Mpi Lab Manual

Dept. of ECE MPI LAB MANUAL

V.K.R, V.N.B & A.G.K ENGINEERING COLLEGE 1

MICRO PROCESSOR & INTERFACING LAB MANUAL

DEPARTMENT OF ELECTRONICS & COMMUNICATION

ENGINEERING

III B.Tech II Semester

(FOR ECE)

V.K.R., V.N.B. & A.G.K. COLLEGE OF ENGINEERING Gudivada-521301, Ph-08674-242189

www.gudivadaenggcollege.co.in



1. Students are required to attend all labs. 2. Students will work in a group of two in hardware laboratories and

individually in computer laboratories.

3. While coming to the lab bring the lab manual cum observation book, record etc.

4. Take only the lab manual, calculator (if needed) and a pen or pencil to the

work area.

5. Before coming to the lab, prepare the prelab questions. Read through the lab experiment to familiarize yourself with the components and assembly sequence.

6. Utilize 3 hours time properly to perform the experiment and noting down

the readings. Do the calculations, draw the graph and take signature from the instructor.

7. If the experiment is not completed in the prescribed time, the pending work

has to be done in the leisure hour or extended hours.

8. You will be expected to submit the completed record book according to the

deadlines set up by your instructor.

9. For practical subjects there shall be a continuous evaluation during the semester for 25 sessional marks and 50 end examination marks.

10. Of the 25 marks for internal, 15 marks shall be awarded for day-to-day work and 10 marks to be awarded by conducting an internal laboratory test.

INSTRUCTIONS TO THE STUDENTS



S.No CONTENTS Page No

1 OVERVIEW OF THE 80X86 FAMILY 6-8

2 INTRODUCTION TO TASM 9-10

3 PROCEDURE TO ENTER A PROGRAM USING TASM

SOFTWARE:

11-13

4 EXAMPLE FOR ASSEMBLE LANGUAGE PROGRAM 14

5 EXECUTION OF PROGRAM THROUGH TRAINER

KIT

15

6 TRAINER KIT PROGRAMMES

1. ADDITION OF „N‟ 8 BIT NUMBERS 16-17

2. MULTI BYTE ADDITION 18-19

3. MULTIBYTE SUBTRACTION 20-21

4. MULTIPLICATION OF 16 BIT NUMBER 22-23

5. DIVISION OF 16 BIT NUMBER 24-25

6. FACTORIAL OF A GIVEN NUMBER 26-27

7 TASM PROGRAMMES

1. MULTI BYTE ADDITION 28-30

2. MULTI BYTE SUBTRACTION 31-33

3. „8‟ BIT MULTIPLICATION (UNSIGNED) 34-35

4. 8-BIT MULTIPLICATION (SIGNED) 36-37

5.„16‟ BIT MULTIPLICATION (UNSIGNED) 38-40

6. „8‟ BIT DIVISION (UNSIGNED) 41-42

7. 8-BIT DIVISION (SIGNED) 43-44

8. „16‟ BIT DIVISION (UNSIGNED) 45-46

9. LARGEST OF A GIVEN NUMBER 47-48

10. ASCII ADDITION 49-50

11. ASCII SUBTRACTION 51-53

8 LOGIC OPERATIONS

1. CHECK WHETHER A GIVEN 16-BIT NUMBER IS

EVEN OR ODD

54-55


POSITIVE OR NEGATIVE

56-57



3. SORTING (ASCENDING) 58-60

4. SORTING (DESCENDING) 61-63

5. PACKED TO UNPACKED BCD 64-66

6. BCD to ASCII CONVERSION 67-68

9 STRING OPERATIONS

1. LENGTH OF THE STRING 69-71

2. BLOCK TRANSFER 72-74

3. STRING COMPARISION 75-77

4. STRING REVERSAL 78-80

5. STRING INSERTION 81-84

6. STRING DELETION 85-87

7. READING KEYBOARD WITH ECHO 88-89

8. READING KEYBOARD WITHOUT ECHO 90-91

9. DISPLAYING A CHARACTER 92-93

10. DISPLAYING A STRING 94-95

INTERFACING

1. 8255 – PPI INTERFACING -1 97-99

2. 8279 – KEYBOARD/DISPLAY INTERFACING 100-102

3. 8259 – PROGRAMMABLE INTERUPT CONTROLLER 103-104

4. 8251 – USART 105-106

APPENDIX

Expected Questions from Microprocessors 108-110

Key to the Questions on Microprocessors 111-113





1.OVERVIEW OF THE 80X86 FAMILY

The 80x86 family was first started in 1981 with the 8086 and the newest member

is the Pentium which was released thirteen years later in 1994. They are all backwards

compatible with each other but each new generation has added features and more speed

than the previous chip. Today there are very few computers in use that have the 8088 and

8086 chips in them as they are very outdated and slow. The number of 286 or 386 based

computers around is declining as today's software becomes more and more demanding.

Even 486's are being replaced by Pentiums. With the Pentium PRO and the MMX based

CPUs Intel keeps increasing performance and features. As computers are used to

manipulate various types of data, there has to be some physical places where these data

can be stored. Some of these places are the registers.

REGISTERS

Registers are places in the CPU where a number can be stored and manipulated. There

are three sizes of registers: 8-bit, 16-bit and on 386 and above 32-bit. There are four

different types of registers:

- General Purpose Registers,

- Segment Registers,

- Index Registers,

- Stack Registers.

GENERAL PURPOSE REGISTERS

First, we are going to describe the main registers. Later, we are going to describe the

stack registers. The main registers are 16-bit registers. There are four general purpose

registers: AX, BX, CX and DX. They are split up into 8-bit registers. AX is split up into

AH which contains the high byte and AL which contains the low byte. On 386's and

above there are also 32-bit registers, these have the same names as the 16-bit registers but

with an 'E' in front i.e. EAX. You can use AL, AH, AX and EAX separately and treat

them as separate registers for some tasks.

SEGMENT REGISTERS

Code segment register – CS- Stores the base address of Code segment

Stack segment register -SS- Stores the Base address of stack

Data segment register-DS- Stores the Base address of data

Extra segment register-ES- Stores the Base address of the extra segment



INDEX REGISTERS

These registers are sometimes called pointer registers. They are 16-bit registers and are

mainly used for string instructions. There are three index registers SI (source index), DI

(destination index) and IP (instruction pointer). On 386's and above there are also 32-bit

index registers: EDI and ESI. You can also use BX to index strings.

STACK REGISTERS

BP and SP are stack registers and are used when dealing with the stack. We will discuss

about them when we will talk about the stack.

The designers of the 8088 decided that the maximum need for memory space will

be one megabyte. So, the chip they build cannot go above that. The problem is to access a

whole megabyte 20 bits are needed. But registers only have 16 bits so 4 more bites are

needed. They came up with what they thought was a clever way to solve this problem:

segments and offsets. This is a another way to do the addressing with two registers but

not using 32 bits. To compute every single address the segment addresses as well as the

offset are needed. A segment is usually 64kBytes long. To address a location inside a

segment, we need the offset (16 bytes long).

THE STACK

As the number of registers used is low, how do they manage to manipulate the data?

There is something called a stack which is an area of memory which you can save and

restore values to. To physically imagine this we can consider a stack of plates. The last

one you put on is the first one that you take off. This is sometimes referred to as Last On

First Off (LOFO) or Last In First Out (LIFO). But, you have to be sure that you do not

put to much data in the stack.

NUMBERS IN BINARY

Before we begin to understand how to program in assembly it is best to try to remember

how numbers are represented in computers. Numbers are stored in binary, base two.

There are several terms which are used to describe different size numbers:

1 BIT: 0 (can have the values 1 or 0)

One bit is the simplest piece of information that exists. Its either a one or a zero.

1 NIBBLE: 0000 (4 BITS)



The nibble is four bits or half a byte. It has a maximum value of 15 (1111 = 15). This is

the basis for the hexadecimal (base 16) number system which is used as it is far easier to

understand. Hexadecimal numbers that begin with a letter are prefixed with a 0 (zero).

1 BYTE: 00000000 (equivalent with 2 NIBBLES or 8 BITS)

A byte is 8 bits or 2 nibbles. A byte has a maximum value of FFh (255 decimal).

A WORD is a 2 byte data. The maximum value that can be represented using a word is

65535. It is the obvious that if bigger number has to be represented, the necessary number

of bites will increase.

TABLE 1: A Byte Diagram

1 (MSD) 1 0 0 1 0 1 1 (LSD)

Above, MSD means Most Significant Digit and LSD means Least Significant Digit. A

word will be represented as below:

TABLE 2: A Word Diagram

Most Significant Byte Least Significant Byte

In memory, bytes are stored one after another. Each one has an associated address. Often

data needs more than one byte (every number greater then 256). Then we are going to use

a word to represent this number. In 8086 programming, the Most Significant Byte is

stored at higher address then LSB.



2. INTRODUCTION TO TASM

INTRODUCTION:

The aim of this experiment is to introduce the student to assembly language

programming and the use of the tools that he will need throughout the lab experiments. This first

experiment let the student use the Dos Debugger and the Turbo Assembler (TASM). Editing,

Assembling, Linking, Execute up can be done using TASM software

OBJECTIVES:

1. Introduction to Turbo Assembler (TASM)

2. General structure of an assembly language program

3. Use of the Dos Debugger program

OVERVIEW: In general, programming of microprocessor usually takes several iterations

before the right sequence of machine code instruction is written. The process, however is

facilitated using a special program called an “Assembler”. The Assembler allows the user to write

alphanumeric instructions. The Assembler, in turn, generates the desired machine instructions

from the assembly language instructions.

