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8 BIT DIVISION Posted by N R Rejin Paul at Wednesday, October 27, 2010 comments (0)

To write an assembly language program in microcontroller to perform division between two 8 bit numbers. ALGORITHM: 1. Start the program. 2. Get the dividend (65) in accumulator. 3. Get the divisor (08) in B reg. 4. Divide the two data. 5. Move the address location to Data Pointer. 6. Store the remainder result in the Data Pointer. 7. Increment the Data Pointer. 8. Move the content of B reg (quotient) to accumulator. 9. Move the content of accumulator to the address location in DPTR. 10. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 4400 74 MOV A, #Data 1 Move the dividend (65) to accumulator. 4401 65 4402 75 MOV B, #Data 2 Move the divisor (08) to B reg. 4403 F0 4404 08 4405 84 DIV A,B Dividend the content of Acc and B reg. 4406 90 MOV DPTR, #4500 Move the address location 4500 to DPTR. 4407 45 4408 00 4409 F0 MOVX @DPTR,A Move the content of accumulator to the address location (4500) in DPTR 440A A3 INC DPTR Increment the content of DPTR. 440B E5 MOV A,B Move the content of B to acc. 440C F0 440D F0 MOVX @DPTR,A Move the content of accumulator to the address location (4501) in DPTR 440E 80 HERE:SJMP HERE

440F FE INPUT: 4401: 65 4404: 08 OUTPUT: 4500: 0C 4501: 00 Labels: MICROPROCESSOR

8 BIT MULTIPLICATION Posted by N R Rejin Paul at comments (0)

ALGORITHM: 1. Start the program. 2. Get the first 8 bit data in accumulator. 3. Get the second 8 bit data in B reg. 4. Multiply the two data. 5. Move the address location to Data Pointer. 6. Store the multiplied result (LSB) in the Data Pointer. 7. Increment the Data Pointer. 8. Move the content of B reg (MSB) to accumulator. 9. Move the content of accumulator to the address location in DPTR. 10. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 4300 74 MOV A, #Data 1 Move the data (0A) to accumulator. 4301 0A 4302 75 MOV B, #Data 2 Move the data (88) to B reg. 4303 F0 4304 88 4305 A4 MUL A,B Multiply the content of Acc and B reg. 4306 90 MOV DPTR, #4500 Move the address location 4500 to DPTR. 4307 45 4308 00 4309 F0 MOVX @DPTR,A Move the content of accumulator to the address location (4500) in DPTR 430A A3 INC DPTR Increment the content of DPTR. 430B E5 MOV A,B Move the content of B to acc. 430C F0 430D F0 MOVX @DPTR,A Move the content of accumulator to the address location (4501) in DPTR 430E HERE 80 SJMP HERE 430F FE

INPUT: 4301: 0A 4304: 88 OUTPUT: 4500: 50 (LSB) 4501: 05 (MSB) Labels: MICROPROCESSOR

8 BIT SUBTRACTION MICROCONTROLLER (8051) Posted by N R Rejin Paul at comments (0)

To write an assembly language program in microcontroller to perform subtraction between two 8 bit numbers. ALGORITHM: 1. Start the program. 2. Get the first 8 bit data in accumulator. 3. Subtract another 8 bit data with the accumulator content. 4. Move the address location to Data Pointer. 5. Store the subtracted result in the Data Pointer. 6. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 4200 C3 CLRC Clear the content 4201 74 MOV A, #Data 1 Move the data (20) to accumulator. 4202 20 4203 94 SUBB A, #Data 2 Subtract the content of acc and the second data (10) 4204 10 4205 90 MOV DPTR, #4500 Move the address location 4500 to DPTR. 4206 45 4207 00 4208 F0 MOVX @DPTR,A Move the content of accumulator to the address location (4500) in DPTR 4209 HERE 80 SJMP HERE 420A FE INPUT: 4202: 20 4204: 10 OUTPUT: 4500: 10 Labels: MICROPROCESSOR

MICROCONTROLLER (8051) 16 BIT ADDITION

Posted by N R Rejin Paul at comments (0)

To write an assembly language program in microcontroller to perform addition between two 16 bit numbers ALGORITHM: 1. Start the program. 2. Get the first 8 bit data (MSB) of the 16 bit in accumulator. 3. Add another 8 bit data (MSB) with the accumulator content. 4. Move the address location to Data Pointer. 5. Store the added result (MSB) in the Data Pointer. 6. Get the first 8 bit data (LSB) of the 16 bit in accumulator. 7. Add another 8 bit data (LSB) with the accumulator content. 8. Increment the Data Pointer. 9. Store the added result (LSB) in the Data Pointer. 10. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 4100 C3 CLRC Clear the content 4101 74 MOV A, #Data M1 Move the data (MSB) 34 to accumulator. 4102 34 4103 24 ADD A, #Data M2 Add the content of acc and the second MSB 78 data. 4104 78 4105 90 MOV DPTR, #4150 Move the address location 4150 to DPTR. 4106 41 4107 50 4108 F0 MOVX @DPTR,A Move the content of accumulator to the address location (4150) in DPTR 4109 A3 INC DPTR Increment the content of DPTR. 410A 74 MOV A, #Data L1 Move the data (LSB) 12 to accumulator 410B 12 410C 34 ADDC A, #Data L2 Add the content of acc and the second (LSB) 56 data. 410D 56 410E F0 MOVX @DPTR,A Move the content of accumulator to the address location (4151) in DPTR 410F HERE 80 SJMP HERE 4110 FE

