input pin enables the 8255 to send data or status information to theDevice controlling the 8255.WR’ – A low on this input pin enables the device controlling the 8255 to write data orControl words to the 8255.A0 and A1 – These input address lines, in conjunction with the RD’ and WR’ signals,allow the selection of one of three ports or the control register, as defined by the tablegiven below:A1 A0 Location0 0 Port A0 1 Port B1 0 Port C1 1 Control RegisterRESET – A high on this input pin clears the control register and all ports are set to inputMode.To program the 8255 PPI, one must write a control word to the control register. ThisControl word will define how the 8255 is going to behave. The table shown below definesthe value of each bit of the control word:Control word bit Function


MICROPROCESSORS LAB MANUAL ECE Dept.D0 1 = Port C (lower) is input, 0 = Port C (lower) is outputD1 1 = Port A is input, 0 = Port A is outputD2 1 = Port B mode 1, 0 = Port B mode 0D3 1 = Port C (upper) is input, 0 = Port C (upper) is outputD4 1 = Port A is input, 0 = Port A is outputD5

D6

Port A00 = Mode 0, 01 = Mode 1, 1X = Mode 2D7 1 = Mode set flag active48Procedure

Write a program to:

1. Reset and initialize the 8255 to mode 0 with ports A, B, and C defined as outputs.

2. Create a menu of options to allow for the selection of which port is going to biased to simulate a three bits up-counter, and to quit the program.

3. When a port is selected from the menu, the screen should be cleared and message indicating which port is in use should be displayed.

The counting sequence should be displayed both on the screen and on the LED s associated withThe selected port until a key is pressed on the keyboard, at which time you start back at step 2.



1. PORT A AND PORT B AS OUT PUT PORT:

ALGORITHM:

Step 1: start

Step 2: move the control word address 0FF46 to register DX

Step 3: move 80 to AL register.

Step 4: locate the contents in AL register to DX register using port out.


Step 6: Initialize port A address.



MICROPROCESSORS LAB MANUAL ECE Dept.Step 8: Call subroutine program.

Step 9: complement the contents of AL register.

Step 10: initialize port B address.

Step 11: locate the contents in AL register to DX register using port “out”.

Step 12: Call subroutine program.

Step 13: Go step 6.

Step 14: Stop.

ALGORITHM FOR SUBROUTINE:

Step1: Load CX register with 0000h.

Step2: Decrement CX and go to step2 until the count CX equal to zero.

Step3: Return to main program.




Start

Initialize the control word register

FLOW CHART

Move the data 80h into AL register &Locate the contents in AL to DX register using port

out

Move the value 55h into AL register

1

Initialize port A address

Locate the contents of AL register to DX register using port out

Move the value 00 into CX register, decrement CX register and jump to main program if CX is equal to zero

Complement the contents of AL register

2

102



1

Stop

2

Locate the contents in AL register into DX register using port out

Initialize port B address

Move the value 00 into CX register, decrement CX register and jump to main program if CX is equal to zero

Jump

103


PROGRAM:MOV DX, 0FF46

MOV AX, 80

OUT DX, AL

MOV AX, 55

RPT: MOB DX, 0FF40

OUT DX, AX

CALL DELAY

NOT AX

MOV DX, 0FF42

OUT DX, AX

CALL DELAY

JMP RPT

DELAY PROGRAM:

MOV CX, 0000H

L1: LOOP L1

RET



CODE TABLE:


Op code operands

Comments

Segment address

Effective address

0000 5000 BA 46 FF MOV DX , FF46

Move FF46 to DX register

5003 B8 80 00 MOV AX, 0080 Load AX with 80

5006 EF OUT DX , AX Move AX content to DX

5007 B8 55 00 MOV AX , 0055 Move 55 to AX register.

500A Rpt BA 40 FF MOV DX , 0FF40

Move 0FF40 to DX register.

500D EF OUT DX,AX Move AX content to DX

500E E8 EE 1F CALL 6000 Go to subroutine

5011 FD D0 NOT AX Complement the AX register

5013 BA 42 FF MOV DX ,0FF42

Initialize the port B address.

5016 EF OUT DX , AX Move the contents of AX into DX.

5017 E8 E9 1F CALL 6000 Go to 6000 location

501A EB F1 JMP RPT Go to 500A location.

RESULT:



2. PORT A AS INPUT PORT AND PORT B AS OUT PUT PORT.

ALGORITHM:

Step 1: Start

Step 2: Move the control word address 0FF40 to register DX

Step 3: Move 90 to AL register.

Step 4: Locate the contents in AL register to DX register using port out.


Step 6: Initialized port A address ie.,, 0FF40 to DX register.

Step 7: Locate the contents in AL register to DX register using port IN.

Step 8: Complement the contents of AL register.

Step 9: Initialize port B address ie,,.0FF42 to DX register.

Step 10: Locate the contents in AL register to DX register using port “out”.

Step 11: Go step 5.

Step 12: Stop.




Start


FLOW CHART


out

Move the value 55h into AL register

1


Locate the contents of AL register to DX register using port out

Complement the contents of AL register

2

107



1

Stop

2

Locate the contents in AL register into DX register using port out


Jump

108


PROGRAM:

MOV DX, 0FF46

MOV AX, 90

OUT DX, AX

RPT: MOV AL, 55

MOV DX, 0FF40

IN AX, DX

NOT AX

MOV DX, 0FF42

OUT DX, AX

JMP RPT



CODE TABLE:


Op code operands

Comments

Segment address

Effective address

0000 4000 BA 46 FF MOV DX , FF46

Move FF46 to DX register



5007 Rpt B8 55 00 MOV AX , 0055 Move 55 to AX register.

500A BA 40 FF MOV DX , 0FF40

Move 0FF40 to DX register.

500D EF IN AX,DX Move DX content to AX

5011 FD D0 NOT AX Complement the AX register

5013 BA 42 FF MOV DX ,0FF42

Initialize the port B address.