Assembly language programming consists of following steps:

STEP PRODUCES

1 Editing Source file

2 Assembling Object file

3 Linking Executable file

4 Executing Results

Table1.1: Assembly Language Programming Phases

ASSEMBLER:

The assembler is used to convert the assembly language instructions to machine

code. It is used immediately after writing the Assembly language program. The assembler starts

by checking the syntax or validity of the structure of each instruction in the source file .if any

errors are found, the assemblers displays a report on these errors along with brief explanation of

their nature. However if the program does contain any errors, the assembler produces an object

file that has the same name as the original file but with the “obj” extension



LINKER: The Linker is used convert the object file to an executable file. A linker is a

program used to join several object files in to one large object file and convert to an executable

file. The linker produces a link file which contains the binary codes for all the combined

modules. The linker however doesn‟t assign absolute address to the program it assigns is said to

be relocatable because it can be put any where in memory to be run.

EXECUTING THE PROGRAM:

The executable contains the machine language code .it can be loaded in the RAM

and executed by the microprocessor simply by typing ,from the DOS prompt ,the name of the file

followed by the carriage Return Key (Enter Key). If the program produces an output on the

screen or sequence of control signals to control a piece of hard ware, the effect should be noticed

almost immediately. However, if the program manipulates data in memory, nothing would seem

to have happened as a result of executing the program.

DEBUGGER:

The debugger can also be used to find logical errors in the program. Even if a

program does not contain syntax errors it may not produce the desired results after execution.

Logical errors may be found by tracing the action of the program. Once found, the source file

should be rewrite to fix the problem, then the re-assembled and re-linked. A special program

called the debugger is designed for that purpose.

The debugger allows the user to trace the action of the program, by single

stepping through the program or executing the program up to a desired point, called break point.

It also allows the user to inspect or change the contents of the microprocessor internal registers or

the contents of any memory location.



3. PROCEDURE TO ENTER A PROGRAM USING TASM

SOFTWARE:

Start

↓

Run

↓

Type CMD

↓

Ok

Display shows

↓

C :\> D:

(Change to D drive because MASM is in D drive)

↓

Press ENTER

↓

D :\> CD TASM

↓

Press ENTER

↓

D: \TASM> EDIT

↓

Press ENTER

↓

Then the display shows the main menu.

↓



Create A file name with .asm extension

Ex: Filename.asm

↓

Write the program in assemble language then Save and Quit

It returns to DOS Prompt

↓

Use the following Commands

↓

D: \TASM> tasm filename.asm

It is for viewing errors and warnings report exist in the asm file(refer Assembler)

↓

D: \TASM> tlink filename.obj

The assembler produces an object file that has the same name as the original file but with the

“obj” extension

↓



D: \TASM> td filename.exe

It is for viewing the output window

↓

Click Run and Click View

It shows the output for Corresponding Inputs



4. EXAMPLE FOR ASSEMBLE LANGUAGE

PROGRAM

The assembly language program can be divided in to two parts

1. Data Segment

2. Code Segment

In the data segment all the input, output and count (for loop programs) have to

specify. This section starts with Data Segment and ends with Data Ends

Example:

DATA SEGMENT

NUM1 DB 04H

NUM2 DB 02H

RESULT DB 00

RESULT1 DB 00

DATA ENDS

In the code segment remaining the entire program with specified instructions are

present

Example:

CODE SEGMENT

START:

MOV AX, DATA

MOV DS, AX

MOV AL, NUM1

MOV BL, NUM2

MUL BL

MOV RESULT, AL

MOV RESULT1, AH

INT 3

CODE ENDS



5. EXECUTION OF PROGRAM THROUGH

TRAINER KIT

Procedure For Executing Assembly Language Programs In 8086 By Using 8086

Trainer Kit:

Procedure:

Step1: After entering the data in to memory locations by using the known procedure then

the display shows – [blinking cursor]

Step2: Type G starting address

Ex: G 2000

Step3: Then [press] enter

Step4: Then the display shows registers with results

Step5: To see results locations, press D 4000 [4000 is the address of the memory location

where the result is stored „D‟ is the command for displaying the data in memory

locations]

Step6: Then press enter.

Step7: Then the display shows the data in 4000 location.

Step8: If we want to substitute Different inputs for the same locations, press S

3000 [3000 is the address of the memory location where the input data is stored

„S‟ is the command for substituting the data in memory locations]



6. TRAINER KIT PROGRAMMES

ARITHMETIC OPERATIONS

1. ADDITION OF „N‟ 8 BIT NUMBERS

AIM: To write a program to add „N‟ 8 bit numbers.

APPARATUS: 8086 microprocessor trainer kit, power supply

ALGORITHM:

Step1: Initialize AX// load SI with 3000h memory location

Step2: Move the contents of SI in to the CL registers [counter]

Step3: Increment SI [first number of n number]

Step 4: Move the first number of n number in to BL

Step 5: Add AL& BL

Step6: If jump on no carry go to the step8 else step7

Step7: Increment AH

Step8: Decrement CL [count]

Step9: If jump on no zero go to the step 3

Step10: Load DI with 4000h memory location {to store the

Result}

Step11: Move the result in to DI

Step 12: End of the program

PROGRAM:

MOV AX, 0000H

MOVSI, 3000H

BACK: MOV CL, [SI]

INC SI

MOV BL, [SI]

ADD AL, BL

JNC GO

INC AH

G0: DEC CL

JNZ BACK

MOV DI, 4000H

MOV [DI], AX



INT 03

OBSERVATIONS:

Input: Input

S3000: 03 S3002:03 S3000: S3002:

S3001: 02 S3003:04 S3001: S3003:

Output: Output

D4000:09 D4000:

RESULT: Thus the addition of n 8 bit numbers is performed and results are observed

using 8086 microprocessor trainer kit.

VIVA QUESTIONS:

1. What is Microprocessors

2. What are the functions of Accumulator?

3. MOV BL, [SI] which Addressing mode?

4. Basic Difference between 8085 and 8086



2. MULTI BYTE ADDITION

AIM: To write a program to add two multibyte numbers.

APPARATUS: 8086 microprocessor trainer kit, power supply.

ALGORITHM:

Step1: initialize AX// load SI with 3000h memory location

Step2: load DI with 4000h memory location

Step3: load the CX register with count 03

Step4: move the first byte of first number [SI] in to AL

Step5: move the first byte of second number [DI] in to DL

Step6: add AL&DL with carry

Step7: move the contents of AL in to DI

Step8: increment SI

Step9: increment DI

Step10: move the second byte of first number [SI] in to AL

Step11: move the second byte of second number [DI] in to DL

Step12: add AL&DL with carry


Step14: decrement CX and jump to the step 8 if the CX register is

Non zero

Step15: end of the program

PROGRAM:

MOV AX, 0000H

MOV SI, 3000H

MOV DI, 4000H

MOV CX, 03

MOV AL, [SI]

MOV DL, [DI]

ADC AL, DL

MOV [DI], AL

BACK: INC SI

INC DI

MOV AL, [SI]

MOV DL, [DI]

ADC AL, DL

MOV [DI], AL



LOOP BACK

INT 03

OBSERVATIONS:

Input:

S3000 S3001 S3002 S3003

10 20 30 40

S4000 S4001 S4002 S4003

20 30 40 50

Output:

D4000 D4001 D4002 D4003

30 50 70 90

Input:

S3000 S3001 S3002 S3003

S4000 S4001 S4002 S4003

Output:

D4000 D4001 D4002 D4003

RESULT:

Multi byte addition is performed and results are observed using TASM.

VIVA QUESTIONS

1. Why 8085 processor is called an 8-bit processor

2. What is the disadvantage of microprocessor?

3. What is the difference between ADD & ADC

4. What is the purpose served by CX register?



3. MULTIBYTE SUBTRACTION

AIM: To write a program to subtract multibyte numbers

APPARATUS: 8086 microprocessor trainer kit, power supply

ALGORITHM:

Step1: initialize AX// load SI with 3000h memory location

Step2: load DI with 4000h memory location

Step3: load the CX register with count 03

Step4: move the first byte of first number [SI] in to AL

Step5: move the first byte of second number [DI] in to DL

Step6: subtract AL from DL with borrow


Step8: increment SI

Step9: increment DI

Step10: move the second byte of first number [SI] in to AL

Step11: move the second byte of second number [DI] in to DL

Step12: subtract AL from DL with borrow


Step14: decrement CX and jump to the step 8 if the CX register is

Non zero


PROGRAM:

MOV AX, 0000H

MOV SI, 3000H

MOV DI, 4000H

MOV CX, 03

MOV AL, [SI]

MOV DL, [DI]

SBB AL, DL

MOV [DI], AL

BACK: INC SI

INC DI

MOV AL, [SI]

MOV DL, [DI]

SBB AL, DL

MOV [DI], AL



LOOP BACK

INT 03

OBSERVATIONS:

Input:

S3000 S3001 S3002 S3003

50 80 50 40

S4000 S4001 S4002 S4003

20 30 40 20

Output:

D4000 D4001 D4002 D4003

30 50 10 20

Input:

S3000 S3001 S3002 S3003

S4000 S4001 S4002 S4003

Output:

D4000 D4001 D4002 D4003

RESULT:

Thus multibyte subtraction is performed and results are observed using 8086

microprocessor kit



4. MULTIPLICATION OF 16 BIT NUMBER

AIM: To write a program to multiply two 16 bit numbers

APPARATUS: 8086 trainer kit, power supply

ALGORITHM:

Step1: initialize AX, DX load SI with 3000h memory location

Step2: move the contents [SI] in to AX

Step3: increment SI

Step4: increment SI

Step5: move the contents [SI] in to BX

Step6: multiply the contents of BX with AX

Step7: move the contents of AX to the 4000h memory location

Step8: move the contents of DX to the 5000h memory location


PROGRAM:

MOV AX, 0000

MOV DX, 0000

MOV SI, 3000

MOV AX, [SI]

INC SI

INC SI

MOV BX, [SI]

