MICRO PROCESSOR AND MICRO CONTROLLER LAB MANUAL

7/25/2019 MICRO PROCESSOR AND MICRO CONTROLLER LAB MANUAL

1/39

DEPARTMENT OF INFORMATION

TECHNOLOGY

IT6411 MICROPROCESSORS AND

MICROCONTROLLERS LAB

LAB MANUALS

FOURTH SEMESTER

REGULATION 2013

[Type the abstract of the document here. The abstract is typically a short summary of the contents of the

document. Type the abstract of the document here. The abstract is typically a short summary of the

contents of the document.]


2/39

cseitquesions.blogspot.in| cseitquesions.blogspot.in|cseitquesions.blogspot.in

IT Dept/KLNCE 2

ANNA UNIVERSITY

CHENNAI-600 025

DEPARTMENT OF INFORMATION TECHNOLOGY

YEAR / SEM : II YEAR / IV SEM REGULATION : AUC / 2013

IT6411 MICROPROCESSORS AND MICROCONTROLLERS LAB

S. No Title of Experiments Page No

8086 programs using kits and MASM

1 Basic arithmetic operationsLogical operations

2

6

2 Move a data block without overlap8086 Programs using Open source

11

13

3 Code conversion and decimal arithmetic andmatrix operation

4 Floating point operationString manipulation

Sorting and searching16

5 Password checking, Print RAM size and systemdate

6 Counters and Time delay

Peripherals and Interfacing Experiments

7 Traffic light control 17

8 Stepper motor Control 18

9 Digital clock 20

10 Keyboard, Display I/F 23

11 Printer status

12 Serial Interface and parallel interface 27

13 A/D Interface and D/A Interface and waveformgeneration

29

8051 Experiments using kits and MASM14 Basic arithmetic and logical operations 31

15 Square and cube ofprogram, Find 2s

complement of a number

16 Unpacked BCD to ASCII

STAFF-IN-CHARGE HOD/IT


3/39


IT Dept/KLNCE 3

EX.NO : 1 A ARITHMETIC OPERATIONS USING 8086

i.Aim:To perform arithmetic operations using 8086.

Apparatus:

1.Microprocessor kit

2.Power chord

3.Keyboard.

Two Byte Addition:

Algorithm:1.Initialize carry register CL with 00.Load AH and AL registers with MSB and LSB

of first data respectively. Load BH and BL registers with MSB and LSB of second

data.

2.Add accumulator register content with base register.

3.Check for carry flag.If carry exixts,increment carry register by one,else go to next

step.

4.Store accumulator content to 1500 and 1501 respectively.

5.Store carry register content to 1502 and stop program.

Program:Address Opcode Operand Label Mnemonics Comments

MOV CL,00

MOV

AX,0FFFEH

MOV BX,1010H

ADD AX,BX

JNC L1

INC CL

L1 MOV [1500],AXMOV [1502],CL

HLT

ii. Multi Byte Addition:

Algorithm:

1. Load SI with address of first byte of first data.

2. Load DI with address of first byte of second data.3. Clear the carry flag and initialize counter register with no.of data

bytes of data to be added.4. Add the content of memory pointed by DI with content of memory pointer by


4/39


IT Dept/KLNCE 4

DS:SI also with carry bit DI pointer acts as destination.

5. Update SI and DI register accordingly using DI and dec counter.6. If counter is not equal to zero,go to step 4,else go to next step.

7. Stop program.

Program

Address Opcode Operand Label Mnemonics Comments

MOV SI,2400

MOV DI,2404

CLC

MOV CL,04

L1 MOV AL,[SI]

ADC[DI],AL

INC SI

INC DI

JN2 L1

HLT

Iii. Two Byte Subtraction:

Algorithm:

1. Move 00 to CL register. Load first and second data to AX and BX.

2. Subtract AX with BXIf carry exists,increase CL count.

3. Store AX register and CL count value to memory location.

4. Stop the program.

Program:Address Opcode Operand Label Mnemonic

s

Comments

MOV

CL,00

MOV

AX,0FFFFH

MOV

BX,1010HSUBAX,BX

JNC L1

INC CL

L1; MOV[1500

],AX

MOV[1502

],CL

HLT


5/39


IT Dept/KLNCE 5

iv.Multibyte Subtraction:

Algorithm:

1. Load SI and DI with first and second data.

2. Clear the carry flag and initialize CL register with no.of data byte.

3. Subtract content of memory pointer by ES:DI from content of memory pointer

by ES:DI along carry bit with DI pointer acts as destination pointer.4. Update SI and DI register with carry bit DI pointer acting as decrement

counter.If counter !=0 then go to step 4,otherwise go to next step.5. Stop the program.

Program:


MOV SI,2400

MOV

DI,2404

CLCMOV CL,04

L1 MOV AL,[SI]

SBB[DI],AL

INC SI

INC DI

DEC CL

JNZ L1

HLT

v.8 Bit Multiplication:

Algorithm:

1. Load accumulator(AL) with 1st data and BL with 2nd data.

2. Multiply and store AX content into memory.

3. Stop the program

Program:


MOV

AL,0AAMOV

BL,0BB

MUL BL

MOV[3100],

AX

HLT

vi.16 Bit Multiplication:

Algorithm:1. Load AX with 1st data and multiply 2nd data with AX content.

