microprocessor Laboratory experiments manual

Laboratory Experiments Manual for 8085 Microprocessor
Experiment No. 1 - Addition of Two 8-bit Numbers and Sum is 8-bit
AIM: Write 8085 assembly language program for addition of two 8-bit numbers and sum is 8 bit.
Instruments Required: 1. 8085 Microprocessor Kit
2. +5V Power supply
Theory : Consider the first number 26H is stored in memory location 8000H and the second number 62H
is stored in memory location 8001H. The result after addition of two numbers is to be stored in the memory
location 8002 H. Assume program starts from memory location 8500H.
Algorithm
1. Initialise the memory location of first number in HL register pair.
2. Move first number/data into accumulator
3. Increment the content of HL register pair to initialise the memory location of second data
4. Add the second data with accumulator
5. Store the result in memory location 8003H
PROGRAM
Memory
address
Machine Codes Labels Mnemonics Operands Comments
8500 21, 00, 80 LXI H, 8000 H Address of first number in
H-L register pair.
8503 7E MOV A,M Transfer first number in
accumulator.
8504 23 INX H Increment content of H-L
register pair
8505 66 ADD M Add first number and
second number
8506 32, 03, 80 STA 8003H Store sum in 8003 H
8509 76 HLT Halt
Experimental Results
Input DATA RESULT
Memory location Data Memory location Data
8000 26H
8001 62H
8003 88H
Conclusion:
The addition of two 8-bit numbers is performed using 8085 microprocessor where sum is 8-bit.
Precautions:
1. Properly connect the 8085 microprocessor kit with power supply terminals.
2. Switch on the power supply after checking connections
3. Handle the Trainer kit carefully.
Viva-Voice Questions:

1. What is the function of LXI H, 8000 H instruction?
2. How you can store a data in a memory location?
3. How you can read a data from a memory location?
4. What are flags available in 8085 ?
5. What is the function of RESET key of a 8085 microprocessor kit
Experiment No. 2 - Addition of two 8 bit numbers and sum is 16-bit
AIM: Write 8085 assembly language program for addition of two 8-bit numbers and sum is 16 bit.
2. +5V Power supply
Theory : The first number F2H is stored in memory location 8501H and the second number 2FH is stored
in memory location 8502H. The result after addition will be stored in the memory location 8503 H and
8504H. Consider program is written from memory location 8000H.
Algorithm:
Initialise the memory location of 1st
data in HL register pair.
Store first data in the memory location
Increment the content of HL register pair for entering next data in the next memory location
Store second data in the memory location
Move second number in accumulator
Decrease the content of HL register pair
Add the content of memory (first data) with accumulator
Store the results in memory location 8503H and 8504H.
PROGRAM
Memory
address
8000 21, 01, 85 LXI H, 8501 H Address of 1st
number in
H-L register pair.
8003 36, F2 MVI M, F2H Store 1st
number in
memory location
represented by H-L
register pair.
8005 23 INX H Increment content of H-L
register pair
8006 36, 2F MVI M, 2FH Store 2nd
number in
memory location
represented by H-L
register pair
8008 7E MOV A, M 2nd
number in accumulator
8009 0E, 00 MVI C,00H Initialise C register with
00H to store MSBs of sum
800B 2B DCX H Address of 1st
number

2501 in H-L pair
800C 66 ADD M Addition of 1st
number
and 2nd
number
800D D2, 11, 85 JNC LEVEL_1 If carry does not
generated , jump to
LEVEL_1 ?
8010 0C INR C When carry is generated,
increment C register.
8011 32, 03, 85 LEVEL_1 STA 8503H Store LSBs of sum in
memory location 8503H
8014 79 MOV A,C Move MSBs of sum in
accumulator
8015 32, 04, 85 STA 2504 H Store MSBs of sum in
memory location 8503H.
8018 76 HLT Halt
Input DATA RESULT
8501 F2H 8503 82H LSBs of sum
8502 2FH 8504 01H MSBs of sum
Conclusion:
Precautions:
Properly connect the 8085 microprocessor kit with power supply terminals.
Switch on the power supply after checking connections
Handle the Trainer kit carefully.
1. What is the function of JNC instruction?
2. What is the difference between conditional and unconditional jump instruction?
3. What is the function of STA 2500$ instruction?
4. What is multi byte?
Experiment No. 3 - Addition of Two 16-Bit Numbers and Sum is 16-bit
AIM: Write 8085 assembly language program for addition of two16-bit numbers and sum is 16 bit.
2. +5V Power supply
Theory : The first 16-bit number is stored in 8501 and 8502 memory locations. The second 16-bit number
is stored in 8503 and 8504 memory locations. After addition result will be stored from 8505 and 8506
memory locations. Assume program starts from memory location 8000H

Algorithm:
1. Store first 16-bit number in H-L pair.
2. Exchange the contents of D-E pair and H-L pair to store first number in D-E registers pair.
3. Store second 16-bit number in HL register pair.
4. Addition of 1st
and 2nd
number
5. Store result in 8505H and 8506H locations.
PROGRAM
Memory
address
8000 2A, 01, 85 LHLD 8501 H Load the content of 8501H
location in L register and H
register is loaded with the
content of 8502H location.
8000 EB XCHG` The contents of HL register
pair are exchanged with
D-E register pair. So that
first data is stored in DE
register pair.
8000 2A, 03, 85 LHLD 8503 H Load second 16-bit number
(data-2) in H-L pair.
8000 19 DAD D The contents of D-E pair
are added with the contents
of H-L pair and result is
stored in H-L pair.
8000 22, 05, 85 SHLD 8505 H Store LSBs of sum in 8505
and MSBs of sum 8506 H.
8000 76 HLT Halt
Input DATA RESULT
Memory location Data Memory
location
Data
8501 05H LSBs of data-1 8505 07H LSBs of sum
8502 01H MSBs of data-1 8506 04H MSBs of sum
8503 02H LSBs of data-2
8504 03H MSBs of data-2
Conclusion:
Precautions:

1. What is the function of XCHG instruction?
2. How you can load 16-bit data in 8500H and 8501H memory locations?
3. What is the difference between LSLD and SHLD instructions?
4. What is the function of DAD D instruction?
5. What is difference between byte and word?
Experiment No. 4 - Decimal Addition of Two 8-Bit Numbers and Sum is 8-bit
AIM: Write 8085 assembly language program for decimal addition of two8-bit numbers and sum is 8 bit.
2. +5V Power supply
Theory : Two decimal numbers are stored in 8000H and 8001H. The result is to be stored in 8002H
location. Consider program starts from memory location 8100H
Algorithm
1. Initialise the memory location of first number in HL register pair.
2. Load the first number in accumulator
3. Increment the content of HL register pair to initialise the memory location of second data
4. Addition of the content of second memory location with first data
5. Decimal adjustment of result
PROGRAM
Memory
address
8100 21, 50, 80 LXI H,8050 Load the address of first
number in H-L register pair
8103 7E MOV A,M Store the first number in
accumulator
8104 23 INX H Increment H-L register pair to
locate second number
8105 86 ADD M Addition of 1st
and 2nd
number
8106 27 DAA Decimal Adjust
8107 32, 02, 80 STA 8002 Store result in 8002H location
810A 76 HLT Halt
Input DATA RESULT
8000 22H
8001 67H
8002 89H

Conclusion:
The decimal addition of two 8-bit numbers is performed using 8085 microprocessor where sum is 8-bit.
Precautions:
1. What is the function of DAA instruction?
Experiment No. 5 - One’s Complement of an 8-bit Number
AIM: Write 8085 assembly language program for one’s complement of an 8-bit numbers
2. +5V Power supply
Theory : The number is stored in memory location 8050H and one’s complement of number will be stored in
location 8051H. Assume the program memory starts from 8000H.
Algorithm
1. Load memory location of data 8050H in H-L registers pair.
2. Move data into accumulator
3. Complement accumulator
PROGRAM
Memory
address
8000 21, 50, 80 LXI H,8050H Load address of number in H-
L register pair
8003 7E MOV A,M Move the number into
accumulator
8004 3F CMA Complement accumulator
8005 32, 51, 80 STA 8051H Store the result in 8051H
8008 76 HLT Stop
In the above program, the first two instructions LXI H,8050H and MOV A,M can be replaced by directly
loading the content of location 8050H in accumulator. For this LDA 8050H can be used.
DATA RESULT

8050 F0H 8051 0FH
Conclusion:
The one’s complement of an 8-bit numbers is performed using 8085 microprocessor.
Precautions:
2. Define one’s complement of an 8-bit numbers
3. What is the function of CMA instruction?
4.
Experiment No. 6 Two’s Complement of an 8-bit Number
AIM: Write 8085 assembly language program for two’s complement of an 8-bit numbers
2. +5V Power supply
Theory : The number is stored in memory location 8500H. The two’s complement will be stored in 8501H. The
program is written from memory location 8510H
Algorithm:
1. Transfer the content of memory location 8500H to accumulator.
2. Complement the content of accumulator
3. Add 01H with accumulator to get two’s complement of number
PROGRAM
Memory
address
Machine
Codes
Labels Mnemonics Operands Comments
8510 3A, 00, 85 LDA 8500H Load the content of memory
location 8500H in
accumulator
8513 2F CMA Complement accumulator
8514 C6, 01 ADI 01H Add 01H with accumulator
to find two’s complement of
number
8516 32, 01, 85 STA 8501H Store result in 8501H
location
8519 76 HLT Stop

DATA RESULT
8500 F0H 8501 10H
Conclusion:
The two’s complement of an 8-bit numbers is performed using 8085 microprocessor.
Precautions:
1. Define two’s complement of an 8-bit numbers
2. What is the function of CMA instruction?
3. Why ADI 01H is used in two’s complement of an 8-bit number.
Experiment No. 7 - Find out Square Root of 0, 1, 4, 9, 16, 25, 36, 49, 64 and 81
Using Look Up Table
AIM: Write 8085 assembly language program to find out square root of 0, 1, 4, 9, 16, 25, 36, 49 , 64 and
81 using look up table
2. +5V Power supply
Theory : Load the number in memory location 9000H and the square root of number will be stored in the
memory location 9001H. The square root of numbers 0, 1, 4, 9, 16, 25, 36, 49, 64 and 81 are stored in
8500H, 8501H, 8504H, 8509H, 8516H, 8525H, 8536H, 8549H, 8564H and 8581H locations respective1y
as given in tabular form. Assume the program is written from memory location 9100H.
Algorithm
1. Store the number in accumulator from memory location 9000H
2. Move the content of Accumulator in L register and store 85H in H register
3. When the number is 16, the content of H and L register are 85 and 16H respectively. Then the H-L
register pair represents 8516H memory location.
4. Copy square root of number in accumulator from memory location which is represented by H-L
register pair
5. Store the result in 9001H
PROGRAM

Memory
address
Machine
Codes
9100 3A, 00, 90 LDA 9000H Load the number in
accumulator from memory
location 9000H
9103 6F MOV L, A Copy the content of
Accumulator in L register
9104 26, 85 MVI H, 85 Load 85H in H register
9106 7E MOV A, M Move square root of decimal
number in accumulator from
memory
9107 32, 01, 90 STA 9001H Store square root value in
9001H
910A 76 HLT
Look –up Table
ADDRESS SUUARE ROOT
8500 00
8501 01
8504 02
8509 03
8516 04
8525 05
8536 06
8549 07
8564 08
8581 09
ADDRESS SQUARE ADDRESS Result
9000 H 16 9001 H 04
Conclusion:
The square root of 0, 1, 4, 9, 16, 25, 36, 49 , 64 and 81 are determined using look up table
in 8085 microprocessor.
Precautions:
1. How look up table is used to determine the square of a number?
2. Write some applications of look up table.