5016 EF OUT DX , AX Move the contents of AX into DX.

501A EB F1 JMP RPT Go to 500A location.

RESULT:

VIVA-VOCE QUESTIONS

72. How data can be transferred in programmed I/O?

73. What is the difference between machine language and assembly language?

74. Expand USART?

75. What are the signals required for memory read operation?

76. What are the signals required for memory write operation?

77. Why 8086 uses hexadecimal number system?

78. On which data stepper motor works?

79. How the stepper motor can be interfaced to 8086 microprocessor?

80. Write the syntax for 32-bit division operation?

81. Give an example for relative based indexed addressing mode?



3.1 a) DIGITAL TO ANALOG CONVERTER

GENERATION OF WAVE FORMS:

AIM: program to generate the following wave forms: Triangular wave forms Saw tooth wave forms Square wave Sine wave

REGISTERS USED: general purpose registers: AL, DX, and CX

PORTS USED: out (port-B)

CONNECTION:J4 of ESA 86/88 to J1 DAC interface.

DESCRIPTIONS: As can be from the circuit only 17 lines from the connector are used totally. The port A and port B of 8255 programmable peripheral interface are used as output ports. The digital inputs to the DAC’s are provided through the port A and port B of 8255.the analog outputs of the DAC’s are connected to the inverting inputs of op-amps µA741 which acts as current to voltage converters. The out puts from the op- amps are connected to points marked Xout and Yout at which the wave forms are observed on a CRO. (port A is used to control Xout port B is used to control Yout).the difference voltage for the DAC’s is derived from an on-board voltage regulator µA 723 .it generates a voltage of about 8V.the offset balancing of the op-amps is done by making use of the two 10k pots provided. The output wave forms are observed at Xout and Yout on an oscillator.THEORY:BASIC DAC TECHNIQUE:

Vo = K VFS (d1 .2-1 + d2 . 2-2 + . . . . . . . .+dn . 2-n )Where d1 = MSB, d2 = LSBVFS = Full scale reading / out put voltageK --- Conversion factor is adjusted to ‘unity’.D/A converters consist of ‘n’ bit binary word ‘D’and is combined with a reference voltage VR to give an analog output. The out put can be either voltage or currentOut put voltage Vo = K VFS (d1 .2-1 + d2. 2-2 + . . . . . . . . +dn. 2-n)MSB weight = ½ VFS if d1 = 1 and all are zero’s, K = 1.LSB weight = VFS/2n if dn = 1 and all are zero’s, K = 1

DUAL DAC INTERFACE: This program generates a square wave or a Triangular wave at points Xout or Yout of

interface. The waveforms may be observed on an oscilloscope. This program can be executed in STAND-ALONE MODE or SERIAL MODE of operation. The program starts at memory location 3000H

ALGORITHM: (FOR TRIANGULAR WAVE):



Step 1: Start

Step 2: Initialize the control word register



Step 5: Move 00FF to CX register.


Step 7: Initialize the port A address.

Step 8 : Locate the contents of AL to DX register.

Step 9: Increment the value in AL by one.

Step 10: Locate AL contents to DX register.

Step 11: Decrement the value of CX register by one and move to step 6 if CX not equal to zero.

Step 11: Other wise move 00FF to CX register.

Step 12: Decrement the value of AL by one.

Step 13: Locate the contents in AL register to DX register.

Step 14: Decrement the value of CX by one and move to step 11 if CX not equal to zero.

Step 15: Otherwise move to step 5.

Step 16: Stop.




Start


FLOW CHART

Copy the contents from AX to 80 &Locate the contents in AL to DX register using port

out Move 0FF to CX register and

1

Locate the contents of AX to DX register&

Increment the value in AX by one

Locate AX contents to DX register

Move 00 to AL register


Decrement CX & CX =0

NO

YES

2

113



1

NO

YES

Move 0FF to CX register

Decrement the value of AL

Locate the contents in AL register to DX register

Stop

Decrement CX &CX=0

Jump

2

114


PROGRAM (FOR TRIANGULAR WAVE):

MOV DX, 0FFE6

MOV AL, 80 OUT DX, AL

MOV DX, 0FFE0

MOV AX, 00

RPT: MOV CX, 0FF

L1: OUT DX, AX

INC AX

LOOP L1

MOV CX, 0FF

L2 : OUT DX,AX

DEC AX

LOOP L2

JMP RPT



CODE TABLE (FOR TRIANGULAR WAVE):


Op code operands

Comments

Segment address

Effective address

0000 2000 BA E6 FF MOV DX , FFE6

Move FFE6 to DX register


2006 EF OUT DX , AX Move DX content to AX

2007 BA E0 FF MOV DX , FFE0 Move FFE0 to DX register

200A B8 00 00 MOV AX , 0000 Load AX with 0000

200D Rpt B9 FF 00 MOV CX , 00FF Move FF to CX register

2010 L1 EF OUT DX , AX Move contents of AX to DX

2011 40 INC AX Increment the contents of AX

2012 E2 FC LOOP L1 Loop go to 2010 location

2014 B9 FF 00 MOV CX , 00FF Move 00FF to CX register

2017 L2 EF OUT DX, AX Move the contents of AX into DX

2018 48 DEC AX Decrement to AX register


201B EB EA JMP RPT Jump to 200D location

RESULT:



ALGORITHM (FOR SAW TOOTH WAVE):

Step 1: Start

Step 2: Move the control word address 0FFE6 to register DX




Step 6: Move 00FF to CX register.

Step 7: Locate or move 0FFE2 to DX register ie, port B address.

Step 8: Locate AL contents to DX register.

Step 9: Increment the value of AL register by one.

Step 10: Decrement the value of CX register by one and move to step 8 if CX not equal to zero.

Step 11: Otherwise move to step 5.

Step 12 Stop.




