RCE MICROPROCESSORS & MICROCONTROLLERS LAB

MICROPROCESSORS & MICROCONTROLLERS

LABORATORY MANUAL

(R13) III – B. Tech., II-Semester

ECE

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

RAMACHANDRA COLLEGE OF ENGINEERING, ELURU – 534 007

Accredited by NAAC with “B++” Grade

(Approved by AICTE, New Delhi & Affiliated to JNTUK: Kakinada)

West Godavari District, Andhra Pradesh

RCE MICROPROCESSORS & MICROCONTROLLERS LAB

Department of Electronics & Communication Engineering 2

Microprocessors & Microcontrollers Lab

(R13) III – B. Tech., ECE II- Semester

Index

PART-I:

MICROPROCESSOR 8086

1. Introduction to MASM/TASM. 2. Arithmetic operation- 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 (Buffered with and without echo) - Display characters, Strings.

PART-II: INTERFACING WITH MICROPROCESSOR

1.8259 – Interrupt Controller-Generate an interrupt using 8259 timer. 2.8279 – Keyboard Display- Write a program to display a string of characters.

3.8255 – PPI-Write ALP to generate sinusoidal wave using PPI.

4.8251 – USART-Write a program in ALP to establish Communication between two processors.

PART-III:

MICROCONTROLLER 8051

1. Reading and Writing on a parallel port.

2. Timer in different modes. 3. Serial communication implementation.

PART-IV: INTERFACING WITH MICROCONTROLLER

1. Simple Calculator using 6 digit seven segment display and Hex Keyboard interface to 8051. 2. Alphanumeric LCD panel and Hex keypad input interface to 8051.

3. External ADC and Temperature control interface to 8051. 4. Generate different waveforms Sine, Square, Triangular, and Ramp, etc. using DAC interface to 8051; change the frequency and Amplitude.



EX: No: 1

1. INTRODUCTION

ESA 86/88-2 is a powerful general purpose micro computer system which can be

operated with 8086 CPU. It is generally supplied with 8086 CPU to 8086 uses has to just

remove 8086 as set 8086 into that socket & set an output switch.8086 & 8088 are 3 rd

generation CPU from Intel that differ primary in thirds external data paths 8086 uses 16 bit

can be generated either 8086 or 8088.

The possible difference would be in the speed of execution in either case the CPU is

operated in the max mode 1.5A: 86/88-2 can be used not only as an industrial, instrumental &

learning aid but also a development tool in R & D labs. It is a versatile system and can be

configured in a number ways as per the user requirements various options the basic system

can be easily expanded through the system bus connection. The built in assembled features

supplies the task. Case of entering assembly language program on board provision for 8087

data process makes ESA 86/88-2 useful for a number of on-board provisions is an added

feature to make a care of the frequent power failure while conducting experiment on the

trainer.

To aid the user further various optional attachments like EARUM programming audio

cassettes interface are available. ESA 86/88-2 is operated from circuit terminals or a

computer system is a connection from a keyboard

1. Examine & optionally modify the contents of memory

2. Examine & optionally modify the contents of process register

3. Assemble & de assemble 8086/88 instruction

4. Perform fast numerical computation using the optional 8087 numerical data

5. Execute the user program at the full speed user program through single step and break

Point facilities.

6. Write/read data to/from input/output parts

7. Move a block of data or program within the memory

8. Download user program into ESA 8086/88-2 from fast counter

9. Transfer code and data to/from audio cassettes is a optional audiocassette interface

10. Read the program verify and branch check EPROM or vice optional PROM

Programmable module

11. Obtain a hard copy via controls complicated parallel printer interface



SPECIFICATIONS:

CONTROL PROCESSOR: 8086 CPU or 8086 operating at 5 MHz with maximum mode

(supplied with 8086 CPU) memory cycles has wait slates and i/o cycles

CO-PROCESSOR: On board 8087 numeric data processor memory EPROM: 4JEDEC

compatible 8 bit EPROM offer the following options: 64k bytes using 27/285 or:128k bytes

using 27/2565 or :256 K bytes using 275/255 System firmware is supplied in 2x27565 the

other two sockets are for user expansion

RAM: 4JEPEC compatible slots after the following 128kb using 622565 64 KB using

2*6022565. The RAM has also a battery backup facility peripherals and controller.

8251A: programmable communications interface for social communication supporting all

standards from 110-19200.

825-3-5: programmable interval time „0‟ is used for generating basic block timer 1 and timer

11 are available for users

8255A: programmable peripherals interface devices provide us programmable I/O devices.

8288: bus controller uses for generating control signal interrupts

EXTERNAL: NMI interrupts for INTR key and keyboard INTR controlled through 8259A

on board interrupts controller provides interrupt vector for B sources complete flexibility for

selecting either off board interrupt source.

8251/JXRD4 and RXRD4

8253/OUT 1 OUT II

8255/PCD and PC3 in hand shake mode

8087/NDDI INT

INTERNAL: Interrupt sector 1-single step &31 break points monitor

CPUSIGNAL: De-multiplexed and fully differentiate signals are available on 2.50 per

ribbon cable connection

PARALLEL I/O: As per programmable I/O line/TTL compatible serial

SERIAL I/O: RS232C on-board b-type formal connection

POWER SUPPLY: +5 V at 3.0A



DEBUG COMMANDS

Assemble A[address]

Compare c[compare]

Dump d[manage]

Enter E[address list]

Fill F[range list]

Ev G[address][address]

Flux H[value][value]

Load L[address]

Input I[port]

Move M[range address]

Proceed N[path name][org list]

Quit Q

Register R[register]

Search S[range list]

True T[address][value]

Unassembled U[range]

Viva Questions

1. What is Microprocessor? 2. Define Compiler 3. Define Interpreter 4. Define Assembler 5. What is Assembly level language? 6. Abbreviate MASM & TASM 7. Differentiate micro & macro 8. Define two pass assembler 9. What are the advantages of modular programming? 10. What is linking and relocation?

11. Explain about .MODEL SMALL?



FLOW CHART:

RESULT:

GIVEN INPUT: OUTPUT:

AL= AL=

BL= BL=

MANUPLATION:

AL=

BL=

START

Copy the immediate

numbers to AX, BX registers

Perform addition on

contents of AX, BX registers

STOP



ARITHMETICOPERATIONS

EX: No: 2

ADDITION

AIM: To write an assembly language program to perform addition operation

REQUIRED SOFTWARE: MASM611

Starting address: Ending address:

ALGORITHM:

1. Start the program

2. Move the input value into the specified register.

3. Move the other value into the other specified register.

4. Perform the arithmetic addition between the two registers & store the result in the

register specified

5. Stop

PROGRAM:

Address Hexadecimal code Label Mnemonic Operands Comments

MOV AL, Copy immediate number to

AL register

MOV BL, Copy immediate number to

BL register

ADD AL,BL Add contents of AL and BL

registers

INT 03 Terminate the program

RESULT:



FLOW CHART:

RESULT:


AL= AL=

BL= BL=

MANUPLATION:

AL=

BL=

Start

Copy the immediate numbers

to AX, BX registers

Perform the subtraction between the contents of AX and BX registers

STOP



SUBTRACTION

AIM: To write an assembly language program to perform subtraction operation.


Starting address:

Ending address:

ALGORITHM:

1. Start the program.

2. Move the specified immediate numbers to AL, BL registers.

3. Subtract the contents of BL from the contents of AL register the result will be stored

in the destination register.

4. Stop the program.

PROGRAM:

Address Hexadecimal code Label Mnemonic Operand Comments

MOV AL,

Copy the immediate number

to AL register

MOV BL,

Copy the immediate number

to BL register

SUB AL,BL

Subtract content of BL with

the content of AL


RESULT:



FLOWCHART:

RESULT:


AL= AL=

BL= BL=

MANUPLATION:

AL=

BL=

START

Copy the immediate

numbers to AL, BL registers

Perform multiplication on contents of AL

and BL registers

STOP



MULTIPLICTION

AIM: To write an assembly language program to perform multiplication operation.


Starting address:

Ending address:

ALGORITHM:


2. Copy the immediate numbers to AL, BL registers.

3. Perform multiplication operation between the contents of AL and BL register the

result will be stored in the destination register specified.

4. Terminate the program.

PROGRAM:


MOV AL,

Copy the immediate number to

AL register

MOV BL,


BL register

MUL BL

Multiply content of AL with the

contents of BL register


RESULT:



FLOW CHART:

RESULT:


AL= AL=

BL= BL=

MANUPLATION:

AL=

BL=

START

Copy the immediate

number to AX, BL register

Divide the content of AL register with content of BL register

STOP



SIGNED MULTIPLICTION

AIM: To write an assembly language program to perform signed multiplication operation.


Starting address:

Ending address:

ALGORITHM:



7. Perform multiplication operation between the contents of AL and BL regis ter the

result will be stored in the destination register specified.


PROGRAM:


MOV AX,


AL register

MOV BX,


BL register

NEG BX

MUL BX

Multiply content of AL with the

contents of BL register

MOV CX,AX

NEG CX


RESULT:



FLOW CHART:

RESULT:


AL= AL=

BL= BL=

MANUPLATION:

AL=

BL=

START

Copy the immediate

number to AX, BL register

Divide the content of AL register with content of BL register

STOP



DIVISION

AIM: To write an assembly language program to perform division operation.


Starting address:

Ending address:

ALGORITHM:



3. Divide the content of AL register with content of BL register & store the result in

specified destination register.


PROGRAM:


MOV AL,

Copy the immediate number to AL

register

MOV BL,

Copy the immediate number to BL

register

DIV BL

Divide the content of AL register

with content of BL register


RESULT:



FLOWCHART:

No

Yes

START

Move the immediate number to CX

register

Add the content of DI and the content of

AX with carry

Increment SI & DI address values by

twice

Decrement CX by

once

STOP

Initialize the SI and DI with desired address

values

Copy the content of SI into AX register

CX=0 =



SIGNED DIVISION

AIM: To write an assembly language program to perform signed division operation.


Starting address:

Ending address:

ALGORITHM:



7. Divide the content of AL register with content of BL register & store the result in

specified destination register.