2. Move MSB in DX & LSB in AX to memory and stop the program.


6/39


IT Dept/KLNCE 6

Program:


MOVAX,0FFFF

MOV

BX,1010

MUL BX

MOV[2902],D

X

MOV[2900],A

X

HLT

Vii.16/8 Bit Division:

Algorithm:1. Load AX with first 16 bit

2. Get 2nd data in BL register.

3. Divide AX with BL content

4. Store AX content in memory

5. Stop the program

Program:

Address Opcode Operand Label Mnemonic

s

Comments

MOVAX,0002

MOV

BL,01

DIV BL

MOV[2300],AX

HLT

Viii.32/16 Bit Division:Algorithm:

1. Load AX with LSB of 1st data and DX with MSB first data

2. Divide DX & accumulator content with 2nd data3. Store quotient (AX) into memory

4. Store remainder(DX) into memory

5. Stop the program.

Program:


s

Comments

MOV

AX,6562


7/39


IT Dept/KLNCE 7

MOV

DX,1232

MOVCX,1F20

DIV CX

MOV

[1200],AX

MOV[1202

],DX

HLT

Result:

Thus arithmetic operations using 8086 is done and program is executed and output is

verified.

Ex.NO:1.B 8086 programs using Logical Operations

i. ONE'S COMPLEMENT OF A 16-BIT NUMBER:

AIM:

To find the one's complement of the data in register pair AX and store the

result at 1400.

Apparatus required:

1.8086 Microprocessor kit

2.Power chord

3.Keyboard.

EXAMPLE:

The example given is to find the one's complement of 1234 and store it inmemory location 1400.

Input:

Data: (AX) = 0001 0010 0011 0100 = 1234

Result: [1400] = 1110 1101 1100 1011 = EDCB

PROGRAM:

MOV AX,1234 ; word in AX register

NOT AXMOV [1400],AX ; One's complement in


8/39


IT Dept/KLNCE 8

; AX and memory

HLT

OBJECT CODES:

Memory Address Object Codes Mnemonics

10001001

1002

1003

1004

1005

10061007

1008

1009

100A

C7C0

34

12

F7

D0

8906

00

14

F4

MOV AX,1234

NOT AX

MOV [1400],AX

HLT

ii. MASKING OFF BITS SELECTIVELY:

AIM:

To clear 8 selected bits,the 2nd

HN and the HN in a 16 bit number.

EXAMPLE:

The 16 bit number is at location 1200 and the result is at location 1400.

Input: [1200] = FF

[1201] = FF

Result: [1400] = 0F

[1401] = 0F

PROGRAM:

MOV AX,[1200] ;data in AX

AND AX,0F0F

MOV [1400],AX ;result in 1400

HLT

OBJECT CODES:


1000 8B MOV AX,[1200]


9/39


IT Dept/KLNCE 9

1001

1002

1003

1004

1005

1006

1007

10081009

100A

100B

100C

06

00

12

81

E0

0F

0F

8906

00

14

F4

AND AX,0F0F

MOV [1400],AX

HLT

iii. SETTING BITS SELECTIVELY IN A 16-BIT NUMBER:

OBJECTIVE:

To set the LN snd the 2nd

LN of a 16-Bit number in AX and store the

result at memory location 1100.

EXAMPLE:

The accumulator AX is initially cleared.Input: (AX) = 0000

Result: [1100] = 0F0F

PROGRAM:

MOV AX,0 ;clear accumulatorOR AX,0F0F ;OR with 0F0F

MOV [1100],AX ;Store result in memory

HLT

OBJECT CODES:


1000

1001

1002

1003

1004

1005

1006

10071008

C7

C0

00

00

81

C8

0F

0F89

MOV AX,0000

OR AX,0F0F

MOV [1100],AX


10/39


IT Dept/KLNCE 10

1009

100A

100B

100C

06

00

11

F4 HLT

iv. SUM OF THE NUMBERS IN A WORD ARRAY:OBJECTIVE:

To obtain the sum of a 16-bit array in memory, using index register SI and store the

result in memory,

EXAMPLE:

Data in array from START =1110

[1110]=0FFF[1112]=0FFF

[1114]=0FFF

[1116]=0FFF

[1118]=0FFF

PROGRAM:MOV CX, 5 ;number of elements in CX

MOV AX, 0MOV SI, AX ;initialise SI to start of array

LOOP1 : ADD AX,START[SI]ADD SI,2 ;decrement CX and check

;if Zero

LOOP LOOP1

MOV [SUM],AX

HLT

OBJECT CODES:


1000

10011002

1003

1004

1005

1006

10071008

1009

100A

100B

100C

C7

C105

00

C7

C0

00

0089

C6

03

06

10

MOV CX, 5

MOV AX, 00

MOV SI, AX

LOOP1 : ADD AX,[1100]


11/39


IT Dept/KLNCE 11

100D

100E

100F

1010

1011

1012

1013

10141015

1016

1017

1018

11

81

C6

02

00

E2

F6

8906

00

12

F4

ADD SI,2

LOOP LOOP1

MOV [1200],AX

HLT

EX.NO 3 STRING MANIPULATION OPERATIONS USING 8086

Aim:

1. To transfer a block of 10 data word from one memory to another in forward &

reverse direction

2. Block transfer in forward direction:

Algorithm

1. Move source index register with necessary offset address

2. Move destination index register with necessary offset address

3.