INPUT: 4102:34 (MSB) 410B:12 (LSB) 4104:78 (MSB) 410D: 56 (LSB) OUTPUT: 4150: AC (MSB) 4151: 68 (LSB) Labels: MICROPROCESSOR

STRING MANIPULATION To write an assembly language program using 8086 to find the number of Characters in a string, to move a byte from the source to destination and to store the string primitive. Posted by N R Rejin Paul at comments (0)

To write an assembly language program using 8086 to find the number of Characters in a string, to move a byte from the source to destination and to store the string primitive. ALGORITHM (LENGTH OF THE STRING): 1. Start the program. 2. Enter the opcodes from 1000. 3. Enter the string starting from the location 1200. 4. After entering the last character in the string enter the hexa value FF. 5. Execute the program and check for result in 1100. PROGRAM ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 1000 BE, 00,12 MOV SI, 1200 Initializing the string and the string starts in SI 1003 BA, FF, FF MOV DX, FFFF Maximum location used for the entry of the string 1006 B4, FF MOV AH, FF Moving the final data of array and also indicates the end of the string 1008 LOOP 42 INC DX Incrementing the string length by 1 1009 8A, 04 MOV AL, [SI] Get the string character to AL 100B 46 INC SI Increment the address location to get the next string 100C 38,C4 CMP AH, AL Check for the end of the string 100E 75,F8 JNZ LOOP Checking for the zero value. 1010 89,16,00,11 MOV [1100], DX Successful increment in the count else stores the

string length. 1014 F4 HLT Stop the program ALGORITHM (STORE THE STRING PRIMITIVE): 1. Start the program. 2. Enter the program and execute it. 3. Check if locations 1100 to 11FF contain 34. 4. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 1000 B9, 00,01 MOV CX, 0100 Moving the total no of location in which the primitive must be stored 1003 BF, 00,11 MOV DI, OFFSET S_ARRAY Moving the starting location value to DI 1006 B8, 34,00 MOV AX, 0034 Moving the primitive value to AX 1009 FC CLD Clearing the data. 100A L AA STOSB Storing the string byte 100B E2, FA LOOP L Checking for the loop 100D F4 HLT Stop the program ALGORITHM (MOVE): 1. Start the program. 2. Enter the program from the location 1000. 3. Fill FF locations as stated above with a particular data, 41. 4. Execute the program. 5. Check if the contents are duplicated to another 255 locations using the compare command. 6. Stop the program. PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 1000 BE, 0E, 10 MOV SI, OFFSET S_ARAY Moving the source location from which the data should be moved 1003 BF, 0E, 11 MOV DI, OFFSET D_ARRAY Moving the ending location to which the data should be moved. 1006 B9, FF, 00 MOV CX, 00FF Total count location within which the data can be moved 1009 FC CLD Clearing the data. 100A MOVE A4 MOVSB Moving the string byte. 100B E2, FD LOOP MOVE Continuous check of data 100D F4 HLT Stop the program Labels: MICROPROCESSOR

SORTING THE NUMBERS IN DESCENDING ORDER To write an assembly language program to sort the numbers in descending order using 8086 microprocessor. Posted by N R Rejin Paul at comments (0)

ALGORITHM: 1. Start the process 2. Move the data to CX and move CX content to D1 3. Move the address to BX and move the count of 2200 to AX 4. Compare the content of AX with 2202` 5. Jump on no borrow to REP 6. Exchange AX and A1 and move AX to BX 7. Add BX to 2 and continue the loop up to CX is zero 8. Perform no operation and move D1 to CX and continue the loop up to CX is zero 9. Stop the program PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 2000 B9, 07, 00 MOV CX, 07 Move data to CX reg 2003 LOOP1 89 CF MOV DI, CX Move CX data to DI reg 2005 BB, 00, 22 MOV BX, 2200 Move 2200 to the BX reg 2008 LOOP2 8B, 07 MOV AX, AI [BX] Move the content of 2200 to memory of AX 200A 3B, 47, 02 CMP AX,AI[BX+2] Compare the content of AX with 2202 200D 73, 05 JNB REP Jump on borrow to rep 200F 87, 47, 02 XCHG AX,AI[BX+2] Move the data in memory to AX 2012 89, 07 MOV AI [BX], AX Move data from AX to AI 201A REP 83, C3, 02 ADD BX,2 Add data to BX 2017 F2, EF LOOP LOOP2 Go to loop2 2019 90 NOP No operation 201A 89, F9 MOV CX,DI Move data of DI to CX 201C E2, E5 LOOP LOOP1 Go to loop1 201E F4 HLT Stop the process DESCENDING ORDER INPUT: 1200 – 0002 1208 - 0078 1202 – 0006 120A - 0012 1204 – 00A1 120C - 0011 1206 – 00C2 OUTPUT: 1200 – 00C2 1208 - 0011 1202 – 00A1 120A - 0006 1204 – 0078 120C - 0002 1206 - 0012