MUL BX

MOV @4000, AX

MOV @5000, DX

INT 03

OBSERVATIONS:

Input: Input:

S3000 S3001 S3002 S3003 S3000 S3001 S3002 S3003

02 01 02 01

Output: Output:

D4000 D4001 D5000 D5001 D4000 D4001 D5000 D5001

04 04 01 00



RESULT:

Thus multiplication of a 16 bit numbers is performed and results are observed

using 8086 microprocessor kit



5. DIVISION OF 16 BIT NUMBER

AIM: To write a program to perform division of two 16 bit numbers


ALGORITHM:

Step1: initialize AX, DX load SI with 3000h memory location

Step2: move the contents [SI] in to AX

Step3: increment SI

Step4: increment SI

Step5: move the contents [SI] in to BX

Step6: divide the contents of BX with AX


Step8: move the contents of DX to the 5000h memory location


PROGRAM:

MOV AX, 0000

MOV DX, 0000

MOV SI, 3000

MOV AX, [SI]

INC SI

INC SI

MOV BX, [SI]

DIV BX

MOV @4000, AX

MOV @5000, DX

INT 03

OBSERVATIONS:

Input: Input:

S3000 S3001 S3002 S3003 S3000 S3001 S3002 S3003

04 00 02 00

Output: Output:

D4000 D4001 D5000 D5001 D4000 D4001 D5000 D5001

02 00 00 00



RESULT:

Thus division of a two 16 bit numbers is performed and results are observed using

8086 microprocessor kit



6. FACTORIAL OF A GIVEN NUMBER

AIM: To write a program to find factorial of a given number


ALGORITHM:

Step1: initialize ax, bx.

Step2: load SI with 3000h memory location

Step3: move the contents [SI] in to AL

Step4: move the contents of AX in to BX

Step5: decrement BX

Step6: multiply the contents of BX with AX

Step7: decrement BX and jump to the step 6 if the BX register is

Non zero



PROGRAM:

MOV AX, 0000H

MOV BX, 0000H

MOV SI, 3000H

MOV AX, [SI]

MOV BX, AX

DEC BX

BACK: MUL BX

DEC BX

JNZ BACK

MOV @4000, AX

INT 03

OBSERVATIONS:

Input: Input:

S3000: 05 S3000:

Output: Output:

D4000: 78 D4000:



RESULT:

Thus factorial of a given number is performed and results are observed using

8086 microprocessor kit.



7. TASM PROGRAMMES

ARITHMETIC OPERATION

1. MULTI BYTE ADDITION

AIM: To write a program to add two multi byte numbers.

APPARATUS: Computer system installed with TASM.

ALGORITHM:

Step1: initialize the data segment.

Step2: load cx with counter

Step3: load the si first number

Step4: load the first byte of num1 (si) into al

Step5: add the first byte of num1 (si) and first byte of num2 (si) with carry

Step6: move the result into num3 (si)

Step7: decrement si

Step8: decrement count

Step9: jump if not zero to step4


PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

NUM1 DB 11H,22H,33H,44H,55H,66H,77H,88H

NUM2 DB 11H,22H,33H,44H,55H,66H,77H,88H

NUM3 DB ?

DATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

CLC

MOV CL, 08

MOV SI, 0007

BACK: MOV AL, NUM1 [SI]

ADC AL, NUM2 [SI]

MOV NUM3 [SI], AL



DEC SI

DEC CL

JNZ BACK

INT 3

CODE ENDS

END START

OBSERVATION: (Theoretical)

INPUT: NUM1=11h,22h,33h,44h,55h,66h,77h,88h

NUM2=11h,22h,33h,44h,55h,66h,77h,88h

OUTPUT: RES=22h,44h,66h,88h,AAH,CCH,EFH,10H

MEMORY LOCATIONS:

Input: DS: 0000h:11H

0001h: 22H

0002h: 33H

0003h: 44h

0004h: 55H

0005H: 66H

0006h: 77H

0007h: 88H

0008h: 11H

0009h: 22H

000Ah:33H

000Bh:44H

000Ch:55H

000DH:66H

000Eh:77H

000Fh:88H

OUTPUT: DS: 0010h: 22H

0011h: 44H

0012h: 66H

0013h: 88H

0014h: AAH

0015H: CCH

0016h: EFH

0017h: 10H

OBSERVATION: (practical):

INPUT: NUM1=

NUM2=

OUTPUT: RES=

Memory Locations:

Input: ds: 0000h: 0001h:



0002h:

0003h:

0004h:

0005H:

0006h:

0007h:

0008h:

0009h:

000Ah:

000Bh:

000Ch:

000DH:

000Eh:

000Fh:

OUTPUT :DS:0010h:

0011h:

0012h:

0013h:

0014h:

0015H:

0016h:

0017H:



2. MULTI BYTE SUBTRACTION

AIM: To write a program to subtract two multi byte numbers.

APPARATUS: Computer system installed with TASM

ALGORITHM:

STEP1: Initialize the data segment

STEP2: Load CX with counter

STEP3: Load the SI first number

STEP4: Load the first byte of num1(SI) into AL

STEP5: subtract the first byte of num1(SI) and first byte of num2(SI)

With borrow

STEP6: move the result into num3(SI)

STEP7: Decrement SI

STEP8: Decrement count

STEP9: Jump if not zero to step4

STEP10: End of the program

PROGRAM:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

NUM1 DB 88H,77H,66H,55H,44H,33H,22H,11H

NUM2 DB 11H,22H,33H,44H,11H,22H,11H,11H

NUM3 DB ?

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

CLC

MOV CL,08

MOV SI,0007

BACK: MOV AL,NUM1[SI]

SBB AL,NUM2[SI]

MOV NUM3[SI],AL

DEC SI

DEC CL



JNZ BACK

INT 3

CODE ENDS

END START

OBSERVATION: (Theoretical)

INPUT: NUM1=88H,77H,66H,55H,44H,33H,22H,11H

NUM2=11H,22H,33H,44H,11H,22H,11H,11H

OUTPUT: RES=77h,55h,33h,11h,33H,11H,11H,00H

Memory Locations:

Input: DS: 0000h:

0001h:

0002h:

0003h:

0004h:

0005:

0006h:

0007h:

0008h:

0009h:

000Ah:

000Bh:

000Ch:

000Dh:

000Eh:

000Fh:

OUTPUT:DS: 0010h:

0011h:

0012h:

0013h:

0014h:

0015h:

0016h:

0017h:


INPUT: NUM1=

NUM2=

OUTPUT: RES=



Memory Locations:

INPUT: DS: 0000h:

0001h:

0002h:

0003h:

0004h:

0005H:

0006h:

0007h:

0008h:

0009h:

000Ah:

000Bh:

000Ch:

000DH:

000Eh:

000Fh:

OUTPUT: DS: 0010h:

0011h:

0012h:

0013h:

0014h:

0015H:

0016h:

0017h:

RESULT:

Multi byte subtraction is performed and results are observed using TASM

EXERCISE:

1. Why is subtract with carry instruction is used in the loop?

2. What is the purpose served by BX register?

3. Why subtraction is done with AL register why not with AX?

4. What is the other instruction which can be used instead of MOV DI, offset NUM2?



3. „8‟ BIT MULTIPLICATION (UNSIGNED)

AIM: To write a program to multiply two 8-bit unsigned numbers.


ALGORITHM:


STEP2: Load the AL with NUM1

STEP3: Load the BL with NUM2

STEP4: Multiply AL & BL (Unsigned)

STEP5: Store the result in AL and AH


PROGRAM:


DATA SEGMENT

NUM1 DB 04H

NUM2 DB 02H

RESULT DB 00

RESULT1 DB 00

DATA ENDS

CODE SEGMENT

START:

MOV AX, DATA

MOV DS, AX

MOV AL, NUM1

MOV BL, NUM2

MUL BL

MOV RESULT, AL

MOV RESULT1, AH

INT 3

CODE ENDS

END START



OBSERVATION: (theoretical)

INPUT: NUM1=04h

NUM2=02h

OUTPUT: RES=08h

RES1=00h

MEMORY LOCATIONS:

INPUT: DS: 0000h: 04H

0001h: 02h

OUTPUT: DS: 0002h: 08H

0003h: 00h

OBSERVATION:

INPUT: NUM1=

NUM2=

OUTPUT: RES=

RES1=

MEMORY LOCATIONS:

INPUT: DS: 0000h:

0001h:

OUTPUT: DS: 0002h:

0003h:

RESULT:

8-bit unsigned multiplication is performed and results are observed using TASM.

EXERCISE:

1. AL and BL are used for multiplying why not AX & BX?

2. Instead of using MOV BL is it not possible to MUL num2?

3. What is the instruction used for signed multiplication?



4. 8-BIT MULTIPLICATION (SIGNED)

AIM: To write a program to multiply two 8-bit signed numbers.


ALGORITHM:


STEP2: Load the AL with NUM1

STEP3: Load the BL with NUM2

STEP4: Multiply AL & BL (signed)

STEP5: Store the result in AL and AH


PROGRAM:

DATA SEGMENT

NUM1 DB 0A5H

NUM2 DB 20H

RES DW ?

DATA ENDS

CODE SEGMENT


START:

MOV AX,DATA

MOV DS,AX

SUB AX,AX

MOV AL,NUM1

MOV BL,NUM2

IMUL BL

MOV RES,AX

INT 3

CODE ENDS

END START




INPUT: NUM1=A5h

NUM2=20h

OUTPUT: RES=A0 F4h


INPUT: NUM1=

NUM2=

OUTPUT: RES=

RESULT:

8-bit signed multiplication is performed and results are observed using TASM.

VIVA QUESTIONS:

1. What is the difference between IMUL and MUL?

2. What is the use of instruction CBW?

3. What is the use of instruction CWD?



5. „16‟ BIT MULTIPLICATION (UNSIGNED)

AIM: To write a program to multiply two 16-bit unsigned numbers.