Start


FLOW CHART


out Move 00 to AX register

Locate the contents of AX to DX register

Move 0FF to CX register



NO YES


Stop

118


PROGRAM (FOR SAW TOOTH WAVE):

MOV DX, 0FFE6

MOV AL, 80

OUT DX,AL

MOV DX, 0FFE0

MOV AX, 00

RPT: MOV CX, 0FF

L1: OUT DX, AX

INC AX

LOOP L1

MOV AX, 00

OUT DX, AX

JMP RPT



CODE TABLE (FOR SAW TOOTH WAVE):


Op code operands

Comments

Segment address

Effective address

0000 3000 BA E6 FF MOV DX , FFE6 Move FFE6 to DX register


3006 EF OUT DX , AX Move DX content to AX


300A B8 00 00 MOV AX , 0000 Load AX with 0000

300D Rpt B9 FF 00 MOV CX , 00FF Move FF to CX register

3010 L1 EF OUT DX , AX Move contents of AX to DX

3011 40 INC AX Increment the contents of AX


3014 B8 00 00 MOV AX , 0000 Move 0000 to AX register

3017 EF OUT DX, AX Move the contents of AX into DX

3018 EB F3 JMP RPT Jump to 300D location

RESULT:



ALGORITHM (FOR SQUARE WAVE):

Step 1: Start




Step 5: Move FF to AL register.

Step 6: Locate the contents in AL register to DX register using port out .

Step 7: Move 91 into CX register.

Step 8: Decrement the value CX by one and jump to step 8 if CX not equal to zero.

Step 9: Other wise move 00 to AL register.

Step 10: Locate the contents in AL register to DX register using port out .

Step 11: Move 91 into CX register.

Step 12: Decrement the value CX by one and jump to step 12 if CX not equal to zero.

Step 13: Jump to location step 5.

Step 14: Stop.




Start


FLOW CHART


out


Move 0FF to AX register



NOYES

Move the data 91h into CX register

12

122



1

NOYES

Move 00h to AX register

Locate the contents in AL register to DX registerMove the data 91h into CX register

Stop

Decrement CX &CX=0

Jump

2

123


PROGRAM (FOR SQUARE WAVE):

MOV DX, 0FFE6

MOV AL, 80

OUT DX, AL

MOV DX, 0FFE0

RPT: MOV AX, 0FF

OUT DX, AX

CALL DELAY

MOV AX, 00

OUT DX, AX

CALL DELAY

JMP RPT

DELAY PROGRAM:

MOV CX, 1E

NOP

NOP

L1: LOOP L1

RET


MICROPROCESSORS LAB MANUAL ECE Dept.CODE TABLE (FOR SQUARE WAVE):


Op code operands

Comments

Segment address

Effective address

0000 5000 BA E6 FF MOV DX , FFE6

Move FFE6 to DX register




500A Rpt B8 FF 00 MOV AX , 00FF Load AX with 00FF

500D EF OUT DX,AX Move AX content to DX

500E E8 FF EF CALL 6000

5011 B8 00 00 MOV AX , 0000 Load AX with 0000

5014 EF OUT DX , AX Move the contents of AX into DX

5015 E8 E8 FF CALL 6000

5018 E8 F0 JMP RPT Jump to 500A location

DELAY PROGRAM:

LOCATION 0000:6000

Physical addressBase adder offset adder

Label Hex codes

Mnemonics Op code Operand

comment

0000 6000 B9 1E 00 MOV CX,001E Move 001E into CX register

6003 NOP

6004 NOP

6005 C3 RET Return to main program

RESULT:



ALGORITHM (FOR SINE WAVE):

Step 1: Start




Step 5: Initialize the SI with 4000h location.

Step 6: Move the number 46 into the counter CX register.

Step 7: Initialize the Port A address ie. 0FFE0

Step 8: Move the contents SI into AL register.

Step 9: Locate the contents in AL register to DX register using port out..

Step 10: Increment the value of SI.

Step 11: Decrement the counter and go to Step 8 until CX =0

Step 12: Jump to location step 5.

Step 13: Stop.



PROGRAM FOR SINE WAVE:

MOV DX, 0FFE6

MOV AL, 80

OUT DX, AL

RPT: MOV SI, 4000

MOV CL, 46

MOV DX, 0FFE0

L1: MOV AL, [SI]

OUT DX, AL

INC SI

LOOP L1

JMP RPT



CODE TABLE:

Base address

Effective address

Label Hex code Mnemonics Operands Comments

0000 2000 BA E6 FF MOV DX, 0FFE6 ;Initialize control word register

2003 B8 80 00 MOV AL, 80 ;Load AL with 802006 EF OUT DX, AL ;Read the data from

AL register using port out

2007 RPT BE 0040 MOV SI, 4000 ;Initialize the SI with 4000 ie,LOOK-UP TABLE

200A 8B 46 00 MOV CX , 46 ;Initialize the counter with 46h

200D BA E0 FF MOV DX, 0FFE0 Initialize port A

2010 L1 B8 C6 MOV AL, [SI] ;move the number in [SI] into al register

2012 EF OUT DX, AL ;read the data from Al register using port out

2013 46 INC SI ;Increment the value of SI

2014 E2 F5 LOOP L1 ;Decrement counter and go to L1 until CX=0

2016 EB EA JMP RPT ;Jump go to rpt with out condition



LOOK – UP TABLE:

EFFECTIVE ADDRESS

HEX CODES

4000 7F 8A 95 A0

4004AA B5 BF C8

4008D1 D9 E0 E7

400CED F2 F7 FA

4010FC FE FF FE

4014FC FA F7 F2

4018ED E7 E0 D9

401CD1 C8 BF B5

4020AA A0 95 8A

40247F 74 69 5F

402853 49 3F 36

402C2D 25 1D 17

403010 0B 07 04

403401 00 01 04

403807 0B 10 17

403C

1D 25 2D 364040

3F 49 53 5F4044

69 74

RESULT:



VIVA-VOCE QUESTIONS

82. How does the 8086 microprocessor instruction set is classified?

83. Which operation is performed by the Negate instruction?

84. What is the difference between RET and IRET instruction?

85. If AH=00, AL=0F then what is the content of AH and AL after AAA instruction?

86. What is the function of SAHF instruction?

87. What is the difference between IN and OUT instructions?

88. Which flags are affected after CMP (compare) instruction?

89. If AL=71, AH=81 then what is the content of AL after DAA instruction?



3.2. KEYBOARD INTERFACING

AIM: To display a string through interfacing 8279.

