mp manual programs

SYLLABUSCS6412 MICROPROCESSOR AND MICROCONTROLLER LABORATOR OBJECTIVES:The student should be made to: Introduce ALP concepts and features Write ALP for arithmetic and logical operations in 8086 and 8051 Differentiate Serial and Parallel Interface Interface different I/Os with Microprocessors Be familiar with MASMLIST OF EXPERIMENTS:8086 Programs using kits and MASM1. Basic arithmetic and Logical operations2. Move a data block without overlap3. Code conversion, decimal arithmetic and Matrix operations.4. Floating point operations, string manipulations, sorting and searching5. Password checking, Print RAM size and system date6. Counters and Time DelayPeripherals and Interfacing Experiments7. Traffic light control8. Stepper motor control9. Digital clock10. Key board and Display11. Printer status12. Serial interface and Parallel interface13. A/D and D/A interface and Waveform Generation8051 Experiments using kits and MASM14. Basic arithmetic and Logical operations15. Square and Cube program, Find 2s complement of a number16. Unpacked BCD to ASCII TOTAL: 45 PERIODSOUTCOMES:At the end of the course, the student should be able to: Write ALP Programmes for fixed and Floating Point and Arithmetic nterface different I/Os with processor Generate waveforms using Microprocessors Execute Programs in 8051 Explain the difference between simulator and Emulator

https://sites.google.com/site/ecerjt/tutorials/the-8051-microcontroller-and-embedded-systems-

LAB EQUIPMENT FOR A BATCH OF 30 STUDENTS:HARDWARE:8086 development kits - 30 nosInterfacing Units - Each 10 nosMicrocontroller - 30 nosSOFTWARE:Intel Desktop Systems with MASM - 30 nos8086 Assembler8051 Cross Assembler

INTRODUCTION TO MASMProgramming of a microprocessor usually takes several iterations before the rightsequence of machine code instruction is written. The process, how machine code isfacilitated casing a special program called an assembler. The assembler allows the userto write alphanumeric instructions (or) mnemonics called assembly language instructions.An assembler takes the written assembly code and converts it into machine code. Often itwill come with a linker that links the assembled file and produces an executable fromit.Window executables have the extension. Here are some of the popular onesMASM:- Microsofts macro assembler(MASM) is an integrated software packagewritten by Microsoft co-operation for professional software developers. It consists of aneditor, an assembler, a linker and a debugger the programmers work bench combinesthese four path into a user friendly programming environment with built in online help.FAMILIARITY WITH MASM:- Available since the beginning of the IBM compatible PCS Works in MS-DOS and windows environments. Its free: Microsoft no longer sells MASM as a standalone product. Bundled with the Microsoft visual studio product Numerous tutorials, books and samples floating around, many are free or low costTASM:- Another popular assembler made by Borland but it is still a commercialproduct.ASSEMBLING THE PROGRAM:-The assembler is used to convert the assembly language instructions to machinecode. It is used immediately after writing the assembly language program. The assemblerstarts by checking the syntax or validity of the structure of each instruction in the sourcefile. If any errors are found the assembler displays a report on these errors along withbrief explanation of their nature. However, if the program does not contain any errors.The assembler produces an object file that the same name as original file but with the.obj extension.LINKING THE PROGRAM:-The linker is used to convert the object file to an executable file the executablefile is the final set of machine code instructions that can directly be executed bymicroprocessor. It is difficult and different than the object file on the sense that it is selfcontained and locatable. An object file may represent on segment of long program. Thisprogram cannot operate by itself and must be integrated with other object filerepresenting the rest of the program in order to produce the final self contained exactablefile.DEBUGGING THE PROGRAM:-The debugger can also be used to find logical errors in the program. Even if aprogram does not contain syntax errors. It may not produce the desired result afterexecution. Logical errors may be found by tracing the action of the program. Once foundthe source file called Debugger is designed for that purpose.The debugger allows the user to trace the action of the program by single steppingthrough the program (or) executing the program up to a desired point called breakpoint. Itallows the user to inspect (or) change the contents of MP internal register (or) thecontents of any memory location.EXECUTING THE PROGRAM:-The executable file contains the machine language code. It can be based on the RAM andbe executed by the micro processor. Simply by typing from the DOS prompt. It theprogram produces an output on the screen or a sequence of control a piece of onhardware, if the programming manipulates data in memory nothing would seem to havehappened as a result of executing the program.THE DOS-DEBUGGER:-THE DOS-Debug program is an example of a simple debugger that comes with MSDOS. Hence it will available on any pc. It was initially designed to give the usercapability to trace logical errors in executable files. It allows the user to take an existingexecutable file and unassembled it i.e., convert into assembly language also it allows theuser to write assembly language instructions directly and then convert them to machinelanguage.MS-PWB:-The PWB allows the user to define a project that means it contains one or more files thenthe user may select and save all the necessary assembly linking and debugging option forthat project the PWB allows the leaving the PWB environment it also allows the user toget help on any keyword by pointing to the keyboard a

Exp No : 1 DATE: Programs for 16 bit Arithmetic operations (Using 8086).(a) ADDITION OF TWO 16-BIT NUMBERS AIM: To write an assembly language program to add the two 16-bit numbers residing in memory and store the result in memory using 8086.APPARATUS REQUIRED:i.8086 Microprocessor kit with key board ii. Power chord.ALGORITHM:1. Load the Addend data to accumulator.2. Get the Augends and Add with Accumulator content.3. Store the result.

FLOW CHART:

STARTGet the Augend and Add with Accumulator contentSTOPStore the resultLoad theAddend data to accumulator

PROGRAM:ADDRESSLABELMNEMONICSOPCODECOMMENT

1000MOV AX,[2000]8B, 06,00,20Move addend to accumulator

1003ADD AX,[2002]03,06,02,20Add addend with augends

1007MOV [2004],AX89,06,04,20Move accumulator content to memory

100AHLTF4Stop

OBSERVATION:16 BIT ADDITIONSINPUTADDRESS BYTESET I (Hex)SET II(Hex)

16-bit Data 12000hLB13FF

2001hHB12D1

16-bit Data22002hLB107F

2003hHB11CF

OUTPUTADDRESS BYTESET I (Hex)Without carrySET II(Hex)With carry

16 bit Sum2004hLB

2005hHB

Carry2006hLB

RESULT: Thus the assembly language program to add the two 16-bit numbers is Written and executed successfully.

(b) SUBTRACTION OF TWO 16-BIT NUMBERS USING 8086AIM: To write an assembly language program to subtract two 16-bit numbers residing in memory and store the result in memory using 8086.APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board,ii. Power chord.ALGORITHM:1. Load the Minuend data to accumulator.2. Get the Subtrahend and subtract with Accumulator content.3. Store the result.

FLOW CHART:

STARTGet the Subtrahend and subtract with Accumulator contentSTOPStore the resultLoad the Minuend data to accumulator


1000MOV AX,[2000]8B,06,00,20Move minuend to accumulator

1003SUB AX,[2002]2B,06,22,20Subtract minuend with subtrahend

1007MOV [2004],AX89,06,04,20Move accumulator content to memory.

100AHLTF4Stop

OBSERVATION:16 BIT SUBTRACTIONSINPUTADDRESS BYTESET I (Hex)SET II(Hex)

Minuend2000hLB15FD

2001hHB13DE

Subtrahend2002hLB136A

2003hHB1418

OUTPUTADDRESS BYTESET I (Hex)Without BorrowSET II(Hex) with Borrow

16 bitDifference2004hLB

2005hHB

Borrow 2006hLB

RESULT: Thus the assembly language program to subtract the two 16-bit Numbers is written and executed successfully.

(C) MULTIPLICATION OF TWO 16-BIT NUMBERS

AIM: To write an assembly language program to multiply two 16 bit numbers andstore the result in memory using 8086.

APPARATUS REQUIRED:i.8086 Microprocessor kit with key board,ii. Power chord.

ALGORITHM:1. Get the multiplier and multiplicand.2. Multiply the given two numbers.3. Store the result in consecutive memory locations.