Experiment No. 8 - Display Digits 0 1 2 3 4 5 6 7 8 9 A B C D E F on the data
field of screen
AIM: Write 8085 assembly language program to display 0 1 2 3 4 5 B C D E F on the data filed.
2. +5V Power supply
Theory : This program will display a 6 7 8 9 A flashing 0 1 2 3 4 5 B C D E F on the data filed. The
flashing rate is 500ms. Consider program is written from memory location 8000H and data is stored from
8C50H.
Algorithm
1. Initialize the stack pointer
2. Clear the display
3. Point to the data which will be displayed in the data field
4. Wait for 500mili second
5. Cear the display and wait for 500 ms
6. Jump to step-3
PROGRAM
Memory
address
Machine
Codes
8000 31, FF, 8C LXI SP,8CFF Initialize the stack pointer
8003 0E, 0F MVI C,0F Load the counter, C register with
0FH
8005 21, 50, 8C LXI H,8C50 Starting address where the display
data is stored
8008 3E, 01 MVI A,01
800A 06, 00 MVI B,00
A is 01 and B is 00 to display
character in the data field
800C E5 LOOP_I PUSH H
800D C5 PUSH B
800E F5 PUSH PSW
Move the content of HL, BC, PSW
and A registers into stack
800F CD, 47, 03 CALL CLEAR Clear the display by using a
subroutine whose address is 0347H
8012 11, 00, 00 LXI D,0000H
8015 CD, BC, 03 CALL DELAY
Generate 500 ms delay
8018 CD, D0, 05 CALL OUTPUT Display character in data field
801B CD, 47, 03 LXI D,0000H
801E 11, 00, 00 CALL DELAY
To display the number for 500 ms
8021 F1 POP PSW
8022 C1 POP B
8023 E1 POP H
POP the content of PSW, A, BC and
HL from stack
8024 23 INX H Increment HL register pair to
display next number
8025 0D DCR C Decrement C register
8026 C2, 0C, 80 JNZ LOOP_I Jump to LOOP_I if the content of C

is not equal to zero
Input
ADDRESS DATA ADDRESS DATA
8C50 00 H 8C58 08 H
8C51 01 H 8C59 09H
8C52 02 H 8C5A 0AH
8C53 03 H 8C5B 0B H
8C54 04 H 8C5C 0C H
8C55 05 H 8C5D 0D H
8C56 06 H 8C5E 0E H
8C57 07 H 8C5F 0F H
Conclusion:
Display 0 1 2 3 4 5 B C D E F on the data filed of 8085 microprocessor.
Precautions:
1. How we can display any number on the data field of 8085 microprocessor.
3. Write some applications of flashing display.
Experiment No. 9 - Rolling Display “HELP 85 up” on the address and data field
of screen
AIM: Write 8085 assembly language program for rolling display “HELP 85 up” in the address and data
filed.
2. +5V Power supply
Theory : This program will display a flashing “HELP 85 up” in the address and data filed. The flashing
rate is 500ms. Assume program is written from memory location 9000H and data is stored from 9C50H.
Algorithm
1. Initialize the stack pointer
2. Clear the display
Point to the data of “HELP 85” and display “HELP 85 ” in the address and data field
Wait for 500 ms
Clear the display and wait for 500 ms
Jump to start

PROGRAM
Memory
address
Machine
Codes
9000 31, FF, 9C LXI SP,9CFF Initialize the stack pointer
9003 CD, 47, 03 CALL CLEAR Clear the display by using a
subroutine whose address is 0347H
9006 0E, 0A START MVI C, 06 Load the counter, C register with 06
9008 AF XRA A
9009 47 MOV B, A
A is 00 to display character in the
address field
900A 21, 50, 9C LXI H, 9C50 Starting address where the display is
to be started
900D E5 LOOP_ I PUSH H
900E C5 PUSH B
900F F5 PUSH PSW
Move the content of HL, BC, PSW
and A registers into stack
9010 CD, D0, 05 CALL OUTPUT Call OUTPUT subroutine to display
characters in address field
9013 3E, 01 MVI A, 01
9015 06, 00 MVI B, 00
Initialize A, B for data to be
displayed in data field
9017 CD, D0, 05 CALL OUTPUT Call OUTPUT subroutine to display
characters in data field
901A 21, 00, 00 LXI D,0000H
901D CD, BC, 03 CALL DELAY
Generate 500 ms delay
9020 F1 POP PSW
9021 C1 POP B
9022 E1 POP H
POP the content of PSW, A, BC and
HL from stack
9023 23 INX H Increment HL register pair to
display next number
9024 0D DCR C Decrement C register
9025 C2, 0D, 90 JNZ LOOP_ I Jump to LOOP_I if the content of C
is not equal to zero.
C3, 06, 90 JMP START Jump to START
Input
ADDRESS DATA ADDRESS DATA
9C50 16 H 9C55 10H (H)
9C51 16 H 9C56 0E H (E)
9C52 16 H 9C57 11H (L)
9C53 16 H 9C58 12H (P)
9C54 16 H 9C59 08 H (8)
9C5A 05 H (5)
Conclusion:
Rolling Display “HELP 85 up” in the address and data filed of 8085 microprocessor is performed.
Precautions:

1. How we can display any number and characters on the data field of 8085 microprocessor.
4. Write some applications of rolling display.
Laboratory Experiments Manual for 8086 Microprocessor
Experiment No. 10 - Program for Addition of Two 8 Bit Numbers and Sum is 16 Bit
AIM: Write 8086 assembly language program for Addition of Two 8 Bit Numbers and Sum is 16 Bit.
2. +5V Power supply
3. Personal Computer
4. UPS
Theory : The first number FFH is stored in AL register and the second number 22H is stored in BL
register. The result after addition will be stored in AX.
Algorithm
1. Store first data in AL register
2. Store second data in BL register
3. Add the content of AL and BL.
4. CY flag will be set, if result is more than 8 bit
PROGRAM
C:>DEBUG
-A 1000
17DA:1000 MOV AL, FF ;Load FFH in AL register
17DA:1002 MOV BL, 22 ;Load 22H in AL register
17DA:1004 ADD AL, BL ;Add content of BL to AL
17DA:1006 HLT
17DA:1007
-U 1000 1006
17DA:1000 B0 FF MOV AL,FF
17DA:1002 B3 22 MOV BL,22
17DA:1004 00 D8 ADD AL,BL
17DA:1006 F4 HLT
-G 1006
AX=0021 BX=0022 CX=0000 DX=0000 SP=0004 BP=20CD SI=0000 DI=0000
DS=17DA ES=17DA SS=9FFF CS=17DA IP=1006 NV UP EI PL NZ AC PE CY
17DA:1006 F4 HLT

Conclusion:
Initially the program is loaded in the memory location 17DA:1000 to 17DA:1006. Then the object codes
of program the program will be visualized after execution of U 1000 1006 command. If the above program
is executed by G1006 command, result will be displayed on the screen. As result is more that 8 bit, the
content of AL is 21 which is LSB and the CY flag is set to detect the MSB.
Precautions:
Properly connect the 8086 microprocessor kit with power supply terminals and Personal computer.
Switch on the power supply after checking connections
Handle the Trainer kit carefully.
What are the registers are used to store data?
At what condition the carry flag is set?.
What are the different flags available in 8086?
Experiment No. 11-Program for Addition of Two 16 Bit Numbers and Sum is 16 Bit
AIM: Write 8086 assembly language program for Addition of Two 16 Bit Numbers and Sum is 16 Bit.
2. +5V Power supply
4. UPS
Theory : The first 16-bit number is stored in AX register. The second 16-bit number is stored in BX
register. After addition result will be stored in AX.
Algorithm
Store first 16-bit number in AX.
Store second 16-bit number in BX.
Addition of 1st
and 2nd
number.
PROGRAM
C:>DEBUG
-A 1000
17DA:1000 MOV AX, 4622 ; 16 bit data in AX.
17DA:1003 MOV BX, 3244 ; 16 bit data in BX.
17DA:1006 ADD AX, BX ; Contents of BX is added to AX.
17DA:1008 HLT
17DA:1009
-U 1000 1008
17DA:1000 B8 22 46 MOV AX,4622
17DA:1003 BB 44 32 MOV BX,3244
17DA:1006 01 D8 ADD AX,BX
17DA:1008 F4 HLT
-G 1008

DS=17DA ES=17DA SS=9FFF CS=17DA IP=1008 NV UP EI PL NZ NA PE NC
17DA:1008 F4 HLT
Conclusion:
The program for addition of two 16 bit numbers is loaded in the memory location 17DA:1000 to
17DA:1008. After editing the program, U 1000 1008 command is used to display the object codes of
program. When the above program is executed by G1008 command, result will be displayed on the screen.
The content of AX is 7866H which is the addition of 4622H and 3244H. As sum is 16 bit, no carry is
generated.
Precautions:
1. Properly connect the 8086 microprocessor kit with power supply terminals and Personal computer.
1. What are the registers are used to store word?
2. At what condition the carry flag is set?.
3. What are the different flags available in 8086?
4. What is the command to execute the above program?
Experiment No. 12 - Program for Addition of a String of Words and Sum is 16 Bit
AIM: Write 8086 assembly language program for Addition of a string of words and sum is 16 Bit.
2. +5V Power supply
4. UPS
Theory : Assume the number of 16-bit data is stored in CX register and a string of words are stored in
17DA:0300 to 17DA:0305. After addition the result is stored in BX. Initially content of BX is 0000H
Algorithm
1. Initialize SI register with 0300H as source address of data.
2. Load number of bytes to be added in CX register.
3. Load a word in AX from the source specified by SI and SI is incremented by 2.
4. Addition of AX content and BX content.
5. Move content of AX to BX.
6. Continue step-3 to step-5 until CX=0.

PROGRAM
C:>DEBUG
-A 0100
17DA:0100 MOV SI,0300 ; Source address in SI.
17DA:0103 MOV CX,0005 ; Count value is loaded in CX.
17DA:0106 MOV AX,[SI] ; Load AX with data which is located by SI
17DA:0108 ADD AX,BX ; Contents of BX in AX.
17DA:010A INC SI ; Increment SI
17DA:010B INC SI ; Increment SI
17DA:010C MOV BX,AX ; Contents of BX in AX.
17DA:010E DEC CX ; Decrement CX
17DA:010F JNZ 0106 ; Jump to 0106 if CX≠0
17DA:0111 HLT
17DA:0112
-U 0100 0111
17DA:0100 BE 00 03 MOV SI,0300
17DA:0103 B9 05 00 MOV CX,0005
17DA:0106 8B 04 MOV AX,[SI]
17DA:0108 01 D8 ADD AX,BX
17DA:010A 46 INC SI
17DA:010B 46 INC SI
17DA:010C 89 C3 MOV BX,AX
17DA:010E 49 DEC CX
17DA:010F 75 F5 JNZ 0106
17DA:0111 F4 HLT
-
-ECS:0300
17DA:0300 00.01 00.01 00.02 00.02 00.03 00.03 00.04 00.04
17DA:0308 00.05 00.05
-G 0111
AX=0F0F BX=0F0F CX=0000 DX=0000 SP=FFEE BP=0000 SI=030A DI=0000
DS=17DA ES=17DA SS=17DA CS=17DA IP=0111 NV UP EI PL ZR NA PE NC
17DA:0111 F4 HLT
-
Conclusion:
The above program is entered in the memory location 17DA:0100 to 17DA:0111. The U 0100 0111
command is used to display the object codes of program. Five 16-bit data 0101, 0202, 0303, 0404 and 0505
are entered by the command -ECS:0300 17DA:0300 00.01 00.01 00.02 00.02 00.03 00.03 00.04
00.04 17DA:0308 00.05 00.05. After addition 0101, 0202, 0303, 0404, and 0505 we get 0F0F.
Therefore, when the program is executed by G0111 command and result will be displayed on the screen as
the content of BX and AX registers, 0F0F.
Precautions:

1. What is a string of words?
2. What are the registers are used to store words?
3. At what condition the carry flag is set?.
4. What is the difference between conditional and unconditional jump instructions?
5. What are the different flags available in 8086?
6. Which register is used as pointer?
7. What is the command to execute the above program?
Experiment No. 13 - Program for 2’s Complement of a 16 Bit Number
AIM: Write 8086 assembly language program for 2’s complement of a 16 bit number.
2. +5V Power supply
4. UPS
Theory : Assume that the 16-bit number is stored in AX register. Find the two’s complement of number
and stored it in BX register.
Algorithm
1. Store the 16-bit number in AX register.
2. Determine 2’s complement of AX .
3. Store two’s complement of AX in BX
PROGRAM
C:>DEBUG
-A 1000
17DA:1000 MOV AX, 2244 ; 16 bit data in AX
17DA:1003 NEG AX ; 2’s complement of 16 bit data
17DA:1005 MOV BX, AX ; Result is stored in BX
17DA:1007 HLT
17DA:1008
-U 1000 1007
17DA:1000 B8 44 22 MOV AX,2244
17DA:1003 F7 D8 NEG AX
17DA:1005 89 C3 MOV BX,AX
17DA:1007 F4 HLT
-G 1007
AX=DDBC BX=DDBC CX=0000 DX=0000 SP=0004 BP=20CD SI=0000 DI=0000
DS=17DA ES=17DA SS=9FFF CS=17DA IP=1007 NV UP EI NG NZ AC PO CY
17DA:1007 F4 HLT
-

Conclusion:
The program for 2’s complement of a 16-bit number is stored in the memory location 17DA:1000 to
17DA:1007. The U 1000 1006 command is used to display the object codes of program. When this
program is executed by G1007 command and result will be available in AX. Therefore the content of AX
is copied into BX.
Precautions:
1. What is 1’s complement?
2. What is 2’s complement?
3. What is the difference between 1’s complement and 2’s complement?
4. What is the application of 2’s complement of a number?
Experiment No. 14 - Program to Multiply Two 8 Bit Numbers
AIM: Write 8086 assembly language program for multiplication of two 8 bit number.
2. +5V Power supply
4. UPS
Theory : Assume the first number 11H is stored in AL register and the second number 22H is stored in BL
register. Multiply contents of AL by BL and the result is to be stored in AX register.
Algorithm:
1. Load first number in AL register.
2. Store the second data in BL register
3. Multiply contents of AL by BL
PROGRAM
C:>DEBUG
-A 1000
17DA:1000 MOV AL,11 ;8 bit multiplicand in AL
17DA:1002 MOV BL,22 ;8 bit multiplier in BL

17DA:1004 MUL BL ;Multiply contents of AL by BL
17DA:1006 HLT
17DA:1007
-U 1000 1006
17DA:1000 B011 MOV AL,11
17DA:1002 B322 MOV BL,22
17DA:1004 F6E3 MUL BL
17DA:1006 F4 HLT
-G 1006
DS=17DA ES=17DA SS=9FFF CS=17DA IP=1006 OV UP EI NG NZ NA PO CY
17DA:1006 F4 HLT
-
Conclusion:
The program for multiplication of two 8 bit numbers is edited in the memory location 17DA:1000 to
17DA:1006. The U 1000 1006command is used to display the object codes of program. This program is
executed by G1006 command and result will be stored in AX register. Hence the result is content of
AX=0242H.
Precautions:
1. Which command is used to multiply two numbers?
2. What are the registers are used for multiplication?
3. Which registers are used to store the result after multiplication?
Experiment No. 15 - Program to Divide Two 8 Bit Numbers.
AIM: Write 8086 assembly language program to divide two 8 bit numbers.
2. +5V Power supply
4. UPS
Theory : Assume the first number 56H is stored in AL register and the second number 02H is stored in CL
register. Divide contents of AL by BL and the result is to be stored in AX register.
Algorithm
1. Load first number in AL register.

Store the second data in CL register
Divide contents of AL by CL and result is AX
PROGRAM
C:>DEBUG
-A 1000
17DA:1000 MOV AL, 56 ; 8 bit dividend in AL
17DA:1002 MOV CL, 02 ; 8 bit divisor in CL
17DA:1004 DIV CL ; Divide contents of AL by CL
17DA:1006 HLT
17DA:1007
-U 1000 1006
17DA:1000 B0 56 MOV AL,56
17DA:1002 B1 02 MOV CL,02
17DA:1004 F6 F1 DIV CL
17DA:1006 F4 HLT
-G 1006
AX=002B BX=0000 CX=0002 DX=0000 SP=0004 BP=20CD SI=0000 DI=0000
DS=17DA ES=17DA SS=9FFF CS=17DA IP=1006 NV UP EI NG NZ NA PO NC
17DA:1006 F4 HLT
-
Conclusion:
The program for division of two 8 bit numbers is stored in the memory location 17DA:1000 to 17DA:1006.
The U 1000 1006command is used to display the object codes of program. When this program is executed
by G1006 command and result will be stored in AX register. Quotient is stored in AL and remainder is
stored in AH. Hence the result is quotient=content of AL= 2B and remainder =content of AH= 00H.
Precautions:
1. Which command is used to divide two numbers?
2. What are the registers are used for division?
3. Which registers are used to store the result after division?
Experiment No. 16 - Program for addition of two string decimal numbers
AIM: Write 8086 assembly language program for addition of two string decimal numbers.
2. +5V Power supply