PROGRAM:


MOV AX,

Copy the immediate number to AL

register

MOV BX,

Copy the immediate number to BL

register

NEG AX

DIV BX

Divide the content of AL register

with content of BL register

MOV CX,AX

NEG CX


RESULT:



FLOWCHART:

No

Yes

START

Move the immediate number to CX

register

Add the content of DI and the content of

AX with carry


twice

Decrement CX by

once

STOP


values


CX=0 =



MULTI BYTE OPERATIONS

Ex: No: 2

MULTIBYTE ADDITION

AIM: To write an assembly language program to perform Multibyte addition using

MASM611.


Starting address:

Ending address:

ALGORITHM:


2. Move the immediate number to CX register.

3. Initialize the source index pointer SI and destination index DI with the specified

address value to point out the desired memory locations.

4. Copy the content of SI into AL register.

5. Perform the addition with carry operation between the content of DI and the content

of AL & store the result in specified destination register.

6. Increment the SI and DI address values by 2 times.

7. Decrement the value of CX register by one.

8. Repeat the step5 to step8 until the condition “CX=0” has been satisfied.

9. When “CX=0” then continue the next step of the main program.




RESULT:

INPUT:

Source index:

Input address Value

Destination index:

Input address Value

OUTPUT:

Destination index:

Output address Value



PROGRAM:


MOV CX,0003 Move the immediate number to CX

register

MOV SI, Initialize the source index pointer SI

with desired address value

MOV DI, Initialize the source index pointer DI


Loop1: MOV AX,[SI] Copy the content of SI into AX

register

ADC [DI],AX Add the content of DI and the content

of AX with carry

INC SI Increment SI

INC SI Increment SI

INC DI Increment DI

INC DI Increment DI

DEC CX Decrement CX

JNZ Loop1 Jump to loop1 address if CX register is

not equal to zero


RESULT:



FLOWCHART:

NO

YES

START

Move the immediate

number to CX register

Subtract the content of AX from the

content of DI with borrow


twice

Decrement CX by

once

STOP


values


CX=0 =



MULTIBYTE SUBTRACTION

AIM: To write an assembly language program to perform Multibyte subtraction using

MASM611.


Starting address:

Ending address:

ALGORITHM:


2. Move the immediate number to CX register.



4. Copy the content of SI into AX register.

5. Perform the subtraction with borrow operation between the content of DI and the

content of AX & store the result in specified destination register.

6. Increment the SI and DI address values by 2 times.

7. Decrement the value of CX register by one.

8. Repeat the step5 to step8 until the condition “CX=0” has been satisfied.

9. When “CX=0” then continue the next step of the main program.




RESULT:

INPUT:

Source index:

Input address Value

Destination index:

Input address Value

OUTPUT:

Destination index:

Output address Value



PROGRAM:


MOV CX,0003 Move the immediate number to CX

register

MOV SI, Initialize the source index pointer SI


MOV DI, Initialize the source index pointer DI


Loop1: MOV AX,[SI] Copy the content of SI into AX

register

SBB [DI],AX Subtract the content of AX from the

content of DI with borrow

INC SI Increment SI

INC SI Increment SI

INC DI Increment DI

INC DI Increment DI

DEC CX Decrement CX

JNZ Loop1 Jump to loop1 address if CX register is

not equal to zero


RESULT:



MULTI BYTE MULTIPLICATION

DATASEGET

BYTESEQU 08H

NUM1DB 01H,02H,03H,04H,05H,06H,07H,08H

NUM2DB 02H,02H,02H,02H,02H,02H, 02H, 02H

NUM3DB 0AH DUP (00)

DATAENDS

CODESEGMENT

ASSUME CS: CODE,

DS: DATA

START: MOVAX, DATA

MOVDS, AX

MOVCX, BYTES

LEASI, NUM1

LEADI, NUM2

LEABX, NUM3

MOVAX,00

NEXT: MOV AL,[SI]

MOV DL,[DI]

MUL DL

MOV [BX], AL

MOV [BX+1], AH

INC SI

INC DI

INC BX

INC BX

DEC CX

JNZ NEXT

INT 3H

CODE ENDS

END START



INPUT:

OUT PUT:

RESULT:



MULTI BYTE DIVISION

DATA SEGMENT

BYTES EQU 08H

NUM2DB 02H,04H,06H,08H,0AH,0CH,0EH,10H

NUM1DB 02H,02H,02H,02H,02H,02H, 02H, 02H

NUM3DB 0AH DUP (00)

DATAENDS

CODE SEGMENT

ASSUME CS: CODE, DS: DATA

START: MOVAX, DATA

MOV DS, AX

MOVCX, BYTES

LEA SI, NUM1

LEA DI, NUM2

LEA BX, NUM3

NEXT: MOVAX, 00

MOV AL,[SI]

MOV DL,[DI]

MUL DL

MOV[BX], AL

MOV [BX+1],AH

INC SI

INC DI

INC BX

INC BX

DEC CX

JNZ NEXT

INT 3H

CODE ENDS

ENDSTART

INPUT:

OUTPUT: RESULT:



FLOWCHART:

RESULT:


AX= AX=

BL= BL=

START

Copy the immediate contents to AL, BL

register

ADD the contents of AL and BL

registers

Perform ASCII Adjust after

addition operation

STOP



ASCII ARITHMETIC OPERATIONS EX: No: 2

ASCII ADDITION

AIM: To write an assembly language program to perform ASCII adjust after addition.


Starting address:

Ending address:

ALGORITHM:


2. Move the input byte to specified register.

3. Move another byte to specified register.

4. Add contents of specified registers.

5. Perform AAA operation and store the result in specified destination register.

6. Stop

PROGRAM:

Address Hexadecimal

Code Label Mnemonic Operand Comments

MOV AL,33h Copy immediate number to AL

register.

MOV BL,32h Copy immediate number to BL

register.

ADD AL,BL Add contents of AL and BL

registers

AAA Perform the operation ASCII

adjust after addition

OR AX,3030H


RESULT:



FLOWCHART:

RESULT:


AX= AX=

BL= BL=

START

Copy the immediate contents to AL, BL

register

Subtract the content of BL from the

content of AL register

Perform the operation ASCII

adjust after subtraction

STOP



ASCII SUBTRACTION

AIM: To write an assembly language program to perform ASCII adjust after subtraction.


Starting address:

Ending address:

Algorithm:


2. Move the input byte to specified registers.


4. Subtract the contents of specified register from the content of another register.

5. Perform AAS operation and store the result in specified register.

6. Stop.

PROGRAM:

Address Hexadecimal


MOV AL,39H Copy immediate number to

AL register.

MOV BL,36H Copy immediate number to

BL register.

SUB AL,BL

Subtract the content of BL

from the content of AL

register.

AAS Perform the operation ASCII

adjust after subtraction

OR AX,3030H


RESULT:



FLOWCHART:

RESULT:


AX= AX=

BL= BL=

START

Copy the immediate

contents to AL, BL register

Multiply the content of BL with the content of AL register

Perform the operation ASCII adjust after multiplication

STOP



ASCII MULTIPLICATION

AIM: To write an assembly language program to perform ASCII adjust after multiplication.


Starting address:

Ending address:

ALGORITHM:




4. Multiply the contents of specified register with the content of another register.

5. Perform AAM operation and store the result in specified register.

6. Stop.

PROGRAM:

Address Hexadecimal


MOV AL,36H Copy immediate number to AL

register

MOV BL,32H Copy immediate number to BL

register

SUB AL,30H

SUB BL,30H

MUL BL Multiply the content of BL

with the content of AL register

AAM Perform the operation ASCII

adjust after multiplication

OR AX,3030H


RESULT:



FLOWCHART:

RESULT:


AX= AX=

BL= BL=

RESULT:

START

Copy the immediate contents to AX, BL

register

Perform the operation ASCII adjust before division

Divide the content of AX with the content of BL

STOP



ASCII DIVISION

AIM: To write an assembly language program to perform ASCII adjust before division.


Starting address:

Ending address:

ALGORITHM:




4. Perform AAD operation.

5. Divide the content of specified register with the content of another register.

6. Stop.

PROGRAM:

Address Hexadecimal


MOV AL,39H Copy immediate number to AL register

MOV BL,33H Copy immediate number to BL register

SUB AL,30H

SUB BL,30H

AAD Perform the operation ASCII adjust before division

DIV BL Divide the content of AX with the content of BL

OR AX,3030H


VIVA QUESTIONS

1. What is an extended accumulator? 2. AL and BL are used for multiplying why not AX & BX? 3. Instead of using MOV BL is it not possible to MUL num2? 4. Why AL & BL are not used in signed multiplication? 5. If result exceeds 32 bit where is it stored? 6. What is the name given to the register combination DX: AX? 7. What happens if DX register contains a nonzero value before DIV instruction? 8. In a program instead of DIV BX is it possible to use DIV num2? 9. Describe about MUL, IMUL, DIV, IDIV instructions 10. What is the purpose of ASCII addition?



FLOW CHART:

RESULT:


AX= AX=

BX= BX=

MANUPLATION:

AX=

BX=

START

Copy values in AX and BX registers

Perform AND operation between the contents of AX &

BX registers

STOP



LOGICAL OPERATIONS Ex: No: 3

Logical AND

AIM: To write an assembly language program to perform logical AND operation.


Starting address:

Ending address:

ALGORITHM:


2. Move the input byte to specified register

3. Move the other byte to specified register

4. Perform logical AND operation between AX and BX registers and store the result in

specified register

5. Stop

PROGRAM:


MOV AX, Copy immediate number to AX

register

MOV BX, Copy immediate number to BX

register

AND AX,BX Perform AND operation between

the contents of AX & BX registers


RESULT:



FLOW CHART:

RESULT:


AX= AX=

BX= BX=

MANUPLATION:

AX=

BX=

START


Perform OR operation between


STOP



Logical OR

AIM: To write an assembly language program to perform logical OR operation.