FLOW CHART:

Get the multiplier and multiplicandSTART [Multiplier] x [multiplicand]STOPStore the result


1000MOV AX,[2000]8B,06,00,20Move the data to accumulator

1003MUL[2002]F7,26,02,11Multiply memory content with accumulator

1007MOV [2004],DX8B,1E,02,20Store MSW

100BMOV[2006],AX89,16,02,20Store LSW

100EHLTF4Stop

OBSERVATION:16 BIT MULTIPLICATIONINPUTADDRESS BYTESET I (Hex)SET II(Hex)

Multiplicand2000hLB1223

2001hHB12FD

Multiplier2002hLB131C

2003hHB141F

OUTPUTADDRESS BYTESET I (Hex)

SET II(Hex)

32 bitResult

2004hLB

2005hHB

2006hLB

2007HB

RESULT: Thus the assembly language program to multiply the two 16-bit numbers is written and executed successfully.

(D) DIVISION OF TWO 16-BIT NUMBERS

AIM: To write an assembly language program to divide two 16 bit numbers and store the result in memory using 8086.

APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board, ii. Power chord.ALGORITHM:

1. Get the dividend and divisor.2. Divide the given two numbers.3. Store the result in consecutive memory locations.

FLOW CHART:

Get the Dividend and Divisor START [Dividend] / [Divisor]STOPStore the result


1000MOV AX,[1100]8B,06,02,00Move the data to accumulator

1003DIV AX, [1102]F7,36,02,11divide memory content with accumulator

1007MOV [1200],AX8D,1E,02,20Store MSW

100BMOV[1202],DXA3,02,12Store LSW

100EHLTF4Stop

OBSERVATION:16 BIT DIVISIONINPUTADDRESS BYTESET I (Hex)SET II(Hex)

Dividend2000hLB23AD

2001hHB43AE

Divisor2002hLB12CB

2003hHB14CA


SET II(Hex)

16 bitQuotient 2004hLB

2005hHB

Remainder2006hLB

RESULT: Thus the assembly language program to divide the two 16-bit numbers is written and executed successfully. (E) SUM OF N NUMBERS IN AN ARRAY

AIM: To write an assembly language program to add a given numbers in an Array and store the result in memory.APPARATUS REQUIRED:i.8086 Microprocessor kit with key board ii. Power chord.ALGORITHM:1. Get the count in register for number of data.2. Initialize the array address.3. Add the given N numbers.4. Continue the addition till the count becomes zero.5. Store the result in consecutive memory locations. 6. Stop the program.

Get the count to count register from first location of sourceSTARTGet the first and second data by increment the pointerIncrement the pointerSTOPStore the resultIs count value 0 ?Add the two numbers and decrement the count value by 1FLOW CHART:

No

yes

Yes


1000MOV SI, 2000C7,C6,00,20Initialize the memory pointer

1003MOV CL,[SI]8A,0CMove the content of 2000 to count register CL.

1007INC SI46increment the memory pointer

100BMOV AH,[SI]8A,24Move the content of SIie 2001 to AH register.

100ENEXT:INC SIFE,09increment the memory pointer

1010ADD AH, [SI]46Add the content of AH and content of 2002.

1014DEC CL02,24Decrement the count value by 1.

1018JNZ :NEXT75,79If the count is non zero go to the address specify by the label NEXT. if count is zero continue with next instruction.

110BINC SI46Increment the SI for store result.

101CMOV [SI],AH88,24Finally sum in AH is store in the memory specify by SI.

101DHLTF4Stop the program.

OBSERVATION: SUM OF N DATA IN AN ARRAY

INPUTADDRESS BYTESET I (Hex)SET II(Hex)

Count (N) value2000h050A

Data 12001hLB7723

Data 22002hLB5653

Data 32003hLB5476

Data 42004hLB12FA

Data 52005hLB09CD

Data 62006hLB23

Data 72007hLB65

Data 82008hLB98

Data 92009hLBFC

Data 10200AhLBBC


SET II(Hex)

Sum of N Numbers2004hLB

2005hHB

RESULT: Thus the assembly language program to add N numbers in a given array is written and executed successfully.

Ex.No : 2 DATE: Programs for Sorting and Searching (Using 8086).

(a)Largest Number in an array

AIM: To write an assembly language program to find a largest value in a given array and store the result in memory.

APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board, ii. Power chord.ALGORITHM:

1. Get the count in register for number of data.1. Initialize the array address.1. Compare the selected two numbers and check carry.1. If bigger, store the largest value in accumulator. 1. Otherwise interchange the contents with register.1. Decrement the count.1. Continue the same operation till the count becomes zero.1. Store the result in memory locations.

FLOW CHART:


1000MOV SI, 2000Initialize the memory pointer.

MOV CL,[SI]Move the content of 2000 to count register CL.

INC SIIncrement the memory pointer as 2001.

MOV AL,[SI]Move the content of 2001 to the AL register.

NEXT:INC SIIncrement the memory pointer

DEC CLAfter getting the first data decrement the count value by 1.

INC SIIncrement the memory pointer

MOV BL,[SI]Move the content of SI to register BL.

CMP AL,BLCompare the content of AL,BL

JNC: SKIPIf no carry occur jump to the address specify by SKIP.if carry occur continue the next instruction.

MOV AL,BLMove the content of BL to AL register.

SKIP:DEC CLDecrement the count value in CL by 1.

JNZ: NEXTIf count is not zero jump to address specify by NEXT. otherwise continue the next instruction.

INC SIIncrement the memory pointer to store the result.

MOV [SI],ALMove the largest no hold by AL to the address hold by SI.

HLTStop the program

OBSERVATION :LARGEST OF N DATA IN AN ARRAYINPUTADDRESS BYTESET I (Hex)SET II(Hex)

Count (N) value2000hLB050A

Data 12001hLB1612

Data 22002hLBFD34

Data 32003hLBDC56

Data 42004hLBCD75

Data 52005hLB1489

Data 62006hLB9F

Data 72007hLBFC

Data 82008hLBAD

Data 92009hLBBC

Data 10200AhLBBB


SET II(Hex)

Largest Value 2006hLB

200BhLB

RESULT: Thus an assembly language program to find the largest value in a given array is written and executed successfully. (b) Smallest Number in an array

AIM: To write an assembly language program to find a smallest value in a given array and store the result in memory.

APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board,ii .Power chord.ALGORITHM:1. Get the count in register for number of data.1. Initialize the array address.1. Compare the selected two numbers and check carry.1. If smaller, store the largest value in accumulator. 1. Otherwise interchange the contents with register.1. Decrement the count.1. Continue the same operation till the count becomes zero.1. Store the result in memory locations.

FLOW CHART:


1000MOV SI, 2000SI 2000

MOV CL,[SI]CL [SI]

INC SIIncrement the content of SI by one.

MOV AL,[SI] AL [SI]

NEXT:INC SIIncrement the content of SI by one.

DEC CLDecrement the content of CL by one.

INC SIIncrement the content of SI one

MOV BL,[SI]BL [SI]

CMP AL,BL Compare the content of AL,BL

JC SKIPIf have carry jump to skip.

MOV AL,BLAL BL

SKIP:DEC CLDecrement the content of CL by one.

JNZ NEXTIf CL is not zero go to next

INC SIIncrement the content of SI by one

MOV [SI],AL[SI] AL

HLTTerminate the program

OBSERVATION:SMALLEST OF N DATA IN AN ARRAYINPUTADDRESS BYTESET I (Hex)SET II(Hex)


Data 12001hLB1612

Data 22002hFD34

Data 32003hDC56

Data 42004hCD75

Data 52005h1489

Data 62006h9F

Data 72007hFC

Data 82008hAD

Data 92009hBC

Data 10200AhBB


SET II(Hex)

Smallest Value 2006hLB

200BhLB

RESULT: Thus an assembly language program to find the smallest value in a given array is written and executed successfully.

(c) Arranging the given numbers in ascending order

AIM: To write and execute an assembly language program to arrange in ascending order in a given array of memory.

APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board,ii. Power chord.ALGORITHM:1. Get the count in register for number of data.2. Initialize the array address.3. Compare the first two numbers and if the first number is larger than second then interchange the number.4. If the first number is smaller, go to step 55. Repeat steps 2 and 3 until the numbers are in required order

FLOW CHART:

PROGRAM:ADDRESSLABELMNEMONICSOPCODE CODECOMMENT


MOV CL,[SI]Move the content SI to CL.

DEC CLDecrement the count value

REPEAT:MOV SI,2000Move 2000 to SI.

MOV CH,[SI]Move the content of to CH.

DEC CHDecrement the CH by one.


RECMP:MOV AL,[SI]Move the content of SI to AL.

INC SIincrement the memory pointer

CMP AL,[SI]Compare the content of AL and content of address hold by SI.

JC :AHEADIf carry occur jump to the address specify by the label AHEAD.

