8085 calicut university lab manual

Laboratory Manual

AE&I Engineering Department, Govt. Engg. College Kozhikode Page 1

SECTION I

ASSEMBLY LANGUAGE PROGRAMMING

BASED ON 8085 KIT

Laboratory Manual


Laboratory Manual


EXPERIMENT NO.1

FAMILIARIZATION OF 8085 TRAINER KIT

AIM:

To familiarize 8085 microprocessor trainer kit

KEY BOARD DESCRIPTION:

The m85-0x LCD kit has 101 ASCII keys and 20*2 Liquid Crystal Display to communicate

with the outside world. As m85-0x LCD is switched on a message 8085 LCD TRAINER

KIT m85 -0x is displayed on the LCD screen and all keys are in command mode.

LIST OF ASCII KEYBOARD COMMANDS:

L - List a memory block

M Examine/modify memory

E Enter a memory block

R Examine/modify register

S Single step

G Go

B Block move

I Insert

D Delete

N Insert data

O Delete data

F Fill

H Relocate

J Memory compare

K - String

COMMAND DESCRIPTION:

RESET:

This key initializes m 85- 0x LCD kit and displays 8085 LCD TRAINER KIT m 85-0x on

the LCD display, indicates that the system is expecting a valid command.

8085 LCD TRAINER KIT

M85-0x

Note: m85-0x means m85-04 or m85-07

Laboratory Manual


Laboratory Manual


LIST A MEMORY BLOCK (L):

This command is used to examine the contents of a block of memory on pressing this key L

is displayed on the upper left end of LCD display. One can now enter the desired address for

examining the contents. Enter the starting address ending address and press I key which is

shown in below example. This displays the contents of memory area in the LCD screen in 20

columns upper half and 20 columns lower half. If we want to see the further press . To

exit from the command mode press key

Format:

L (starting address), (ending address)

EXAMINE/MODIFY MEMORY (M)

This command is used to examine the contents of any memory location and modify the

contents of a RAM area on pressing this key m- is displayed on the upper left of the LCD

display. One can now enter the desired address of any location. For changing the contents,

enter the address and press , it shows the present content of the address is displayed

followed by a-to modify the contents of the location displayed. Type the new data and press

press 1 key to exit from command mode.

Format:

M [Starting address],

EXAMINE/MODIFY REGISTER (R):

This command is used to examine or modify any internal register of the CPU. One wants to

examine the contents of all registers press R display R on the upper left end of the LCD

display. One can start from A register, another examine all the registers by pressing .

Where-as it some specific register is to examine then the key for that register can be entered

directly. The contents of any register can be changed. To exit from this command mode 1

key

Format:

R (Register identifier)

Laboratory Manual


Laboratory Manual


The Register identifier values for various CPU registers are given below.

REGISTER IDENTIFIER REGISTER

A Register A

B Register B

C Register C

D Register D

E Register E

F Flag byte

I Interrupt mask

H Register H

L Register L

S Stack point MSB and LSB

P Program Counter MSB&LSB

GO COMMAND (G):

This command is used to execute the program in full clock speed. On pressing this key the

program counter contents are displayed. Enter the address of the program and press 1 key.

The CPU will start executing the operation.

Format:

G (Starting address)

Pressing the G key will display the PC content and the first byte of the instruction. To modify

it enter the desired address and then press key. CPU start executing the program.

ASSEMBLER AND DISASSEMBLER:

In m 85-0x LCD kit there is an on board facility of assembler or disassemble on pressing 1

key. M 85 -0x LCD kit comes in assembler or disassembler mode.

A ASSEMBLER MODE

C DISASSEMBLER MODE

ASSEMBLER MODE

On pressing the key A m 85 -0x LCD kit comes in to the assembler mode. As soon as the A

key is pressed kit asks RAM address. This will be the starting address of the program to be

entered. After entering the starting address press key. It displays the entered starting

address in the upper line of the LCD screen. Now it waits for mnemonics to entry.

One can enter all the valid mnemonics of 8085 and the pseudo commands. If the entered

alphabets do not form a valid mnemonic of a pseudo command, the carriage goes to some

line prints the address of previous line. Hence the entry of wrong mnemonics is indicated by

giving the same line to the users.