ALGORITHM:


STEP2: Load the AX with NUM1

STEP3: Load the BX with NUM2

STEP4: Multiply AX & BX(Unsigned)

STEP5: Store the result in AX and DX


PROGRAM:


DATA SEGMENT

NUM1 DW 0004H

NUM2 DW 0002H

RESULT DW 00

RESULT1 DW 00

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

MOV AX,NUM1

MOV BX,NUM2

MUL BX

MOV RESULT,AX

MOV RESULT1,DX

INT 3

CODE ENDS

END START




INPUT: NUM1=0004h

NUM2=0002h

OUTPUT: RES=0008h

RES1=0000h

MEMORY LOCATIONS:

INPUT: DS: 0000h:04H

0001h:00h

0002h:02H

0003h:00h

OUTPUT: DS: 0004:08h

0005:00h

0006:00h

0007:00h

OBSERVATION: (PRACTICAL):

INPUT: NUM1=

NUM2=

OUTPUT: RES=

RES1=

MEMORY LOCATIONS:

INPUT: DS: 0000h:

0001h:

0002h:

0003h:

OUTPUT: DS: 0004h:

0005h:

0006h:

0007h:

RESULT:

16-bit Multiplication is performed and results are observed using TASM.



EXERCISE:

1. Why AL & BL are not used in this program?

2. If result exceeds 32 bit, where is it stored?



6. „8‟ BIT DIVISION (UNSIGNED)

AIM: To write a program to perform division on two 8-bit unsigned numbers.


ALGORITHM:

STEP1: Initialize the data segment and AX, BX registers

STEP2: Load the AL with NUM1(Divisor)

STEP3: Load the BL with NUM2(Dividend)

STEP4: divide AL & BL(Unsigned)

STEP5: Store the result in AL(quotient) and AH(remaider)


PROGRAM:


DATA SEGMENT

NUM1 DB 04H

NUM2 DB 02H

QUOTIENT DB ?

REMAINDER DB ?

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

MOV AX,0000h

MOV DX,0000h

MOV AL,NUM1

MOV BL,NUM2

DIV BL

MOV QUOTIENT,AL

MOV REMAINDER,AH

INT 3

CODE ENDS

END START



OBSERVATION: (THEORETICAL)

INPUT: NUM1=04h

NUM2=02h

OUTPUT: QUEO=02h

REM=00h

MEMORY LOCATIONS:

INPUT: DS: 0000h: 04H

0001h: 02h

OUTPUT: DS: 0002:02h

0003:00h


INPUT: NUM1=

NUM2=

OUTPUT: RES=

RES1=

MEMORY LOCATIONS:

INPUT: DS: 0000h:

0001h:

OUTPUT: DS: 0002:

0003:

RESULT:

Division of two 8-bit numbers is performed and results are observed using

TASM.

EXERCISE:

1. Why is the registers DX & AX made zero in the above program?

2. Where is the remainder in 8 bit division?

3. Where is the quotient in 8 bit division?

4. Which interrupt is used when a divide overflow error occurs?



7. 8-BIT DIVISION (SIGNED)

AIM: To write a program to perform division on two 8-bit signed numbers.


ALGORITHM:

STEP1: Initialize the data segment and AX, BX registers

STEP2: Load the AL with NUM1(Divisor)

STEP3: Load the BL with NUM2(Dividend)

STEP4: divide AL & BL(signed)

STEP5: Store the result in AL(quotient) and AH(remainder)


PROGRAM:

DATA SEGMENT

NUM1 DB 0D5H

NUM2 DB 20H

QUO DB ?

REM DB ?

DATA ENDS

CODE SEGMENT


START:

MOV AX,DATA

MOV DS,AX

XOR AX,AX

MOV AL,NUM1

MOV BL,NUM2

IDIV BL

MOV QUO,AL

MOV REM,AH

INT 3

CODE ENDS

END START

OBSERVATION: (THEORETICAL)



INPUT: NUM1=D5h

NUM2=20h

OUTPUT: QEO= FFh

REM= F5h


INPUT: NUM1=

NUM2=

OUTPUT: QEO=

REM=

RESULT:

8-bit signed division is performed and results are observed using TASM.

EXERCISE:

1. What is the purpose of XOR AX,AX?

2. What is the difference between IDIV and DIV?

3. What is the use of instruction CBW & CWD?



8. „16‟ BIT DIVISION (UNSIGNED)

AIM: To write a program to perform division on two 16-bit unsigned numbers.


ALGORITHM:

STEP1: Initialize the data segment and AX, DX registers

STEP2: Load the AX with NUM1(Divisor)

STEP3: Load the BX with NUM2(Dividend)

STEP4: divide AX & BX(Unsigned)

STEP5: Store the result in AX(quotient) and DX(remainder)


PROGRAM:


DATA SEGMENT

NUM1 DW 0004H

NUM2 DW 0002H

QUOTIENT DW ?

REMAINDER DW ?

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

MOV AX,0000

MOV DX,0000

MOV AX,NUM1

MOV BX,NUM2

DIV BX

MOV QUOTIENT,AX

MOV REMAINDER,DX

INT 3

CODE ENDS

END START




INPUT: NUM1=0004h

NUM2=0002h

OUTPUT: QUEO=0002h

REMA=0000h


INPUT: NUM1=

NUM2=

OUTPUT: RES=

RES1=

RESULT:

16-bit division is performed and results are observed using TASM.

EXERCISE:

1. What happens if DX register contains a nonzero value before DIV instruction?

2. What is the instruction used for signed division?

3. In the above program instead of DIV BX is it possible to use DIV num2?



9. LARGEST OF A GIVEN NUMBER

AIM: To write a program to find largest number from a given unordered array of 16 bit

numbers.


ALGORITHM:

Step1: initialize ax// load si with 3000h memory location

step2: load the cx register with count 03

step3: move the contents [si] in to al

step3: compare the contents of [si] with al

step4: jump if not less to the step6

step5: move the contents [si] in to al

step6: increment si

step7: increment si

step8: decrement cl

step9: jump on non zero to the step3

step10: move the contents of ax to the 4000h memory location

step11: end of the program

PROGRAM:

Assume cs:code, ds:data

Data segment

Count db 01h,03h,02h,04h

Data ends

Code segment

Start: Mov ax,data

Mov ds,ax

Mov ax,0000h

Mov si,0000h

Mov cl,04h

Mov al,num1[si]

Back: Inc si

Cmp al,[si]

Jnl next

Mov ax,[si]

Next: Dec cl

Jnz back



Mov num1[si],al

Int 03

Code ends

End start

OBSERVATIONS:

Input: Input:

Output: Output:

RESULT:

Thus largest number from given numbers are found and results are observed using

trainer kit.



10. ASCII ADDITION

AIM: To write a program to perform ASCII addition.


ALGORITHM:


STEP2: Load the DL with the NUM1

STEP3: Load the DH with NUM2

STEP4: ANDed DL & DH with 0FH

STEP5:Move the DX into AX

STEP6:Add the AH & AL

STEP7:Ascii Adjust After ADDITION

STEP8:ORed AL,AH with 30H and store the result


PROGRAM:

ASSUME CS: CODE,DS:DATA

DATA SEGMENT

NUM1 DB 36H

NUM2 DB 32H

RES DW 00H

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

XOR AX,AX

MOV DL,NUM1

MOV DH,NUM2

AND DL,0FH

AND DH,0FH

MOV AX,DX

ADD AL,AH

MOV AH,00

AAA



OR AL,30H

OR AH,30H

MOV RES,AX

INT 03H

CODE ENDS

END START

OBSERVATIONS:(Theoretical)

INPUT: NUM1=36H

NUM2=32H

OUTPUT: RES=38H 30H

OBSERVATION: (practical)

INPUT: NUM1=

NUM2=

OUTPUT: RES=

RESULT:

ASCII addition is performed and results are observed using TASM.

EXERCISE:

1. What is the purpose of ASCII addition?

2. What is the instruction used for ASCII addition?

3. Why do we make use of instruction ORL AX, 3030H ?

4. Why is AAA after addition?



11. ASCII SUBTRACTION

AIM: To write a program to perform ASCII subtraction.


ALGORITHM:


STEP2: Load the DL with the NUM1

STEP3: Load the DH with NUM2

STEP4: ANDed DL & DH with 0FH

STEP5:Move the DX into AX

STEP6:Subtract the AH & AL

STEP7:Ascii Adjust After subtraction

STEP8:ORed AL,AH with 30H and store the result


PROGRAM:


DATA SEGMENT

NUM1 DB 36H

NUM2 DB 32H

RES DW 00H

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

XOR AX,AX

MOV DL,NUM1

MOV DH,NUM2

AND DL,0FH

AND DH,0FH

MOV AX,DX

SBB AL,AH

MOV AH,00

AAS

OR AL,30H



OR AH,30H

MOV RES,AX

INT 03H

CODE ENDS

END START


INPUT: NUM1=36H

NUM2=32H

OUTPUT: RES=34H 30H

MEMORY LOCATIONS:

Input: ds: 0000h:36H

0001h:32H

output: ds: 0002:34H

0003:30H


INPUT: NUM1=

NUM2=

OUTPUT: RES=

MEMORY LOCATIONS:

Input: ds: 0000h:

0001h:

output: ds: 0002h:

0003h:

RESULT:

ASCII subtraction is performed and results are observed using TASM.



EXERCISE:

1. What is the purpose of ASCII subtraction?

2. What is the instruction used for ASCII subtraction?

3. Why do we make use of instruction ORL AX,3030H ?

4. Why is AAS after addition?



8. LOGIC OPERATIONS


EVEN OR ODD

AIM: To write a program to check whether a given 16-bit number is even or odd.