XCHG AL,[SI]Exchange the content of AL and the content hold by the address specify in SI.

XCHG AL,[SI-1]Exchange the content of AL and the content hold by the address specify in SI-1.

AHEAD:DEC CHDecrement the CH register by 1.

JNZ: RECMPIn the count value not reach to zero jump to address specify by RECMP.

DEC CLDecrement the content of CL.

JNZ: REPEATIn the count value not reach to zero jump to address specify by REPEAT.

HLTStop the program.

OBSERVATION:ASCENDING ORDERINPUTADDRESS BYTESET I (Hex)OUTPUTAscendingSETII(Hex)OUTPUTAscending

Count(N) value2000hLB050A

Data 12001hLB34AD

Data 22002h56DC

Data 32003h78EB

Data 42004h87EE

Data 52005hDB12

Data 62006h76

Data 72007h31

Data 82008h58

Data 92009h89

Data 10200Ah90

RESULT: Thus an assembly language program to arrange the numbers of an array in ascending order is written and executed successfully.

(d) Arranging the given numbers in descending order

AIM: To write and execute an assembly language program to arrange in descending order in a given array of memory.

APPARATUS REQUIRED:i.8086 Microprocessor kit with key board,ii. Power chord.ALGORITHM:

1. Get the count in register for number of data.2. Initialize the array address.3. Compare the first two numbers and if the first number is smaller than second then interchange the number.4. If the first number is smaller, go to step 55. Repeat steps 2 and 3 until the numbers are in required order

FLOW CHART:



MOV CL,[SI]Move the content SI to CL.

DEC CLDecrement the count value

REPEAT:MOV SI,2000Move 2000 to SI.

MOV CH,[SI]Move the content of to CH.

DEC CHDecrement the CH by one.


RECMP:MOV AL,[SI]Move the content of SI to AL.

INC SIincrement the memory pointer

CMP AL,[SI]Compare the content of AL and content of address hold by SI.

JNC :AHEADIf carry occur jump to the address specify by the label AHEAD.

XCHG AL,[SI]Exchange the content of AL and the content hold by the address specify in SI.

XCHG AL,[SI-1]Exchange the content of AL and the content hold by the address specify in SI-1.

AHEAD:DEC CHDecrement the CH register by 1.

JNZ :RECMPIn the count value not reach to zero jumps to address specify by RECMP.

DEC CLDecrement the content of CL.

JNZ: REPEATIn the count value not reach to zero jump to address specify by REPEAT.

HLTStop the program.

OBSERVATION :DESCENDING ORDERINPUTADDRESS BYTESET I (Hex)OUTPUTAscendingSETII(Hex)OUTPUTAscending


Data 12001hLB3316

Data 22002h4567

Data 32003hDB56

Data 42004hFD88

Data 52005hEA90

Data 62006h64

Data 72007h32

Data 82008h11

Data 92009hAD

Data 10200AhEB

RESULT: Thus an assembly language program to arrange the numbers of an array in descending order is written and executed successfully.

Ex.no.3 Date:Programs for String manipulation operations (Using 8086). (a) 8086 STRING MANIPULATION FILL A STRING

AIM: To fill a given string byte to a given destination location.

APPARATUS REQUIRED:i.8086 Microprocessor kit with key board,ii. Power chord.

ALGORITHM:1. Load the destination index register with starting and the ending address respectively.2. Initialize the counter with the total number of words to be copied.3. Clear the direction flag for auto incrementing mode of transfer.4. Use the string manipulation instruction STOSB with the prefix REP to fill a string to destination.

FLOW CHART:


1000MOV CX,0100Move 0100 to CX register.

1003MOV DI,1FFFMove 1FFF to DI register.

1007MOV AH,FFMove FF to AH register.

100BNOPNo operation

100EREPEAT:CLDClear the direction flag.

100FSTOSBStore the string byte

1010LOOP: REPEATUnconditional loop to address specified by the label REPEAT.

1013HLTstop the program

OBSERVATION:ADDRESSDATA

1FFFFF

2000FF

2001FF

2002FF

2003FF

2005FF

2006FF

RESULT: Thus the assembly language program to fill a given string in a selected array is written and executed successfully.

(b) 8086 COPY A STRINGAIM: To copy a string of data words from one location to the other.

APPARATUS REQUIRED:i.8086 Microprocessor kit with key board, ii. Power chord.

ALGORITHM:1. Load the source and destination index register with starting and the ending address respectively.2. Initialize the counter with the total number of words to be copied.3. Clear the direction flag for auto incrementing mode of transfer.4. Use the string manipulation instruction MOVSW with the prefix REP to copy a string from source to destination.

FLOW CHART:


1000MOV SI,2000HInitialize the source address.

1003MOV DI,2100HInitialize the destination address.

1007MOV CX,0006 HInitialize count value to the count register.

100BMOV AH,55HMove 55 to AH register.

100EREPEAT:CLDClear the direction flag.

100FMOVSBMove the string byte.

1010LOOP:REPEATUnconditional loop to address specified by the label REPEAT.

1013HLTStop the program.

OBSERVATION:INPUTSOUTPUTS

ADDRESSDATAADDRESSDATA

200055210055

200155210155

220255210255

220355210355

220455210455

220555210555

RESULT: Thus the assembly language program to copy a string of data byte from one location to other is written and executed successfully. (c) 8086 FINDING STRING LENGTHAIM: To write & execute an ALP to find the length of a given string using 8086 Microprocessor kit.

APPARATUS REQUIRED:i. 8086 Microprocessor kit with key board, ii. Power chordALGORITHM:1. Load the source and destination index register with starting and the ending address respectively.2. Load the terminate data value as FF.3. Load the content of source index to the AL register.4. Increment SI register and compare with register AH.5. If it is non- zero value, increment the memory location by using the control instruction to store the data.6. Compare the string byte with FF, if it is not equal, repeat.7. Count the string byte till zero flag is set. If zero flag is set then store the count value to the memory.8. Terminate the program when FF is matched.

FLOW CHART:

PROGRAM:ADDRESSLABELMNEMONICSOP CODECOMMENT

1000MOV SI,2000hC7,C6,00,12SI 1200

1004MOV DX,FFFFhC7,C7,FF,FFDXFFFF

1008MOV AH,FFhC6,C4,FFAH FF, check FF

100BNOENDINC DX42Increment the DX reg. by 1

100CMOV AL,[SI]8A,04AL [SI]

100EINC SI46Increment the SI reg. by 1

100FCMP AH,AL38,C4Compare the content of AH & AL

1011JNZ : NOEND75,FBIf the Compared result is not 0 go to by the address specified label NOEND.

1013MOV [1100],DX89,16,00,11[1100] DXie. store the result

1017HLTF4Terminate the program

OBSERVATION: INPUT ADDRESS (Hex)Data (Hex)

1200E3

1201F4

120254

120366

120488

120599

120610

1207FF

OUTPUT :ADDRESS (Hex)String length (Hex)

110008

110100

RESULT: Thus the assembly language program to find length of a given string of data bytes in an array is written and executed successfully.EX. NO.4 DATE:PROGRAMS FOR DIGITAL CLOCK (USING 8086)ADDRESSLABELMNEMONICSOP-CODECOMMANDS

