MICROPROCESSOR LAB MANUAL - Study materialssulavpaudel.com.np/wp-content/uploads/2017/08/LAB-Manual.pdf · LAB MANUAL ON 8085 MICROPROCESSOR BSC-CSIT: II SEM Prepared By: Sulav Paudel

MICROPROCESSOR LAB MANUAL July 20, 2017

S. PAUDEL. MSc | LECTURER, DEPARTMENT OF INFORMATION TECHNOLOGY (BSC CSIT)

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Hetauda City College Affiliated to Tribhuwan University

LAB MANUAL

ON

8085 MICROPROCESSOR

BSC-CSIT: II SEM

Prepared By:

Sulav Paudel (MSc)

Lecturer



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PERFORM ALL PROGRAMS AND PREPARE THE LAB REPORT INCLUDING OBJECTIVE,

ALGORITHMS, PROGRAMS, OBSERVATION AND RESULT OR COMCLUSION

LAB 1

1.1 STORE 8-BIT DATA IN MEMORY LOCATION

AIM: To perform the 8085 assembly program to store 8-bit data 2AH in a memory location

8050H.

ALGORITHM 1:

Start the program by loading the first data into an accumulator.

Store accumulator contents at given memory address.

Terminate program.

PROGRAM 1:

MVI A, 2AH

STA 8050H

HLT

ALGORITHM 2:

Start the program by loading the first data into an accumulator.

Load HL pair register with given memory address.

Store accumulator contents in memory location pointed by HL register pair.

Terminate program.

PROGRAM 2:

LXI H, 8050H

MVI M, 2AH

HLT

1.2 EXCHANGE THE CONTENTS OF MEMORY LOCATIONS

AIM: To perform the 8085 assembly program exchange the contents of memory locations

5000H and 4000H.

ALGORITHM 1:

Start the program by loading the first data of memory location 4000H into

accumulator.

Move the data to a register (B register).

Load the second data of memory location 5000H into accumulator.

Store the data of accumulator at address 4000H

Move the data back into accumulator that is saved in a register (B register).

Store the data of accumulator at address 5000H

Terminate program.



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PROGRAM 1:

LDA 4000H

MOV B, A

LDA 5000H

STA 4000H

MOV A, B

STA 5000H

HLT

ALGORITHM 2:

Start the program by initializing register pair (HL) as a pointer to memory location

4000H.

Initializing register pair (DE) as a pointer to a memory location 5000H.

Move the data of memory location 4000H into a register (B register).

Move the data of memory location 5000H into an accumulator.

Move the data of accumulator into memory location pointed by register pair (HL).

Move the data from register (B) into accumulator.

Store the data of accumulator into memory location pointed by register pair (DE).

Terminate program.

PROGRAM 2:

LXI H, 4000H

LXI D, 5000H

MOV B, M

LDAX D

MOV M, A

MOV A, B

STAX D

HLT



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LAB 2: Use 8085 KIT (MPS 85-3)

RAM USAGE

The system monitor utilizes RAM locations 8F90H to 8FFFH for Storing the System Stack

and variables. User programs should not disturb this area; otherwise the results are

unpredictable.

2.1 EXAMINE/MODIFY MEMORY COMMAND

FUNCTION: This command is used to examine the contents of selected memory

locations. The contents can be optionally modified if the memory location is in

RAM area.

FORMAT: EXAM MEM <address> NEXT [[<data>] NEXT/PREV]* [EXEC]

Example 1: Examining a series of memory locations starting from 0000H (the start of the

keyboard monitor).

Key Pressed Display

Comments Address Field Data Field

RESET -UPS 85 System Reset

EXAM MEM . Examine Memory command

0 0000 First memory location to be

examined 0000H

NEXT 0000 F3 Contents of this location.

NEXT 0001 3E Next location and its

contents.

NEXT 0002 0F Next location and its

contents.

PREV 0001 3E Previous location and its

contents.

EXEC - Command

termination/prompt.


EXAM MEM . Examine Memory command

8 0008.

D 008D. Memory location to be

examined & modified.

0 08D0. Memory location to be

examined & modified.

0 8D00.

NEXT 8D00 XX. Contents of this location

A 8D00 0A. New data to be entered

F 8D00 AF.

EXEC - Command termination

prompt



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2.2 EXAMINE/MODIFY REGISTER COMMAND

FUNCTION: The Examine Register Command is used to examine and optionally

modify the contents of any of 8085s’ registers.