Starting address:

Ending address:

ALGORITHM:




4. Perform logical OR operation between AX and BX registers and store the result in

specified register

5. Stop

PROGRAM:



register


register

OR AX,BX Perform OR operation between



RESULT:



FLOW CHART:

RESULT:


AX= AX=

BX= BX=

MANUPLATION:

AX=

BX=

START


Perform XOR operation between the contents of AX & BX registers

STOP



LOGICAL XOR

AIM: To write an assembly language program to perform logical XOR operation.


Starting address:

Ending address:

ALGORITHM:




4. Perform logical OR operation between AX and BX registers and store the result in

specified register

5. Stop

PROGRAM:



register


register

XOR AX,BX Perform X OR operation between



RESULT:



FLOW CHART:

RESULT:


AX= AX=

MANUPLATION:

AX=

STAR

T

Copy the values in AX register

Perform NOT operation on the

content of AX register

STOP



LOGICAL NOT

AIM: To write an assembly language program to perform logical NOT operation.


Starting address:

Ending address:

ALGORITHM:


2. Move the input byte to specified register.

3. Perform logical NOT operation on the content of AX register.

4. Stop

PROGRAM:

Address Hexadecimal

code Label Mnemonic Operand Comments


register

NOT AX Perform NOT operation on the

content of AX register


RESULT:



FLOW CHART:

RESULT:


AX= AX=

CL= CX=

MANUPLATION:

AX=

CL=

START

Copy the immediate contents to AX, CL

register

Perform shift left operation on

the content of AX register

STOP



SHIFT AND ROTATE OPERATIONS

EX: No:3

SHIFT LEFT OPERATION

AIM: To write an assembly language program to perform the shift left operation.


Starting address:

Ending address:

ALGORITHM:


2. Move the immediate number to AX register.

3. Move the count value to CL register.

4. Perform the shift left operation on the content of AX register.

5. Stop

PROGRAM:



register

MOV CL, Copy immediate number to CL

register

SHL AX,CL Perform shift left operation on



RESULT:



FLOW CHART:

RESULT:


AX= AX=

CL= CX=

MANUPLATION:

AX=

CL=

TART

Copy the immediate

contents to AX, CL register

Perform shift right operation on


STOP



SHIFT RIGHT OPERATION

AIM: To write an assembly language program to perform the shift right operation.


Starting address:

Ending address:

ALGORITHM:




4. Perform the shift right operation on the content of AX register.

5. Stop

PROGRAM:



register


register

SHR AX,CL Perform shift right operation

on the content of AX register


RESULT:



FLOW CHART:

RESULT:


AX= AX=

CL= CX=

MANUPLATION:

AX=

CL=

START


register

Perform rotate left operation on


STOP



ROTATE LEFT OPERATION

AIM: To write an assembly language program to perform the rotate left operation.


Starting address:

Ending address:

ALGORITHM:




4. Perform the rotate left operation on the content of AX register.

5. Stop

PROGRAM:



register


register

ROL AX,CL Perform rotate left operation



RESULT:



FLOW CHART:

RESULT:


AX= AX=

CL= CX=

MANUPLATION:

AX=

CL=

START


register

Perform rotate right operation on


STOP



ROTATE RIGHT OPERATION AIM: To write an assembly language program to perform the rotate right operation.


Starting address:

Ending address:

ALGORITHM:




4. Perform the rotate right operation on the content of AX register.

5. Stop

PROGRAM:



register


register

ROR AX,CL Perform rotate right operation



RESULT:



FLOW CHART:

START

Clear the contents of AX BX

register

Initialize the SI with desired address

value

Copy the content of SI into AX

register

Copy the content of AL to

BL register

Perform AND operation between the content of AL

&0F

Move the count value to CL

register

Perform rotate right operation on the content of BL register

Copy the content of BL in AH register & the content of AX in SI address value

STOP

Perform AND operation between the content of BL &0F0



EX:No:3

CONVERSIONS

PACKED BCD TO UNPACKED BCD

AIM:

To write an assembly language program to convert packed BCD to unpacked BCD.


Starting address:

Ending address:

ALGORITHM:


2. Clear the contents of AX,BX registers

3. Initialize the source index pointer (SI) with the specified address value to point out the

desired memory location.

4. Copy the content specified memory location to AL register

5. Copy the content of AL register to BL register

6. Perform AND operation between the content of AL register and the immediate value

0F.

7. Perform AND operation between the content of BL register and the immediate value

0F0.

8. Initialize the CL register with the desired count value.

9. Perform rotate right operation on the content of BL register.

10. Copy the content of BL register to AH register.

11. Copy the content of AX register to the specified memory location.




RESULT:

GIVEN INPUT:

3000-

OUTPUT:

3000-

3001-

AX=



PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 2000H

MOV AX, 0000H

MOV BX, 0000H

MOV SI,3000H

MOV AL,[SI]

MOV BL,AL

AND AL,0FH

AND BL,0F0H

MOV CL,04H

ROR BL,CL

MOV AH,BL

MOV [SI],AX

INT 03

.EXIT

END

RESULT:



FLOWCHART:

Now copy the content of SI in BL register

Initialize CLregister with the required count

value

START

Clear the registers AX and

BX

Initialize SI with some address value

Copy the content of Si in AL

register

Increment the memory location

pointed by SI by one time

The content of AL register is shifted left by 4 times

ADD content of AL register with the content of BL register

STOP



UNPACKED BCD TO PACKED BCD

AIM:

To write an assembly language program to convert unpacked BCD format to packed BCD

format.


Starting address:

Ending address:

ALGORITHM:


2. Clear the contents of AX and BX registers.



4. Move the value of SI memory location to the specified AL register.

5. Increment the value of SI.

6. Move the value of memory location pointed by SI BL register.

7. Initialize CL register with the required count value.

8. Perform shift left operation on the content of AL register.

9. Add the content of AL register with the content of BL register.




RESULT:

GIVEN INPUT:

4000-

4001-

OUTPUT:

AX=



PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 2000H

MOV AX,0000H

MOV BX,0000H

MOV SI,4000H

MOV AL,[SI]

INC SI

MOV BL,[SI]

MOV CL,04H

SHL AL,CL

ADD AL,BL

INT 03

.EXIT

END

RESULT:



FLOW CHART:

START

Clear the register AX & BX

Initialize the specified memory location with some

value

Initialize the CL register with some count

value

Copy the content of specified memory location to AL register

Copy the content of AL to BL register

Perform AND operation between specified

values

Shift the content of BL register to right with specified number of bits

ADD the content of AL register with 30

ADD the content of BL register with 30

STOP



PACKED BCD TO ASCII FORMAT

AIM:

To write an assembly language program to convert packed BC D format to ASCII format.


Starting address:

Ending address:

ALGORITHM:


2. Clear the contents of AX, BX registers.



4. Initialize CL register with some count value.

5. Copy the content of specified memory location to AL register.

6. Copy the content of AL to BL register.

7. Perform AND operation between the content of AL and immediate number 0F.

8. Shift the content of BL register to left with specified number of bits.

9. Add the content of AL register with the immediate number 30.

10. Add the content of BL register with the immediate number 30.




RESULT:

GIVEN INPUT:

3000-

OUTPUT:

AX=

BX=



PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 3000H

MOV AX,0000H

MOV BX,0000H

MOV CL,04H

MOV SI,3000H

MOV AL,[SI]

MOV BL,AL

AND AL,0FH

SHR BL, CL

ADD AL,30H

ADD BL,30H

INT 03

.EXIT

END

RESULT:



FLOWCHART:

START

Clear the contents of AX, BX registers

Initialize the specified memory location with some address values

Initialize CL registers with some value

Copy the contents of specified memory locations to AL, BL register

Perform AND operation between the contents of BL register

and 0F

Shift the content of BL to left with specified number of

bits

Perform OR operation between the contents of AL, BL and store result in specified register

STOP



ASCII TO PACKED BCD

AIM:

To write an assembly language program for the conversion of numbers in ASCII format

to packed BCD format.


Starting address:

Ending address:

ALGORITHM:


2. Clear the contents of AX,BX registers

3. Initialize CL register with some count value.



5. Copy the contents of specified memory location to AL and BL register.

6. Perform AND operation between the content BL register and immediate number 0F.

7. Shift left the content of AL register with the specified number of bits.

8. Perform OR operation between the contents AL & BL registers.

9. Stop the program



RESULT:

GIVEN INPUT:

6000-

6002-

OUTPUT:

AX=



PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 2000H

MOV AX,0000H

MOV BX,0000H

MOV CL,04H

MOV SI,6000H

MOV DI,6002H

MOV AL,[SI]

MOV BL,[DI]

AND BL,0FH

SHL AL,CL

OR AL,BL

INT 03

.EXIT

END

RESULT:

VIVA QUESTIONS

1. What is the purpose of the instruction ROR AL, CL? 2. What is the purpose of the instruction AND AL, 0FH & AND AL, 0F0H in the

program? 3. What is the expansion of UPBCD? 4. What is the use of DAA instruction? 5. What is the reason for packing unpacked BCD? 6. What is common between unpacked BCD and ASCII? 7. What is the difference between adding 30h and OR 30H to a BCD number to conversion

to ASCII? 8. Why unpacking is necessary during the conversion? 9. What is the ASCII character for symbol A?

10. What is the ASCII character for symbol zero „0‟?



FLOWCHART:

YES

NO

YES

NO YES

NO

Decrement the value of CH by one

START

Copy the content of SI into CL register

Decrement the value of CL

by one

Copy the content of SI into CH

register Decrement the value of CH

by one Increment the value of SI memory location by one

Copy the content of SI address into

AL register

Increment the value of SI memory location by one

Compare the values of AL and the specified

memory location

If AL<SI Exchange the contents of AL & SI and also the contents of AL & SI-

01 memory location

If CH!=

0

Initialize the specified memory

location with some address

values

Decrement the value of CL by one

If CL!=

0 TOP



SORTING & STRING OPERATIONS

EX: No:4

ASCENDING ORDER

AIM:

To write an assembly language program to perform ascending order on the given string.


Starting address:

Ending address:

ALGORITHM:




3. Copy the content of SI memory location into CL register.

4. Decrement the value of CL register.

5. Copy the content of SI memory location into CH register.

6. Decrement the value of CH register.

7. Increment the value of memory location pointed by SI by once.

8. Now copy the content of SI memory location into AL register.


10. Compare the content of AL with the content specified memory location pointed by SI.

11. If the content of AL is less than the content of SI jump to step-14 otherwise continue with

the next step.