1000STARTCALL CONVERTE8 00 60

1003CALL DISPLAYE8 00 51

1006DELAYMOV AL,B0B0 B0

1008OUT 16H,ALE6 16

100AMOV CL,07HB1 07

100CS2MOV AL,88HB0 88

100EOUT 14H,ALE6 14

1010MOV AL,80HB0 80

1012OUT 14H,ALE6 18

1014S1MOV AL,80HB0 80

1016OUT 16H,ALE6 16

1018NOP90

1019NOP90

101ANOP90

101BNOP90

101CIN AL,14HE4 14

101EMOV DL,AL8A D0

1020IN AL,14HE4 14

1022OR AL,DL0A C2

1024JNZ S175 F2

1026DEC CLFE C9

1028JNZ S275 E6

102AMOV SI,1500HBE 00 15

102DMOV AL,[SI]8A 04

102FINC AL FE C0

1031MOV [SI],AL88 04

1033CMP AL,3CH3C 3C

1035JNZ START75 CD

1037MOV AL,00B0 00

1039MOV [SI],AL88 04

103BINC SI46

103CMOV AL,[SI]8A 04

103EINC ALFE C0

1040MOV [SI],AL88 04

1042CMP AL,3CH3C 3C

1044JNZ START75 BE

1046MOV AL,00B0 00

1048MOV [SI],AL88 04

104AINC SI46

104BMOV AL,[SI]8A 04

104DINC ALFE C0

104FMOV [SI],AL88 04

ADDRESSLABELMNEMONICSOP-CODECOMMANDS

1051CMP AL,18H3C 18

1053JNZ START75 AF

1055MOV AL,00B0 00

1057MOV [SI],AL88 04

1059JMP STARTEB A9

105BDISPLAYMOV AH,06HB4 06

105DMOV DX,1600HBA 00 16

1060MOV CH,01HB5 01

1062MOV CL,00HB1 00

1064INT 5CD 05

1066RETC3

1067CONVERTMOV SI,1500HBE 00 15

106AMOV BX,1608HBB 08 16

106DMOV AL,24HB0 24

106FMOV[BX],AL88 07

1071MOV AL,[SI]8A 04

1073MOV AH,00B4 00

1075MOV DH,0AHB6 0A

1077DIV DHF6 F6

1079ADD AH,30H80 C4 30

107CDEC BX4B

107DMOV[BX],AH88 27

107FDEC BX4B

1080ADD AL,30H04 30

1082MOV[BX],AL88 07

1084DEC BX4B

1085MOV AL,3AHB0 3A

1087MOV[BX],AL88 07

1089DEC BX 4B

108AINC SI (minutes)46

108BMOV AL.[SI]8A 04

108DMOV AH,00B4 00

108FMOV DH,0AHB6 0A

1091DIV DHF6 F6

1093ADD AH,30H80 C4 30

1096MOV [BX],AH88 27

1098DECBX4B

1099ADD AL,30H04 30

109BMOV[BX],AL88 07

109DDEC BX4B

109EMOV AL,3AHB0 3A

10A0MOV [BX],AL88 07

10A2DEC BX4B

10A3INC SI(hours)46

ADDRESSLABELMNEMONICSOP-CODECOMMANDS

10A4MOV AL,[SI]8A 04

10A6MOV AH,00B4 00

10A8MOV DH,0AHB6 0A

10AADIV DH F6 F6

10ACADD AH,30H80 C4 30

10AFMOV[BX],AH88 27

10B1DEC BX4B

10B2ADD AL,30H04 30

10B4MOV [BX],AL88 07

10B6RETC3

10B7GETCIN AL,02HE4 02

10B9AND AL,FFH24 FF

10BBCMP AL,F0H3C F0

10BDJNE GETC75 F8

TIMING DISPLAYADDRESSSPECIFICATION VALUE

1500SECONDS 32

1501MINUTES0F

1502HOURS05

Result: Thus the ALP for digital clock was written and executted succesfully

BIOS/DOS CALLS2.a.i 16 bit Addition/Subtraction .MODEL TINY .CODE MOV BX,1234h MOV CX,7698h MOV AL,BL ; for subtraction replace with ADD AL,CL ; SUB AL,CL DAA ; DAS MOV DL,AL MOV AL,BL ADC AL,CH ; SBB AL,CH DAA ; DAS MOV DH,AL MOV AH,4Ch INT 21h END

2.c.i 16 bit Multiplication/Division .MODEL TINY .CODE MOV AX,1234h MOV BX,7698h ADD AL,31h ; for Division replace with MUL BX ; DIV BX MOV AH,4Ch INT 21h END1.(A).DISPLAY A MESSAGE

AIM: To display a message on CRT screen of computer using DOS calls

PROCEDURE:Step1: Switch on computerStep2: Go to start menu and select run prompt, type cmdStep3: Then type edit and open MASM windowStep4: Type your programme and save filename.asmStep5: Go to command prompt and then compile the programmeStep6: Run the programme and see the output in the memory window and register windows

PROGRAMME:ASSUME CS:CODE,DS:DATADATA SEGMENTMSG DB 0DH,0AH,GOOD MORNING,0DH,0AH,$DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV AH,09HMOV DX,OFFSET MSGINT 21HMOV AH,4CHINT 21HCODE ENDSEND START

OUTPUT: GOOD MORNING

RESULT: Thus the above bios/dos-> display a message program has been successfully executed & the output is verified.

1(B).FILE CREATION

AIM: To create a file using DOS calls

PROCEDURE:

Step1: Switch on computerStep2: Go to start menu and select run prompt, type cmdStep3: Then type edit and open MASM windowStep4: Type your programme and save filename.asmStep5: Go to command prompt and then compile the programmeStep6: Run the programme and see the output in the memory window and register windows

PROGRAMME:ASSUME CS: CODE, DS:DATADATA SEGMENTFILENAME DB SAMPLE.DAT,$DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV DX,OFFSET FILENAMEMOV AH,3CHINT 21HMOV AH,4CHINT 21HCODE ENDSEND START

OUTPUT:

SAMPLE.DAT file was created( in documents and settings folder )

RESULT: Thus the above program has been successfully executed and the output is verified

1( C).ASCII EQUIVALENT OF A TEXT

AIM: To display ASCII equivalent of a text


PROGRAMME:

ASSUME CS:CODE, DS:DATADATA SEGMENTMESSAGE DB 0DH,0AH,GOOD MORNING,0DH,0AH,$DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV AH,36HMOV DX,OFFSET MSGINT 21HMOV AH,4CHINT 21HCODE ENDSEND START

OUTPUT:

GOOD MORNING

RESULT: Thus the above program has been executed successfully and the output is verified.

1(D) .LARGEST NUMBER IN ARRAY USING MASM

AIM: To implement sorting and searching of Largest number in an array using MASM software

PROCEDURE:


PROGRAMME:

ASSUME CS:CODE,DS:DATADATA SEGMENTLIST DW 52H, 23H, 56H, 45HCOUNT EQU 04LARGEST DB 01H DUP (?)DATA ENDSCODE SEGMENTSSTART: MOV AX,DATA MOV DS,AX MOV SI,OFFSET LISTMOV CL,COUNTMOV AL,[SI]AGAIN:CMP AL,[SI+1]JNL NEXTMOVAL,[SI+1]NEXT:INC SIDEC CLJNZ AGAINMOV SI,OFFSET LARGESTMOV [SI],ALMOV AH,4CHINT 21HCODE ENDSEND STAR

RESULT: Thus the above program has been successfully executed and the output is verified

1(E).DESCENDING ORDER OF AN UNSORTED ARRAY USING MASM

AIM: To implement sorting and searching of Largest number in an array using MASM software

PROCEDURE:


PROGRAMME:

ASSUME CS:CODE,DS:DATADATA SEGMENTLIST DW 53H,25H,19H,02HCOUNT EQU 04DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV DX,COUNT-1AGAIN0:MOV CX,DXMOV SI,OFFSET LISTAGAIN1:MOV AX,[SI]CMP AX,[SI+2]JNL PR1XCHG [SI+2],AXXCHG [SI],AXPR1:ADD SI,02LOOP AGAIN1DEC DXJNZ AGAIN0MOV AH,4CHINT 21HCODE ENDSEND START

RESULT: Thus the above program has been successfully executed and the output is verified.

1(F).DISPLAY THE STRING

AIM: To implement string display using MASM software and verify result


PROGRAMME:

ASSUME CS: CODE, DS:DATADATA SEGMENTMESSAGE DB 0DH, 0AH,MICROPROCESSOR,$DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV AH,09HMOV DX,OFFSET MESSAGEINC 21HMOV AH,4CHINT 21HCODE ENDSEND START

OUTPUT: MICROPROCESSOR

RESULT: Thus the MASM programme is executed successfully and output verified

1.8086 STRING MANIPULATION SEARCH A WORD

AIM:To search a word from a string.

ALGORITHM:5. Load the source and destination index register with starting and the ending address respectively.6. Initialize the counter with the total number of words to be copied.7. Clear the direction flag for auto incrementing mode of transfer.8. Use the string manipulation instruction SCASW with the prefix REP to search a word from string.9. If a match is found (z=1), display 01 in destination address. Otherwise, display 00 in destination address. RESULT:A word is searched and the count of number of appearances is displayed.