REGISTERS USED: AX, BX, CX, DX, SI

PORTS USED: command port, data port

CONNECTIONS: 8279 study card P1 to J2 of ESA 86/88 study card adapter.

DESCRIPTION:8279 study card provides keyboard as well as display section. The display section features

size 8 digit seven segment displays while the keyboard connections comprises a 4x 4 matrix hex key pad and associated circuitry. The options for using shift and controls keys during key scanning are also provided. The interface has 3 connectors, P1, J1, & J4 to interface the card with ESA 86/88E trainers. Connect the 50 pin FRC connectors P1 to connector J2 to ESA 86/88E study card adapter.

INTRODUCTION:In many microprocessors-based systems, calculator keypad is used as an input device. A

calculator keypad can be interfaced to a microprocessor using a dedicated peripheral controller like INTEL 8279Akeyboard/display controller. In this case, the controller can handle the interface problems like key debounce, 2-key lock-out, N-key roll-over etc,. Further such an alternative approach, the calculator keypad interface is passive and software is used for encoding the key positions and for handling problems like key debounce, roll-over etc.

The present interface module provides a calculator style calculator keypad consisting of the key 0 to 9 , + ,- , ×,= ,% , . , C, CE and two spare keys. These 20 keys are arranged in a 3×8 matrix (the third row has only four keys). The row lines can be driven through port C and the status of column lines can be read through port A. this interface allows the user to study a number of techniques generally used in calculator keypad interfacing. User can write programs for software debouncing of key closures, two key understanding of keyboard interface. Further , user can become familiar with the arithmetic group of processor instructions by implementing the calculator functions like addition, subtraction, multiplication , diversion, percentage etc..



THEORY: A programmable keyboard and display interfacing chip.

o Scans and encodes up to a 64-key keyboard. o Controls up to a 16-digit numerical display.

Keyboard has a built-in FIFO 8 character buffer

The display is controlled from an internal 16x8 RAM that stores the coded display information

PIN OUT DEFINITION 8279

A0: Selects data (0) or control/status (1) for reads and writes between micro and 8279. BD: Output that blanks the displays. CLK: Used internally for timing. Max is 3 MHz. CN/ST: Control/strobe, connected to the control key on the keyboard. CS: Chip select that enables programming, reading the keyboard, etc. DB7-DB0: Consists of bidirectional pins that connect to data bus on micro. IRQ: Interrupt request, becomes 1 when a key is pressed, data is available.


MICROPROCESSORS LAB MANUAL ECE Dept. OUT A3-A0/B3-B0: Outputs that sends data to the most significant/least significant nibble

of display. RD (WR): Connects to micro’s IORC or RD signal, reads data/status registers. RESET: Connects to system RESET. RL7-RL0: Return lines are inputs used to sense key depression in the keyboard matrix. Shift: Shift connects to Shift key on keyboard. SL3-SL0: Scan line outputs scan both the keyboard and displays.

8279 Interfaced to the 8088



Keyboard Interface of 8279

KEYBOARD INTERFACE OF 8279:

The keyboard matrix can be any size from 2x2 to 8x8.

Pins SL2-SL0 sequentially scan each column through a counting operation. o The 74LS138 drives 0’s on one line at a time. o The 8279 scans RL pins synchronously with the scan. o RL pins incorporate internal pull-ups, no need for external resistor pull-ups.

Unlike the 82C55, the 8279 must be programmed first.

D7 D6 D5 Function Purpose

0 0 0 Mode set Selects the number of display positions, type of key scan…

0 0 1 Clock Programs internal clk, sets scan and debounce times.



0 1 0 Read FIFO Selects type of FIFO read and address of the read.

0 1 1 Read Display Selects type of display read and address of the read.

1 0 0 Write Display Selects type of write and the address of the write.

1 0 1 Display write inhibit Allows half-bytes to be blanked.

1 1 0 Clear Clears the display or FIFO

1 1 1 End interrupt Clears the IRQ signal to the microprocessor.

o The first 3 bits of # sent to control port selects one of 8 control words.


First three bits given below select one of 8 control registers (opcode).

000DDMMM o Mode set: Opcode 000.

DD sets displays mode. MMM sets keyboard mode.

o DD field selects either: 8- or 16-digit display Whether new data are entered to the rightmost or leftmost display position.

DD Function

00 8-digit display with left entry

01 16-digit display with left entry

10 8-digit display with right entry

11 16-digit display with right entry




o MMM field:

DD Function

000 Encoded keyboard with 2-key lockout

001 Decoded keyboard with 2-key lockout

010 Encoded keyboard with N-key rollover

011 Decoded keyboard with N-key rollover

100 Encoded sensor matrix

101 Decoded sensor matrix

110 Strobed keyboard, encoded display scan

111 Strobed keyboard, decoded display scan

o Encoded: SL outputs are active-high, follow binary bit pattern 0-7 or 0-15. o Decoded: SL outputs are active-low (only one low at any time).

Pattern output: 1110, 1101, 1011, 0111. o Strobed: An active high pulse on the CN/ST input pin strobes data from the RL pins

into an internal FIFO for reading by micro later.

o 2-key lockout/N-key rollover: Prevents 2 keys from being recognized if pressed simultaneously/Accepts all keys pressed from 1st to last.

INTERFACE OF 8279

001PPPPP o The clock command word programs the internal clock driver. o The code PPPPP divides the clock input pin (CLK) to achieve the desired operating

frequency, e.g. 100KHz requires 01010 for a 1 MHz CLK input.