FORMAT: EXAM REG <reg key> [[<data>] NEXT]* [EXEC]

Example 1: Examining a series of processor register contents

Key Pressed Display



EXAM REG . Examine Register Command

8/H H XX. Contents of register.

8 H 08. New data.

NEXT L XX. H register updated. Next

register’s contents displayed.

EXEC - Command termination

prompt.

2.3 GO COMMAND

FUNCTION: The GO command is used to transfer control of the system from the

monitor to user’s program.

FORMAT: GO address EXEC

Example 1: Suppose the following program is entered in the memory by EXAM MEM

command.

Location Opcode Mnemonic

8000 3E MVI A, 42

8001 42

8002 4F MOV C, A

8003 EF RST 5

To run this program, press the keys according to the following sequence.

Key Pressed Display



GO XXXX. XX GO command; current PC

and the byte at this PC are

displayed.

8 0008. New starting address.

0 0080.

0 0800.

0 8000.

EXEC -UPS 85 Program is executed from

8000H control returned to

the monitor.



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Example 2: Fill the registers A and C with 00H. Execute the above program from 8000H.

Later examine the contents of reg A & C. Both the registers must contain 42H.

Key Pressed Display



EXAM REG . Examine Register command

A A XX. Content of reg A

0 A 00. Reg A altered.

NEXT B XX. Content of reg B

NEXT C XX. Content of reg C

0 C 00. Reg C altered.

EXEC - Command terminated

GO XXXX. XX GO Command; Current PC

and the byte at this PC are

displayed.

8 0008. New value of the PC

0 0080.

0 0800.

0 8000.

EXEC -UPS 85 Control returns to monitor

EXAM REG . Examine Reg command

A A 42. Contents of reg A

NEXT B XX. Contents of reg B

NEXT C 42. Contents of reg C

EXEC - Control returns to monitor

due to the break-point.

2.4 SINGLE STEP COMMAND

FUNCTION: This command is used to execute a program, one instruction at a time.

With each instruction executed, control is returned to the monitor. Thus, this

command is an extremely useful debugging tool.

FORMAT:

SINGLE STEP [<Start address> NEXT <Start address> NEXT]* EXEC

Example 1: Assume the program given in Example 1 of GO Command illustration is entered

in the memory. Now this program can be single-stepped as follows.

Key Pressed Display



SINGLE STEP XXXX. XX Current value of PC and the

byte at this PC

8 0008.

0 0080.

0 0800.

0 8000.

NEXT 8002 4F Instruction at 8000 is

executed. Next instruction to



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be executed is displayed.

NEXT 8003 EF Next instruction

EXEC - Control returns to monitor

due to the break-point.

2.5 BLOCK MOVE COMMAND

FUNCTION: This command is used to move a block of data from one area of the

memory to another area.

FORMAT:

BLK MOVE <Start address> NEXT <End address> NEXT <Destination address>

EXEC

Example 1: Moving the block of program code listed in the example for the GO command.

Key Pressed Display



BLK MOVE . Block move command

8 0008. Start address of source block

0 0080.

0 0800.

0 8000.

NEXT .

8 0008. End address of source block.

0 0080.

0 0800.

3 8003.

NEXT .

8 0008. Destination address

8 0088.

0 0880.

0 8800.

EXEC - Block move operation

successful. Command

prompt.

2.6 INSERT COMMAND

FUNCTION: This command is used to insert one or more instructions into the user

program.

FORMAT:

INSERT <Low-Limit> NEXT <High-Limit> NEXT <Low Insert Address> NEXT

<No. of Bytes> NEXT <data> NEXT <data> NEXT .... [EXEC]



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Example 1: Assume that the following program has already been entered into memory using

EXAM MEM Command.

LOCATION VALUE INSTRUCTION

8000 C5 PUSH B

8001 0B DELAY: DCX B

8002 78 MOV A,B

8003 B1 ORA C

8004 C2 JNZ DELAY

8005 01

8006 80

8007 C1 POP B

8008 D1 POP D

8009 E1 POP H

800A EF RET 5

Now, suppose, you discovered that you did not initialize BC register pair before starting the

DELAY. You can insert one instruction, say LXI B, 3030H starting at the location 8001 by

the following key sequence.

Key Pressed Display



INSERT XXXX. XX Current Low-Limit

8 0008. The Low Limit of user

program

0 0080.

0 0800.

0 8000.