PROGRAM:ASSUME CS: CODE, DS: DATA DATA SEGMENTLIST DW 53H, 15H, 19H, 02H DEST EQU 3000HCOUNT EQU 05HDATA ENDSCODE SEGMENT START:MOV AX, DATAMOV DS, AXMOV AX, 15HMOV SI, OFFSET LISTMOV DI, DESTMOV CX, COUNTMOV AX, 00CLDREPSCASWJZ LOOPMOV AX, 01LOOPMOV [DI], AXMOV AH, 4CHINT 21HCODE ENDSEND START

INPUT:LIST: 53H, 15H, 19H, 02H

OUTPUT:300001

2.8086 STRING MANIPULATION FIND AND REPLACE A WORD

AIM:To find and replace a word from a string.

ALGORITHM:1. Load the source and destination index register with starting and the ending address respectively.2. Initialize the counter with the total number of words to be copied.3. Clear the direction flag for auto incrementing mode of transfer.4. Use the string manipulation instruction SCASW with the prefix REP to search a word from string.5. If a match is found (z=1), replace the old word with the current word in destination address. Otherwise, stop.

RESULT:A word is found and replaced from a string.

PROGRAM:ASSUME CS: CODE, DS: DATA DATA SEGMENTLIST DW 53H, 15H, 19H, 02H REPLACE EQU 30HCOUNT EQU 05HDATA ENDSCODE SEGMENT START:MOV AX, DATAMOV DS, AXMOV AX, 15HMOV SI, OFFSET LISTMOV CX, COUNTMOV AX, 00CLDREPSCASWJNZ LOOPMOV DI, LABEL LISTMOV [DI], REPLACELOOPMOV AH, 4CHINT 21HCODE ENDSEND START


OUTPUT:LIST: 53H, 30H, 19H, 02H

3. 8086 STRING MANIPULATION COPY A STRING

AIM:To copy a string of data words from one location to the other.

ALGORITHM:10. Load the source and destination index register with starting and the ending address respectively.11. Initialize the counter with the total number of words to be copied.12. Clear the direction flag for auto incrementing mode of transfer.13. Use the string manipulation instruction MOVSW with the prefix REP to copy a string from source to destination.RESULT:A string of data words is copied from one location to other.

PROGRAM:ASSUME CS: CODE, DS: DATA DATA SEGMENTSOURCE EQU 2000HDEST EQU 3000HCOUNT EQU 05HDATA ENDSCODE SEGMENT START:MOV AX, DATAMOV DS, AXMOV ES, AXMOV SI, SOURCEMOV DI, DESTMOV CX, COUNTCLDREPMOVSWMOV AH, 4CHINT 21HCODE ENDSEND START

INPUT:OUTPUT:2000 483000482001843001842002 67300267200390300390200421300421

4.8086 STRING MANIPULATION SORTING

AIM:To sort a group of data bytes.

ALGORITHM: Place all the elements of an array named list (in the consecutive memory locations). Initialize two counters DX & CX with the total number of elements in the array. Do the following steps until the counter B reaches 0. Load the first element in the accumulator Do the following steps until the counter C reaches 0.1. Compare the accumulator content with the next element present in the next memory location. If the accumulator content is smaller go to next step; otherwise, swap the content of accumulator with the content of memory location.2. Increment the memory pointer to point to the next element.3. Decrement the counter C by 1. Stop the execution.

RESULT: A group of data bytes are arranged in ascending order.

PROGRAM:ASSUME CS: CODE, DS: DATA DATA SEGMENTLIST DW 53H, 25H, 19H, 02H COUNT EQU 04HDATA ENDSCODE SEGMENT START:MOV AX, DATAMOV DS, AXMOV DX, COUNT-1LOOP2:MOV CX, DXMOV SI, OFFSET LISTAGAIN:MOV AX, [SI]CMP AX, [SI+2]JC LOOP1XCHG [SI +2], AXXCHG [SI], AXLOOP1:ADD SI, 02LOOP AGAINDEC DXJNZ LOOP2MOV AH, 4CHINT 21HCODE ENDSEND START


OUTPUT:LIST: 02H, 19H, 25H, 53H

1(G).ARITHMETIC OPERATIONS USING MASM

AIM: To Implement the programme for arithmetic operations using MASM software

PROCEDURE:


PROGRAMME:

ASSUME CS:CODE,DS:DATADATA SEGMENTOPR1 EQU 98HOPR2 EQU 49HSUM DW 01 DUP(00)SUBT DW 01 DUP(00)PROD DW 01 DUP(00)DIVS DW 01 DUP(00)DATA ENDSCODE SEGMENTSTART:MOV AX,DATAMOV DS,AXMOV BL,OPR2XOR AL,ALMOV AL,OPR1ADD AL,BLDAAMOV BYTE PTR SUM,ALJNC MSB0INC [SUM+1]MSB0:XOR AL,ALMOV AL,OPR1SUB AL,BLDASMOV BYTE PTR SUBT,ALJNB MSB1INC [SUBT+1]MSB1:XOR AL,ALMOV AL,OPR1MUL BLMOV WORD PTR PROD,AXXOR AH,AHMOV AL,OPR1DIV BLMOV WORD PTR DIVS,AXMOV AH,4CHINT 21HCODE ENDSEND START

RESULT: Thus the MASM programme is executed successfully and output is verified.

8086/Masm Program To Multiply Two Matrices .DATA SEGMENTAR1 DB 1H,2H,-3HAR2 DB 4H,5H,6HAR3 DB 2H,-1H,3HBC1 DB 2H,4H,-4HBC2 DB 3H,-2H,5HBC3 DB 1H,5H,2HC DB 9 DUP (?)L2 DB (?)L1 DB (?)DATA ENDSCODE SEGMENTASSUME CS:CODE,DS:DATASTART: MOV AX,DATAMOV DS,AXMOV ES,AXMOV BP,0HMOV L2,3HLEA SI,AR1REPEAT2: LEA DI,BC1MOV L1,3HREPEAT1: MOV CX,3HMOV BX,0HMOV DL,0HAGAIN: MOV AL,[SI][BX]IMUL BYTE PTR[DI][BX]ADD DL,ALINC BXLOOP AGAINMOV DS:C[BP],DLINC BPADD DI,3HDEC L1JNE REPEAT1ADD SI,3HDEC L2JNE REPEAT2MOV AH,4CHINT 21HCODE ENDSEND START

2.f.i Binary to BCD conversion .MODEL SMALL .DATA Binary DB 63h Ans DB 00h, 00h, 00h .CODE MOV AX,@DATA MOV DS, AX MOV AX, 00h MOV AL, Binary MOV CL, 64h DIV CL MOV BCD, AL MOV AL, AH MOV AH, 00h MOV CL, 0Ah DIV CL MOV Ans+ 1, AL MOV Ans+2, AH OR Ans, 30h OR Ans+ l,30h OR Ans+2,30h MOV AH, 4Ch INT 21h END

BCD to Binary conversion .MODEL SMALL .DATA BCD DB 15h Ans DB 00h .CODE MOV AX,@DATA MOV DS, AX MOV AL, BCD AND AL, 0Fh MOV BL, AL MOV AL, BCD AND AL, 0F0h MOV CL, 04h ROR AL, CL MOV CL, 0Ah MUL CL ADD AL, BL MOV Ans, AL MOV AH, 4Ch INT 21h END

Bit manupilation to check if the data is positive or negative .MODEL SMALL .DATA Msg1 DB ENTERED NUMBER IS POSITIVE. $ Msg2 DB ENTERED NUMBER IS NEGATIVE. $ Input DB ? .STACK .CODE MOV AX, @Data MOV DS, AX MOV AL, Input ROL AL, 01h JC NEXT LEA DX, Msg1 MOV AH, 09h INT 21h JMP LAST NEXT: LEA DX, Msg2 MOV AH, 09h INT 21h LAST: MOV AH, 4Ch INT 21h END

Bit manupilation to check if the data is odd or even .MODEL SMALL .DATA Msg1 DB ENTERED NUMBER IS ODD. $ Msg2 DB ENTERED NUMBER IS EVEN. $ Input DB ? .STACK .CODE MOV AX, @Data MOV DS, AX MOV AL, Input SAR AL, 01h JC NEXT LEA DX, Msg2 MOV AH, 09h INT 21h JMP LAST NEXT: LEA DX, Msg1 MOV AH, 09h INT 21h LAST: MOV AH, 4Ch INT 21h END

Bit manupilation to count the number of 1s and 0s in given data .MODEL SMALL .CODE MOV CX, 0008h MOV AL, 24h MOV BL, 00h MOV DL, BL NEXT: SAR AL, 01h JC DOWN INC BL LOOP NEXT JMP LAST DOWN: INC DL LOOP NEXT LAST: MOV AH, 4Ch INT 21h END