Load the count value in CX register4. Move string byte from source to destination address through accumulator

5. Increment the SI & DI registers

6. Decrement CX,if it is zero ,go to next step else step 4

7. Stop the program

Program


s

Comments

MOV

SI,1500

MOV

DI,1600

MOV

CX,0005

L1 MOV

AX,[SI]

L1 MOV

[DI],AX

INC SI

INC SI

INC DI


12/39


IT Dept/KLNCE 12

INC DI

L2 DEC CX

JNZ L1

HLT

Block transfer in reverse directionAlgorithm

1. Move source index register with necessary offset address2. Move destination index register with necessary offset address

3. Load the count value in CX register

4. Move content of memory pointed by source index reg to accumulator

5. Move the accumulator content to memory pointed by destination index register

6. Increment DI and decrement SI register

7. Decrement CX,if zero go to next step,else go to step 4

8.

Stop the programProgram


s

Comments

MOV

SI,1500

MOV

DI,1600

MOV

CX,0005L1 MOV

AX,[SI]

MOV

[DI],AX

DEC DI

DEC DI

INC SI

INC SI

DEC CXJNZ L1

HLT

Array addition

Algorithm

1. Load the counter reg CX with the number of data bytes to be added & also

clear AX.

2. Load SI reg with the starting address of the data

3.

Initialize the carry reg with zero.Add the numbers in the array pointed by SIwith content of AX along carry

4. Check for carry.If carry exists then increment the carr reg by 1 , then go to next


13/39


IT Dept/KLNCE 13

step

5. Update the SI content to point to the next data & decrement the CX regcount.Check for zero flag.If not equal to zero go to next step

6. Move the result in AX and carry reg to memory

7. Stop the program

Program


s

Comments

MOV

CX,0005

MOV

AX,0000

MOVBX,0000

MOVSI,1700

CLC

L2 ADC

AX,[SI]

JNZ L1

INC BX

L1 INC SI

INC SI

DEC CXJNZ L2

MOV[1600

],AX

MOV[1602

],BX

HLT

Even and odd number generation

1. Load count value to CX register

2.

Initialize DI register to a address & clear AL reg3. Load accumulator with 00 for even number generation and 01 for odd number

generation4. Move the content of accumulator to memory location pointed by DI

5. Add 02 to accumulator . Increment DI register

6. Decrement CX reg . If zero go to next step,else step 4

7. Stop the program

Even number generation


s

Comments

MOV


14/39


IT Dept/KLNCE 14

CX,0005

MOV

DI,1500

MOV

AL,00

L1 MOV

[DI],AL

ADD

AL,02

INC DI

DEC CX

JNZ L1

HLT

Odd number generation:Address Opcode Operand Label Mnemonics Comments

MOV

CX,0005

MOV

DI,1500

MOV

AL,01

L1 MOV

[DI],ALADD AL,02

INC DI

DEC CX

JNZ L1

HLT

Arithmetic operations using 8086 simulator(emuOpen source)

Aim:

To perform arithmetic operations using 8086.

Apparatus Required:

1.PC with 8086 emulator

Program: 16 bit addition

ORG 1000hMov ax,0002h


15/39


IT Dept/KLNCE 15

Mov bx,0013h

Mov cx,0000hAdd ax,bx

Jnc l1

Inc cxMov si,1400h

Mov[si],ax

Add si,0002hMov [si],cx

Hlt

SAMPLE DATAS:ADDITION

INPUT OUTPUT

Address/Reg Data Address Data

Ax

BxCX

004E

00FF0000

Ax

BxCX

01 4D

OOFFOOOO

16 bit subtraction

Mov ax,0005hMov bx,0003h

Mov cx,0000hsub ax,bx

Jnc l1

Inc cxMov si,1400h

Mov[si],ax

Add si,0002hMov [si],cx

Hlt

SAMPLE DATAS:SUBTRACTION

INPUT OUTPUT


Ax

Bx

1020

20 30

Ax

Bx

EFF0

2030


16/39


IT Dept/KLNCE 16

CX 0000 CX 0001

16 bit Multiplication

Mov ax,1020hMov bx,2020h

Mov cx,0000h

Mov si,1400hMul bx

Mov [si],ax

Add si,0002hMov[si],cx

hlt

SAMPLE DATAS:MULTIPLICATION

INPUT OUTPUT


AxBx

CX

102020 20

0000

AxBx

CX

DX

04002020

0000

0206

16 bit divisionMov ax,1020h

Mov bx,2020h

Mov cx,0000h

Mov si,1400

Div bx

Mov[si],axAdd si,0002h

Mov[si],cx

hlt

SAMPLE DATAS:DIVISION

INPUT OUTPUT


17/39


IT Dept/KLNCE 17


Ax

Bx

CX

0025

0005

0000

Ax

Bx

CX

DX

0007

2025

0000

0002

SORTING IN 8086:

org 1000h

mov ax,0505hL3:mov bx,1400h

L2:mov ax,[bx]cmp ax,0002h[bx]jc l1

xchg ax,0002[bx]

mov [bx],ax

L1:add bx,0002h

dec di

jnc L2

mov cl,chdec ch

jmp L3hlt

BLOCK TRANSFER [Forward ] :

mov si,1400h

mov di,1500h

mov cx,0005hL1:mov ax,[si]

mov [di],ax

add si,0002hadd di,0002h

dec cx

jnz l1mov [si],1500h

mov [si],ax

hlt


18/39


IT Dept/KLNCE 18

BLOCK TRANSFER [Reverse] :

mov si,1500hmov di,1400h

mov cx,0005h

L1:mov ax,[si]mov [di],ax

add di,0002h

add si,0002hdec cx

jnz l1

mov [si],1400hmov [si],ax

hlt

Ex.No .7 TRAFFIC LIGHT CONTROLLER

AIM:

To write a Program for Microprocessor based Traffic light system using

8086LCD mnemonics

APPARATUS REQUIRED:

8086 MP kit, 50 core cable,Traffic light controller interface

Procedure:

Enter the program starting from location 1000. Execute the program

The traffc light controller Led glows as per the given control word.Direction can be varied by entering the data in the look uptable in the reverse order.


1000

1003

1007

100A

100C

100E

1010

10121016

C6 C0 80

E6 26

C7 C3 73 10

C7 C6 7F 10

E8 33 00

8A 04

E6 20

E8 4D 0046

MOV AL,80H

MOV CL,04

MOV AL,[DI]

MOV C0,AL

MOV DX,1010

DEC DX

JNZ 100F

INC DILOOP 1007


19/39


IT Dept/KLNCE 19

1017

1019

101A

101C

101F

1020

43

E8 27 00

8A 04

E6 22

E8 41 00

46

43

E8 1B 00

JMP 1000

TABLE 09 05 06 0A

Ex.No:8

Stepper motor Interface with 8086AIM:

To run a stepper motor at different speed.

APPARATUS REQUIRED:

8086 MP kit, RS 232 cable, Stepper motor interface

Procedure:

Enter the program starting from location 1000. Execute the program

The stepper motor rotates. Speed can be varied by varying the count at dx register pair.

Direction can be varied by entering the data in the lookuptable in the reverse order.

PROGRAM


1000

1004

10071009

100B

100F1010

1012

1013

1015

1018

C7 C7 18 10

C6 C1 04

8A 05E6 C0

C7 C2 1010

4A75 FD

47

E2 F2

E9 E8 FF

09 05 06 0A

MOV DI,1018

MOV CL,04

MOV AL,[DI]MOV C0,AL

MOV DX,1010

DEC DXJNZ 100F

INC DI

LOOP 1007

JMP 1000

TABLE 09 05 06 0A

AIM:To run a stepper motor for required angle within 360 which is equivalent to 256

steps


20/39


IT Dept/KLNCE 20

Procedure:

Enter the program and execute it

By converting the requiring steps in decimal to hex and entering the hex data at 1001 the

motor rotates for so much steps and stops.

PROGRAM


1000

1003

1007

100A

100C

100E1010

1012

1016

1017

1019

101A

101C

101F

1020

C7 C3 45

C7 C7 20 10

C6 C1 04

8A 05

E6 C0

FE CB74 OD

C7 C2 10 10

4A

75 FD

47

E2 EE

E9 E4 FF

F4

09 05 06 0A

MOV DI,1018

MOV CL,04

MOV AL,[DI]

MOV C0,AL

MOV DX,1010

DEC DXJNZ 100F

INC DI

LOOP 1007

JMP 1000

TABLE 09 05 06 0A

AIM: To run a stepper motor in both forward and reverse direction with delay

APPARATUS REQUIRED:

8086 MP kit, RS 232 cable, Stepper motor interface

Procedure:

Enter the program and execute it after connecting the stepper motor in port1. Execute the

program. Now stepper motor runs in both forward and reverse direction continuously with a

delay

By converting the requiring steps in decimal to hex and entering the hex data at 1001 the

motor rotates for so much steps and stops.

PROGRAM



21/39


IT Dept/KLNCE 21

1000

1003

1007

100A

100C

100E

1011

10141018

101B

101D

101F

1022

1025

1028102A102C

1030

1031

1033

1034

1036

1037

103B103C

103E

103F

1043

C6 C3 20

C7 C7 3F10

E8 1B 00

FE CB

75 F5

E8 2A 00

C6 C3 20

C7 C7 43 10E8 0A 00

FE CB

75 F5

E8 19 00

E9 DB FF

C6 C1 04

8A 05E6 C0

C7 C2 10 10

4A 75 FD

47

E2 F2

C3

C7 C2 FF FF

4A

75 FDC3

09 05 06 0A

0A 06 05 09

MOV BL,20

MOV DI,103F

CALL 1025

DEC BL

JNZ 1003

CALL 103B

MOV BL,20

MOV DI,1043CALL 1025

DEC BL

JNZ 1014

CALL 103B

JMP 1000

MOV CL,04

MOV AL,[DI]OUT C0,AL

MOV DX,1010

DEC DX

JNZ 1030

INC DI

LOOP 1028

RET

MOV DX,0FFFF

DEC DXJNZ 103B

RET

FORWARD DATA

REVERSE DATA

RESULT:

EX.NO:9

DIGITAL CLOCKAim

To generate a digital clock in 8086 kit.