Labels: MICROPROCESSOR

MICROPROCESSOR 8086 LAB MANUAL SORTING THE NUMBERS IN DESCENDING ORDER Posted by N R Rejin Paul at comments (0)

AIM: To write an assembly language program for the sorting in ascending order using 8086 microprocessor. ALGORITHM: (ASCENDING ORDER) 1. Start the program. 2. Move data to CX and move CX data to D1 3. Move 2200 to the BX and move the content of 2200 to memory of AX. 4. Compare the content of AX with 2202 5. Jump on borrow to rep 6. Move the data in memory to AX and move Data from AX to AI 7. Add Data to BX and go to loop2 8. Perform no operation and move data & D1 to CX 9. Go to the loop 10. Stop the program PROGRAM: ADDRESS LABEL OPCODE MNEMONICS OPERAND COMMENTS 2000 B9, 07, 00 MOV CX, 07 Move data to CX reg 2003 LOOP1 89 CF MOV DI, CX Move CX data to direg 2005 BB, 00, 22 MOV BX, 2200 Move 2200 to the BX reg 2008 LOOP2 8B, 07 MOV AX, AI [BX] Move the content of 2200 to memory of AX 200A 3B, 47, 02 CMP AX,AI[BX+2] Compare the content `of ax with 2202 200D 72, 05 JB REP jump on borrow to rep 200F 87, 47, 02 XCHG AX,AI[BX+2] Move the data in memory to AX 2012 89, 07 MOV AI[BX],AX Move data from AX to AI 201A REP 83, C3, 02 ADD BX,2 Add data to BX 2017 F2, EF LOOP LOOP2 Go to loop2 2019 90 NOP No operation 201A 89, F9 MOV CX, DI Move data of DI to CX 201C E2, E5 LOOP LOOP1 Go to loop1 201E F4 HLT Stop the process INPUT: 1200 – 0007 1208 – 0003 1202 – 0006 120A – 0002 1204 – 0005 120C – 0001 1206 – 0004 OUTPUT: 1200 – 0001 1208 – 0005 1202 – 0002 120A – 0006

1204 – 0003 120C - 0007 1206 – 0004 Labels: MICROPROCESSOR

MICROPROCESSOR 8086 SORTING THE NUMBERS IN ASCENDING ORDER Posted by N R Rejin Paul at comments (0)

ALGORITHM: (ASCENDING ORDER) 1. Start the program. 2. Move data to CX and move CX data to D1 3. Move 2200 to the BX and move the content of 2200 to memory of AX. 4. Compare the content of AX with 2202 5. Jump on borrow to rep 6. Move the data in memory to AX and move Data from AX to AI 7. Add Data to BX and go to loop2 8. Perform no operation and move data & D1 to CX 9. Go to the loop 10. Stop the program PROGRAM: ADDRESS LABEL

OPCODE

MNEMONICS

OPERAND

B9, 07, 00 89 CF

MOV MOV

CX, 07 DI, CX

BB, 00, 22

MOV

8B, 07

MOV

200A

3B, 47, 02

CMP

200D

72, 05

JB

2000 2003

LOOP1

2005 2008

LOOP2

COMMENTS

Move data to CX reg Move CX data to direg Move 2200 to the BX BX, 2200 reg Move the content of AX, AI [BX] 2200 to memory of AX Compare the content AX,AI[BX+2] `of ax with 2202 REP jump on borrow to rep

200F

87, 47, 02

XCHG

2012

89, 07

MOV

83, C3, 02 F2, EF 90 89, F9 E2, E5 F4

ADD LOOP NOP MOV LOOP HLT

201A 2017 2019 201A 201C 201E

REP

INPUT: 1200 – 0007

1208 – 0003

1202 – 0006

120A – 0002

1204 – 0005

120C – 0001

Move the data in memory to AX Move data from AX to AI[BX],AX AI BX,2 Add data to BX LOOP2 Go to loop2 No operation CX, DI Move data of DI to CX LOOP1 Go to loop1 Stop the process

AX,AI[BX+2]

1206 – 0004 OUTPUT: 1200 – 0001 1202 – 0002

1208 – 0005

120A – 0006 1204 – 0003

120C -

0007 1206 – 0004 Labels: MICROPROCESSOR Newer Posts Older Posts

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