ALGORITHM:


STEP2: Load the AX with the NUM1

STEP3: Load the BX with 0002H(NUM2)

STEP4: Divide AX & BX

STEP5:Compare CX & DX

STEP6:Jump on even step9

STEP7:Load AX with 1111H(ODD NUMBER)

STEP8:Jump to step 10

STEP9: Load the AX with FFFFH(EVEN NUMBER)

STEP10:End of the program.

PROGRAM:


DATA SEGMENT

NUM1 DW 0008H

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV AX,NUM1

MOV BX,0002H

MOV CX,0000H

DIV BX

CMP CX,DX

JE EVEN

MOV AX,1111H

JMP NEXT

EVEN: MOV AX,FFFFH



NEXT: INT 03H

CODE ENDS

END START

OBSERVATIONS:(Theoritical)

INPUT: NUM1=0008H

OUTPUT: AX=FFFFH

BX=0002H


INPUT: NUM1=

OUTPUT: AX=

BX=

RESULT: Checking whether a given 16-bit number is even or odd is performed and

results are observed using Tasm.




POSITIVE OR NEGATIVE

AIM: To write a program to check whether a given 16-bit number is positive or

negative.


ALGORITHM:


STEP2: Load the AX with the NUM1

STEP3: Rotate the contents of AX by one through carry

STEP4:Jump on carry (negative number) go to step6

STEP5:Load the AX with FFFFH (if positive number)

STEP6:Jump to step8

STEP7:Load AX with 1111H(negative number)

STEP8:End of the program.

PROGRAM:


DATA SEGMENT

NUM1 DW 0002H

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

MOV AX,NUM1

ROL AX,01H

JC NEG_NUM

MOV AX,FFFFH

JMP EXIT

NEG_NUM:MOV AX,1111H

EXIT: INT 3

CODE ENDS

END START




INPUT: NUM1=0008H

OUTPUT: AX=FFFFH

BX=0000H


INPUT: NUM1=

OUTPUT: AX=

BX=

RESULT:

Checking whether a given 16-bit number is positive or negative is performed and

results are observed using TASM.

EXERCISE:

1. How can we differentiate between a positive number and a negative number?



3. SORTING (ASCENDING)

AIM: To write a program to sort a given list of numbers in ascending order.


ALGORITHM:


STEP2: Initialize the count in DX Register

STEP3: Load DX with count

STEP4: move DX into the CX Register

STEP5: Load the SI with first byte of List

STEP6:Move the content of SI into AX register

STEP7:Compare AXwith second byte of SI

STEP8: Jump if Less to the step11

STEP9:Exchage AX and second byte of SI

STEP10: Exchage AX and first byte of SI

STEP11:Add SI with 02

STEP12:continue the loop until to get the result and go to the step7

STEP13:Decrement DX

STEP14:Jump if non zero to the step 4

STEP15:END of the program

PROGRAM:


DATA SEGMENT

LIST DW 53H,25H,89H,02H

COUNT EQU 04H

DATA ENDS

CODE SEGMENT

START:MOV AX,DATA

MOV DS,AX

MOV DX,COUNT

AGAIN0: MOV CX,DX

MOV SI,OFFSET LIST

AGAIN1: MOV AX,[SI]

CMP AX,[SI+02]

JL NEXT

XCHG [SI+02],AX



XCHG [SI],AX

NEXT: ADD SI,02H

LOOP AGAIN1

DEC DX

JNZ AGAIN0

MOV AH,4CH

INT 03

CODE END

END START


INPUT: LIST=53H,25H,89H,02H

OUTPUT: LIST=02H,25H,53H,89H

MEMORY LOCATIONS:

OUTPUT:DS: 0000H:00h

0001H:00h

0002H:02h

0003H:00h

0004H:25h

0005H:00h

0006H:53h

0007H:00h

0008H:89h

OBSERVATIONS:(practical)

INPUT: LIST=

OUTPUT: LIST=

MEMORY LOCATIONS:

OUTPUT:DS: 0000H:

0001H:

0002H:

0003H:

0004H:

0005H:

0006H:

0007H:

0008H:



RESULT:

Sorting of a given list of numbers is performed and results are observed using

TASM.

EXERCISE:

1. What is the sorting technique used in above program?

2. Are there any other ways to sort a list?



4. SORTING (DESCENDING)

AIM: To write a program to sort a given list of numbers in descending order.


ALGORITHM:


STEP2: Initialize the count in DX Register

STEP3: Load DX with count

STEP4: move DX into the CX Register

STEP5: Load the SI with first byte of List

STEP6:Move the content of SI into AX register

STEP7:Compare AXwith second byte of SI

STEP8: Jump if Greater to the step11

STEP9:Exchage AX and second byte of SI

STEP10: Exchage AX and first byte of SI

STEP11:Add SI with 02

STEP12:continue the loop until to get the result and go to the step7

STEP13:Decrement DX

STEP14:Jump if non zero to the step 4

STEP15:END of the program

PROGRAM:


DATA SEGMENT

LIST DW 53H,25H,89H,02H

COUNT EQU 04H

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV DX,COUNT

AGAIN0: MOV CX,DX

MOV SI,OFFSET LIST

AGAIN1: MOV AX,[SI]

CMP AX,[SI+02]

JG NEXT



XCHG [SI+02],AX

XCHG [SI],AX

NEXT:ADD SI,02H

LOOP AGAIN1

DEC DX

JNZ AGAIN0

MOV AH,4CH

INT 03

CODE END

END START


INPUT: LIST=53H,25H,89H,02H

OUTPUT: LIST=89H,53H,25H,02H

MEMORY LOCATIONS:

OUTPUT:DS: 0000H:00h

0001H:00h

0002H:89h

0003H:00h

0004H:53h

0005H:00h

0006H:25h

0007H:00h

0008H:02h


INPUT: LIST=

OUTPUT: LIST=

MEMORY LOCATIONS:

OUTPUT:DS: 0000H:

0001H:

0002H:

0003H:

0004H:

0005H:

0006H:



0007H:

0008H:

RESULT:

Sorting of a given list of numbers is performed and results are observed using

TASM.

EXERCISE:

1. What is the sorting technique used in above program?

2. Are there any other ways to sort a list?



5. PACKED TO UNPACKED BCD

AIM: To write a program to convert a packed BCD number to unpacked BCD

number.


ALGORITHM:


STEP2: Load AL with BCDIP & store into the DL

STEP3: ANDed the AL with 0F0H

STEP4:Load the 04 into into CL

STEP5:Rotate accumulator contents by 04 through carry ane moved into the

BH

STEP6: ANDed the DL with 0F0H and moved into BL

STEP7:Store the result

STEP8:End of the program

PROGRAM:

DATA SEGMENT

BCDIP DB 56H

UBCDOP DW 0

DATA ENDS

CODE SEGMENT


START:

MOV AX,DATA

MOV DS,AX

XOR AX,AX

MOV AL, BCDIP

MOV DL,AL

AND AL,0F0H

MOV CL,4

ROR AL,CL

MOV BH,AL

AND DL,0FH

MOV BL,DL

MOV UBCDOP,BX



INT 3

CODE ENDS

END START


INPUT: BCDIP=56H

OUTPUT: UBCDOP=06H 05H

MEMORY LOCATIONS:

Input: ds: 0000h:56h

output: ds: 0001h:06h

0002h:05h


INPUT: BCDIP=

OUTPUT: UBCDOP=

MEMORY LOCATIONS:

Input: ds: 0000h:

output: ds: 0001h:

0002h:

RESULT:

Conversion of packed BCD to unpacked BCD is performed and results are

observed using TASM.

EXERCISE:

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?



6. BCD to ASCII CONVERSION

AIM: To write a program to convert a BCD number to its equivalent ASCII number.


ALGORITHM:


STEP2: Load AL with BCDIP & store into the DL

STEP3:ORed the AL with 30H

STEP4:Store the result


PROGRAM:

ASSUME CS:CODE, DS:DATA

DATA SEGMENT

BCDIP DB 04H

ASCOP DB 00

DATA ENDS

CODE SEGMENT

START:

MOV AX,DATA

MOV DS,AX

XOR AX,AX

MOV AL, BCDIP

OR AL,30H

MOV ASCOP,AL

INT 3

CODE ENDS

END START


INPUT: BCDIP=04H

OUTPUT: ASCOP=34H



MEMORY LOCATIONS:

Input: ds: 0000h:04h

output: ds: 0001h:34H


INPUT: BCDIP=

OUTPUT: ASCOP=

MEMORY LOCATIONS:

Input: ds: 0000h:

output: ds: 0001h:

RESULT:

Conversion of BCD to ASCII is performed and results are observed using TASM.

EXERCISE:

1. What is the difference between adding 30h and OR 30H to a BCD number to

conversion to ASCII?

2. Why unpacking is necessary during the conversion?

3. What is the ASCII character for symbol A?

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



9. STRING OPERATIONS

1. LENGTH OF THE STRING

AIM: To write a program to find the length of the given string.


ALGORITHM:

Step1:Initialize the data segment

Step2: Initialize si and di

Step:3 Take the count as 0000h

Step4: Move the character of the string to al reg.

Step5: compare the al contents with dl ( the data in dl is 00h)

If zero flag is set goto step6 else goto step 8

Step6: increment the count value and increment the si

Step7: jump to step4

Step8: move the count value to output

Step9: end of the program.

PROGRAM:


Data segment

String1 db “VKR”

String2 db ?