NEXT XXXX. Current High-Limit

8 0008. The High-Limit of the user

program

0 0080.

0 0800.

A 800A.

NEXT .

8 0008. Low Insert Address

0 0080.

0 0800.

1 8001.

NEXT .

3 0003. Insert one instruction of 3

Bytes

NEXT 8001 XX. Start entering the instruction

1 8001 01.

NEXT 8002 XX.

3 8002 03.

0 8002 30.

NEXT 8003 XX.



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3 8003 03.

0 8003 30.

EXEC - Instruction inserted return to

command prompt.

To see that the instruction has indeed been inserted, you can use the EXAM MEM command

to observe the contents of the locations 8000H to 800DH (800A+3=800D). Also note that the

address reference for the JNZ instruction has changed from 8001 to 8004.

2.7 DELETE COMMAND

FUNCTION: This command is used to delete one or more instructions from the

user program.

FORMAT:

DELETE <Low-Limit> NEXT <High-Limit> NEXT <Low Delete Address> NEXT

<High Delete Address> EXEC

Example 1: After inserting the instruction and after examining the program in 2.6 INSERT

COMMAND, Example 1, suppose you discover that you have two extra instructions POP D

and POP H. You can delete these instructions by the following key sequence.

Key Pressed Display



DELETE 8000. Current Low-Limit. You do

not want to change it. So

press NEXT.

NEXT 800A. Current Hight-Limit.

8 0008. New High-Limit

0 0080.

0 0800.

D 800D.

NEXT . Enter Low Delete Address.

Note that POP D is now at

800BH and not at 8008H

8 0008.

0 0080.

0 0800.

B 800B.

NEXT . Enter High Delete Address.

8 0008.

0 0080.

0 0800.

C 800C.

EXEC - The two bytes are deleted.

Command prompt appears.

To check the operation, use EXAM MEM key to observe the contents location 8000H to 800BH.

They should be C5, 01, 30, 0B, 78, B1, C2, 04, 80, C1, C9.



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LAB 3

3.1 STORE 8-BIT DATA IN MEMORY LOCATION

AIM: Loads the registers B and C with specific values and stores a specific value in a RAM

location. Assume the program is to be loaded from 8000 H. The program and the hand

assembly are shown below.

LOCATION CONTENTS MNEMONIC COMMENTS

8000 06 22 MVI B, 22H Initialize Reg B

8002 0E 82 MVI C, 82H Initialize Reg C

8004 78 MOV A, B

8005 32 00 81 STA 8100H Load the RAM

location

8008 EF RST 5 Return control to the

monitor

3.2 ADDITION OF TWO 8 BIT NUMBERS

AIM: To perform the subtraction of two 8 bit numbers using 8085 Kit (MPS 85-3) and also

run the program in 8085 Simulator.

ALGORITHM:

Start the program by loading the first data into Accumulator.

Move the data to a register (Say B Register).

Get the second data and load into Accumulator.

Add the two register contents.

Check the carry.

Store the value of sum and carry in memory location.

Terminate the program.


8000 0E 00 MVI C, 00H Initialize register C

to 00H

8002 21 00 81 LXI H, 8100 HL points 8100H

memory location

8005 7E MOV A, M Move data

8006 23 INX H

8007 86 ADD M Add contents

8008 82 0C 80 JNC LOOP Jump control

800B 0C INR C

800C 23 LOOP: INX H

800D 77 MOV M, A

800E 23 INX H

800F 71 MOV M, C


monitor



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LAB 4

4.1 SUBTRACTION OF TWO 8 BIT NUMBERS

AIM: To perform the subtraction of two 8 bit numbers using 8085 Kit (MPS 85-3) and also

run the program in 8085 Simulator.

ALGORITHM:

Start the program by loading the first data into Accumulator.

Move the data to a register (Say B Register).

Get the second data and load into Accumulator.

Subtract the two register contents.

Check the carry.

If carry is present take 2’s complement of Accumulator

Store the difference value and value of borrow in memory location.



8000 0E 00 MVI C, 00H Initialize register C

to 00H

8002 21 00 81 LXI H, 8100 HL points 8100H

memory location

8005 7E MOV A, M Move data

8006 23 INX H Increment HL pair

8007 96 SUB M Subtract contents

8008 82 0E 80 JNC LOOP Jump control

800B 2F CMA Complement Acc

800C 3C INR A Increment Acc

800D 0C INR C Increment C Reg

800E 23 LOOP: INX H Increment HL pair

800F 77 MOV M, A Move content


8011 71 MOV M, C Move content


monitor



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LAB 5

5.1 MULTIPLICATION OF TWO 8 BIT NUMBERS

AIM: To perform the multiplication of two 8 bit numbers using 8085 Kit (MPS 85-3) and

also run the program in 8085 Simulator.