4. UPS
Theory : The first string decimal numbers 11 22 33 44 55 are stored in memory locations 17B3:0100 to
17B3:0104. The second string decimal numbers 22 33 55 66 77 are stored in memory locations 17B3:0400
to 17B3:0404. After addition of two string decimal numbers, the result will be stored in memory locations
17B3:0500 to 17B3:0504.
Algorithm
1. Initialize SI and BX as offset address of first and second string decimal numbers.
2. Initialize DI as offset address of result and number of decimal numbers in a string is loaded in
CX
3. Move first decimal number from first string to AL
4. Add first decimal number from second string decimal numbers with AL
5. Decimal adjustment of AL after addition and store AL content in destination address represented
by DI Decrement CX
6. Increment SI, BX and DI
7. Move next decimal number from first string to AL
8. Next decimal number from second string decimal numbers is added with AL and carry
9. Decimal adjustment of AL after addition and store AL content in destination address represented
by DI
10. Decrement CX. If CX≠0, jump to step 6
PROGRAM
C:>DEBUG
-A0100
17B3:0100 MOV SI,0300 ;Load 0300H in SI as offset address of first string
17B3:0103 MOV BX,0400 ;Load 0400H in BX as offset address of second string
17B3:0106 MOV DI,0500 ;Load 0500H in DI as offset address of destination to store result
17B3:0109 MOV CX,0005 ;Number of decimal numbers in a string is loaded in CX
17B3:010C MOV AL,[SI] ; Load decimal number from first string
17B3:010E ADD AL,[BX] ; Add decimal number of second string to AL
17B3:0110 DAA ;Decimal adjustment after addition
17B3:0111 MOV [DI],AL ; Store the content of AL into destination memory location
17B3:0113 DEC CX ; decrement CX by1
17B3:0114 INC SI ; increment SI by 1
17B3:0115 INC BX ; increment BX by 1
17B3:0116 INC DI ; increment DI by 1
17B3:0117 MOV AL,[SI] ; Load next decimal number from first string
17B3:0119 ADC AL,[BX] ; Add next decimal number of second string to AL with carry
17B3:011B DAA ;Decimal adjustment after addition
17B3:011C MOV [DI],AL ; Store the content of AL into destination memory location
17B3:011E DEC CX ; decrement CX by1
17B3:011F JNZ 0114 ; if CX≠0, jump to offset address 0114
17B3:0121 HLT
17B3:0122

-U 0100 0121
17B3:0100 BE 00 03 MOV SI,0300
17B3:0103 BB 00 04 MOV BX,0400
17B3:0106 BF 00 05 MOV DI,0500
17B3:0109 B9 05 00 MOV CX,0005
17B3:010C 8A 04 MOV AL,[SI]
17B3:010E 02 07 ADD AL,[BX]
17B3:0110 27 DAA
17B3:0111 88 05 MOV [DI],AL
17B3:0113 49 DEC CX
17B3:0114 46 INC SI
17B3:0115 43 INC BX
17B3:0116 47 INC DI
17B3:0117 8A 04 MOV AL,[SI]
17B3:0119 12 07 ADC AL,[BX]
17B3:011B 27 DAA
17B3:011C 88 05 MOV [DI],AL
17B3:011E 49 DEC CX
17B3:011F 75 F3 JNZ 0114
17B3:0121 F4 HLT
-ECS:0300
17B3:0300 00.11 00.22 00.33 00.44 00.55
-ECS:0400
17B3:0400 00.22 00.33 00.55 00.66 00.77
-G0121
AX=0033 BX=0404 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0304 DI=0504
DS=17B3 ES=17B3 SS=17B3 CS=17B3 IP=0121 NV UP EI PL ZR NA PE CY
17B3:0121 F4 HLT
-ECS:0500
17B3:0500 33. 55. 88. 10. 33.
Conclusion:
The above program can be used for addition of two string decimal numbers. The U 0100 0121 command is
used to display the object codes of program. The ECS:0300 and ECS:0400 command are used to enter first
string 11 22 33 44 55 and second string 22 33 55 66 77 respectively. The program is executed by G0121
command and after addition result 33 55 88 10 33 will be stored in memory location 17B3:0500 to
17B3:0404.
Precautions:
1. How a string of decimal numbers is called in 8086?
2. Why DAA instruction is used after addition?
3. Which register is used to store the address of the source?
4. Which register is used to store the address of the destination?

Experiment No. 17 - Program to Find out the Largest Number from a String of
Words
AIM: Write 8086 assembly language program to find out the largest number from a string of words.
2. +5V Power supply
4. UPS
Theory : The count value of number of words 05H is stored in CX register. A string of words is stored in
the memory locations starting from17B3:0300 to 17B3:0309. The largest word will be stored in memory
location 17B3:0400.
Algorithm
1 Initialize SI as source offset address of word and load numbers of words in CX register.
2. Initialize AX register with 0000H
3. Compare the word from memory with content of accumulator AX. If the word is above and equal
to AX, jump to step-5
4. Move word from memory to AL
5. Increment SI by 2
6. Decrement CX. If CX≠0, jump to step- 3
7. Store result in memory location 17B3:0400.
PROGRAM
C:>DEBUG
-A0100
17B3:0100 MOV SI,0300 ; Initialize SI with 0300H as source offset address of word
17B3:0103 MOV CX,0005 ; Count value of words in CX
17B3:0106 MOV AX,0000 ; Initialize AX with 0000H
17B3:0109 CMP AX,[SI] ; compare word from memory with AX
17B3:010B JAE 010F ; Jump to 010F if above and equal
17B3:010D MOV AX,[SI] ; Load word from memory
17B3:010F INC SI ; Increment SI
17B3:0110 INC SI ; Increment SI
17B3:0111 LOOPNZ 0109 ; If CX≠0, jump to0109
17B3:0113 MOV [0400],AX ; Store the content of AX at destination address
17B3:0116 HLT
-U 0100 0116
17B3:0100 BE 00 03 MOV SI,0300
17B3:0103 B9 05 00 MOV CX,0005
17B3:0106 B8 00 00 MOV AX,0000
17B3:0109 3B 04 CMP AX,[SI]
17B3:010B 73 02 JNB 010F
17B3:010D 8B 04 MOV AX,[SI]
17B3:010F 46 INC SI
17B3:0110 46 INC SI
17B3:0111 E0 F6 LOOPNZ 0109