010Z0AAA o The read FIFO control word selects the address (AAA) of a keystroke from the

FIFO buffer (000 to 111). o Z selects auto-increment for the address.


MICROPROCESSORS LAB MANUAL ECE Dept. 011ZAAAA

o The display read control word selects the read address of one of the display RAM positions for reading through the data port.

100ZAAAA o Selects write address – Z selects auto-increment so subsequent writes go to

subsequent display positions.

INTERFACE OF 8279

1010WWBB o The display write inhibit control word inhibits writing to either the leftmost 4 bits of

the display (left W) or rightmost 4 bits. o BB works similarly except that they blank (turn off) half of the output pins.

1100CCFA o The clear control word clears the display, FIFO or both o Bit F clears FIFO and the display RAM status, and sets address pointer to 000.

If CC are 00 or 01, all display RAM locations become 00000000. If CC is 10, 00100000, if CC is 11, 11111111.

1110E000 o End of Interrupt control word is issued to clear IRQ pin in sensor matrix mode.

1) Clock must be programmed first. If 3.0 MHz drives CLK input, PPPPP is programmed to 30 or 11110.

INTERFACE OF 8279

2) Keyboard type is programmed next. o The previous example illustrates an encoded keyboard, external decoder used to

drive matrix.

3) Program the FIFO.

Once done, a procedure is needed to read data from the keyboard. o To determine if a character has been typed, the FIFO status register is checked.


MICROPROCESSORS LAB MANUAL ECE Dept.o When this control port is addressed by the IN instruction, the contents of the FIFO

status word is copied into register AL:

INTERFACE OF 8279

Code given in text for reading keyboard.

Data returned from 8279 contains raw data that need to be translated to ASCII:

o Row and column number are given the rightmost 6 bits (scan/return).

o This can be converted to ASCII using the XLAT instruction with an ASCII code lookup table.

o The CT and SH indicate whether the control or shift keys were pressed.

o The Strobed Keyboard code is just the state of the RLx bits at the time a 1 was `strobed’ on the strobe input pin.

ALGORITHM:

Step1: Move the value 00 to BX register.

Step2: Move the value 00 to AX register.

Step3: Move the 0FF42 to DX & out AX to DX.

Step4: Move the 90 to AL , and out value AL to DX.

Step5: Initialize counters CX with the value 08h

Step6: Move the 0ff40 to DX register and Load AL with 00

Step7: Write the value in AL register into DX register using port OUT.

Step8: Go to step 6 until equal to zero.

Step9: Move the address 2050 into SI register.

Step10: Move the value 0FF42 to DX register.

Step11: Read the value in DX register into AL register using port IN.

Step12: Add the contents of AL register with the value 07h.


MICROPROCESSORS LAB MANUAL ECE Dept.Step 13: Go to step 8 if the result not equal to zero.

Step14: Get the value 40 to AL register.


Step 16: move the 0ff40 to DX register.

Step 17: read the value in DX register into AL register using port IN.

Step 18: Logical AND ed the contents of AL with 01F.

Step19: Move the value of AL into BL register.

Step 20: Add the contents of SI with BX register.

Step21: Initialize data port and move the value 94 into AL register.

Step22: Write the data in AL register into DX register using port OUT.

Step 23: Initialize Command port and move the value F3 into AL register.


Step25: Initialize the data port and move the value 95 into AL register.

Step 26: Write the value in AL register into DX register using port OUT.

Step 27: Initialize the Command port.

Step 28: Move the contents of SI into AL register.


Step 30: Go to step 8.

Step31: Stop.

PROGRAM: MOV BX, 00H

MOV DX, 0FF42

MOV AL, 00

OUT DX, AL

MOV AL, 90


MICROPROCESSORS LAB MANUAL ECE Dept.OUT DX, AL

MOV CX, 08

MOV AL, 00

MOV DX, 0FF40

RPT: OUT DX, AL

LOOP RPT

BACK: MOV SI, 2050

MOV DX, 0FF42

KEY: IN AL, DX

AND AL, 07

JZ KEY

MOV AL, 40

OUT DX, AL

MOV DX, 0FF40

IN AL, DX

AND AL, 1F

MOV BL, AL

ADD SI, BX

MOV DX, 0FF42

MOV AL, 94

OUT DX, AL

MOV DX, 0FF40

MOV AL, 0F3

OUT DX, AL


MICROPROCESSORS LAB MANUAL ECE Dept.MOV DX, 0FF42

MOV AL, 95

OUT DX, AL

MOV DX, 0FF40

MOV AL, [SI]

OUT DX, AL

JMP BACK

CODE TABLE:


Op code operands

Comments

Segment address

Effective address

0000 2000 BB 00 00 MOV BX,00 Load 00 with BX register

2003 BA 42 FF MOV DX,0FF42 Initialize control port

2006 B0 00 MOV AL,00 Load 00 with Al register

2008 EE OUT DX,AL Routine to clear

2009 B0 90 MOV AL, 90 All display LEDs

200B EE OUT DX,AL

200C B9 08 00 MOV CX,08 Set the count

200F B0 00 MOV AL,00 Load 00 with AL register

2011 BA 40 FF MOV DX,0FF40 Initialize data port

2014 RPT EE OUT DX,AL Read the data from port

2015 E2 FD LOOP RPT Continuous loop

2017 BACK

BE 50 20 MOV SI, 2050 Initialize SI with 2050

201A BA 42 FF MOV DX,0FF42 Read 8279 status to check if any character is available