12. Perform exchange operation between the contents of AL and SI memory location.

13. Perform exchange operation between the contents of AL and [SI-01] memory location.

14. Decrement the content of CH register by one.

15. If the content of CH!=0 then go to step-8.

16. Decrement the content of CL register by one.

17. If the content of CL!=0 then go to step-5.

18. Stop the program



RESULT:

INPUT:

Source Index

Address Values

OUTPUT:

Source index:

Address Values



PROGRAM:

.MODEL SMALL

.286

.CODE

ORG 2000H

MOV SI,0500H

MOV CL,[SI]

DEC CL

LOOP1: MOV SI,0500H

MOV CH,[SI]

DEC CH

INC SI

LOOP2: MOV AL,[SI]

INC SI

CMP AL,[SI]

JB LOOP3

XCHG AL,[SI]

XCHG AL,[SI-01]

LOOP3: DEC CH

JNZ LOOP2

DEC CL

JNZ LOOP1

INT 03

.EXIT

END

RESULT:



FLOWCHART:

YES

NO

NO

YES NO

YES NO

Copy the content of SI address into AL

register

Increment the value of SI memory location by

one Compare the values of AL and the specified memory

location

If AL>SI

Exchange the contents of AL & SI and also the contents of AL & SI-01 memory location

Decrement the value of CH by one

If CH!=0

Decrement the value of CL by

one

If CL!=0

STOP

START

Copy the content of SI into CL register

Decrement the value of CL by one

Copy the content of SI into CH register

Decrement the value of CH by

one

Increment the value of SI memory location by one

Initialize the specified memory location with some address

values



DESCENDING ORDER

AIM:

To write an assembly language program to perform descending order on the given string.


Starting address:

Ending address:

ALGORITHM:




3. Copy the content of SI memory location into CL register.


5. Copy the content of SI memory location into CH register.

6. Decrement the value of CH register.

7. Increment the value of memory location pointed by SI by once.

8. Now copy the content of SI memory location into AL register.


10. Compare the content of AL with the content specified memory location pointed by SI.

11. If the content of AL is greater than the content of SI jump to step-14 otherwise continue

with the next step.

12. Perform exchange operation between the contents of AL and SI memory location.

13. Perform exchange operation between the contents of AL and [SI-01] memory location.

14. Decrement the content of CH register by one.

15. If the content of CH!=0 then go to step-8.

16. Decrement the content of CL register by one.

17. If the content of CL!=0 then go to step-5.

18. Stop the program



RESULT:

INPUT:

Source Index

Address Values

OUTPUT:

Source index:

Address Values



PROGRAM:

.MODEL SMALL

.286

.CODE

ORG 2000H

MOV SI,0500H

MOV CL,[SI]

DEC CL

LOOP1: MOV SI,0500H

MOV CH,[SI]

DEC CH

INC SI

LOOP2: MOV AL,[SI]

INC SI

CMP AL,[SI]

JNB LOOP3

XCHG AL,[SI]

XCHG AL,[SI-01]

LOOP3: DEC CH

JNZ LOOP2

DEC CL

JNZ LOOP1

INT 03

.EXIT

END

RESULT:

Microprocessor & Microcontroller Lab

Dept. of ECE RCE

FLOW CHART:

YES

NO

START

T

Clear the contents of AX & BX registers


values

Initialize CL register by a

count value

Copy the content of specified memory location

into AL register

Copy the content of AL

register into specified memory location [DI]

Increment SI & DI memory locations

by once


by one

If CL!=0

STOP


Dept. of ECE RCE

MOVE BLOCK OF DATA

AIM:

To write an assembly language program to move a block of data from one memory location

to

another memory location.


Starting address:

Ending address:

ALGORITHM:


2. Clear the contents of AX & BX registers.



4. Load the contents in specified memory locations of SI & DI before execution.

5. Copy the immediate number in the CL register.

6. Copy the content of specified memory location [SI] into AL.

7. Copy the content of AL register into specified memory location [DI].

8. Increment the address value of the specified memory locations SI and DI by once.


10. If CL!=0 goto step-6 until CL=0.



Dept. of ECE RCE

RESULT:

INPUT:

OUTPUT:

Source index Values

Destination index Values


Dept. of ECE RCE

PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 2000H

MOV AX,0000H

MOV BX,0000H

MOV SI,

MOV DI,

MOV CL,

LOOP1: MOV AL,[SI]

MOV [DI],AL

INC SI

INC DI

DEC CL

JNZ LOOP1

INT 03

.EXIT

END

RESULT:


Dept. of ECE RCE

FLOWCHART:

YES NO

NO

If

CL!=0

START

Initialize the AX

register

Initialize the specified memory location to SI & DI

Initialize CL register with some count

value

Copy the content of SI and DI into AL and BL

registers

Compare the contents of AL & BL registers

If

AL!=BL

Increment SI & DI locations by

once


register

Copy the immediate number to AX

register

STOP

Copy content in the AX register

STOP


Dept. of ECE RCE

STRING COMPARISON

AIM:

To write an assembly language program to compare two strings.


Staring address:

Ending address:

ALGORITHM:


2. Initialize the AX register .



4. Load the contents in specified memory locations of SI & DI before execution.

5. Copy the immediate number in the CL register.

6. Copy the content of specified memory location [SI] into AL and [DI] into BL.

7. Compare the contents of AL & BL register.

8. If the content of AL is not equal to the content of BL then go to step-13.

9. Increment the value of SI & DI by once.


11. If CL!=0 goto step-6.

12. Copy the immediate number to AX register and terminate the program.

13. Copy the given immediate content to AX register.



Dept. of ECE RCE

RESULT:

INPUT:

Source Index: Destination Index:

OUTPUT:

When Strings are equal: AX=

When Strings are not equal: AX=

Address Values

Address Values


Dept. of ECE RCE

PROGRAM:

.MODEL SMALL

.286

.DATA

ORG 2000H

MOV AX, 0000H

MOV SI,

MOV DI,

MOV CL,

LOOP 2: MOV AL,[SI]

MOV BL,[DI]

CMP AL,BL

JNE LOOP1

INC SI

INC DI

DEC CL

JNZ LOOP2

MOV AX,0000H

INT 03

LOOP1: MOV AX,0001H

INT 03

.EXIT

END

RESULT:


Dept. of ECE RCE

FLOWCHART:

YES

NO

START

Clear the contents of AX and BX

registers

Initalize the SI and DI with some address value

Copy the contents of SI and DI to AL and BL

registers

Exchange the contents of AL and BL registers

Copy the contents of AL and BL to [SI] and [DI]

Increment the value of

[SI]

Decrement the value of

[DI]

Compare the address values of SI and

DI

If [SI]<[DI]

STOP


Dept. of ECE RCE

STRING REVERSAL

AIM:

To write an assembly language program to perform string reversal operation


Starting address:

Ending address:

ALGORITHM:


2. Initialize the AX & BX register.



4. Copy the content of SI into AL register.

5. Copy the content of DI into BL register.

6. Exchange the contents of AL and BL registers.

7. Copy the contents of AL and BL registers to SI and DI memory locations.

8. Increment the value of SI memory location.

9. Decrement the value of DI memory location.

10. Compare the address values of SI and DI.

11. If [SI]<[DI] goto step-4.



Dept. of ECE RCE

RESULT:

INPUT:

ADDRESS VALUES

OUTPUT:

ADDRESS VALUES


Dept. of ECE RCE

PROGRAM:

.MODEL SMALL

.286

.CODE

ORG 2000H

MOV AX,0000H

MOV BX,0000H

MOV SI,

MOV DI,

LOOP1: MOV AL,[SI]

MOV BL,[DI]

XCHG AL,BL

MOV [SI],AL

MOV [DI],BL

INC SI

DEC DI

CMP SI,DI

JB LOOP1

INT 03

.EXIT

END

RESULT:

VIVA QUESTIONS 1. What is the purpose of string length? 2. What does „equ‟ stands for? 3. What is the purpose of label start after the end directive? 4. What is the operation performed by the instruction cmp al, $? 5. Why SI is not been incremented in the program? 6. What is the significance of CLD? 7. How does CMPSB perform the comparison? 8. After executing the instructions mov ah,02h / int 21h, what are contents of AL ? 9. What are the contents of AH for displaying a sting in the screen? 10. What are the contents of AH for accepting a number from keyboard?

11. Which register contains the address of the string to be displayed on screen?


Dept. of ECE RCE

MICROPROCESSOR (8086) INTERFACING PROCEDURE WITH

DIFFERENT INTERFACING MODULES

1) Switch on the PC.

2) Connect ESA 86/88E Trainer to the Host PC using 9 to 9 pin serial cable

supplied by kit manufacturer.

3) Give power to ESA 86/88E Trainer.

4) Make the Dipswitches on ESA 86/88E Trainer for 9600 baud rate and for

serial

Monitor (Select Dip switches 3,4 and 7 should be ON for Serial Mode).

5) Start Win86E application on Host PC. You should then see the following

Dialog Box.

6) If the Trainer is connected to COM1 port on your Host PC & configured for

9600-baud rate, click No. Click Yesotherwise. If yes, you will see the following

dialog box.

7) Select the Baud Rate as per the dipswitch configuration on ESA 86/88E

Trainer. Select the Com Port where the trainer is connected. Click Apply. You

should then see the following window.


Dept. of ECE RCE

8) Click on the „download’ option in the above window and browse ‘.hex’ file

of the current program and click ok. After downloading the file click on „cmd‟

option in the above window and then to execute the current program give the

command ‘g 0000:2000’ (i.e., starting address of the program)from

keyboard. Check out the output on the interface or study card kit which has

connected to the 8086 microprocessor kit.

9) In some cases if the communication is not established you will see the

following

Dialog box. In this case please check the Power on ESA 86/88E Trainer. Check

whether Serial cable is connected to the PC &ESA 86/88E Trainer and check

the parameter by clicking Settings.