7(b) .CUBE OF NUMBERAIM: WRITE AN ASSEMBLY LANGUAGE PROGRAM TO FIND THE CUBE OFTHE GIVEN NUMBER.SOFTWARE REQUIRED: MASM 611ASSEMBLY LANGUAGE PROGRAM:ASSUME CS: CODE, DS: DATADATA SEGMENTNO1 DB 06HAD1 DW 5000HDATA ENDSCODE SEGMENTSTART:ORG 600HMOV AX, DATAMOV DS, AXXOR AX, AXMOV SI, AD1MOV CL, NO1ABC: MOV AL, [SI]MOV BL, ALMUL BLMUL BLMOV [SI], AL,INC SIDEC CLJNZ ABCINT 21HCODE ENDSEND STARTRESULT: CUBE OPERATION IS PERFORMED USING MASM SOFTWARE.7(c).AVERAGE OF NUMBERAIM: WRITE AN ASSEMBLY LANGUAGE PROGRAM TO FIND THE CUBE OFTHE GIVEN NUMBER.SOFTWARE REQUIRED: MASM 611ASSEMBLY LANGUAGE PROGRAM:ASSUME CS: CODE, DS: DATADATA SEGMENTNO1 DW 7000HDATA ENDSCODE SEGMENTSTART:ORG 600HMOV AX, DATAMOV DS, AXXOR CX, CXMOV AX,[BX]MOV CL, 06HDEC CXABC: INC SIADD AX,[BX]DEC CXJNZ ABCDIV BXINT 21HCODE ENDSEND STARTRESULT: AVERAGE OF NUMBERS OPERATION IS PERFORMED USING MASMSOFTWARE.

7(a) SQUARE OF NUMBERAIM: WRITE AN ASSEMBLY LANGUAGE PROGRAM TO FIND THE SQUARE OFTHE GIVEN NUMBER.SOFTWARE REQUIRED: MASM 611ASSEMBLY LANGUAGE PROGRAM:ASSUME CS: CODE, DS: DATADATA SEGMENTNO1 DB 06HAD1 DW 5000HDATA ENDSCODE SEGMENTSTART:ORG 600HMOV AX, DATAMOV DS, AXXOV AX, AXMOV SI, AD1MOV CL, NO1ABC: MOV AL, [SI]MOV BL, ALMUL BLMOV [SI], AL,INC SIDEC CLJNZ ABCINT 21HCODE ENDSEND STARTRESULT: SQUARING OPERATION IS PERFORMED USING MASM SOFTWARE.

1.b.i Block transfer without overlap .MODEL SMALL .DATA Array1 DW 1111h,2222h,3333h,4444h,5555h Array2 DW 5 DUP (0) Count DW 0005H .CODE MOV AX,@DATA MOV DS,AX LEA SI,Array1 LEA DI,Array2 MOV CX,Count NEXT: MOV AX,[SI] MOV [DI],AX INC SI INC SI INC DI INC DI LOOP NEXT MOV AH,4Ch INT 21h END

1.b.ii Block transfer with overlap .MODEL SMALL .DATA Array DB 11h,22h,33h,44h,55h Count DW 0005h .CODE MOV AX,@DATA MOV ES,AX MOV DS,AX LEA SI,Array ADD SI,Count MOV CX,Count DEC SI MOV DI,SI ADD DI,2h STD REP MOVSB MOV AH,4Ch INT 21h END

5.i String transfer .MODEL SMALL .DATA String1 DB 'BMSCE DEPT OF ECE$' Length EQU ($-String1) String2 DB LEN DUP (0) .CODE MOV AX, @DATA MOV DS, AX MOV ES, AX MOV CX, Length CLD LEA SI, String1 LEA DI, String2 REP MOVSB MOV AH, 4Ch INT 21h END

String reverse .MODEL SMALL .DATA String DB 'BMSCE$' Length EQU ($-String) Rvrs DB Length DUP (0) .CODE MOV AX,@DATA MOV DS, AX MOV ES, AX MOV CX, Length LEA SI, String+Length-1 LEA DI, Rvrs REPEAT: MOV AL, [SI] MOV [DI], AL DEC SI INC DI LOOP REPEAT MOV AH, 4Ch INT 21h END

Character search in a string .MODEL SMALL .DATA String DB 'BMS COLLEGE' Length EQU ($-String) Key DB 'X' Dis1 DB '-IS PRESENT IN GIVEN STRING$' Dis2 DB '-IS NOT PRESENT IN GIVEN STRING$' .CODE MOV AX,@DATA MOV DS, AX MOV ES, AX MOV DL, Key MOV AH, 02h INT 21h LEA DI, String MOV AL, Key MOV CX, Length REPNE SCASB JE PRESENT LEA DX, Dis2 CALL Display JMP OVER PRESENT: LEA DX, Dis1 CALL Display OVER: MOV AH, 4Ch INT 21h Display PROC NEAR MOV AH, 09h INT 21h RET Display ENDP END

Matrix Keyboard Interfacing CODE SEGMENT ASSUME CS:code,DS:code,ES:code,SS:code CWR EQU 46h PORTA EQU 40h PORTB EQU 42h PORTC EQU 44h ORG 0400h MOV AL, 88h ; port a and port c high as output OUT CWR,AL ; port b and port c low as output READKEY: MOV DL,0 ; clear e/dl register MOV AL,0F0h ; output all one's to pc high OUT PORTC,AL LP: IN AL,PORTC AND AL,0F0h CMP AL,0F0h JNZ LP CALL FAR PTR ONEMS KEYCHK: IN AL,PORTC AND AL,0F0h CMP AL,0F0h JZ KEYCHK ;wait for key press CALL FAR PTR ONEMS MOV AL,7Fh MOV BH,04h NXTCOLM: ROL AL, 01h ; scan each column MOV DH,AL ; and wait for the data OUT PORTC,AL ; in any of the four IN AL,PORTC ; rows AND AL,0F0h MOV CL,04h NXTROW: ROL AL,01h ; scan each column JNC CODEN ; scan each column INC DL ; in any of the four DEC CL ; rows JNZ NXTROW MOV AL,DH DEC BH JNZ NXTCOLM JMP KEYCHK

CODEN: MOV AL,DL MOV DH,0h MOV BX,OFFSET LOOKUP+8000h ADD BX,DX MOV AL,BYTE PTR[BX] OUT PORTB,AL JMP READKEYONEMS: ; delay routine PUSH AX MOV AL,0FFh LOP: DEC AL JNZ LOP POP AX RETFLOOKUP: DB 00h,04h,08h,0Ch,01h,05h,09h,0Dh DB 02h,06h,0Ah,0Eh,03h,07h,0Bh,0Fh CODE ENDS END

1. ADDITION OF TWO 8-BIT BCD

DATA SEGMENT MESS1 DB 0AH,0DH,'ENTER FIRST NUMBER:','$' MESS2 DB 0AH,0DH,'ENTER SECOND NUMBER:','$' MESS3 DB 0AH,0DH,'SUM OF TWO 8-BIT NUMBER IS:','$'DATA ENDS

CODE SEGMENT ASSUME CS:CODE,DS:DATA START: MOV AX,DATA MOV DS,AX LEA DX,MESS1 MOV AH,09H INT 21H CALL READ MOV BL,DL LEA DX,MESS2 MOV AH,09H INT 21H CALL READ MOV CL,00H MOV AL,BL ADD AL,DL DAA JNC NEXT INC CL NEXT:MOV BL,AL CALL DISP MOV AH,4CH INT 21H

READ PROC NEAR PUBLIC READ

MOV AH,01H INT 21H MOV DL,AL MOV CL,04H SUB DL,30H CMP DL,0AH JC R1 SUB DL,07H R1:SHL DL,CL MOV AH,01H INT 21H SUB AL,30H CMP AL,0AH JC R2 SUB AL,07H R2:AND AL,0FH OR DL,AL RET READ ENDP

DISP PROC NEAR PUBLIC DISP LEA DX,MESS3 MOV AH,09H INT 21H MOV DL,CL ADD DL,30H MOV AH,06H INT 21H MOV CL,04H MOV DL,BL SHR DL,CL CMP DL,0AH JC L1 ADD DL,07H L1:ADD DL,30H MOV AH,06H INT 21H AND BL,0FH CMP BL,0AH JC L2 ADD BL,07H L2:ADD BL,30H MOV DL,BL MOV AH,06H INT 21H RET