201D KEY EC IN AL,DX Write the data from port IN



201E 24 07 AND AL,07 Logical AND between [AL] and 07h

2020 74 FB JZ KEY Jump if condition

2022 B0 40 MOV AL,40 Load AL register with 40

2024 EE OUT DX,AL Read the data from port

2025 BA 40 FF MOV DX,0FF40 Read FIFO RAM

2028 EC IN AL,DX Mask SIFT

2029 24 1F AND AL,1F AND AL with 1F

202B 8A D8 MOV BL,AL Move the contents of AL into BL

202D 03 F3 ADD SI,BX Add SI with BX

202F BA 42 FF MOV DX,0FF42 Address to display 0

2032 B0 94 MOV AL,94 Load AL register with 94


2035 BA 40 FF

MOV DX,0FF40 Addressed control port

2038 B0 F3 MOV AL,F3 Move the value 0F3 into AL register

203A EE OUT DX,AL Read the data from port

203B BA 42 FF MOV DX,0FF42 Addressed data port

203E B0 95 MOV AL,95 Set 95 into AL register


2041 BA 40 FF MOV DX,0FF40 Initialize control port

2044 8A 04 MOV AL,[SI] Move the contents of SI into the AL register


2047 E9 CD FF JMP BACK Unconditional jump



INPUT DATA:

Segment address

Offset address

DATA INPUT

2050 F3 60 B5 F4

2054 66 D6 D7 70

2058 F7 76 77 C7

205C 93 E5 97 17

RESULT:

VIVA-VOCE QUESTIONS

90. What is the difference between opcode and operand?

91. What is the function of instruction pointer (IP) register?

92. What does assembler mean?

93. What is the difference between RCL and ROL instructions?

94. Which signal is used to restart the 8086 microprocessor?

95. What is the power supply required for 8086 microprocessor?



ADDITIONAL PROGARMS

DUAL SLOPE ANALOG TO DIGITAL CONVERTERS

AIM: To Convert digital data into Analog voltage.

INTRODUCTION:The analog to digital conversion can be done in many ways. One of the methods is the Dual

Slope method, used to achieve high noise immunity.The input voltage is integrated for a fixed time T1. Also a known reference voltage VR is

integrated for a time TX. Now the input voltage VX is given by

But =K (a constant)

Hence VX = K . TX I.e. the input voltage is proportional to the measured time TX. If integrated time TC is chosen as 20 m sec.the microprocessor measures time as the number of counts, namelyTX = NX. TC where

NX is number of counts. TX is counter period.By proper choice of the counter period TC scaling can be incorporated into the counter such that a software multiplication can be avoided.

THEORY: Voltage,current,temperature,pressure,time,etc, are available in analog form.ot is difficult to

process ,store or transmit the analog signal with out introducing considerable error because of the superimposition of noise as in the case of amplitude modulation. Therefore, for processing, transmission and storage purpose. It is often convenient to express these variables in digital form, it gives better accuracy system is based.

The operation of any digital communication system is based upon analog to digital and digital to analog conversion.

A/D converter requires sample and hold(S/H) circuit. The ADC output is a sequence of binary digit.D/A converter is to convert digital signal to analog signal and function of DAC is exactly opposite to that of ADC. The D/A converter is usually operated at the same frequency as the ADC.the output of D/A converter is commonly a stair case. This stair case like digital output is passed through a smoothing filter to reduce the effect of quantization noise.

ADC’s are classified broadly into two groups according to their conversion technique:1. Direct type ADC’s and2. Integrating type ADC’s.

Direct type ADC’s compare a given analog signal with the internally generated equivalent signal. This group includes:1. Flash (comparator) type converter2. Counter type converter.


MICROPROCESSORS LAB MANUAL ECE Dept.3. Tracking or servo converter4. Successive approximation type converter.

Integral type ADC’s perform conversion in an indirect manner by first changing the analog input signal to a linear function of time or frequency and then to a digital code .the two most widely used integrating type converters are:1. Charge balancing ADC.2. Dual slope ADC

The flash (comparator) type is expensive for high of accuracy the integrated type converter is used in application such as digital meter, panel meter and so on.

PARALLEL COMPARATOR (FLASH) A/D CONVERTER:It is simplest possible converter and the fastest and most expensive technique.

3-BIT A/D CONVERTER:Circuit consists of resistive divider network, 8 op-amp comparators and 8-5line encoder. At each node of the resistive divider, a comparison voltage is available .since all the resistors are of equal value, the voltage levels available at the nodes are equally divided between the reference voltage VR and the ground.

The purpose of circuit is to compare the analog input voltage VA with each of the voltages.The circuit has the advantage of high speed as the conversion take place

simultaneously rather than sequentially. Typical conversion time is loons or less. Conversion time is limited only by the speed of the comparator and of the priority encoder by using advanced micro devices AMA 686A comparator and a T1147 priority encoder ,conversion delays of the order of 20 no’s can be obtained.

This type of ADC has the disadvantage that the number of comparators required almost doubles for each added bit. A 2-bit ADC requires 3 comparators .in general, the number of comparators required are 2n-1 where n is the desired number of bits. Hence the number of comparators approximately doubles for each added bit. Also the larger the value of n the more complex is the priority encoder.

INTEGRATING TYPE OF ADC’S: the generating type of ADC’s do not require a S/H circuit at the input. If the input changes during conversion, the ADC’s output code will be proportional to the value of the input averaged over the integration period.

DUAL SLOPE ADC:

DUAL SLOPE OR DUAL RAMP CONVERTERS:

The analog part of the circuit consists of a high input impedance buffer A1, precision integrator A2 and a voltage comparator. The converter fort integrates the analog input signal VA for a fixed duration of 2n clock periods. Then to integrate an internal reference voltage VR of opposite polarity until the integrator output is zero. The number N of clock cycles required to return the integrated period. Hence N represents the desired output code. Since VR and n are constant, the analog voltage VA is proportional to the count reading N and is independent of R,C and T.

The dual-slope ADC integrates the input signal for a fixed time, hence to provides excellent noise rejection of ac signals whose periods are integral multiples of the integration time T1.

The main disadvantage of the dual-slope ADC is the long conversion time. For instance, if 2n-T =1 / 50 is used to reject line pick-up , the conversion time will be 20 m sec.


MICROPROCESSORS LAB MANUAL ECE Dept.Dual-slope converters are particularly suitable for accurate measurement of slowly varying

signal, such as thermo couples and weighing scales. Dual slope ADC’s also form the basics of digital panel meters and millimeters.