Data ends

Code segment

Start: Mov ax,data

Mov ds,ax

Mov ax,0000h

Mov si,0000h

Mov cl,00h

Mov dl,00h

Up: Mov al,string1[si]

Mov dl,00h

Cmp al,dl

Je Up1

Inc cl



Inc si

Jmp Up

Mov String2[si],al

Up1: Int 03

Code ends

End start


INPUT: STRING=KAKINADA

OUTPUT:

K=6BH

A=61H

K=6BH

I=69H

N=6EH

A=61H

D=64H

A=61H

MEMORY LOCATIONS:

output: ds: 0000h:6BH

0001h:61H

0002h:6BH

0003h:69H

0004h:6EH

0005h:61H

0006h:64H

0007h:61H


OUTPUT: STRING=

MEMORY LOCATIONS:

output: ds: 0000h:

0001h:



0002h:

0003h:

0004h:

0005h:

0006h:

0007h:

RESULT:

Length of a string is performed and results are observed using TASM.

EXERCISE:

1. What is the operation performed by the instruction cmp al,$ ?

2. What is function 09h / int 21h performed?

3. Why SI is not been incremented is the program?



2 BLOCK TRANSFER

AIM: To write a program to transfer a block of characters from one place to

another place using string instructions.


ALGORITHM:

STEP1: Initialize the data segment and ES

STEP2: Load SI with first number (num1)

STEP3: Load DI with first number (num2)

STEP4: Load the count into CL

STEP5: clear the direction flag and transfer the string of bytes into

Respective memory locations


PROGRAM:


DATA SEGMENT

NUM1 DB 11H,22H,33H,44H,55H,66H,77H,88H

NUM2 DB 01H,02H,03H,04H,05H,06H,07H,08H

COUNT EQU 08H

DATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV ES, AX

MOV SI, OFFSET NUM1

MOV DI, OFFSET NUM2

MOV CL, COUNT

CLD

REP MOVSW

INT 03H

CODE ENDS

END START




INPUT: NUM1:11H,22H,33H,44H,55H,66H,77H,88H

NUM2:01H,02H,03H,04H,05H,06H,07H,08H

OUTPUT: NUM1: 11H,22H,33H,44H,55H,66H,77H,88H

MEMORY LOCATIONS:

OUTPUT: DS: 0000h:11H

0001h:22H

0002h:33H

0003h:44H

0004h:55H

0005h:66H

0006h:77H

0007h:88H

OBSERVATIONS:(PRACTICAL)

INPUT: NUM1:

NUM2:

OUTPUT: NUM1:

MEMORY LOCATIONS:

OUTPUT: DS: 0000h:

0001h:

0002h:

0003h:

0004h:

0005h:

0006h:

0007h:

RESULT:

Block transfer is performed and results are observed using TASM.



EXERCISE:

1. If the DF=1, will the SI and DI register decremented?

2. The destination memory is pointed by which register combination?

3. The source is pointed to by which register combination?



3 STRING COMPARISION

AIM: To write a program to Compare the two strings.


ALGORITHM:


STEP2: Load CL with first count (string1)

STEP3: Load CL with first count (string2)

STEP4: Compare CL &CH (string1 & string2)

STEP5: Jump if non zero to the step18

STEP6:load the first string character into AL

STEP7:load the second string character into BL




STEP11:Compare the strings


STEP13:Increment SI

STEP14:Decrement CL &CH


STEP17:load the AX with 0000h

STEP18: Jump to the step20

STEP19:load the AX with1111h


PROGRAM:


Data segment

Count1 db 04h

Count2 db 04h

String1 db “JNTU”

String2 db “JNTU”

Data ends

Code segment

Start: Mov ax,data

Mov ds,ax



Mov cl,count1

Mov ch,count2

Cmp ch,cl

Jnz Below

Mov si,0000h

Up: Mov al,string1[si]

Mov bl,string2[si]

Cmp al,bl

Jnz Below

Inc si

Dec cl

Dec ch

Jnz Up

Jmp Below1

Below: Mov ax,0000h

Jmp exit

Below1: Mov ax,1111h

Exit: Int 03

Code ends

End start


INPUT: STRING1 =JNTU

STRING2=KAKINADA

OUTPUT: AX:1111H

MEMORY LOCATIONS:

OUTPUT: DS: 0000h:04H

0001h:04H

0002h:4AH:J

0003h:4EH:N

0004h:54H:T

0005h:55H:U

0006h:4BH:K

0007h:41H:A

0008h:4BH:K

0009h:49H:I

000Ah:4EH:N

000Bh:41H:A



000Ch:44H:D

000Dh:41H:A

OBSERVATIONS:(Practical)

INPUT: STRING1 =

STRING2=

OUTPUT :AX:1111H

MEMORY LOCATIONS:

OUTPUT: DS: 0000h:

0001h:

0002h:

0003h:

0004h:

0005h:

0006h:

0007h:

0008h:

0009h:

000Ah:

000Bh:

000Ch:

000Dh:

RESULT: compare the two strings are performed and results are observed using TASM.



4 STRING REVERSAL

AIM: To write a program to reverse a given string.


ALGORITHM:


STEP2: Initialize SI &DI

STEP3: Load BX with count

STEP4: move the BL into DL







PROGRAM:


Data segment

Count dw 0003h

String1 db “VKR”

String2 db ?

Data ends

Code segment

Start : Mov ax,data

Mov ds,ax

Mov si,0000h

Mov di,0000h

Mov bx,count

Mov cl,04h

Up: Mov al,String1[si+bx]

Mov String2[di],al

Inc di

Dec bx

Jnz Up

Int 03



Code ends

End start


INPUT: STRING1=KAKINADA

OUTPUT: STRING2=ADANIKAK

MEMORY LOCATIONS:

INPUT:DS: 0000H:K=6BH OUTPUT:DS: 0008H:A=61H

0001H:A=61H 0009H:D=64H

0002H:K=6BH 000AH:A=61H

0003H:I=69H 000BH:N=6EH

0004H:N=6EH 000CH:I=69H

0005H:A=61H 000DH:K=6BH

0006H:D=64H 000EH:A=61H

0007H:A=61H 000FH:K=6BH


INPUT: STRING1=

OUTPUT: STRING2=

MEMORY LOCATIONS:

INPUT:DS: 0000H =

0001H =

0002H =

0003H =

0004H =

0005H =

0006H =

0007H =

OUTPUT:DS: 0008H =

0009H =

000AH =

000BH =

000CH =

000DH =

000EH =

000FH =

RESULT:

String reversal is performed and results are observed using TASM.



EXERCISE:

1. Why BX register is added with „5‟?

2. Why MOVS instruction is not used?

3. What is the function of LODS and STOS instructions?



5. STRING INSERTION

AIM: To write a program to insert a word in the given sentence.


ALGORITHM:


STEP2: Initialize the SI Register with 0000

STEP3: Load DL with Location

STEP4: Initialize the DI Register with 0000

STEP5: Move the content of LEN into the CL register

STEP6:Compare DX with SI Register

STEP7: Jump if Equal to the step18

STEP8: Move the content of String1 into the AL register

STEP9:Move the content of AL into the String2 of DI Register

STEP10: Decrement CL and increment SI,DI

STEP11:Jump to STEP6

STEP12:Move the content of ELEMENT into the AL register

STEP13:Move the AL into DI

STEP14:Increment DI

STEP15:Move the SI into AL and AL into DL

STEP16: Increment SI

STEP17: Continue the loop until to get result


PROGRAM:


Data segment

String1 db “GUDVADA”

String2 db ?

Element db “ I “

Len db 08h

Location db 03h

Data ends

Code segment

Start: Mov ax,data



Mov ds,ax

Mov si,0000h

Mov dl,Location

Mov di,0000h

Mov cl,len

Up: Cmp dx,si

Je down

Mov al,string1[si]

Mov string2[di],al

Dec cl

Inc si

Inc di

Jmp Up

Down: Mov al,element

Mov string2[di],al

Up1: inc di

Mov al,string1[si]

Mov string2[di],al

Inc si

Loop Up1

Int 03

Code ends

End start


INPUT: STRING1=RAJASKAR

ELEMENT=E

OUTPUT: STRING2=RAJASEKAR

MEMORY LOCATIONS:

INPUT:DS: 0000H:R=52H

0001H:A=41H

0002H:J=4AH

0003H:A=41H

0004H:S=53H

0005H:K=4BH

0006H:A=41H

0007H:R=52H



OUTPUT:DS: 0008H:R=52H

0009H:A=41H

000AH:J=4AH

000BH:A=41H

000CH:S=53H

000DH:E=45H

000EH:K=4BH

000FH:A=41H

0010H:R=52H


INPUT: STRING1=

ELEMENT=

OUTPUT: STRING2=

MEMORY LOCATIONS:

INPUT:DS: 0000H =

0001H =

0002H =

0003H =

0004H =

0005H =

0006H =

0007H =

OUTPUT:DS: 0008H =

0009H =

000AH =

000BH =

000CH =

000DH =

000EH =

000FH =



RESULT:

String insertion is performed and results are observed using TASM.

EXERCISE:

1. Why register „DI‟ is loaded with 5?

2. What is the function of rep movsb?

3. What is the purpose of mov ah,01h / int 21h?



6. STRING DELETION

AIM: To delete a word from the given sentence.


ALGORITHM:


STEP2: Initialize the SI Register with 0000

STEP3: Load CL with counter

STEP4: Load DH with counter 1

STEP5: Move the content of String1 into the AL register

STEP6: Compare AL and DL

STEP7 Jump on zero to step15

STEP8:Increment CH and SI

STEP9: Compare CL and CH

STEP10:Jump if non zero to step5

STEP11:Subtract DH and CL

STEP12:Move the SI into AL and AL into DL

STEP13: Increment SI and decrement DH

STEP14:Jump if non zero to step12


PROGRAM:


Data segment

String1 db “GUDIVADA”

String2 db ?