Store time value in memory location 1500-SECONDS

1501-MINUTES

1502-HOURS

Address Opcode Label Mnemonics Comments

1000 E8 00 60 START CALL

CONVERT

1003 E8 00 51 CALLDISPLAY


22/39


IT Dept/KLNCE 22

1006 B0 B0 DELAY: MOV

AL,0B0H

1008 E6 16 OUT 16H,AL

100A B1 07 MOV CL,07H

100C B0 88 S2: MOV AL,88H

100E E6 14 OUT 14H,AL

1010 B0 80 MOV A;,80H

1012 E6 14 OUT 14H,AL

1014 90 S1: NOP

1015 90 NOP

1016 90 NOP

1017 90 NOP

1018 E4 14 IN AL,14H

101A 8A D0 MOV DL,AL

101C E4 14 IN AL,14H

101E 0A C2 OR AL,DL

1020 75 F2 JNZ S1

1022 FE C9 DEC CL

1024 75 E6 JNZ S2

1026 BE 15 00 MOV

SI,1500H

1029 8A 04 MOV AL,[SI]

102B FE C0 INC AL

102D 88 04 MOV [SI],AL

102F 3C 3C CMP AL,3CH

1031 75 CD JNZ START

1033 B0 00 MOV AL,00H

1035 88 04 MOV [SI],AL

1037 46 INC SI

1038 8A 04 MOV AL,[SI]

103A FE C0 INC AL

103C 88 04 MOV [SI],AL

103E 3C 3C CMP AL,3CH

1040 75 BE JNZ START

1042 B0 00 MOV AL,0

1044 88 04 MOV [SI],AL

1046 46 INC SI

1047 8A 04 MOV AL,[SI]

1049 FE C0 INC AL

104B 88 04 MOV [SI],AL

104D 3C 18 CMP AL,18H

104F 75 AF JNZ START

1051 B0 00 MOV AL,0


23/39


IT Dept/KLNCE 23

1053 88 04 MOV [SI],AL

1055 EB A9 JMP START

1057 B4 06 DISPLAY: MOV

AH,06H1059 BA 1600 MOV

DX,1600H

105C B5 01 MOV

CH,01H

105E B1 00 MOV CL,0H

1060 CD 05 INT 5

1062 C3 RET

1063 BE 1500 CONVERT: MOVSI,1500H

1066 BB 1608 MOV

BX,1608H

1069 B0 24 MOV AL,24H

106B 88 07 MOV[BX],AL

106D 8A 04 MOV AL[SI]

;SECONDS

106F B4 00 MOV AH,01071 B6 0A MOV

DH,0AH

1073 F6 F6 DIV DH

1075 80 C4 30 ADD AH,30H

1078 4B DEC BX

1079 88 27 MOV[BX],AL

107B 4B DEC BX

107C 04 30 ADD L,30H107E 88 07 MOV

[BX]AL

1080 4B DEC BX

1081 B0 3A MOV

AL,3AH

1083 88 07 MOV

[BX],AL

1085 4B DEC BX

1086 46 INC SI


24/39


IT Dept/KLNCE 24

;MINUTES

1087 8A 04 MOV AL,[SI]

1089 B4 00 MOV AH,0

108B B6 0A MOV

DH,0AH

108D F6 F6 DIV DH

108F 80 C4 30 ADD AH,30H

1092 88 27 MOV

[BX],AH

1094 4B DEC BX

1095 04 30 ADD AL,30H

1097 88 07 MOV

[BX],AL

1099 4B DEC BX

109A B0 3A MOV

AL,3AH

109C 88 07 MOV

[BX],AL

109E 4B DEC BX

109F 46 INC SI

;HOURS

10A0 8A 04 MOV AL,[SI]

10A2 B4 00 MOV AH,010A4 B6 0A MOV

DH,0AH

10A6 F6 F6 DIV DH

10A8 80 C4 30 ADD AH,30H

10AB 88 27 MOV

[BX],AH

10AD 4B DEC BX

10AE 04 30 ADD AL,30H

10B0 88 07 MOV[BX],AL

10B2 C3 RET

10B3 E4 02 GETC: IN AL,02H

10B5 24 FF AND

AL,0FFH

10B7 3C F0 CMP

AL,0F0H

10B9 75 F8 JNE GETCEND


25/39


IT Dept/KLNCE 25

Result:

EX.NO:10 INTERFACING & PROGRAMMING WITH 8279Aim

A.To display a letter in display interfacing with 8279

B.To roll the display in 8279 using 8086

i)TO DISPLAY A IN THE FIRST DIGIT:

Program

Address Opcode

Operand

Label Mnemonics Comments

1000 B0 00 MOV AL,00

1002 E6 C2 OUT C2,AL

1004 B0 CC MOV

AL,0CC


1008 B0 90 MOV AL,90

100A E6 C2 OUT C2,AL

100C B0 88 MOV AL,88

100E E6 C0 OUT C0,AL

1010 B0 FF MOV AL,0FF

1012 B9 05 00 MOVCX,0005

1015 E6 C0 NEXT OUT C0,AL

1017 E2 FC LOOP NEXT

1019 F4 HLT

II) ROLLING DISPLAY

PROGRAM:

Address Opcode

Operand


1000 BE 00 12 START MOV SI,1200

1003 B9 0F MOV

CX,000F

1006 B0 10 MOV AL,10


100A B0 CC MOV AL,CC

100C B0 90 OUT C2,AL

100E E6 C2 MOV AL,90


1012 8A 04 NEXT MOV AL,[SI]



26/39


IT Dept/KLNCE 26

1016 E8 E7 04 CALL

DELAY

1019 46 INC SI

101A E2 F6 LOOP NEXT

101C EB E2 JMP START

1500 BA,FF,A0 DELAY MOV

DX,A0FF

1503 4A LOOP1 DEC DX

1504 74,FD JNZ LOOP1

1506 C3 RET

1200 FF,FF,FF,FF

FF,FF,FF,FF

98,68,7C,C8FF,1C,29,FF

Ex.No :12a. Serial Interface (8251) Interface with 8086AIM:

To send the data serially to another serial port.