Laboratory Manual


Laboratory Manual


Entry of a space completes are field of entry and processing of that field, we mean is done

immediately by the command. By field we mean, mnemonics as one field operands or

operand labels as another field. On pressing A key m 85-0x LCD kit waits for the RAM

address.

Format:

RAM ADR: STARTING ADDRESS

ADDRESS: MNEMONICS

SINGLE BYTE INSTRUCTION:

NNNN mnemonics registered 1 or RST. NO [if needed] . If the mnemonics entered is a

single operand the command wait for operand entry. No message is given by the common in

this field. The users enters the required operand which is normally a register name of the two

register when single register or RST number operand is loaded.

DISASSEMBLER:

C command dis assemble the program as specified by the starting address and end address.

In this case one wants to proceed further press key. Otherwise key will exit from

the disassembly mode on pressing C key m85-ox LCD kit comes into the disassembly

mode.

FORMAT:

PROG ADR [starting address] @ [ending address]

LIST ADR

RESULT:

Familiarized the 8085 microprocessor trainer kit

*****

Laboratory Manual


Laboratory Manual


EXPERIMENT NO.2

ARITHMETIC OPERATION-I

AIM:

a) Write assembly language program for addition of two 8-bit numbers

b) Write assembly language program for subtraction of two 8-bit numbers

EQUIPMENTS REQUIRED:

8085 Trainer Kit, Keyboard, Power Chord

ALGORITHM:

a) 8-bit addition

Step 1: Read the first 8-bit number from memory location address 2500H to register A

Step 2: Read the second 8-bit number from memory location address 2501H to register B

Step 3: Add the two numbers

Step 4: Write the result to memory location address 2502H

Step 5: If carry exists, write the 01H value to memory location address 2503H

b) 8-bit subtraction

Step 1: Read the first 8-bit number from memory location address 2600H to register A

Step 2: Read the second 8-bit number from memory location address 2601H to register B

Step 3: Subtract the two numbers

Step 4: Write the result to memory location address 2602H

Step 5: If borrow exists, write the 01H value to memory location address 2603H

Laboratory Manual


OBSERVATION:

a) 8-bit addition

1. First 8-bit number: FF@2500

Second 8-bit number:FE@2501

Expected result: FD@2502, 01@2503

Observed result:

2. First 8-bit number:

Second 8-bit number:

Expected result:

Observed result:



Expected result:

Observed result:


1. First 8-bit number: FF@2600

Second 8-bit number:FE@2601

Expected result: 01@2602, 00@2603

Observed result:



Expected result:

Observed result:



Expected result:

Observed result:

Laboratory Manual


PROGRAM:

a) 8-bit addition

Memory

address

Hex code Label Mnemonics Comments

2000 3A 01 25 LDA 2501 Read the second 8-bit number

2003 MOV B,A Save the second number in B

register

2004 LDA 2500 Read the first 8-bit number

2007 ADD B Perform the addition

2008 STA 2502 Save the result

200B JC AHEAD Carry exist? If yes, go to AHEAD

200E MVI A,00 Else represent 0 value for carry

2010 STA 2503 Save no carry condition

2013 RST 5 Stop execution

2014 AHEAD MVI A,01 Represent carry using value 01

2016 STA 2503 Save the carry



AIM:

Memory

address


2100 3A 01 26 LDA 2601 Read the second 8-bit number

2103 MOV B,A Save the second number in B

register

2104 LDA 2600 Read the first 8-bit number

2107 SUB B Perform the subtraction

2108 STA 2602 Save the result

210B JC AHEAD Borrow exist? If yes, go to AHEAD

210E MVI A,00 Else represent 0 value for borrow

2110 STA 2603 Save no borrow condition


2114 AHEAD MVI A,01H Represent borrow using value 01

2116 STA 2603 Save the carry


RESULT:

Following programs assembled, executed using 8085 trainer kit and outputs were verified

a) Addition of two 8-bit numbers

b) Subtraction of two 8-bit numbers

*****

Laboratory Manual


Laboratory Manual


EXPERIMENT NO.3

ARITHMETIC OPERATION-II

AIM:

a) Write assembly language program for addition of two 16-bit numbers

b) Write assembly language program for subtraction of two 16-bit numbers

c) Write a assembly language program for decimal addition using DAA instruction



ALGORITHM:

a) 16-bit addition

Step 1: Load first 16-bit number in HL register pair

Step 2: Load second 16-bit number in DE register pair

Step 3: Add the lower bytes and store in L register

Step 4: Add the higher byte with previous carry and store in H register

Step 5: Store the 16-bit result in memory location 3000H


Step 1: Load first 16-bit number in HL register pair

Step 2: Load second 16-bit number in DE register pair

Step 3: subtract the lower bytes and store in L register

Step 4: Subtract the higher byte with previous borrow and store in H register

Step 5: Store the 16-bit result in memory location 3800H

c) Decimal addition

Step 1: Initialize counter with number of bytes to be added stored in memory 3500H

Step 2: Add two bytes starting from memory 3501 through accumulator

Step 3: Update the carry register if carry generated in step 2

Step 4: Decrement the counter and if not zero go to step 2

Step 5: Store the final sum result in memory 3600H and final carry in memory 3601H

Laboratory Manual


OBSERVATION:

a) 16-bit addition

1. First 16-bit data: 2050

Second 16-bit data: 2030

Expected result: 4080@2100

Observed result:

2. First 16-bit data:

Second 16-bit data:

Expected result:

Observed result:


1. First 16-bit data: 2030

Second 16-bit data: 2050

Expected result: 0020@3800

Observed result:

2. First 16-bit data:

Second 16-bit data:

Expected result:

Observed result:

Laboratory Manual


PROGRAM:

a) 16-bit addition

Memory

address


2000 21 50 20 LXI H 2050 Load 2050 in HL pair

2003 LXI D 2030 Load 2030 in DE pair

2006 MOV A,E Move lower value from E to A

2007 ADD L Add values in L to A

2008 MOV L,A Move values from A to L

2009 MOV A,D Move values from D to A

200A ADC H Add value in H with A along with

carry

200B MOV H,A Move value in A to H

200C SHLD 2100 Store value in HL pair to location

2100H

200F RST 5 Stop execution


AIM:

Memory

address


3002 21 30 20 LXI H 2030 Load 2030H in the HL pair

3005 LXI D 2050 Load 2050H in the DE pair

3008 MOV A,E Move value from E to A

3009 SUB L Subtract value in L from A

300A MOV L,A Move value from A to L

300B MOV A,D Move value from D to A

300C SBB H Subtract with borrow from A

300D MOV H,A Move value from A to H

300E SHLD 3800 Store result in HL pair to location

3800H


Laboratory Manual


c) Decimal addition

1. No of bytes to be added: 03@3500

First 8-bit data in 3501: 50

Second 8-bit data in 3502: 60

Third 8-bit data in 3503: 70

Expected result: 80@3600, 01@3601

Observed result:

2. No of bytes to be added: 05@3500

First 8-bit data in 3501:

Second 8-bit data in 3502:

Third 8-bit data in 3503:

Fourth 8-bit data in 3504:

Fifth 8-bit data in 3505:

Expected result:

Observed result:

Laboratory Manual


c) Decimal addition using DAA instruction

Memory

address


2300 AF XRA A Clear accumulator

2301 MOV D,A Clear register D

2302 LXI H 3500 Load 3500H in HL pair

2305 MOV C,M Copy the contents from M to C

2306 LOOP INX H Increment HL

2307 ADD M Add A and M

2308 DAA Decimal adjust accumulator

2309 JNC AHEAD Jump to AHEAD if no carry

230C INR D Increment D

230D AHEAD DCR C Decrement counter

230E JNZ LOOP Jump to LOOP if non zero

2311 STA 3600 Store 3600 with content of A

2314 MOV A,D Copy from D to A

2315 STA 3601 Store 3601 with content of A


RESULT:


a) Addition of two 16-bit numbers

b) Subtraction of two 16-bit numbers

c) Decimal addition using DAA instruction

*****

Laboratory Manual


Laboratory Manual


EXPERIMENT NO.4

ARITHMETIC OPERATION-III

AIM:

a) Write Assembly Language Program to perform 16x8 multiplication

b) Write Assembly Language Program to perform 16x8 division

c) Write Assembly Language Program to find the square of a 8-bit number



ALGORITHM:

a) 16x8 Multiplication

Step 1: Read 16 bit multiplicand from memory location

Step 2: Read 8 bit multiplier from memory location

Step 3: Initialize count register with multiplier value

Step 4: Add multiplicand value multiplier times to get the result

Step 5: Carry is accumulated if exists on each addition

Step 6: Store the final result in memory location

b) 16x8 Division

Step 1: Read 16 bit dividend from memory location

Step 2: Read 8 bit divisor from memory location

Step 3: Initialize a borrow register

Step 4: Initialize a register pair to hold the quotient

Step 5: Subtract divisor from dividend

Step 6: Check for borrow, if no borrow go to step 5 else proceed to step 7

Step 7: Quotient accumulated in register pair stored in memory location

c) Square of a 8-bit number

Step 1: Read 8 bit number to be squared

Step 2: Store the number temporarily in a register

Laboratory Manual


OBSERVATION:

a) 16X8 multiplication

1. 16-bit multiplicand: FF@2100

01@2101

8-bit multiplier: 22@2102

Expected result: DE@2103 43@2104 00@2105

Observed result:

2. 16-bit multiplicand:

8-bit multiplier:

Expected result:

Observed result:

Laboratory Manual


Step 3: Read the same number as a counter in another register

Step 4: Add the same number counter times and the result is stored in the accumulator

Temporarily

Step 5: Carry is checked to store in another register

Step 6: 8 bit result is stored in one memory location

Step 7: 8 bit carry result is stored in another memory location

PROGRAM:

a) 16X8 Multiplication

Memory

address


2000 24 00 21 LHLD 2100 Read the 16 bit multiplicand in HL register

pair from memory location 2100H

XCHG Save the 16 bit number in DE register

LDA 2102 Copy multiplier into accumulator from

2102 memory location

MOV C,A Initialise the counter

LXI H 0000 Load HL with zero result initially

BACK DAD D Add content in HL with DE pair and store

the result in HL pair

JNC SKIP Jump to SKIP if there is no carry

INR B Increment register B for carry

SKIP DCR C Decrement counter

JNZ BACK Jump to BACK if counter value is non zero

SHLD 2103 Store 16 bit result in HL pair to memory

location 2103

MOV A,B Move value from B to A carry value

STA 2104 Store carry in accumulator to memory

location 2104

RST 5 Stop execution

Laboratory Manual


OBSERVATION:

b) 16X8 division

1. 16-bit dividend: FF@3100

: FF@3101

8-bit divisor: FF@3102

Expected result: 01@3103 01@3104

Observed result:

2. 16-bit dividend:

8-bit divisor:

Expected result:

Observed result:

Laboratory Manual


b) 16X8 division

Memory

address


3000 2A 00 31 LHLD 3100 Read dividend in HL from memory

location 3100

LDA 3102 Read 8 bit divisor from memory

location 3102 to A

MOV C,A Copy value to C

MVI B 00 Initialize Register B to store borrow

MVI E FF Initialize DE pair for quotient

MVI D FF

BACK INX D Increment DE register pair

MOV A,L Copy LSB of dividend from L to A

SUB C Subtract value in C from A

MOV L,A Move result from A to L

MOV A,H Copy MSB of dividend to

accumulator

SBB B Borrow from LSB and subtract

from MSB

MOV H,A Copy result from A to H

JNC BACK If no borrow jump to BACK

MOV A,E Copy the value from E to A

STA 3103 LSB of quotient is stored in

memory location 3103

MOV A,D Copy result from D to A

STA 3104 MSB of quotient is stored in

memory location 3104


Laboratory Manual


OBSERVATION:


1. 8-bit number to be squared: 05@2500

Expected result: 19@2501 00@2502

Observed result:

2. 8-bit number to be squared:

Expected result:

Observed result:

Laboratory Manual



Address Hex code Label Mnemonics Comments

2300 LXI H 2500 Load HL register pair

MOV A,M Copy the register in A

MOV B,A Copy result from A to B

MOV C,B Set C as a counter

MVI A 00 A is loaded with D

MVI D 00 D is initialised to store carry

BACK ADD B Add content in B with A

JNC SKIP Jump to SKIP if no carry

INR D Increment D

SKIP DCR C Decrement counter C

JNZ BACK If non zero jump to BACK

STA 3000 Store LSB

MOV A,D Move MSB from D to A

STA 3501 Store MSB in 3501H


RESULT:


a) 16X8 multiplication

b) 16X8 division


*****

Laboratory Manual


Laboratory Manual


EXPERIMENT NO.5

LOGICAL OPERATIONS

AIM:

a) Write Assembly Language Program to study rotation, logical operators and complement

operations

b) Write Assembly Language Program to find the largest of an array



ALGORITHM:

a) Program to study rotation, logical and complement operations

Step 1: The two 8bit values are stored in memory location

Step 2: Find the 1s complement and 2s complement of the first byte and store the result

Step 3: Use rotate instructions for left and right rotation through carry flag

Step 4: Load the second byte and perform logical AND, OR and XOR operations

Step 5: Store the results in consecutive memory locations

b) Program to find largest of an array

Step 1: Initialize a count register for the number of data to be compared

Step 2: Initialize a register pair as address pointer to the data array

Step 3: Increment the address pointer

Step 4: compare the contents of memory location with accumulator

Step 5: Keep the largest number in accumulator on each comparison

Step 6: Decrement the count register and go to step 3 if not zero

Step 7: Else store the final largest number in accumulator in memory location

Laboratory Manual


OBSERVATION:

a) Rotation, logical and complement operations

1s complement operation

1 Input data: 21@3000 2 Input data: 36@3000

Expected output: DE@3500 Expected output: C9@3500

Observed output: Observed output:

2s complement operation


Expected output: DF@3501 Expected output: CA@3501


Rotate left operation


Expected output: 42@3502 Expected output: 6C@3502


Rotate right operation


Expected output: 90@3503 Expected output: 1B@3503


AND operation


Input data: 22@3001 Input data: 12@3001

Expected output: 20@3504 Expected output: 12@3504


Laboratory Manual


PROGRAM:

a) Program to study rotation, logical and complement operations

Memory

address


2000 LXI H 3000 Initialize address pointer

MOV A,M Load the first byte

MOV B,A Temporarily store the data

CMA 1s complement

STA 3500 Store the result

ADI 01 To find 2s complement

STA 3501 The 2s complement is stored

MOV A,B Restore the data

RLC Rotate content in A to left

STA 3502 The result in A is stored

MOV A,B Copy first operand to A

RRC Rotate accumulator right

STA 3503 Store the result

INX H Increment HL pair.

MOV C,M Copy the second byte

MOV A,B Copy 1st operand to A

ANA C Logical AND operation

STA 3504 Result is stored in memory

MOV A,B 1st operand is restored

ORA C Logical OR operation

STA 3505 The result is stored

MOV A,B 1st

operand is copied back

XRA C Logical XOR operation

STA3506 Store the result


Laboratory Manual


OBSERVATION:

OR operation





XOR operation





b) Largest of an array

1 No. of data: 04@2500

I/P data (1): 82@2501

I/P data (2): 19@2502

I/P data (3): 20@2503

I/P data (4): 06@2504

Expected output: 82@3500

Observed output:

2 No. of data: 05@2500

I/P data (1): 10@2501

I/P data (2): 30@2502

I/P data (3): 60@2503

I/P data (4): 70@2504

I/P data (4): 80@2505

Expected output: 80@3500

Observed output:

Laboratory Manual


b) Program to find the largest of an array

Memory

address


2600 LXI H 2500 Initialize the HL pair

MOV C,M Load number of operands

XRA A Clear accumulator

BACK INX H Point to first number

CMP M Compare the first number with

accumulator

JNC SKIP Jump to SKIP if no carry

MOV A,M Transfer the large number to A

SKIP DCR C Decrement the count register

JNZ BACK Check if zero flag set

STA 3500 Store the largest number


RESULT:


a) Study rotation, logical and complement operations

b) Find the largest of an array

*****

Laboratory Manual


8085 calicut university lab manual

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