17B3:0113 A3 00 04 MOV [0400],AX
17B3:0116 F4 HLT
-ECS:0300
17B3:0300 00.11 00.11 00.22 00.22 00.33 00.33 00.FF 00.FF
17B3:0308 00.99 00.99
-G0116
AX=FFFF BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=030A DI=0000
DS=17B3 ES=17B3 SS=17B3 CS=17B3 IP=0116 NV UP EI PL NZ NA PE NC
17B3:0116 F4 HLT
-ECS:0400
17B3:0400 FF. FF.
Conclusion:
The five 16-bit data or word 1111, 2222, 3333, FFFF, and 9999 are stored in the memory location starting
from 17B3:0300 to 17B3:0309. Therefore the largest number is FFFFH. After execution of above
program, the largest number is stored in memory location 17B3:0400 and it can be displayed by ECS:0400
command as given above.
Precautions:
1. What is the use of CMP AX, [SI] instruction?
2. What is the function LOOPNZ 0109H instruction?
3. How you can store the largest number in memory?
Experiment No. 18 - Program to Arrange a String of Word in Descending Order
AIM: Write 8086 assembly language program to arrange a string of words in descending order.
2. +5V Power supply
4. UPS
Theory : A series of five words 1111H, 5555H, 3333H, 2222H, and 4444H are stored in memory
locations from 17B3:0302 to 17B3:030B and number of words is stored in memory location 17B3:0300.
Arrange the above words in Descending Order.
Algorithm
1. Store 0005H, number of words to be arranged in DX register from memory and store number of
comparisons in CX register
2. Load the 1st
word in accumulator from memory

3. Increment SI register by 2 for addressing next word
4. Compare next word from memory with accumulator. Store the smallest word in accumulator and
largest word in memory
5. Then next number (word) is compared with accumulator and Store the largest number in memory
and smallest number in accumulator.
6. This process will continue, till comparisons of all numbers have been completed. After completion
of comparison of all numbers, the smallest number in accumulator and store it in memory. In this
way first process will be completed.
7. At the starting of second process, DX register is decremented by one and store number of
comparisons in CX register. Then repeat step-2 to step-6. After completion of this process,
smallest number in 17B3:030A and second smallest number in 17B3:0308.
8. DX register is decremented by one and the next process starts, if the content of DX register is not
zero. Then repeat step-2 to step-6.
PROGRAM
C:>DEBUG
-A0100
17B3:0100 MOV SI,0300 ; SI loaded with 0300 as source address
17B3:0103 MOV DX,[SI] ; DX loaded with the content of SI
17B3:0105 MOV CX,[SI] ; CX loaded with the content of SI
17B3:0107 DEC CX ; Decrement CX
17B3:010A MOV AX,[SI] ;AX is loaded with word from memory represented by SI
17B3:010C INC SI ; Increment SI
17B3:010D INC SI ; Increment SI
17B3:010E CMP AX,[SI] ; Compare AX with the content of memory represented by SI
17B3:0110 JNB 0118 ; Jump to 0118
17B3:0112 XCHG AX,[SI] ; Exchange AX and word stored at memory represented by SI
17B3:0114 DEC SI ; Decrement SI
17B3:0115 DEC SI ; Decrement SI
17B3:0116 MOV [SI],AX ;Move AX to memory represented by SI
17B3:0118 LOOP 010A ; CX decrement by 1. If CX≠0, jump to 010A
17B3:011A DEC DX ; Decrement DX
17B3:011B MOV SI,0300 ; SI loaded with 0300 as source address
17B3:011E JNZ 0105 ; Jump not zero to 0105
17B3:0120 HLT
-U 0100 0120
17B3:0100 BE 00 03 MOV SI,0300
17B3:0103 8B 14 MOV DX,[SI]
17B3:0105 8B 0C MOV CX,[SI]
17B3:0107 49 DEC CX
17B3:0108 46 INC SI
17B3:0109 46 INC SI
17B3:010A 8B 04 MOV AX,[SI]
17B3:010C 46 INC SI
17B3:010D 46 INC SI
17B3:010E 3B 04 CMP AX,[SI]

17B3:0110 73 06 JNB 0118
17B3:0112 87 04 XCHG AX,[SI]
17B3:0114 4E DEC SI
17B3:0115 4E DEC SI
17B3:0116 89 04 MOV [SI],AX
17B3:0118 E2 F0 LOOP 010A
17B3:011A 4A DEC DX
17B3:011B BE 00 03 MOV SI,0300
17B3:011E 75 E5 JNZ 0105
17B3:0120 F4 HLT
-ECS:0300
17B3:0300 00.05 00.00 00.11 00.11 00.55 00.55 00.33 00.33
17B3:0308 00.22 00.22 00.44 00.44
-G0120
DS=17B3 ES=17B3 SS=17B3 CS=17B3 IP=0120 NV UP EI PL ZR NA PE NC
17B3:0120 F4 HLT
-ECS:0300
17B3:0300 05. 00. 55. 55. 44. 44. 33. 33.
17B3:0308 22. 22. 11. 11.
Conclusion:
The above program is used to arrange a string of words in descending order and this program is stored into
memory 17B3:0100 to 17B3:0120. The U 0100 0120 command is used to display the object codes of the
above program. The 5 words (1111H, 5555H, 3333H, 2222H, and 4444H) are entered by the command
ECS:300. Thereafter G0120 is used to execute the program and words will be stored in descending order
(5555H, 4444H 3333H, 2222H, and 1111H) as given above.
Precautions:
1. What do you mean by descending order?
2. What is the function of LOOP and XCHG AX,[SI] instructions?
3. How you can store the smallest number in memory?
Experiment No. 19 - Program to Find the Transpose of a 3x3 Matrix
AIM: Write 8086 assembly language program to find the transpose of a 3 x 3 matrix.
2. +5V Power supply