Dual slope converters are available in monolithic form and are available both in microprocessors compatible and in display oriented versions.

REGISTERS USED: AX, BX, CX, DX.

PORTS USED: Port A and Port B.

ALGORITHM:

Step1: Initialize 8255 Control Word Register and Port A as O/P port and Port B as I/P port.

Step2: Load the number 02 in AL register.

Step3: Initialize Port A address and read the data from port out.

Step4: Perform NOP five times.

Step5: Move the number 01h into AL register.

Step6: Read the data from port out.

Step7: Load CX register with 1000h.

Step8: Continue the loop if CX is not equal to zero.

Step9: Load AL with 04 and read the data from port out.

Step10: Initialized port A with 0FFE0 and initialized BX register with 0000h.

Step11: Initialized port B with 0FFE2 and write the data from register AL using port IN.

Step12: Logical AND operation AL register with 01h value.

Step13: Jump the corresponding condition to specify in instruction.

Step14: After the increment the BL register jump step 11.


MICROPROCESSORS LAB MANUAL ECE Dept.Step15: Move the contents of BL register into the AL register.

Step16: Initialize SI with 3000h and move the value in AL register into the SI location.

Step 17: Load AL register with 02h

Step 18: Initialize port A and read the data from port out.

Step19: Get the value 03h into BX register.

Step20: Initialize CX with 0FFFFh and continue the loop until CX is equal to zero.

Step21: Decrement the value of BX register and jump step 20 if BX equal to zero.

Step22: Jump step 2 Unconditional

Step23: Stop.

PROGRAM:MOV DX, 0FFE6

MOV AL, 82

OUT DX, AL

START: MOV AL, 02H

MOV DX, 0FFE0

OUT DX, AL

NOP

NOP

NOP

NOP

NOP


MICROPROCESSORS LAB MANUAL ECE Dept.MOV AX, 01

OUT DX, AL

MOV CX, 1000H

L2: LOOP L1

MOV AL, 04H

OUT DX, AL

MOV DX, 0FFE0

MOV BL, 00H

L1: MOV DX, 0FFE2

IN AX, DX

AND AL, 01

JE DISP

INC BL

JMP L1

DISP: MOV AL, BL

MOV SI, 3000H

MOV [SI], AL

MOV AL, 02H

MOV DX, 0FFE0

OUT DX, AL

MOV BX, 20

DELAY: MOV CX, 0FFFF

$: LOOP $


MICROPROCESSORS LAB MANUAL ECE Dept. JNZ DELAY

JMP START

CODE TABLE:

Physical address Label Hex code

Mnemonic

Op code operands

Comments

Segment address

Effective address

0000 4000 BA E6 FF MOV DX,0FFE6 Initialize the Control word

register4003 B0 82 MOV AL,82 Port A as output &

port B as input4004 EE OUT DX,AL

4006 START B0 02 MOV AL,02 Reset integrator

4008 BA E0 FF MOV DX,0FFE0 Addressed port A ie initializes.

400B EE OUT DX,AL Read the data from port out

400C 90 NOPNo operation

4011 90 NOP

4014 90 NOP

4015 90 NOP

4018 90 NOP

401A B8 01 00 MOV AX,01 Load AX with 01h

401C EE OUT DX,AL Read the data

401D B9 00 10 MOV CX, 1000H Set the count with 1000h

4020 L1 E2 FE LOOP L1 Continuous loop

4022 B0 04 MOV AL,04H Load AL with 04number

4025 EE OUT DX,AL Read the data

4026 BA E0 FF MOV DX,0FFE0 Initialized port A addressed

4028 B3 00 MOV BL,00 Clear BL register



402A L2 BA E2 FF MOV DX,0FFE2 Initialize port B

402C EC IN AL,DX Write the data from port IN

402E 24 01 AND AL, 01H AND ed operation between AL and 01

4030 74 04 JE DISP Jump conditionally4033 FE C3 INC BL Increment value of

BL4035 EB F4 JMP L2 Jump

Unconditionally4037 DSP 8A C3 MOV AL,BL Move the contents

from AL into BL403A B8 00 30 MOV SI,3000 Initialize SI with

3000h403B MOV [SI],AL Move the value in

AL register into contents of SI

403E B0 02 MOV AL,02 Load AL with the value 02h

4041 BA E0 FF MOV DX,0FFE0 Initialize port A

4043 EE OUT DX,AL Read the data from port out

4044 BB 20 00 MOV BX,20 Set the 20h in BX register

4046 DY B9 FF FF MOV CX,0FFFH Set the count with 0FFFFh value into

CX register4049 L3 F2 EE LOOP L3 Continuous loop

404B 4B DEC BX Decrement the value of BX

register404C 75 F8 JNZ DY Jump

Conditionally404E EB AD JMP START Jump

Unconditionally.

RESULT:



STEPPER MOTOR INTERFACING

AIM: program to design a stepper motor to rotate shaft of a 4 phase stepper motor in clockwise 15 rotations.

REGISTERS USED: General purpose registers: AL , DX , CX

PORTS USED: Port B, port C (out)

CONNECTIONS: J4 of ESA 86/88E to J1 of stepper motor.OPERATING PRINCIPLE OF PERMANENT MAGNET STEPPER MOTOR:

It consists of two stator windings A,B and a motor having two magnetic poles N and S. when a voltage +v is applied to stator winding A, a magnetic field Fa is generated. The rotor positions itself such that its poles lock with corresponding stator poles.

With the winding ‘A’ excited as before ,winding ‘b’ is now to Fa. the resulting magnetic field F makes an angle of 450. the rotor consequently moves through 450 in anti clockwise direction,again to cause locking of rotor poles with corresponding stator poles.


MICROPROCESSORS LAB MANUAL ECE Dept.While winding ‘B’ has voltage +V applied to it, winding ‘A’ is switched off. The rotor then

moves through a further 450 in anti-clockwise direction to aligne itself with stator field Fb. with voltage +V on winding B, a voltage –V is applied to winding A. then the stator magnetic field has two components Fa , Fb and their resultant F makes an angle of 1350 position.