APPARATUS REQUIRED:

8086 MP kit, RS 232 cable

Procedure:

Connect the two serial ports with RS232C Cable. First make the receiver ready and

Then execute the program at transmitter end.

EXAMPLE:

Data in Transmitter :41H

Receiver [1500]:41H

OBJECT CODES:TRANSMITTER END



27/39


IT Dept/KLNCE 27

1000

1002

1004

1006

1008

100A

100C

100E1010

1012

1014

1016

1018

101A

101C101E

B0 36

E6 CE

B0 10

E6 C8

B0 00

E6 C8

B0 4E

E6 C2B0 37

E6 C2

E4 C2

24 04

74 FA

B0 41

E6 C0CD 02

MOV AL,36

OUT CE,AL

MOV AL,10

OUT C8,AL

MOV AL,00

OUT C8,AL

MOV AL,4E

OUT C2,ALMOV AL,37

OUT C2,AL

LOOP1: IN AL,C2

AND AL,04

JZ LOOP1

MOV AL,41

OUT C0,ALINT2

Receiver end


1200

12021204

1206

1208120A

120C

120E

1210

12121214

1216

1218

121A121C

B0 36

E6 CEB0 10

E6 C8

B0 00E6 C8

B0 4E

E6 C2

B0 37

E6 C2E4 C2

24 04

74 FA

B0 41E6 C0

MOV AL,36

OUT CE,ALMOV AL,10

OUT C8,AL

MOV AL,00OUT C8,AL

MOV AL,4E

OUT C2,AL

MOV AL,37

OUT C2,ALLOOP 2: IN AL,C2

AND AL,02

JZ LOOP2

IN AL,C0MOV BX,1500


28/39


IT Dept/KLNCE 28

121E CD 02 MOV [BX],AL

INT2

RESULT:

______________________________________

Ex.No:12 b.

Parallel Interface(8255) Interface with 8086AIM:

To initialize Port A as an input port in mode -0.

APPARATUS REQUIRED:

8086 MP kit, 50 core cable, 8255 kit

Procedure:

1. Connect the 8086 kit with 8255 using core cable.

2. Enter the program starting from user RAM address 1000H set a known data in SPDT switch

3.

Type and execute the program in 8086 kit

4. Verify the result at 1500H as data data given through SPDT switch

OBJECT CODES:To initialize PORT A as input port in mode 0.



29/39


IT Dept/KLNCE 29

1000

1003

1005

1007

1009

100B

BE 00

B0 90

E6 C6

E4 C0

88 04

F4

MOV SI,1500H

MOV AL,90H

OUT C6,AL

IN AL,C0

MOV [SI],AL

HLT

AIM:To initialize Port A as an input port and port B as output port in mode 0

APPARATUS REQUIRED:


Procedure:


5. Set a known data in SPDT switch

6. Type and execute the program in 8086 kit

7. Verify visually that the data output at the LEDs is same as that set by the SPDT switch

settings

OBJECT CODES:To initialize PORT A as input port and port B as output port in mode-0


1000

1003

1005

1007

1009

100B

BE 00

B0 90

E6 C6

E4 C0

88 04

F4

ORG 1000H

MOV AL,90H

OUT C6,AL

IN AL,C0

OUT C2,AL

HLT

AIM:To initialize Port C as an Output port and port B in mode 0 and to explain the BIT

set and reset features of PORT C

APPARATUS REQUIRED:



30/39


IT Dept/KLNCE 30

PROCEDURE:


8.

Enter the program starting from user RAM address 1000H

9.

Type and execute the program in 8086 kit

10.

Verify the result at LEDS after changing the datas

OBJECT CODES:To initialize PORT A as input port and port B as output port in mode-0


1000

10021004

10061008

100A

100B

100C

B0 80

E6 C6B0 01

E6 C4B0 07

E6 C6

F4

ORG 1000H

MOV AL,80OUT C6,AL

MOV AL,01OUT C4,AL

MOV AL,07

OUT C6,AL

HLT

Ex.No 13.i. INTERFACING & PROGRAMMING WITH ADC

Aim:

To design an ADC converting using 8086

Algorithm:

1. Select the channel and insert the ALE by sending corresponding data C8

2. Assert the SOC(Start of Conversion) signal high and provide some delay

3. After the delay make the SOC signal low and wait for the EOC(End of

Conversion)

4.