DISP ENDPCODE ENDSEND STARTOUTPUT:

ENTER FIRST NUMBER : 97ENTER SECOND NUMBER : 56SUM OF TWO 8-BIT NUMBER IS : 153

ENTER FIRST NUMBER : 82ENTER SECOND NUMBER : 19SUM OF TWO 8-BIT NUMBER IS : 101

2. SUBTRACTION OF TWO 8-BIT BCD NUMBERS

DATA SEGMENT MESS1 DB 0AH,0DH,'ENTER FIRST NO:','$' MESS2 DB 0AH,0DH,'ENTER SECOND NO:','$' MESS3 DB 0AH,0DH,'SUBTRACTION OF TWO 8-BIT NO IS:','$'DATA ENDS

CODE SEGMENT ASSUME CS:CODE,DS:DATA

START: MOV AX,DATA MOV DS,AX LEA DX,MESS1 MOV AH,09H INT 21H CALL READ MOV BL,DL LEA DX,MESS2 MOV AH,09H INT 21H CALL READ MOV CL,00H MOV AL,BL SUB AL,DL DAS JNC NEXT MOV CH,99H SUB CH,AL MOV AL,CH ADD AL,01H DAA MOV CL,'-' NEXT: MOV BL,AL CALL DISP MOV AH,4CH INT 21H


MOV AH,01H INT 21H MOV DL,AL MOV CL,04H SUB DL,30H CMP DL,0AH JC R1 SUB DL,07H R1:SHL DL,CL MOV AH,01H INT 21H SUB AL,30H CMP AL,0AH JC R2 SUB AL,07H R2:AND AL,0FH OR DL,AL RET READ ENDP

DISP PROC NEAR PUBLIC DISP

LEA DX,MESS3 MOV AH,09H INT 21H MOV DL,CL ADD DL,30H MOV AH,06H INT 21H MOV CL,04H MOV DL,BL SHR DL,CL CMP DL,0AH JC L1 ADD DL,07H L1:ADD DL,30H MOV AH,06H INT 21H AND BL,0FH CMP BL,0AH JC L2 ADD BL,07H L2:ADD BL,30H MOV DL,BL MOV AH,06H INT 21H RET

DISP ENDPCODE ENDSEND START

OUTPUT:

ENTER FIRST NUMBER : 98ENTER SECOND NUMBER : 54SUBTRACTION OF TWO 8-BIT NUMBER IS : 044

ENTER FIRST NUMBER : 87ENTER SECOND NUMBER : 98SUBTRACTION OF TWO 8-BIT NUMBER IS : -11

23.TRANSPOSE OF THE MATRIX

DATA SEGMENT M1 DB 'ENTER THE ORDER OF THE MATRIX:','$' M2 DB 0AH,0DH,0AH,'ENTER THE ORDER OF SECOND MATRIX:','$' M3 DB 0AH,0DH,0AH,'TRANSPOSE MATRIX:',0AH,0DH,'$' MAT1 DB 10 DUP(0) MAT2 DB 10 DUP(0) ROW DB 00H COL DB 00H DATA ENDS

MESSAGE MACRO MESS LEA DX,MESS MOV AH,09H INT 21H ENDM

BSPCE MACRO ASC MOV DL,ASC MOV AH,06H INT 21H ENDM

CODE SEGMENT ASSUME CS:CODE,DS:DATA START:MOV AX,DATAMOV DS,AXMESSAGE M1CALL READOMOV BX,DXMOV ROW,DHMOV COL,DLMESSAGE M2LEA SI,MAT1CALL READ1LEA DI,MAT1LEA SI,MAT2MOV DH,00HMOV DL,COLMOV AH,COLL3:MOV AL,ROWMOV BX,DIL1:MOV CL,[BX]MOV [SI],CLADD BX,DXINC SIDEC ALJNZ L1INC DIDEC AHJNZ L3LEA SI,MAT2CALL DISPMOV AH,4CHINT 21H

READO PROC NEAR PUBLIC READO MOV AH,01H INT 21H MOV DH,AL SUB DH,30H BSPCE ' ' MOV AH,01H INT 21H MOV DL,AL SUB DL,30H RET READO ENDP

READ1 PROC NEAR BSPCE 0AH MOV CH,ROW N2:MOV BH,COL N1:CALL READ MOV [SI],DL INC SI BSPCE ' ' DEC BH JNZ N1 DEC CH JZ N3 BSPCE 0AH BSPCE 0DH JMP N2 N3:RET READ1 ENDP


MOV AH,01HINT 21HMOV CL,04HMOV DL,ALSUB DL,30HCMP DL,0AHJC R1SUB DL,07HAND DL,0FHR1:SHL DL,CLMOV AH,01HINT 21HSUB AL,30HCMP AL, 0AHJC R2SUB AL,07HAND AL,0FHR2:OR DL,ALRET READ ENDP

DISP PROC NEAR PUBLIC DISP MESSAGE M3BSPCE 0AHMOV DH,COLD4:MOV BH,ROWLOP:MOV CL,04HMOV DL,[SI]SHR DL,CLCMP DL,0AHJC D1ADD DL,07HD1: ADD DL,30HMOV AH,06HINT 21HMOV DL,[SI]AND DL,0FHCMP DL,0AHJC D2ADD DL,07HD2: ADD DL,30HMOV AH,06HINT 21HINC SIBSPCE ' 'DEC BHJNZ LOPDEC DHJZ D3BSPCE 0AHBSPCE 0DHJMP D4 D3: RET DISP ENDP CODE ENDS END START

OUTPUT :

ENTER THE ORDER OF THE MATRIX : 2 3

ENTER THE ELEMENTS OF THE MATRIX:11 55 9922 74 51

TRANSPOSE MATRIX :11 2255 7499 51

3. SUBTRACTION OF TWO 16-BIT BCD NUMBERS

DATA SEGMENT MESS1 DB 0AH,0DH,'ENTER FIRST NO:','$' MESS2 DB 0AH,0DH,'ENTER SECOND NO:','$' MESS3 DB 0AH,0DH,'SUBTRACTION OF TWO 16-BIT NO IS:','$'DATA ENDS

CODE SEGMENT ASSUME CS:CODE,DS:DATASTART: MOV AX,DATAMOV DS,AXLEA DX,MESS1MOV AH,09HINT 21HCALL READMOV BX,DXLEA DX,MESS2MOV AH,09HINT 21HCALL READMOV AX,BXMOV CL,' 'MOV AL,BLSUB AL,DLDASMOV BL,ALMOV AL,BHSBB AL,DHDASMOV BH,ALJNC NEXTMOV AX,0000HSUB AL,BLADD DL,01HDASMOV BL,ALMOV AL,AHSBB AL,BHDASMOV BH,ALMOV CL,'-'NEXT: CALL DISPMOV AH,4CHINT 21H

READ PROC NEAR PUBLIC READMOV CH,02HR3: MOV AH,01HINT 21HMOV CL,04HMOV DL,ALSUB DL,30HCMP DL,0AHJC R1SUB DL,07HR1:SHL DL,CLMOV AH,01HINT 21HSUB AL,30HCMP AL,0AHJC R2SUB AL,07HR2: AND AL,0FHOR DL,ALDEC CHJZ R4MOV DH,DLJMP R3R4:RETREAD ENDP

DISP PROC NEAR PUBLIC DISPLEA DX,MESS3MOV AH,09HINT 21HMOV DL,CLMOV AH,06HINT 21HMOV CH,02HL3:MOV CL,04HMOV DL,BHSHR DL,CLCMP DL,0AHJC L1ADD DL,07HL1:ADD DL,30HMOV AH,06H INT 21H AND BH,0FH CMP BH,0AH JC L2 ADD BH,07HL2:ADD BH,30H MOV DL,BH MOV AH,06H INT 21H DEC CH JZ L4 MOV BH,BL JMP L3 L4:RET DISP ENDPCODE ENDSEND START

OUTPUT:

ENTER FIRST NUMBER : 9873ENTER SECOND NUMBER : 8642SUBTRACTION OF TWO 8-BIT NUMBER IS : 01231

ENTER FIRST NUMBER : 2431ENTER SECOND NUMBER : 9247SUBTRACTION OF TWO 8-BIT NUMBER IS : -681640.TO DISPLAY CURRENT TIME.DATA SEGMENT MESS1 DB 0AH,0DH,'TIME IS:'0AH,0DH,'$'DATA ENDS


START: MOV AX,DATA MOV DS,AX LEA DX,MESS1 MOV AH,09H INT 21H L1:MOV AH,2CH INT 21H MOV AL,CH CALL CONV CALL DISP MOV DL,':' MOV AH,06H INT 21H MOV AL,CL CALL CONV CALL DISP MOV DL,':' MOV AH,06H INT 21H MOV DL,0DH MOV AH,06H INT 21H JMP L1 MOV AH,4CH INT 21H


MOV BL,AH MOV DL,AL ADD DL,30H MOV AH,06H INT 21H MOV DL,BL ADD DL,30H MOV AH,06H INT 21H RET

DISP ENDP

CONV PROC NEAR PUBLIC CONV

MOV AH,00H MOV BH,0AH DIV BH RET

CONV ENDP CODE ENDS END START

OUTPUT:

THE CURRENT TIME IS: 23:37:23.85

41.TO DISPLAY CURRENT DATE.