Dept. of ECE RCE

10) If the communication is established properly, from the displayed window

you can work with ESA 86/88E Trainer. You can view the Registers, Memory

Dumps, Download the hex images, Upload the hex dump to the Host PC, Run

the downloaded hex images, single stepping, breakpoints, watch windows,

Memory modification etc., These options are explained in detail in online help

of Win86E (Win86E.hlp).


Dept. of ECE RCE

OUT PUT:

RESULT:


Dept. of ECE RCE

READING FROM KEYBOARD WITH ECHO

EX: No: 5

AIM: To write an assembly language program to read the string from keyboard with echo by using DOS functions REQUIRED SOFTWARE: MASM611

Starting address: Ending address:

PROGRAM:

.MODEL SMALL

.286

ASSUME DS: DATA, CS: CODE DATA SEGMENT STRING DB "ROY BHURCHANDI"

LEN EQU 0EH BUFFER DB 0FH DUP(?)

MESSAGE DB 0AH, 0DH, "STRING DISPLAYED", 0AH, 0DH, "$" DATA ENDS CODE SEGMENT

ORG 2000H START: MOV AX, DATA

MOV DS, AX RAJ: MOV CX, LEN MOV DI, OFFSET BUFFER

NXTCHAR: MOV AH, 01H INT 21H

CMP AL, 0DH JE STOP MOV [DI], AL

INC DI DEC CX

JNZ NXTCHAR STOP: MOV CX, LEN MOV SI, OFFSET STRING

MOV DI, OFFSET BUFFER MOV AX, SEG BUFFER

MOV ES, AX CMPSB JNZ SORRY

MOV AH, 4CH INT 21H

SORRY: MOV DX, OFFSET MESSAGE MOV AH, 09H INT 21H

JMP RAJ CODE ENDS

END START .EXIT END


Dept. of ECE RCE

OUTPUT:

RESULT:


Dept. of ECE RCE

READING FROM KEYBOARD WITHOUT ECHO

AIM: To write an assembly language program to read the string from keyboard without echo

by using DOS functions

REQUIRED SOFTWARE: MASM611 Starting address: Ending address

PROGRAM:

.MODEL SMALL

.286 ASSUME CS: CODE, DS: DATA

DATA SEGMENT STG DB "RAJESHNAIDUARE"

LEN DW 0EH BUFFER DB 0FH DUP(?) MESSAGE DB 0AH,0DH,"SORRY!",0AH,0DH,"$"

DATA ENDS CODE SEGMENT

ORG 2000h START: MOV AX,DATA MOV DS,AX

RAJ : MOV CX,LEN MOV DI,OFFSET BUFFER

NXTCHAR: MOV AH,08H INT 21H CMP AL,0DH

JE STOP MOV [DI],AL

INC DI DEC CX JNZ NXTCHAR

STOP: MOV CX,LEN MOV SI,OFFSET STG

MOV DI,OFFSET BUFFER MOV AX,SEG BUFFER MOV ES,AX

CMPSB JNZ SORRY

MOV AH,4CH INT 21H SORRY: MOV DX,OFFSET MESSAGE

MOV AH,09H INT 21H

JMP RAJ CODE ENDS END START

.EXIT END


Dept. of ECE RCE

OUT PUT

RESULT:


Dept. of ECE RCE

DISPLAY STRING

AIM: To write an assembly language program to display the string by using DOS functions

REQUIRED SOFTWARE: MASM611 Starting address:

Ending address

PROGRAM

.MODEL SMALL

.286 ASSUME CS: CODE, DS: DATA DATA SEGMENT

MESSAGE DB 0DH, 0AH,"STUDY OF MICROPROCESSORS IS INTERFACING", 0DH, 0AH, "$"

DATA ENDS CODE SEGMENT ORG 2000H

START : MOV AX,DATA MOV DS, AX

MOV AH, 09H MOV DX, OFFSET MESSAGE INT 21H

MOV AH, 4CH INT 21H

CODE ENDS END START .EXIT

END


Dept. of ECE RCE

VIVA QUESTIONS

1. Abbreviate DOS & BIOS 2. Why INT21H is used in the Program? 3. Why BIOS interrupt is called as video service interrupt? 4. Why INT10H is used in the program? 5. What is the use of BIOS CALLS? 6. In echo program AH register should loaded with which DOS function? 7. In without echo program AH register should loaded with which DOS function? 8. Why INT09H is used in DOS calls? 9. What happened If AH is loaded with 4CH in program 10. Why INT11, INT12 and INT13H are used?


Dept. of ECE RCE

EX: No:

8259 INTERRUPT CONTROLLER

Aim: To write an assembly language program to generate interrupts by using 8259 interrupt

controller which is interfacing with 8086 microprocessor. Apparatus:Hardware: soft ware 8086 Microprocessor Kit ESA 86

8259 interrupt controller kit FRC, USB cables

Power adapter Theory:

INTERRUPT SEQUENCE

The powerful features of the 8259A in a microcomputer system are its programmability and the interrupt routine addressing capability. The latter allows direct or indirect jumping to the

specific interrupt routine requested without any polling of the interrupting devices. The normal sequence of events during an interrupt depends on the type of CPU being used. The events occur as follows in an MCS-80/85 system:

1. One or more of the INTERRUPT REQUEST lines (IR7±0) are raised high, setting the

corresponding IRR bit(s). 2. The 8259A evaluates these requests, and sends an INT to the CPU, if appropriate. 3. The CPU acknowledges the INT and responds with an INTA pulse.

4. Upon receiving an INTA from the CPU group, the highest priority ISR bit is set, and the corresponding IRR bit is reset. The 8259A will also release a CALL instruction

code (11001101) onto the 8-bit Data Bus through its D7±0 pins. 5. This CALL instruction will initiate two more INTA pulses to be sent to the 8259A

from the CPU group.

6. These two INTA pulses allow the 8259A to release its preprogrammed subroutine address onto the Data Bus. The lower 8-bit address is released at the first INTA pulse

and the higher 8-bit address is released at the second INTA pulse. 7. This completes the 3-byte CALL instruction released by the 8259A. In the AEOI

mode the ISR bit is reset at the end of the third INTA pulse. Otherwise, the ISR bit

remains set until an appropriate EOI command is issued at the end of the interrupt sequence.

The events occurring in an 8086 system are the same until step 4.

4. Upon receiving an INTA from the CPU group, the highest priority ISR bit is set and the corresponding IRR bit is reset. The 8259A does not drive the Data Bus during this cycle.


Dept. of ECE RCE

5. The 8086 will initiate a second INTA pulse. During this pulse, the 8259A releases an 8-bit

pointer onto the Data Bus where it is read by the CPU. 6. This completes the interrupt cycle. In the AEOI mode the ISR bit is reset at the end of the

second INTA pulse. Otherwise, the ISR bit remains set until an appropriate EOI command is issued at the end of the interrupt subroutine.

If no interrupt request is present at step 4 of either\ sequence (i.e., the request was too short in duration) the 8259A will issue an interrupt level 7. Both the vectoring bytes and the

CAS lines will look like an interrupt level 7 was requested.

When the 8259A PIC receives an interrupt, INT becomes active and an interrupt

acknowledge cycle is started. If a higher priority interrupt occurs between the two INTA pulses, the INT line goes inactive immediately after the second INTA pulse. After an

unspecified amount of time the INT line is activated again to signify the higher priority interrupt waiting for service. This inactive time is not specified and can vary between parts. The designer

should be aware of this consideration when designing a system which uses the 8259A. It is recommended that proper asynchronous design techniques be followed.

PROGRAM: DISPM EQU 0B5BH ; DISPLAY ROUTINE ADDRESS TO DISPLAY

; MESSAGE ON SERIAL ORG 8000H

START: MVI A, 12H ; single, edge triggerd mode, ; call address interval 8. OUT 90H

MVI A,81H OUT 91H

MVI A, 20H OUT 90H EI

SSS: JMP SSS UP: RST 3

ORG 8100H LXI H, MES0 JMP SUBRT

NOP NOP

LXI H, MES1 JMP SUBRT NOP

NOP LXI H, MES2

JMP SUBRT NOP NOP

LXI H, MESG3 JMP SUBRT

NOP NOP


Dept. of ECE RCE

LXI H, MESG4

JMP SUBRT NOP

NOP LXI H, MESG5 JMP SUBRT

NOP NOP

LXI H, MESG6 JMP SUBRT NOP

NOP LXI H, MESG7

JMP SUBRT NOP NOP

SUBRT: CALL DISPM JMP UP

RESULT:

VIVA QUESTIONS:

1. What is an Interrupt? 2. What is the need for interrupt controller? 3. List some of the features of INTEL 8259 (Programmable Interrupt Controller) 4. Differentiate software and hardware interrupts 5. What is meant by NMI? 6. Explain all initialization and operational command word format of 8259? 7. With single 8259 how many interrupts are possible? 8. With cascading mode of 8259 how many interrupts are possible? 9. How many 8259 ICs will be required for cascading mode? 10. What is the difference between polled command mode and vectored (interrupt driven)

command mode of 8259?


Dept. of ECE RCE

EX: No: 8279 KEY BOARD INTERFACE Aim: To write an assembly language program to blink Port C bit 0 of the 8279 by interfacing

with 8086 microprocessor.

Apparatus: Hardware: Software 8086 Microprocessor Kit ESA 86 8279 key board interface kit

FRC, USB cables Power adapter

Theory:

A programmable keyboard and display interfacing chip. Scans and encodes up to a 64-key keyboard. Controls up to a 16-digit numerical display. Keyboard section has a built- in

FIFO 8 character buffer. The display is controlled from an internal 16x8 RAM that stores the coded display information. 1. The section has eight lines(RL0 - RL7) that can be connected to eight coluns of a keyboard

matrix. 2. It has two additional input : shift and control/strobe. The keys are automatically

debounced. 3. The two operating modes of keyboard section are 1) 2-key lockout 2) N-key rollover 4. In the 2-key lockout mode, if two keys are pressed simultaneously, only the first key is

recognized. 5. In the N-key rollover mode simultaneous keys are recognized and their codes are stored in

FIFO. 6. The keyboard section also have an 8 x 8 FIFO (First In First Out) RAM.The FIFO can store eight key codes in the

scan keyboard mode. The status of the shift key and control key are also stored along with key code.