ALGORITHM:

Start the program by loading the 00H data into Accumulator.

Get the first data to a register (Say B Register).

Get the second data to a register (Say C Register) act as a counter loop.

Add the first data with accumulator content.

Check the carry.

If carry is present, increment the value of carry.

Decrement the value of second data. (C Register)

Check the zero flag.

If zero flag is reset, then repeat addition step else store the value of product and carry

in memory location.



8000 16 00 MVI D, 00H Initialize register D

to 00H

8002 3E 00 MVI A, 00H Initialize register A

to 00H

8004 21 00 81 LXI H, 8100 HL points 8100H

memory location

8007 46 MOV B, M Move data


8009 4E MOV C, M Move data

800A 80 LOOP: ADD B Add content

800B 2F 0F 80 JNC NEXT Jump control

800E 3C INR D Increment D Reg

800F 0C NEXT: DCR C Decrement C Reg

8010 D2 0A 80 JNZ LOOP Jump control


8014 77 MOV M, A Move content


8016 72 MOV M, D Move content


monitor



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LAB 6

6.1 DIVISION OF TWO 8 BIT NUMBERS

AIM: To perform the division of two 8 bit numbers using 8085 Kit (MPS 85-3) and also run

the program in 8085 Simulator.

ALGORITHM:

Start the program by loading the 00H data into Accumulator.

Get the first data to a register (Say B Register).

Get the second data to a register (Say C Register) act as a counter loop.

Add the first data with accumulator content.

Check the carry.

If carry is present, increment the value of carry.

Decrement the value of second data. (C Register)

Check the zero flag.

If zero flag is reset, then repeat addition step else store the value of product and carry

in memory location.



8000 16 00 MVI D, 00H Initialize register D

to 00H

8002 3E 00 MVI A, 00H Initialize register A

to 00H

8004 21 00 81 LXI H, 8100 HL points 8100H

memory location

8007 46 MOV B, M Move data


8009 4E MOV C, M Move data

800A 80 LOOP: ADD B Add content

800B 2F 0F 80 JNC NEXT Jump control

800E 3C INR D Increment D Reg

800F 0C NEXT: DCR C Decrement C Reg

8010 D2 0A 80 JNZ LOOP Jump control


8014 77 MOV M, A Move content


8016 72 MOV M, D Move content


monitor



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LAB 7

7.1 LARGEST NUMBER IN AN ARRAY OF DATA

AIM: To find the largest number in an array of data using 8085 instruction set

ALGORITHM:

Load the address of the first element of the array in HL pair.

Move the count to a register (Say B Register).

Increment the HL pair pointer

Get the first data in a register (Say A Register)

Decrement the count.

BACK: Increment the HL pair pointer

Compare the content of memory addressed by HL pair with that of A reg.

If carry is reset, go to step: NEXT or if carry is set, proceed sequentially.

Move the content of memory addressed by HL to A reg.

NEXT: Decrement the count.

Check for Zero of the count. If ZF is reset, go to step: BACK, else proceed

sequentially.

Store the largest data in memory.




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LAB 8

8.1 SMALLEST NUMBER IN AN ARRAY OF DATA

AIM: To find the smallest number in an array of data using 8085 instruction set

ALGORITHM:

Load the address of the first element of the array in HL pair.

Move the count to a register (Say B Register).

Increment the HL pair pointer

Get the first data in a register (Say A Register)

Decrement the count.

BACK: Increment the HL pair pointer

Compare the content of memory addressed by HL pair with that of A reg.

If carry is set, go to step: NEXT or if carry is reset, proceed sequentially.

Move the content of memory addressed by HL to A reg.

NEXT: Decrement the count.

Check for Zero of the count. If ZF is reset, go to step: BACK, else proceed

sequentially.

Store the smallest data in memory.


Documents

MICROPROCESSOR LAB MANUAL - Study materialssulavpaudel.com.np/wp-content/uploads/2017/08/LAB-Manual.pdf · LAB MANUAL ON 8085 MICROPROCESSOR BSC-CSIT: II SEM Prepared By: Sulav Paudel