Count1 db 04h

Data ends

Code segment

Start: Mov ax,data

Mov ds,ax

Mov si,0000h

Mov ch,00h

Mov cl,count1

Mov dh,count1



Mov dl,00h

Up1: Mov al,string1[si]

Cmp al,dl

Jz Below

Inc ch

Inc si

Cmp cl,ch

Jnz Up1

Sub dh,cl

Up2: Mov al,string1[si]

Mov string2[si],al

Inc si

Dec ch

Jnz Up2

Below: int 03h

Code ends

End start


INPUT: STRING1=KAKINADA

OUTPUT: STRING2=NADA

MEMORY LOCATIONS:

INPUT:DS: 0000H:K=4BH

0001H:A=41H

0002H:K=4BH

0003H:I=49H

0004H:N=4EH

0005H:A=41H

0006H:D=44H

0007H:A=41H

OUTPUT:DS:

000EH:N=4EH

000FH:A=41H



0010H:D=44H

0011H:A=41H


INPUT: STRING1=

OUTPUT: STRING2=

MEMORY LOCATIONS:

INPUT:DS: 0000H:

0001H:

0002H:

0003H:

0004H:

0005H:

0006H:

0007H:

OUTPUT:DS:

000EH:

000FH:

0010H:

0011H:

RESULT:

String deletion is performed and results are observed using TASM.

EXERCISE:

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?



7. READING KEYBOARD WITH ECHO

AIM: To write a program to read the keyboard with echo.


ALGORITHM:


STEP2: Move 01 into AL

STEP3: Initialize DOS

STEP4: Move AL into address


PROGRAM:

DATA SEGMENT

CHAR DB 0

DATA ENDS

CODE SEGMENT


START: MOV AX,DATA

MOV DS,AX

MOV AH,01H

INT 21H

MOV CHAR, AL

INT 3

CODE ENDS

END START

OBSERVATIONS:

INPUT: N

OUTPUT:N



RESULT:

Reading a keyboard is performed and results are observed using TASM.

EXERCISE:

1. Write a brief note on DOS interrupts.

2. What does AH register hold when INT 21H is called?



8. READING KEYBOARD WITHOUT ECHO

AIM: To write a program to read the keyboard without echo.


ALGORITHM:






PROGRAM:

DATA SEGMENT

CHAR DB 0

DATA ENDS

CODE SEGMENT


START:

MOV AX,DATA

MOV DS,AX

MOV AH,08H

INT 21H

MOV CHAR, AL

INT 3

CODE ENDS

END START

OBSERVATIONS:

INPUT:

OUTPUT:

RESULT:

Reading a keyboard is performed and results are observed using TASM.



EXERCISE:





9. DISPLAYING A CHARACTER

AIM: To write a program to displaying a given string.


ALGORITHM:






PROGRAM:

DATA SEGMENT

CHAR DB „*‟

DATA ENDS

CODE SEGMENT


START:

MOV AX, DATA

MOV DS, AX

MOV AH, 02H

MOV DL,CHAR

INT 21H

MOV AX, 4C00H

INT 21H

CODE ENDS

END START

OBSERVATIONS:

INPUT:

OUTPUT:

RESULT:



Displaying a character on screen is performed and results are observed using

TASM.

EXERCISE:





10. DISPLAYING A STRING

AIM: To write a program to displaying a given string.


ALGORITHM:


STEP2: Load DX with the string



PROGRAM:


DATA SEGMENT

STRING DB 'HAPPY NEW YEAR', 0DH, 0AH,'$'

DATA ENDS

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

LEA DX, STRING

MOV AH, 09H

INT 21H

CODE ENDS

END START

OBSERVATIONS:

INPUT: HAPPY NEW YEAR

OUTPUT: HAPPY NEW YEAR

RESULT:

Displaying a string on screen is performed and results are observed using TASM.



EXERCISE:







1. 8255 – PPI INTERFACING -1

AIM: Configure 8255A such that port A and port B as an output ports .Execute the

program at 0000:2000h

APPARATUS:

1. 8086 TRAINER KIT

2. 8255 STUDY CARD

3. ADAPTER, KEYBOARD, CABLES, ETC…

PROGRAM:

// DAC FOR ADC INTERFACE

// Connect the interface over J4 of the trainer

// This program illustrates the use of counter method for A/D conversion

// The program can be executed in Standalone or Serial Mode

// Execute the program from 2000H

OUTPUT 2500AD

ORG 2000H

MOV AX,00H //INITIALISE SEGMENT REGISTERS

MOV CS,AX

MOV ES,AX

MOV SP,3000H //INITIALISE STACK POINTER

CALL FAR 0FE00:0031H //NEWLINE

MOV DX,0FFE6H //INITIALISE 8255 AS FOLLOWS

MOV AL,81H //PORT A = OUTPUT

OUT DX,AL //PORT C = INPUT

JMP SHORT START

TEST: DB 0DH,20H,'CONVERTING... ', 00H

MES: DB 0AH,0DH,20H, 'DIGITAL VALUE = ',00H

START: MOV DX,0FFE4H //wait for start of conversion(SOC)

IN AL,DX

CALL DELAY

CALL DELAY

AND AL,02H

JZ START



MOV AX,0FFH

PUSH AX

CONT: MOV DX,0FFE4H //check for SOC

IN AL,DX

CALL DELAY

AND AL,02H

JNZ START //If true start again

POP AX //else continue

MOV DX,0FFE0H

INC AL //increment count and

OUT DX,AL //output to port

PUSH AX

PUSH DX

LEA DX,TEST //display 'converting...'

MOV AX,DX

POP DX

CALL FAR 0FE00:0013H

CALL DELAY

MOV DX,0FFE4H //read comparator output

IN AL,DX

AND AL,00000001B

JNZ CONT //if PC0=1,continue else

LEA DX,MES //conversion is over

MOV AX,DX

CALL FAR 0FE00:0013H //display message

POP AX

CALL FAR 0FE00:0052H //DISPLAY OUTPUT

CALL FAR 0FE00:0031H

JMP SHORT START

DELAY: MOV CX,8000H //delay routine

HERE: LOOP HERE

RET

END



OBSERVATIONS:

INPUT:

OUTPUT:

RESULT:

8086 Interfacing with 8255 is done and results are observed.



2. 8279 – KEYBOARD/DISPLAY INTERFACING

AIM: To interface keyboard/display with 8086 microprocessor and display a given

sentence on the display board.(Left Entry)

APPARATUS:

1. 8086 TRAINER KIT

2. 8279 STUDY CARD


PROGRAM:

CMDREG EQU 0FFEBH

DATAREG EQU 0FFE9H

.8086

.MODEL TINY

.STACK 32

.DATA

.CODE

ORG 2000H

START:

MOV AX,CS

MOV DS,AX

MOV AL,10H // ENCODED SCAN-8CHAR 8 BIT

LEFT ENTRY

MOV DX,CMDREG

OUT DX,AL

MOV AL,090H //WRITE RAM AUTO INCREMENT

OUT DX,AL

MOV CX,08

CLEAR:MOV AL,00

MOV DX,DATAREG

OUT DX,AL

LOOP CLEAR

BACK: MOV DX,CMDREG

IN AL,DX // READ THE COMMAND R4EGISTER TO GET THE 8279

STATUS



AND AL,07 // NUMBER OF KEYS PRESSED IS IN THE LOWER 3 BITS

MASK //THESE AND CHECK IF NOT ZERO. IF ZERO NO

KEY PRESSED

JZ BACK

MOV BX,OFFSET SSCHARLUT

MOV AL,040H // READ THE FIRST RAM ADDRESS OF

FIFO

OUT DX,AL

MOV DX,DATAREG //

IN AL, DX

AND AL,01FH

XLAT // CONVERT THE KEY READ TO THE

SS CODE

MOV DX,DATAREG

OUT DX,AL // DISPLAY AT THE CURRENT DIGIT

POSITION

JMP BACK

SSCHARLUT:

SS0 DB 3FH //0

SS1 DB 6H //1

SS2 DB 5BH //2

SS3 DB 4FH //3

SS4 DB 66H //4

SS5 DB 6DH //5

SS6 DB 7DH //6

SS7 DB 07H //7

SS8 DB 7FH

SS9 DB 6FH

SSA DB 77H

SSB DB 7CH

SSC DB 39H

SSD DB 5EH

SSE DB 79H

SSF DB 71H

END START

OBSERVATIONS:

INPUT:



OUTPUT:

RESULT:




3. 8259 – PROGRAMMABLE INTERUPT CONTROLLER

AIM: To interface Programmable Interrupt Controller with 8086 microprocessor.

APPARATUS:

1. 8086 TRAINER KIT

2. 8259 STUDY CARD


PROGRAM:

ORG 1000H

MOV AL,17H

OUT 0C0H, AL

MOV AL,08H

OUT 0C2H, AL

MOV AL,01H

OUT 0C2H, AL

MOV AL,FEH

OUT 0C2H, AL

STI

HERE: JMP HERE

MOV AL,20H

LOOP 0C0H,AL

INT 02H

Interrupt Vector:-

0000:0200 00h

0000:0021 12h

0000:0022 00h

0000:0023 00h

OBSERVATIONS:

INPUT:

OUTPUT:



RESULT:




4. 8251 – USART

AIM: To interface USART (Universal Synchronous Asynchronous Receiver

Transmitter) with 8086 microprocessor.