4. UPS
Theory :
The transpose of a matrix A = is A










333231
232221
131211
aaa
aaa
aaa
T
=










332313
322212
312111
aaa
aaa
aaa
The matrix A is stored in the memory location 17B3:0200 to17B3:0209. After transpose, AT
matrix will be
stored in the memory location 17B3:0300 to17B3:0309.
Algorithm
1. Store source address of matrix A in SI and store destination address of AT
in DI
2. Load number of rows of a matrix in CL register
3. Load element of matrix A from memory to AL
4. Store element into destination address
5. Increment SI by three and DI by one
6. Load element of matrix A from memory to AL and store into destination address
7. Increment SI by three and DI by one
8. Load element of matrix A from memory to AL and store into destination address
9. Subtract 05 from SI
10. Decrement CL, If CL≠0, continue step-3-step-10
PROGRAM
C:>DEBUG
-A0100
17B3:0100 MOV SI,0200 ; SI loaded with 0200 as source address of matrix A
17B3:0103 MOV DI,0300 ; DI loaded with 03200 as destination address of matrix A
17B3:0106 MOV CL,03 ; Number of rows in CL
17B3:0108 MOV AL,[SI] ; Load element of matrix A
17B3:010A MOV [DI],AL ; Store element of A in destination address
17B3:010C ADD SI,+03 ; Add 03 with SI
17B3:010F INC DI ; Increment DI
17B3:0112 MOV [DI],AL ; Store element of A in destination address
17B3:0114 ADD SI,+03 ; Add 03 with SI
17B3:0117 INC DI ; Increment DI
17B3:011A MOV [DI],AL ; Store element of A in destination address
17B3:011C SUB SI,+05 ; Subtract 03 from SI
17B3:011F INC DI ; Increment DI
17B3:0120 DEC CL ; Decrement CL
17B3:0122 JNZ 0108 ; If CL≠0, jump to 0108C
17B3:0124 HLT
-U100 124
17B3:0100 BE 00 02 MOV SI,0200

17B3:0103 BF 00 03 MOV DI,0300
17B3:0106 B1 03 MOV CL,03
17B3:0108 8A 04 MOV AL,[SI]
17B3:010A 88 05 MOV [DI],AL
17B3:010C 83 C6 03 ADD SI,+03
17B3:010F 47 INC DI
17B3:0110 8A 04 MOV AL,[SI]
17B3:0112 88 05 MOV [DI],AL
17B3:0114 83 C6 03 ADD SI,+03
17B3:0117 47 INC DI
17B3:0118 8A 04 MOV AL,[SI]
17B3:011A 88 05 MOV [DI],AL
17B3:011C 83 EE 05 SUB SI,+05
17B3:011F 47 INC DI
17B3:0120 FE C9 DEC CL
17B3:0122 75 E4 JNZ 0108
17B3:0124 F4 HLT
-ECS:200
17B3:0200 11. 22. 33. 44. 55. 66. 77. 88.
17B3:0208 99. 00.
-G124
DS=17B3 ES=17B3 SS=17B3 CS=17B3 IP=0124 NV UP EI PL ZR NA PE NC
17B3:0124 F4 HLT
-ECS:0300
17B3:0300 11. 44. 77. 22. 55. 88. 33. 66.
17B3:0308 99.
-
Conclusion:
This program can be used to find transpose of a 3x3 matrix. The ECS : 200 command is used to enter the
matrix elements of A such as 11, 22, 33, 44, 55, 66, 77, 88 and 99. When we execute the program using
G0124 command, AT
of matrix A will be stored in the destination address. To display the result, ECS :
0300 will be used and AT
will displayed on screen as 11, 44, 77, 22, 55, 88, 33, 66 and 99
Precautions:
1. What is the use of transpose of a matrix ?
2. What is the function MOV AL,[SI] instruction?
3. How you can determine the transpose of a matrix?

Experiment No. 20 - Program to Find Factorial of BCD number
AIM: Write 8086 assembly language program to find the factorial of BCD number.
2. +5V Power supply
4. UPS
Theory : The program to determine the factorial of one digit BCD number is developed based on the
algorithm as given below:
Algorithm
1. If N=1, then factorial=1
2. If N>1, the factorial =N×(factorial of N-1)
PROGRAM
C:>DEBUG
-A0100
17DA:0100 MOV CX,0005 ; Store number in CX register
17DA:0103 MOV AX,CX ; Copy the content of CX in AX register
17DA:0105 DEC CX ; Decrement CX
17DA:0106 MUL CX ; Multiply CX with AX
17DA:0108 DEC CX ; Decrement CX
17DA:0109 JNZ 0106 ; Jump not zero to 0106
17DA:010B MOV BX,AX ; Store factorial value in BX register
17DA:010D HLT
-U100 10D
17DA:0100 B9 05 00 MOV CX,0005
17DA:0103 89 C8 MOV AX,CX
17DA:0105 49 DEC CX
17DA:0106 F7 E1 MUL CX
17DA:0108 49 DEC CX
17DA:0109 75 FB JNZ 0106
17DA:010B 89 C3 MOV BX,AX
17DA:010D F4 HLT
-G010D
DS=17DA ES=17DA SS=17DA CS=17DA IP=010D NV UP EI PL ZR NA PE NC
17DA:010D F4 HLT
Conclusion:
The above program can be used to determine factorial 5 or 5!. After editing the program, G010D is used to
execute. Then factorial 5 will be stored in BX register i.e. 78H.
Precautions:

1. What is the use of MUL CX and DEC CX instructions?
2. What are the registers used to determine the factorial of a number?
3. How you can determine the factorial of a number?

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