Read the converted digital data and stores it in the memory location.Procedure:

Place jumper J2 in A position

Place jumper J5 in A position

Enter and execute the program

Vary the analog input (using trimpot ) and verify the digital data displayed with

that data stored in memory location 1100H

Program:

Address Opcode/Oper

and


1000 C6 C0 10 MOV AL,10


31/39


IT Dept/KLNCE 31


1005 C6 C0 18 MOV AL,18


100A E6 C0 01 MOV AL,01

100D E6 D0 OUT D0,AL

100F C6 C0 00 MOV AL,00

1012 C6 C0 00 MOV AL,00

1015 C6 C0 00 MOV AL,00

1018 C6 C0 00 MOV AL,00

101B E6 D0 OUT D0,AL

101D E4 D8 LOOP IN AL,D8

101F 80 E0 01 AND AL,01

1022 80 E0 01 CMP AL,01

1025 75 F6 JNZ LOOP

1027 E4 C0 IN AL,C0H

1029 C7 C3 00 11 MOV

BX,1100

102D 88 07 MOV

[BX],AL

102F F4 HLT

Result:

Ex.No 13.ii. INTERFACING & PROGRAMMING WITH DACAim

To generate the square wave at the output of DAC 2.

Algorithm

1. Initialize AL and move 00 to AL

2. The accumualator content of AL is then sent to C8

3. After calling 1010 the content of FF is moved to accumulator content AL

4. Again the content of AL is sent to C8 and 1010 is called

5. The 05FF is moved to the CX and looped the content to1013

6.

Return the programProgram

Address Opcode Mnemonics Comments

1000 C6 C0 00 MOV AL,00


1004 E8 09 00 CALL 1010

1007 B0 FF MOV AL,FF


100B E8 02 00 CALL 1010

100E EB F0 JMP 1000

1010 B FF 05 MOV CX,05FF


32/39


IT Dept/KLNCE 32

1013 E2 FE LOOP 1013

1015 C3 RET

2.AIM: To create a saw tooth wave at the output of DAC

Algorithm:

1.

Initialize AL and move the first data 00 to AL2. After moving to AL,the accumulator content of AL is sent to c0

3. The accumulator content of AL is ncremented by 1

4. After incrementing ,if it is non zero ,jump to 1002

5. After checking for non zero , jump to 1000

Program:

Address Opcode Label Mnemonics Comments

1000 B0 00 L2 MOV AL,00

1002 E6 C0 L1 OUT C8,AL

1004 FE C0 INC AL1006 75 FA JNZ L1

1008 EB F6 JMP L2

3.Aim:

To generate the triangular waveform at DAC 2 output

Algorithm

1. Initialize BL and move 00 to BL2. Move the content of BL to AL as its content

3.

The content moved to accumulator content AL is then sent to C84. The content of BL is incremented by 1

5. If the data is m=non zero , then jump to STEP2

6. Move the content of FF to BL & then to AL

7. After moving content to AL,it is sent to C88. Decrement BL by 1

9. If it is non zero , then jump on to 100C

10.Jump to step 1

Program

Address Opcode Label Mnemonics Comments1000 B3 00 L2 MOV BL,00

1002 88 D8 MOV AL,BL


1006 FE C3 INC BL

1008 75 F8 JNZ 1002

100A B3 FF MOV BL,FF

100C 88 D8 L1 MOV AL,BL

100E E6 C8 OUT C8,AL

1010 FE CB DEC BL1012 75F8 JNZ L1


33/39


IT Dept/KLNCE 33

1014 EB EA JMP L2 JUMP TO L2

EX.NO 14 ARITHMETIC OPERATIONS USIGN 8051AIM

To write an ALP to perform

Add 2 given 8 bit numbers

Subtract 2 given 8 bit numbers

Multiply 2 8 bit numbers

Divide 2 8 bit numbers

Array addition

APPARATUS REQUIRED

1.8051 Microprocessor kit2.Eliminator

Algorithm

8 Bit Addition

1.Load the first data into accumulator & 2nd data in RO register.

2.Initialize the carry register(Rl) with OoH and initialize the data pointer with

address of the result location.

3.Add the accumulator content with the second data.

4.Check for carry flag if carry exits then increment carry register by one otherwise

5. go to next step.

6.Store the accumulator and Rl content into memory whose address pointed by

data pointer.

7.Stop the program.

Program

Address Opcodes Mnemonics Comments

MOV R1,#00H

MOV A,#FEH

MOV R0,#10H

MOV DPTR,#4500

ADD A,R0

JNC L1

INC R1

L1 MOVX @DPTR,A

INC DPTR

MOV A,R1

MOVX @DPTR,A


34/39


IT Dept/KLNCE 34

L2 SJMP L2

8 BIT SUBTRACTION

ALGORTITHM

1.Load the first data into accumulator & 2nd data in r0 register

2.Initialize the carry register (R1) with 00H and initialize the data pointer withaddress of the result location

3.Subtract the second data from accumulator content

4.Check for carry flag if carry exists then increment carry register by one by one

otherwise go to next step

5.Store the accumulator and R1 content into memory whose address pointed by datapointer

6.Stop the program

Address Opcode Operand Label Mnemonics CommentsMOV R1,#00H

MOV A,#FEH

MOV R0,#10H

MOV DPTR,#4500

SUB A,R0

JNC L1

INC R1

L1 MOVX @DPTR,A

INC DPTRMOV A,R1

MOVX @DPTR,A

L2 SJMP L2

8 BIT MULTIPLICATION

ALGORITHM

1. Load the accumulator and B register with first and second data

2. Multiply the B register with content of A register

3. Initialize the data pointer with address of the result location

4.