APPARATUS:

1 8086 TRAINER KIT

2 8251 STUDY CARD

3 ADAPTER, KEYBOARD, CABLES, ETC…

PROGRAM:

TRANSMITTER END:

ORG 1000H

MOV AL,36H

OUT 0CEH, AL

MOV AL,10H

OUT 0C8H, AL

MOV AL,00H

OUT 0C8H, AL

MOV AL,4EH

OUT 0C2H, AL

MOV AL,37H

OUT 0C2H, AL

LOOP1:IN AL,0C2H

AND AL,04H

JZ LOOP1

MOV AL,41H

OUT 0C0H,AL

INT 02H

RECEIVER END:

ORG 1200H

MOV AL,36H

OUT 0CEH, AL

MOV AL,10H

OUT 0C8H, AL

MOV AL,00H



OUT 0C8H, AL

MOV AL,4EH

OUT 0C2H, AL

MOV AL,37H

OUT 0C2H, AL

LOOP2:IN AL,0C2H

AND AL,04H

JZ LOOP2

IN AL,0C0H

MOV BX,I500H

MOV [BX],AL

INT 02H

OBSERVATIONS:

INPUT:

OUTPUT:

RESULT:






Expected Questions from Microprocessors

1. Who Invented 8086 microprocessor?

2. In which year 8086 microprocessor is invented?

3. How many address lines are available on 8086?

4. How many data lines are available on 8086?

5. Why address bus is unidirectional?

6. Whether data bus is unidirectional or bidirectional?

7. What is a bus?

8. If AL=0FFh and ADD AL,01h instruction is executed, what is the status of status

flags?

9. Under what condition bus contention occurs?

10. How can you avoid bus contention?

11. The segment 1234h and offset is 0022h, then the physical address is?

12. If an instruction is accessed by 8086 is 20ms, the time taken by 8088 is?

13. The no. of cycles required by 8086 for writing/reading to/from memory/IO ports is?

14. Whether 8086 contains internal clock generator?

15. 8086 requires external clock signal with a duty cycle of?

16. In 8086 the lower bank of 8086 contains?

17. When ready pin of 8086 is at logic 0 then wait states are

18. The vectored interrupt in 8086 is?

19. Non vectored interrupt in 8086 is?

20. To restart from the initial state when „1‟ is placed on 8086 RESET pin, the duration

for which the signal must present is?

21. The rate of transmission in serial mode is?

22. In serial I/O, data transmission takes place using which code?

23. USART converts data from which form to which from?

24. The inputs and outputs of 8251 are compatible with which logic?

25. Whether RS232C is compatible with TTL Logic?

26. If TxD output remains high the 8251 contains which characters?

27. In RS232C, RS stands for?

28. In RS 232C standard uses positive logic or negative logic?

29. Which transmission is used for high speed serial data transmission?

30. Expand DCE and DTE?

31. Which type of ADC is the fastest one?

32. Which type of ADC is the most expensive one?

33. Which type of ADC used DAC inside?

34. How many comparators are required by an 8-bit flash type ADC?



35. In 8259 if IR3, IR6, IR7 are activated, which interrupt has the highest priority?

36. Which register in 8259 is used to disable interrupt requests?

37. Which ports are used in BSR mode of 8255?

38. Which port provides hand shaking signals in Mode1 & Mode2 of 8255?

39. What is the IC used for DMA transfer?

40. How many peripherals can be connected to 8237 at any point of time?

41. For bidirectional data transfer, 8255 is operated in which mode?

42. DMA will send a request signal to which pin of 8086?

43. Which instruction is used to return from the interrupt service procedure?

44. What are the ICs MC 1488 and MC 1489?

45. When accessing a memory location shared with a coprocessor, the signal that

prevents the coprocessor from accessing that location is?

46. There are two no. of 2Kx8 memory chips. How to obtain 4Kx8 memory?

47. When the processor has 32 address lines, how much memory can be interfaced to

that processor?

48. Whether Stepper Motor is an AC Motor or DC Motor?

49. Give any application of Stepper Motor in our real life?

50. What do you mean by interfacing?

Expected Questions from Microcontrollers

1. Who has introduced MCS-51 family of microcontrollers?

2. MCS-51 family of microcontrollers contains which ICs?

3. What is the difference between MP and MC?

4. Give any Example of Microcontrollers?

5. Is 8051 an 8-bit controller?

6. Why 8051 is an 8-bit controller?

7. What is the on chip ROM capacity of 8051?

8. What is the on chip RAM capacity of 8051?

9. How many IO ports are available on 8051 on chip?

10. How many timers are available on 8051 on chip?

11. Which version of 8051 doesn‟t have on chip ROM?

12. Which version of 8051 contains EPROM on chip?

13. 8051 contains which type of ROM?

14. What is the purpose of ROM in 8051?

15. What is the purpose of RAM in 8051?

16. Whether ROM is accessible randomly or not?

17. What is the capacity of on chip EPROM on 8751?

18. Can you connect external program memory to 8051?

19. Can you connect external data memory to 8051?



20. What is the capacity of external program memory that can be connected to 8051?

21. What is the capacity of external data memory that can be connected to 8051?

22. How can you access internal program memory?

23. How can you access external program memory?

24. How is it possible to access both internal & external ROM using PC?

25. How can you access internal data memory?

26. How can you access external data memory?

27. Do you require any signal to access external data memory?

28. Do you require any signal to access external program memory?

29. What is the reason behind connecting only 64KB of external program memory?

30. What is the reason behind connecting only 64KB of external data memory?

31. Expand DPTR?

32. Is a DPTR an 8-bit register?

33. Can we access PC as two 8-bit registers?

34. Is there any register in 8086 similar to PC in 8051?

35. Expand SFR?

36. Is PC a special function register?

37. Whether external interrupts are edge triggered or level triggered type?

38. Expand ALE?

39. Expand PSEN?

40. What is the function of PSEN pin?

41. For serial communication the pins of which IO port is used?

42. Which register is used to operate 8051 in different modes?

43. What are the modes of operation of 8051?

44. Whether PCON is a bit addressable SFR?

45. Which port will hold the higher order address information during memory access?

46. Which Io port of 8051 doesn‟t have dual function?

47. What is the Architecture of 8051?

48. What is stack?

49. What is the function of SP register?

50. The number of clock cycles in one machine cycle of 8051 is?



Key to the Questions on Microprocessors

1. Intel Corporation

2. 1974

3. 20 address lines

4. 16 data lines

5. The address bus is unidirectional because the information on the address

bus is always provided by the microprocessor.

6. Data bus is bidirectional.

7. Bus is a collection of signal lines. Generally they carry address, data and

control signals.

8. CF=1, PF=1, AF=1, ZF=1, SF=0 and OF=0

9. Bus contention occurs when two outputs trying to drive the same line.

10. Bus contention can be avoided by using tri-state condition which isolates one of

the outputs.

11. 12362h.

12. 40ms.

13. 4 clock cycles.

14. No

15. 33% duty cycle for its optimal functioning.

16. Even address locations.

17. Between T3 and T4 states.

18. NMI.

19. INTR.

20. 4 clock cycles.

21. Baud.

22. ASCII.

23. Parallel form to Serial.

24. TTL Logic.

25. No.

26. No data (Stop bits).

27. Recommended Standard.

28. Negative logic.

29. Synchronous transmission.

30. DCE stands for Data Communications Equipment and DTE stands for Data

Terminal Equipment.

31. Flash type ADC.

32. Successive Approximation ADC.

33. Successive Approximation ADC.

34. 8 Comparators.



35. IR3.

36. Interrupt Mask Register.

37. Only Port C bits.

38. Port C provides hand shaking signals.

39. 8237 or 8257

40. Four

41. Mode 2

42. HOLD pin.

43. IRET.

44. Voltage translators.

45. LOCK

46. By connecting address lines in serial and data lines parallel

47. 232

= 4GB

48. DC Motor

49. Printers, X-Y plotters, Hard disk/Floppy Disk Drives, etc.,

50. Interfacing means not only connecting the devices but also communicating

between them.

Key to the Questions on Microcontrollers

1. Intel Corporation

2. 8031, 8051, 8052, 8751 and 8752 Microcontrollers.

3. MP contains ALU, Control Unit and On chip memory where as

Microcontroller contains MP, ROM, RAM, IO Ports, Counters/Timers on chip.

4. 8051

5. Yes

6. Because all the internal registers and accumulator are of 8-bit wide and they can

process 8-bit data.

7. 4KB

8. 128 bytes

9. 4 no. of 8-bit IO Ports

10. 2 no. of 16-bit Timers/Counters

11. 8031

12. 8751

13. Simple Read Only Memory

14. ROM is used for storing Programs hence it is called as Program Memory.

15. RAM is nothing but Data Memory and is used to store data

16. Yes

17. 4KB

18. Yes



19. Yes

20. 64KB

21. 64KB

22. Program Counter

23. Program Counter

24. Whenever we access external program memory, we use MOVX instruction which

distinguish the internal & external ROM.

25. By means of its address in the program instructions.

26. Data Pointer.

27. Yes. We require Logic „0‟ on External Access pin.

28. No signal is required to access external program memory.

29. Because PC is a 16-bit register, the maximum memory that can be

interfaced externally is 216

= 64KB only.

30. DPTR is a 16-bit register// the maximum memory that can be interfaced

externally is 216

= 64KB only.

31. Data Pointer

32. DPTR is a 16-bit register, but we can access it as two 8-bit registers DPH & DPL.

33. No, we can‟t access PC as two 8-bit registers.

34. PC is very similar to IP register in 8086.

35. SFR stands for Special Function register.

36. No, PC is the only register which doesn‟t come under SFR.

37. They can be programmed as either edge triggered or level triggered.

38. Address Latch Enable

39. PSEN stands for Program Store ENable.

40. PSEN is used to store program in external memory.

41. Port 3 pins are used for serial communication.

42. PCON (Power CONtrol) SFR is used.

43. Power Down Mode and Idle Mode.

44. No. PCON is not bit addressable.

45. Port 2 will hold the higher order address information during memory access.

46. Port 1 pins have no dual functions.

47. The Architecture of 8051is called Harvard Architecture.

48. Stack is an area in internal RAM that is used in conjunction with certain opcodes

to store and retrieve data quickly.

49. SP holds an internal RAM address that is called top of the stack.

50. 12 clock cycles.

Mpi Lab Manual

Documents

mpi lab manual

computer system

hand shaking

0000h mov

0000h mov

0000h mov

al mov result1

ah mov ah