7. The 8279 generate an interrupt signal(IRQ) to MPU when the FIFO is not empty. Scan section:

1. The scan section has a scan counter and four scan lines, SL0 to SL3.

2. These four scan lines cab be decoded as 16 scan lines by using 4-to-16 decoder. 3. These scan lines are connected to the rows of a matrix keyboard and digit drivers of a

multiplexed display Display section:

1. The display section has eight output lines divided into two groups A0-A3 and B0-B3.

2. The output lines can be used either as a single group of eight lines or as two groups of four

lines, in conjunction with the scan lines for a multiplexed display.

3. The display can be blanked by BD (low) line.

4. The display section consists of 16 x 8 display RAM. The CPU can read from or write into

any location of the display RAM.

PROGRAM:

DEMONSTRATION PROGRAM FOR 86/88E TRAINER ; WRITEN BY SIRI 06.08 ; PORTA INTPUT 0FFE0

; PORTB OUTPUT 0FFE2


Dept. of ECE RCE

; PORTC OUTPUT 0FFE4

; CONTROL PORT 0FFE6

OUTPUT 2500AD ORG 2000H

MOV AX,0000H ; Clear AX MOV AL,90H ; Initiasilse 8255 ports as above

MOV DX,0FFE6H OUT DX,AL CALL CLR ; Clear LEDs

BACK: CALL KESCN ; Identify the pressed key MOV AL,BH ; Store ASCII value of pressed key

MOV SI,2100H ; Load MSG address MOV AH,00H ADD AX,SI

MOV SI,AX MOV CL,[SI] ; Get display code

CALL CLR ; Clear LEDs MOV CH,08H ; No of segments / LED D0: RCL CL,01H ; Get 1 data bit

JNC D1 MOV AL,02H ; Output the data bit

MOV DX,0FFE4H OUT DX,AL MOV AL,03H ; Out put the clock to shift Register

MOV DX,0FFE4H OUT DX,AL

JMP D2 D1: MOV AL,00H ; Output the data bit MOV DX,0FFE4H

OUT DX,AL MOV AL,03H ; Out put the clock to shift Register

MOV DX,0FFE4H OUT DX,AL D2: DEC CH ; All bits over?

JNE D0 ; No, continue JMP BACK ; yes, go to BACK

CLR: MOV BL,06H ; No of LEDs S2: MOV BH,08H ; 8 segments / LED

S1: MOV AL,02H ; Output the data bit MOV DX,0FFE4H

OUT DX,AL MOV AL,03H ; Out put the clock to shift Register MOV DX,0FFE4H

OUT DX,AL DEC BH ; All sigments over?

JNE S1 ; No jump to S1 DEC BL ; All LEDs over?


Dept. of ECE RCE

JNE S2 ; No, jump to S2

RET

KESCN: MOV BL,03H ; No of Scan lines MOV BH,0FH ; Reference key value MOV CL,08H ; Initialize scan line high

NXTGRP: MOV AL,CL

MOV DX,0FFE2H OUT DX,AL ; writing High scan line to portB RCR AL,01H

MOV CL,AL ; Updating the next scan line MOV DX,0FFE0H

IN AX,DX ; Read portA AND AL,1FH ; Masking upper 3 bits CMP AL,00H ; Compare key pressed

JNE NXTKEY ; If pressed go to identification part

SUB BH,05H ; Else change the reference key value & scan line DEC BL ; come to next scan line CMP BL,FFH ; repeat until 4 scan lines are checked

JNE NXTGRP JMP KESCN

NXTKEY: RCR AL,01H JNB NZ ; Key code is OK, return

RET

NZ: ADD BH,01H ; If not, try next key code JMP NXTKEY

ORG 2100H

MSG: DB 0C0H, 0F9H, 0A4H, 0B0H,99H DB 92H, 82H, 0F8H, 80H, 90H

DB 7FH, 0B7H, 88H, 0C8H,0AFH DB 89H, 0BFH, 86H, 0A1H, 0C6H

RESULT:

VIVA QUESTIONS: 1. What are the tasks involved in keyboard interface? 2. How a keyboard matrix is formed in keyboard interface using 8279? 3. What is scanning in keyboard and what is scan time? 4. What is scanning in display and what is the scan time? 5. What is meant by De-Bouncing? 6. What is meant by Polling? 7. What is meant by mode set in 8279? 8. What is the data bus size in 8279? 9. What are the display modes supported by the 8279 chip? 10. Give the format of program clock word of 8279 and mention its purpose.


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3. 8255- PROGRAMMABLE PERIPHERAL INTERFACE (PPI)

AIM:

To write an assembly language program to generate sine wave by using 8255 interfacing with 8086 microprocessor.

APPARATUS:

1. ESA 86/88-2 Microprocessor kit.

2. C.R.O.

3. F R C Cable.

4. Power supply.

5. Dual DAC kit.

THEORY:

The parallel input-output port chip 8255 is also called as Programmable peripheral input-output

Port. The Intel’s 8255 are designed for use with Intel’s 8 -bit, 16-bit and higher capability microprocessors. It has

24 input/output lines which may be individually programmed in 2-groups of 12 lines each, or 3 groups of 8 lines .The two groups of I/O pins are named as GROUP A and GROUP B. Each of these two groups contain a sub group of 8 I/O lines called as 8-bit Port and other sub group of 4 I/O lines are a 4-bit port. Thus GROUP A

Contains an 8-bit port A along with a 4-bit port, C upper. Port lines are identified by symbols PAO-PA7,

While the port C lines are identified as PC4-PC7 .Similarly group B contains an 8-bit port B, containing lines

PB0-PB7 and a 4-bit port C with lower bits PC0-PC3 .The port C upper and port C lower can be used in recombination as an 8-bit port-C .Both the port Cs are assigned the same address. Thus one may have either

Three 8-bit I/O ports are two 8-bit and two 4-bit I/O ports from 8255.All of these ports can function independently either as input or as output ports. This can be achieved by programming the bits of an internal register of 8255 called as Control word register. (CWR).

PORT ADDRESSES OF 8255

Control word register = FFE6

Port A = FFE0

Port B = FFE2


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Generation of a Sine Waveform:

Program:

Memory address

Op codes labels Mnemonics

Operands

Comments

2000

2002

2005

2006

2009

200C

2010

2012

2013

2014

2015

2017

B0,80

BA,E6,FF

EE

BA,E0,FF

B9, 24,00

8B,36,00,30

8A,04

EE

46

49

75,F9

EB,F0

BACK:

L1:

MOV

MOV

OUT

MOV

MOV

MOV

MOV

OUT

INC

DEC

JNZ

JMP

AL,80H

DX,FFE6H

DX,AL

DX,FFE0H

CX,24H

SI,3000

AL,[SI]

DX,AL

SI

CX

L1

BACK

;Initialize 8255 all ports o/p

;Keep on writing the sine wave co ordinate data values to port A

Lookup table:

Memory address

Corresponding values

3000 75,95,AA,BF,D1,E0,ED,F6,FC,FE,FC,F6,ED,E0,D1,BF,AA,95,75

69,54,40,2D,1E,11,08,02,00,02,08,11,1E,2D,40,54,69

After Execution: Output Sine wave Amplitude 29.4V,Frequency 50Hz.

RESULT:

8086 MICRO

PROCESSOR

8255

PPI

8-BIT

DAC

CRO


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VIVA QUESTIONS:

1. Give the BSR format?

2. What is the angle of a stepper motor in full step mode?

3. 8255 has how many pins?

4. If the Control word 09bH is given to control register of the 8255 ppi then explain what is the condition of ports?

5. If the 8255 is selected for addresses 0F800H-0F806H what is the address of port C?

6. Mode 1 of 8255 is used for which of the I/O methods?

7. 8255 has how many ports?

8. What are the applications of the ports?

9. Why we need interfacing devices?


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Ex: No:

INTERFACING 8051& 8253 USART TO 8086

AIM:

To interface 8251& 8253 Interface board to 8086 p and verify the operation of receiver

APPARATUS:Hardware: soft ware: 8086 Microprocessor Kit ESA 86

8251 PPI FRC, USB cables

Power adapter THEORY:

The 8251 is a USART (Universal Synchronous Asynchronous Receiver Transmitter) for

serial data communication. As a peripheral device of a microcomputer system, the 8251

receives parallel data from the CPU and transmits serial data after conversion. This device

also receives serial data from the outside and transmits parallel data to the CPU after

conversion

RXD (input terminal)

This is a terminal which receives serial data.

RXRDY (Output terminal)

This is a terminal which indicates that the 8251 contains a character that is ready to READ. If

the CPU reads a data character, RXRDY will be reset by the leading edge of RD signal.

Unless the CPU reads a data character before the next one is received completely, the

preceding data will be lost. In such a case, an overrun error flag status word will be set.

RXC (Input terminal)

This is a clock input signal which determines the transfer speed of received data. In

"synchronous mode," the baud rate is the same as the frequency of RXC. In "asynchronous

mode," it is possible to select the baud rate factor by mode instruction. It can be 1, 1/16, 1/64

the RXC.

SYNDET/BD (Input or output terminal)

This is a terminal whose function changes according to mode. In "internal synchronous

mode." this terminal is at high level, if sync characters are received and synchronized. If a

status word is read, the terminal will be reset. In "external synchronous mode, "this is an

input terminal. A "High" on this input forces the 8251 to start receiving data characters.

In "asynchronous mode," this is an output terminal which generates "high level"output upon

the detection of a "break" character if receiver data contains a "low-level" space between the


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stop bits of two continuous characters. The terminal will be reset, if RXD is at high level.

After Reset is active, the terminal will be output at low level.

DSR (Input terminal)

This is an input port for MODEM interface. The input status of the terminal can be

recognized by the CPU reading status words.

DTR (Output terminal)

This is an output port for MODEM interface. It is possible to set the status of DTR by a

command.