Store the accumulator content into the memory location pointed bythe data pointer and then increment data pointer by one

5. Move the B register content to A

6. Store A's content into memory location pointed by DPTR

7. Stop the program

Program


MOV A,#05H

MOV B,#03HMOV DPTR,#4500


35/39


IT Dept/KLNCE 35

MUL A,B

MOVX @DPTR,A

INC DPTR

MOV A,B

MOVX @DPTR,A

L2 SJMP L2

DIVISION OF 8 BIT NUMBERS

ALGORITHM

1. Load accumulator and B register with divided and divisor data

respactively

2. Divide the accumulator content by the content of B register

3. Initialize the data pointer with the address of the result location

4. Store the accumulator content into the memory location pointed by

the pointer and then increment the data pointer by one5. Move the B register content to the accumulator

6. Store the accumulator content into the memory location pointed

out by the data pointer7. Stop the program

Program


MOV A,#05HMOV B,#03H

MOV DPTR,#4500

DIV A,B

MOVX @DPTR,A

INC DPTR

MOV A,B

MOVX @DPTR,A

L1 SJMP L1

8 BIT ARRAY ADDITION

ALGORITHM1. Initialize the data pointer with the starting address of the source

array , also initialize the countert (R0) with number of bytes going

to be added

2. Initialaize the sum register (R2) and carry register (R1) with 00H3. Load accumulator with data byte pointed by the data pointer

4. Add the accumulator content with content of sum register(B) and

store the result in the same sum register

5.

Check for carry flag status if carry is set,then incrementcarry(R1)by one,else go to next step


36/39


IT Dept/KLNCE 36

6. Increment tha data pointer by one to point to the next data byte

7. Decrement the count register (R0) by one and check whether thedecremented value equal to zero or not,if not go to step 4 else go

to next step

8. Store the content of sum register(B) and carry (R1) into the 2

consecutive memory locations on pointed by data pointer9. Stop the program

Program:Array addition


MOV DPTR,#4700

MOV R0,#05H

MOV R2,#00H

MOV R1,#00H

L2 MOVX A,@DPTR

ADD A,R2MOV R2,A

JNC L1

INC R1

L1 INC DPTR

DJNZ R0,L2

MOV A,R2

MOV @DPTR,A

INC DPTR

MOV A,R1MOVX @DPTR,A

L3 SJMP L3

Result

EX.NO 14. LOGICAL AND BIT MANIPULATION IN 8051

(B) i LOGICAL OPERATORS

AIM:

To perform logical & but manipulation in 8051ALOGRITHM

1. Get 1stbyte in R1

2. Get 2nd

operand in R2

3. Move R1 to Accumulator

4. And the Accumulator with R2

5. Move the content of the accumulator to R0.

6. Move R1 to Accumulator

7. Or the Accumulator with R2

8.

Move Accumulator content to R39. Move R1 to Accumulator


37/39


IT Dept/KLNCE 37

10.Take compliment for the Accumulator content

11.Move R4 to Accumulator

Program:


MOV R1,#20H First BYTE

MOV R2 #10H Second BYTE

MOV A,R1 ANDING

R0=R1 AND R2ANL A, R2

MOV R0, A

MOV A,R1 ORING

R3=R1 OR R2ORL A, R2

MOV R3,A

MOV A, R1 NEGATION

R4=~R1CPL A

MOV R4,A

MOV A,R1 XORING

R5=R1 XOR R2XRL A,R2

MOV R5,A

LCALL 0003H

11. (B)ii. 8051HEXADECIMAL TO DECIMAL CONVERSION

AIM:

To perform hexadecimal to decimal conversionALGORITHM:

1.Load the number to be converted into the accumulator

2.If the number is less than 100(64H), go to next step; otherwise, subtract

100 (64H) repeatedly until the remainder is less than 100(64H). Have thecount (100s value) in separate register which is the carry.

3.If the number is less than 10 (0AH), go to next step; otherwise, subtract

10 (0AH) repeatedly until the remainder is less than 10(0AH). Have thecount (tens value) in separate register.

4.The accumulator now has the units5.Multiply the tens value by 10 and add it with the units. 6.Store the result and carry in the specified memory location.


MOV DPTR, #4500

MOVX A, @DPTR

MOV B, #64

DIV A, B

MOV DPTR, #4501

MOV @DPTR, AMOV A,B


38/39


IT Dept/KLNCE 38

MOV B, #0A

DIV A, B

INC DPTR

MOVX @DPTR, A

INC DPTR

MOV A,B

MOVX @DPTR,A

HLT: SJMP HLT

Result:

The given hexadecimal number is converted into decimal number

11.(b)iii. 8051DECIMAL TO HEXADECIMAL CONVERSION

AIM:

To perform decimal to hexadecimal conversion

ALGORITHM:1.Load the number to be converted in the accumulator

2.Separate the higher order digit from lower order.

3.Multiply the higher order digit by 10 and add it with the lower order digit

4.Store the result in the specified memory location


MOV DPTR, #4500

MOVX A, @DPTR

MOV B, #0AMUL A,B

MOV B, A

INC DPTR

MOVX A ,@DPTR

ADD A,B

INC DPTR

MOVX @DPTR, A

MOV DPTR, #4500

MOVX A, @DPTRHLT: SJMP HLT

RESULT:The given decimal number is converted to hexadecimal number


39/39


MICRO PROCESSOR AND MICRO CONTROLLER LAB MANUAL

Documents

Transcript of MICRO PROCESSOR AND MICRO CONTROLLER LAB MANUAL