In this way it can be seen that ,as the pattern of excitation of the state of winding is changed, the rotor moves successively through450 steps. And completes one full revolution in anti clock-wise direction. A practical PM stepper motor will have 1.80 step angle and 50 tooth on it’s rotor;there are 8 main poles on the stator, each having five tooth in the pole face. The step angle is given by

A = 360 / (N * K) degrees

Where N = number of rotor tooth.

K = execution sequence factor.

PM stepper motors have three modes of excitation i,e..

Single phase mode

Two phase mode

Hybrid mode

Single phase mode: in this mode only one of the motor winding is excited at a time. There are four steps in the sequence, the excitation sequence factor K=2 ,so that step angle is 900.

Two phase mode: Here both stators phase are excited at a time. There are four steps in the excitation sequence, K = 2 and the step angle is 900. However, the rotor positions in the two phase mode are 450 way from those in single phase mode.

Hybrid mode: this is a combination of single and two phase modes. There are 8 steps in excitation sequence=2 and step angle = 450. a voltage +V is applied to a stator winding during some steps, which voltage V is applied during certain other steps. This requires a bipolar regulated power supply capable of yielding +V,-V and zero outputs and a air of SPDT switches, which is quite cumbersome. Consequently each of the two stator windings is split into two sections A1-A2 , B1-B2. these sections are wound differentially. These winding sections can now be excited from a univocal regulated power supply through switcher S1 to S4. this type of construction is called bipolar winding construction. Bipolar windingesults in reduced winding inductance and consequently improved torque stepping rate.Description: the stepper motor interfaces uses four transistor pairs (SL 100 and 2N 3055) in a Darlington pair configuration. Each Darlington pair is used to excite the particular winding of the motor connected to 4 pin connector on the interface. The inputs to these transistors are from the 8255 PPI I/O lines of the microprocessor kit or from digital I/O card plugged in the PC. “port A” lower nibble PA0 , PA1, PA2 , PA3 are the four lines brought out to the 26 pin FRC male connector(J1) on the interface module. The freewheeling diodes across each winding protect transistors from switching transients.Theory:

A motor used for moving things in small increments is known as stepper motor. Stepper motor rotate from one fixed position to next position rather than continuous rotation as in case of other ac or dc motor stepper motors are used in printers to advance the paper from one position to advance the paper from one position to another in steps. They are also used to position the read/write head on the desired track of a floppy disk. To rotate the shaft the stepper motor a


MICROPROCESSORS LAB MANUAL ECE Dept.sequence of pulses are applied to the windings in a predefined sequence. The number of pulses required for one complete rotation per pulse is given by 3600/NT. where “NT” is the number of teeth on rotot. Generally the stepper motor is available with 10 to 300 rotation. They are available with two phase and four phase common field connections.

Instead of rotating smoothly around and around as most motors, stepper motors rotate or step one fixed position to next. Common step size range from 0.90 to 300. it is stepped from one position to next by changing the currents through the fields in the motor.

The two common field connections are referred to as two phase and four phase. The drive circuitry is simpler in 4 phase stepper. The figure shows a circuitry that can interface a small 4 stepper motor to four microcomputer port lines.

The 7406 buffers are inverting , so. A high on ah output port pin turns on current to a winding. The purpose of clamp diodes across each winging is to save transistors from inductive kick. Resistors R1 and R2 are current limiting resistors.Typical parameters of stepper motor:1. Operating voltage - 12 volts2. Current rating - 1.2 Amp3. Step angle - 1.80

4. Step for revolution - 200(No. of teeth on rotor)5. Torque - 3 kg/cmWorking of stepper motor:

Suppose that SW1 and SW2 are turned ON. Turning OFF SW2 and turning ON SW4 cause the motor to rotate one step of 1.80 clockwise. Changing to SW4 and SW3 ON will cause the motor to rotate 1.80 clockwise another. Changing SW3 and SW2 ON will cause another step. To step the motor in counter clock wise direction simply work through the switch sequence in the reverse direction.

The switch pattern for changing from one step to another step in clockwise direction is simply rotated right one position. For counter clockwise direction rotated left one position.

ALGORITHM:

Step 1: Start

Step 2: move the control word address 0FFE6 to register DX



Step 5: move port A address ie.,,0FFE0 to DX register.



Step 8: move 300 to CX register.

Step 9: repeat step 8 until the content in CX register becomes equal to zero.

Step 10: Rotate carry left through bit.

Step 11: jump to location / step 7.



Step 12: stop.


Start


FLOW CHART


out


out


1

Load CX register with 300

2

154



1

NO YES

Stop

Decrement CX &CX=0

Jump

2

Rotate carry left through bit

155


PROGRAM:

MOV DX, 0FFE6

MOV AX, 80

RPT: OUT DX, AX

MOV DX, 0FFE0

MOV AX, 0011

RPT: OUT DX, AX

MOV CX, 0300

L1: LOOP L1

RCL AL, 01H

JMP RPT



CODE TABLE:


Op code operands

Comments

Segment address

Effective address

0000 5000 BA E6 FF Mov DX , FFE6 Move FFE6 to DX register

5003 B8 80 00 Mov AX, 0080 Load AX with 80

5006 EF Out DX , AX Move AX content to DX

5007 BA E0 FF Mov DX , FFE0 Move FFE0 to DX register

500A Rpt B8 FF 00 Mov AX , 00FF Load AX with 00FF

500D EF Out DX,AX Move AX content to DX

500E E8 FF EF Call 6000

5011 B8 00 00 Mov AX , 0000 Load AX with 0000

5014 EF Out DX , AX Move the contents of AX into DX

5015 E8 E8 FF Call 6000

5018 E8 F0 Jmp Rpt Jump to 500A location

RESULT:


Documents

R05 Mp Lab Journal