CTS (Input terminal)

This is an input terminal for MODEM interface which is used for controlling a transmit

circuit. The terminal controls data transmission if the device is set in "TX Enable" status by a

command. Data is transmitable if the terminal is at low level.

RTS (Output terminal)

This is an output port for MODEM interface. It is possible to set the status RTS by a

command.

PROGRAM:

ORG 8000H LXI D, 000AH ;TIMER 0 for

;9600 BAUD MVI A, 36H ; TIMER0 IN ;MODE 3

OUT CMD53 MOV A,E ; LOAD LSB VALUE

OUT 80H MOV A, D ; LOAD MSB VALUE OUT 80H

XRA A ; DUMMY MODE OUT CMD51 ; WRITE DUMMY

; WORD OUT CMB51 ; IN MODE ; REGISTER

OUT CMD51 OUT CMD51

MVI A,40H ;RESET 8251 OUT CMD51 MVI A,0CEH ;ASYNCHRONOUS

;16XBAUD, ;2 STOP BITS, 8

;DATA ;BITS, NO PARITY MVI A,27H ;WRITE COMMAND

;WORD OUT CMD51


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LXI H,8500H ;MEMORY TO STORE

;THE BACK: CALL GETCH ; CHARACTER TO

;MEMORY MOV A, C CPI 27

JZ DOWN CALL SOUTPT

MOV M, C INX H

JMP BACK

DOWN: RST 5 GETCH: IN CMD51 ;GET USART

; STATUS ANI 02H ; RECEIVER READY JZ GETCH ; NO, WAIT

IN DAT51 ; GET CHARACTER MOV C, A ; FROM USART

RET SOUTPT: IN CMD51 ANI 81H

CPI 81H JNZ SOUTPT

MOV A, C OUT DAT51 RET

RESULT:

VIVA QUESTIONS: 1. What is synchronous data transfer scheme? 2. What is asynchronous data transfer scheme?

3. What is baud rate? 4. What is USART?

5. What are the functions performed by INTEL 8251A? 6. What are the control words of 8251A and what are its functions? 7. What are the information that can be obtained from the status word of 8251?

8. How the microprocessor is synchronized with peripherals? 9. What is parity error?

10. Write applications of 8251?


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EMBEDDED C PROGRAMS FOR

8051-MICROCONTROLLER Ex: No: 1

EMBEDDED C PROGRAMS

AIM:

To Execute the Following Embedded C Programs On Kiel Software 1. Reading and Writing on a parallel port. 2. Timer in different modes.

3. Serial communication implementation. 4. Using external interrupts

5. Understanding three memory areas of 00 – FF (Programs using above areas).

APPARATUS: Hardware: soft ware: 1. PC 1.KIEL S/W

THEORY:

The AT89C51 is a low-power, high-performance CMOS 8-bit microcomputer with 4K bytes

of

Flash programmable and erasable read only memory (PEROM). The device is manufactured

using Atmel‟s high-density nonvolatile memory technology and is compatible with the

industry standard MCS-51 instruction set and pin out. The on-chip Flash allows the program

memory to

be reprogrammed in-system or by a conventional nonvolatile memory programmer. By

combining a versatile 8-bit CPU with Flash on a monolithic chip, the Atmel AT89C51 is a

powerful microcomputer which provides a highly-flexible and cost-effective solution to many

embedded control applications.

INTPUT:


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PROCEDURE:

Project creation:

Project ---New---Project name--- select Atmel---select 89c51R02.

File---New---write the program --- source with .c extension.

Right click on target --- select options for target set the crystal frequency as 11.0592--

- select output --- click ON create HEX file option---click OK.

Right click on source group and select Add files to group.

Compile the file--- Build and rebuild all.

Select Debug--- start or stop debug.

Select step or step over.

Go to peripherals and select Timers/ports/serial window.


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OUTPUT:


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S1. READING AND WRITING ON A PARALLEL PORT

Embedded c program

#include<reg51.h>

sbit in0=P1^0;

sbit in1=P1^1;

sbit in2=P1^2;

sbit in3=P1^3;

sbit in4=P1^4;

sbit in5=P1^5;

sbit in6=P1^6;

sbit in7=P1^7;

void main()

{

P1=0xff;

P2=0x00;

while(1)

{

P2=P1;

}

}

VIVA QUESTIONS:

1. How many parallel ports are there in 8051? 2. What about special functions of port-3?

3. What are the special functions of port-0 and port-1? 4. What is difference between microprocessor and microcontroller?

5. What is the use of volatile keyword? 6. What is IDE? 7. What is the size of parallel port in 8051?

8. Why we need embedded-c? How it will differ from normal „c‟?


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OUTPUT:


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2. TIMERS IN DIFFERENT MODES

THEORY:

Thebasic8051hastwo on-chip timers that can be used for timing durat ions or for

counting external events Interval t iming allows the programmer to perform operations at

specific instants in time. Forexample,inour LED flashing program the LED was turned on

for a specific length of time and then turned off for a specific length of time. We achieved

this through the use of time delays. Since the microcontroller operates at aspecific

frequency, we could work out exact ly how many iterations of the time delay was needed to

give us the desired delay. However, this is cumbersome and prone to error. And there is

another disadvantage; the CPU is occupied, stepping through the loops. If we use the on-

chip timers, the CPU could be o ff doing something more useful while the timers take on

the menial task of keeping track of time

The 8051 has two16-bit timers. The high byte for timer1(TH1) is at address 8DH

while the low byte (TL1) is at 8BH The high byte for timer0 (TH0) is at 8CH while the low

byte (TL0) is at 8AH. Both timers can be used in a number o f different modes. The

programmer sets the t imers to a specific mode by loading the appropriate 8-bit number in

to the Timer Mode Register (TMOD) which is at address 89H.

C PROGRAM:

#include<reg51.h>

void T0delay();

sbit out=P1^7;

void main()

{

TMOD=0X21;

TH1=0XFD;

SCON=0X50;

TR1=1;

while(1)

{

out=~out;

T0delay();


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}

}

void T0delay() //TO GENERATE 50 m sec delay using timer-0 mode1

{

TL0=0XFD;

TH0=0X4B;

TR0=1;

while(TF0==0);

TR0=0;

TF0=0;

}

VIVA QUESTIONS:

1. What is meant by timer? 2. Differentiate timer and counter? 3. How many timers are there in 8051?

4. What is meant by auto reload mode in timers? 5. Draw the format of TMOD register of 8051

6. List the modes of Timer in 8051 7. What is the significance of C/T bit in TMOD register of 8051?

8. What is the significance of TRx bit in TCON register of 8051?


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3. SERIAL COMMUNICATION IMPLEMENTATION

THEORY:

In Parallel communicat ion, data being transferred between one locat ion and another

(R0 to the accumulator, for example) travel alo ng the 8-bit data bus. Because o f this data

bus, data bytes can be moved about the microcontroller at high speed. However, parallel

communicat ion has the disadvantage of requiring at least eight separate lines (inan8-

bitsystem) and in most cases extra lines to synchronize the data transfer (in the case of the

microcontroller, the control bus).

Serial co mmunication has the advantage of requiring only one line for the data, a

second line for ground and possibly a third line for the clock. Therefore, because seria l

communicat ion requires less physical wires, it is more suitable for transmitting data over

longer distances.

#include<reg51.h>

voidInit_Uart(void);

voidsend_char(unsigned char c);

void MSDELAY(unsigned int time);

voidsend_string(unsigned char *ptr);

unsigned char RX_CHAR();

//---------------------------

//============================================

voidInit_Uart(void)

{

SCON=0x50;

TMOD=0x20;

TH1=0XFD;

TR1=1;

}

//============================================

voidsend_char(unsigned char c)

{


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SBUF=c;

while(!TI);

TI=0;

}

//============================================

unsigned char RX_CHAR()

{

unsigned char c;

while(!RI);

c=SBUF;

RI=0;

return SBUF;

}

//=============================================

void MSDELAY(unsigned int time)

{

unsignedinti,j;

for(i=0;i<time;i++)

for(j=0;j<1275;j++);

}

//============================================

voidsend_string(unsigned char *ptr)

{

while(*ptr)

send_char(*ptr++);

}

//==============================================

void main()

{

Init_Uart();

send_char('E');

MSDELAY(100);

send_char('C');


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MSDELAY(100);

send_char('E');

MSDELAY(100);

send_char('@E');

MSDELAY(100);

send_string(" RCE, ELURU");

}

VIVA QUESTIONS:

1. What are the types of serial communication?

2. Draw the Frame format of Asynchronous serial communication 3. Draw the format of SCON register of 8051 4. What is the importance of SMOD bit in SCON register?

5. What are the modes of asynchronous serial communication in 8051? 6. Explain how baud rate is calculated for serial data transfer in mode 1?

7. Explain how baud rate is set in 8051 for serial data transfer using timer register? 8. Why we are using START/STOP bit?


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OUTPUT:


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INPUT:


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OUT PUT:


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Interfacing with Micro Controller

Hardware Interfacing

1. Waveform Generation using Dual DAC

2. 4X4 hexadecimal Keyboard interface

3. simple Calculator using Keyboard and Seven segment display.

4. Temperature sensor.

Features of Embedded C

• C is a simple programming language and so very easy to code.

• Embedded C has most features of C-language with more stress on certain bit manipulative instructions.

• This feature makes it easy to write program for μC and μP.

• Keil is a versatile software with a cross compiler that will convert the C program to assembly language and thus the program can be executed on the desired target (say 8051).

Some of the bit manipulative instructions used are

Symbol Operation

& Bitwise AND

| Bitwise OR

~ Bitwise NOT

>> Shift Right

<< Shift Left

^ P0.0

1.DualDac Interface to generate

a.Square waveform

b.Triangular Waveform

c.Ramp waveform

d.Sine waveform

8

0

5 P0 1 P1

μC

Dual

DAC

CRO

Xout


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1. a).Algorithm for Square wave generation

Let initial, amplitude of the square wave be 2.5v(7F) and frequency count 100.

Output the values 00h (0ff) and 7fh (on) Values through P0.

If amplitude key is pressed then increase the voltage in steps of 0.15v (8).

If the frequency key is pressed increment the count in steps of 50. If the count exceeds 1000 reset it back to 100.

Every time amplitude and frequency changes output the value through P0 and note the waveform on CRO.

Program for square wave

#include <REG51xD2.H>

sbit Amp = P3^3; /* Port line to change amplitude */

sbitFre = P3^2; /* Port line to change frequency */

void delay(unsigned int x) /* delay routine */

{

for(;x>0;x--);

}

main()

{

unsigned char on = 0x7f,off=0x00;

unsignedintfre = 100;

while(1)

{

if(!Amp) /* if user choice is to change amplitude */

{

while(!Amp); /* wait for key release */

on+=0x08; /* Increase the amplitude */

}

if(!Fre) /* if user choice is to change frequency */

{


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if(fre> 1000) /* if frequency exceeds 1000 reset to default */

fre = 100;

while(!Fre); /* wait for key release */

fre += 50; } /* Increase the frequency */

P0=on; /* write amplitude to port */

P1=on;

delay(fre);

P0 = off; /* clear port */

P1 = off;

delay(fre);

}}

1.b).Algorithm for Triangular wave generation

Output the initial value 00 through P0.

Increment it in steps of 1 until a count value of FFh (5V) is reached. Every time repeat step 1.

Decrement it in steps of 1 until a zero value is reached and repeat step 1.

Program for triangular wave:


main()

{

unsigned char i=0;

P0 = 0x00; /* P0 as Output port */

while(1)

{

for(i=0;i<0xff;i++){ /* Generate ON pulse */

P1 = i;

P0 = i;

}

for(i=0xfe;i>0x00;i--) /* Generate OFF pulse */

{P0 = i;

P1 = i;}


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}

}

1.cAlgorithm for Ramp wave generation

Output the initial value 00 through P0.

Increment it in steps of 1 until a count value of FFh (5V) is reached. Every time repeat step 1.

Repeat step 1 & 2 continuously.

Program for Ramp waveform


main ()

{ Unsigned char i=0;


while (1)

{

for (i=0;i<0xff;i++) /* Generate ON pulse */

{

P1 = i;

P0 = i; }

}

}


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1.d). Algorithm for Sine wave

Compute different step values (θ = 20o,15o…) of sine using the equation

V= 2.5V +2.5Vsinθ. . Output the values thro P0.

More the steps smoother will be sine wave.

E.g.: θ = 0o

V= 2.5V +2.5Vsinθ = 2.5V

The value sent to DAC is 25.6X5V= 128.

Program for sine wave


main()

{

static int a[13]={128,192,238,255,238,192,128,64,17,0,17,64,128};

unsigned char i=0;


while (1)

{

for(i=0;i<13;i++) /* Output different values */

{ P0 = a[i]; }

}

}


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2. 4X4 HEX Keyboard

Algorithm for Keyboard Interface

• Configure P1 as output port to scan the rows and P0 as input port to read the column values.

• First select the last row by grounding that row. Scan the columns of entire row if a key is pressed in that row then one of the column reads ‘0’.

• If now key is pressed in the selected row all 1’s is returned. So scan next row. Repeat the action until all rows are scanned.

//Program for 4X4 hex keypad.

#include<reg51.h> #include"lcd.h" #include"keypad.h" void main() { unsigned char s; lcd_init(); lcdcmd(0x80); msgdisplay("WELCOME"); delay(200); lcdcmd(1); lcdcmd(0xc0); while(1) { s=keypad(); lcddata(s); delay(10); } }


Dept. of ECE RCE

3. Calculator using Keyboard and LCD

Algorithm

• Read the numbers n1 and n2 from keyboard and display them on seven segment.

• Read the operand from the keypad if key pressed is B (+), C(-),D(*),E(/) then respective operation is performed.

• Result is displayed on 2 digit seven segment display.

• If any time the key pressed value returned as 10h then clear the LCD.

8051µC

PORT 3

PORT

1

PS

Keypad

PS

LCD Display


Dept. of ECE RCE

Program for calculator

#include<reg51.h> #include"lcd.h" #include"keypad.h" #include<string.h> void main() { floata,b; unsigned char k; unsignedint temp; lcd_init(); start: lcdcmd(0x80); msgdisplay("Welcome To"); lcdcmd(0xc0); msgdisplay("Calculator "); while(1) { delay(200); lcdcmd(1); lcdcmd(0x80); msgdisplay("enter val1: "); lcdcmd(0xc0); a=b=0; while(1) { temp=keypad(); delay(100); if(temp=='#') { break; } else if(temp=='*') { if(a) { lcdcmd(0x10); lcddata(' '); delay(100); a=a/10; lcdcmd(0x10); }


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} else { a=a*10+temp-48; lcddata(temp); delay(100); temp++; } } lcdcmd(1); lcdcmd(0x80); msgdisplay("enter val2: "); lcdcmd(0xc0); b=0; while(1) { temp=keypad(); delay(100); if(temp=='#') { break; } else if(temp=='*') { if(b) { lcdcmd(0x10); lcddata(' '); delay(100); lcdcmd(0x10); b=b/10; } } else { b=b*10+temp-48; lcddata(temp); delay(100); temp++; } } lcdcmd(1); lcdcmd(0x80);


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msgdisplay("Select operation"); lcdcmd(1); msgdisplay("1:ADD 2:SUB"); lcdcmd(0xc0); msgdisplay("3:MUL 4:DIV"); while(1) { k=keypad(); delay(100); if(k=='1') { lcdcmd(1); lcdcmd(0x80); msgdisplay("a="); convert1(a); lcddata(' '); lcdcmd(0x87); msgdisplay("b="); convert1(b); lcddata(' '); lcdcmd(0xc0); msgdisplay("a+b="); convert1(a+b); delay(200); again1: k=keypad(); if(k=='#') { goto start; } else { goto again1; } } if(k=='2') { lcdcmd(1); lcdcmd(0x80); msgdisplay("a="); convert1(a); lcddata(' ');


Dept. of ECE RCE

msgdisplay("b="); convert1(b); lcddata(' '); if(a<b) { lcdcmd(0xc0); msgdisplay("a-b="); lcddata('-'); convert1(b-a); } else { lcdcmd(0xc0); msgdisplay("a-b="); convert1(a-b); } delay(300); again2: k=keypad(); if(k=='#') { goto start; } else { goto again2; } } if(k=='3') { lcdcmd(1); lcdcmd(0x80); msgdisplay("a="); convert1(a); lcddata(' '); msgdisplay("b="); convert1(b); lcddata(' '); lcdcmd(0xc0); msgdisplay("a*b="); convert1(a*b); delay(100); again3: k=keypad();


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if(k=='#') { goto start; } else { goto again3; } } if(k=='4') { lcdcmd(1); lcdcmd(0x80); msgdisplay("a="); convert1(a); lcddata(' '); msgdisplay("b="); convert1(b); lcddata(' '); lcdcmd(0xc0); msgdisplay("a/b="); convert1(a/b); delay(200); again4: k=keypad(); if(k=='#') { goto start; } else { goto again4; } } } } }


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4. Temperature Sensor

The interface card has a DAC to convert the actual temperature to digital this is compared with reference temperarture. Realay also a part of interface card will turn on and off to indicate if the actual temperature is above or below reference.

Algorithm for Temperature sensor

1. Configure P0 and P1 as o/p, P3 as input port.

8051µC P0,P2,P3

PS

Temp Sensor

Interface

PS

Heat

Source


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2. Set up a counter with intial value 0xff send it to dacthro P0 after a delay check if comparator o/p has gone low.

3. If low, compare with set value if actual greater than set turn on the relay else turn off.

Program for temperature sensor.

#include<reg51.h> #include"lcddisplay.h" #include<intrins.h> // connection between 8051 and ADC MCP3202 sbit ADC_CS = P2^0; sbit ADC_CLK = P2^3; sbit ADC_DO = P2^2; sbit ADC_DI = P2^1; voidwrite_adc_byte(char data_byte); unsignedintReadADC(unsigned char channel); //void DelayMs(unsigned int count); void convert(unsigned int value); void main() { unsignedint temperature ; float x; // buz=1; lcd_init(); // LCD Initilize lcdcmd(0x85); msgdisplay("WELCOME"); delay(1000); lcdcmd(0x01); msgdisplay("TEMPERATURE"); lcdcmd(0xC0); msgdisplay("MONITOR..."); delay(1000); delay(1500); lcdcmd(0x1); msgdisplay("TEMP:- "); while(1) { delay(100); temperature=ReadADC(0); x=(temperature/8); lcdcmd(0x87); convert(x); msgdisplay(" deg"); delay(1000); }


Dept. of ECE RCE

} unsignedintReadADC(unsigned char channel) { unsigned char i,k; unsignedintAdcResult; // 12 bit ADC_CS=0; // Active chip select k++; // Delay about 1 uS ADC_CLK=0; // make clock low first k++; k++; channel = channel? 0xF0 : 0xD0; k++; k++; // delay about 2 uS //--- write command 4 bit ---------- for(i=0; i< 4;i++) { ADC_DI = (channel & 0x80) != 0; channel<<=1; ADC_CLK=1; //DelayMs(20); k++; k++; // delay about 2 uS ADC_CLK=0; // DelayMs(20); // k++;k++; } k++; k++; // delay about 2 uS ADC_CLK=1; //DelayMs(20); k++; k++; // delay about 2 uS ADC_CLK=0; //DelayMs(20); k++; k++; // delay about 2 uS //--- read ADC result 12 bit -------- AdcResult=0; for(i=0;i<12;i++) { ADC_CLK=1; k++; k++;


Dept. of ECE RCE

// delay about 2 uS AdcResult<<=1; AdcResult=AdcResult|(ADC_DO); ADC_CLK=0; k++; k++; // delay about 2 uS } ADC_CS=1; return(AdcResult); } void convert(unsigned int temp1_value) { unsignedint d1,d2,d3; d3=temp1_value%10; temp1_value=temp1_value/10; d2=temp1_value%10; d1=temp1_value/10; lcddata(d1+48); lcddata(d2+48); lcddata(d3+48); }

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Transcript of RCE MICROPROCESSORS & MICROCONTROLLERS LAB