DATA SEGMENT MESS1 DB 0AH,0DH,'THE CURRENT DATE IS:'$'DATA ENDS


START: MOV AX,DATA MOV DS,AX LEA DX,MESS1 MOV AH,09H INT 21H MOV AH,2AH INT 21H MOV SI,CX MOV BL,DH MOV BH,DL CALL DISP MOV DL,'-' MOV AH,06H INT 21H MOV BL,BH CALL DISP MOV DL,'-' MOV AH,06H INT 21H XOR DX,DX MOV AX,SI MOV CX,0064H DIV CX MOV DH,DL MOV BL,AL CALL DISP MOV BL,DH MOV AH,4CH INT 21H

CONV PROC NEAR PUBLIC CONV

MOV CL,04H XOR AH,AH MOV AL,BL MOV BL,0AH DIV BL SHL AL,CL ADD AL,AH DAA MOV BL,AL RET

CONV ENDP


MOV CL,04H CALL CONV MOV DL,BL SHR DL,CL ADD DL,30H MOV AH,06H INT 21H AND BL,0FH ADD BL,30H MOV DL,BL MOV AH,06H INT 21H RET

DISP ENDP CODE ENDS END START

OUTPUT:

THE CURRENT DATE IS: MON 07/11/2011

14. AVERAGE OF 8-BIT NUMBERS (BCD)

DATA SEGMENT MESS1 DB 0AH,0DH,'ENTER THE LIMIT: ''$' MESS2 DB 0AH,0DH,'ENTER THE NUMBER:','$' MESS3 DB 0AH,0DH,'AVERAGE IS : ', '$' QUO DB 00H DATA ENDS

CODE SEGMENT ASSUME CS:CODE,DS:DATA START :MOV AX,DATA MOV DS,AX LEA DX,MESS1 MOV AH,09H INT 21H CALL READ MOV CH,DL LEA DX,MESS2 MOV AH,09H INT 21H CALL READ MOV BL,DL MOV DH,CH MOV BH,00H N1:MOV AL,CH SUB AL,01H DAS MOV CH,AL JZ N2 MOV DL,',' MOV AH,06H INT 21H CALL READ MOV AL,BL ADD AL,DL DAA MOV BL,AL JNC N3 MOV AL,BH ADD AL,01H DAA MOV BH,AL N3:JMP N1 N2:MOV CH,02H MOV DL,DH MOV DH,00H MOV CL,00H N4: MOV AX,BX SUB AL,DL DAS MOV BL,AL MOV AL,BH SBB AL,00H DAS MOV BH,AL MOV AL,CL ADD AL,01H DAA MOV CL,AL CMP BX,DX JAE N4 DEC CH JZ N5 MOV QUO,CL MOV CL,00H MOV BH,BL JMP N4 N5: MOV BL,QUO MOV BH,CL CALL DISP MOV AH,4CH INT 21HREAD PROC NEARPUBLIC READ

MOV AH,01H INT 21H MOV CL,04H SHL AL,CL MOV DL,AL MOV AH,01H INT 21H AND AL,0FH OR DL,AL RET

READ ENDP


LEA DX,MESS3 MOV AH,09H INT 21H MOV CH,02H LOP:MOV CL,04H MOV DL,BL SHR DL,CL ADD DL,30H MOV AH,06H INT 21H AND BL,0FH ADD BL,30H MOV DL,BL MOV AH,06H INT 21H DEC CH JZ RT CMP CH,01H JNZ L3 MOV DL,'.' MOV AH,06H INT 21H L3:MOV BL,BH JMP LOP RT:RET

DISP ENDP

CODE ENDS END START

OUPTUT:

ENTER THE LIMIT: 05ENTER THE NUMBERS: 34,45,56,67,78THE AVERAGE IS : 56

19.ADDITION OF TWO MATRICES (HEXADECIMAL)

DATA SEGMENT M1 DB 'ENTER THE ORDER OF FIRST MATRIX:','$' M2 DB 0AH,0DH,0AH,'ENTER THE ORDER OF SECOND MATRIX:','$' M3 DB 0AH,0DH,0AH,'ENTER THE ELEMENTS OF FIRST MATRIX:',0AH,0DH,'$' M4 DB 0AH,0DH,0AH,'ENTER THE ELEMENTS OF SECOND MATRIX:',0AH,0DH,'$' M5 DB 0AH,0DH,0AH,' SUM OF TWO MATRICES: ',0AH,0DH, '$' M6 DB 0AH,0DH,0AH,' MATRICES CANNOT BE ADDED: ','$' MAT1 DB 10 DUP(0) MAT2 DB 10 DUP(0) MAT3 DB 30 DUP(0) ROW DB 00H COL DB 00H DATA ENDS

MESSAGE MACRO MESS LEA DX,MESS MOV AH,09H INT 21H ENDM

BSPCE MACRO ASC MOV DL,ASC MOV AH,06H INT 21H ENDM

CODE SEGMENT ASSUME CS:CODE,DS:DATA START:MOV AX,DATA MOV DS,AX MESSAGE M1 CALL READO MOV BX,DX MOV ROW,DH MOV COL,DL MESSAGE M2 CALL READO CMP BX,DX JZ L1 MESSAGE M6 JMP L7 L1:MESSAGE M3 LEA SI,MAT1 CALL READ1 MESSAGE M4 LEA SI,MAT2 CALL READ1 MESSAGE M5 BSPCE 0AH LEA DI,MAT1 LEA SI,MAT2 LEA BX,MAT3 MOV CH,ROW L6:MOV CL,COL L2:MOV DH,00H MOV DL,[SI] ADD DL,[DI] JNC L5 INC DH L5:MOV [BX],DH INC BX MOV [BX],DL INC DI INC SI INC BX DEC CL JNZ L2 DEC CH JZ L3 JMP L6 L3: CALL DISP L7: MOV AH,4CH INT 21H

READO PROC NEAR PUBLIC READO MOV AH,01H INT 21H MOV DH,AL SUB DH,30H BSPCE ' ' MOV AH,01H INT 21H MOV DL,AL SUB DL,30H RET READO ENDP READ1 PROC NEARPUBLIC READ1 BSPCE 0AH MOV CH,ROW N2:MOV BH,COL N1:CALL READ MOV [SI],DL INC SI BSPCE ' ' DEC BH JNZ N1 DEC CH JZ N3 BSPCE 0AH BSPCE 0DH JMP N2 N3:RET READ1 ENDP

READ PROC NEAR PUBLIC READMOV AH,01HINT 21HMOV CL,04HMOV DL,ALSUB DL,30HCMP DL,0AHJC R1SUB DL,07HR1:SHL DL,CLMOV AH,01HINT 21HSUB AL,30HCMP AL, 0AHJC R2SUB AL,07HAND AL,0FHR2:OR DL,ALRET READ ENDP

DISP PROC NEAR PUBLIC DISP LEA BX,MAT3 MOV CH,ROW D4:MOV DH,COL LOP:MOV CL,04H MOV DL,[BX] ADD DL,30H MOV AH,06H INT 21H INC BX MOV CL,04H MOV DL,[BX] SHR DL,CL CMP DL,0AH JC D1 ADD DL,07H D1: ADD DL,30H MOV AH,06H INT 21H MOV DL,[BX] AND DL,0FH CMP DL,0AH JC D2 ADD DL,07H D2: ADD DL,30H MOV AH,06H INT 21H INC BX BSPCE ' ' DEC DH JNZ LOP DEC CH JZ D3 BSPCE 0AH BSPCE 0DH JMP D4 D3: RETDISP ENDP CODE ENDS END START

OUTPUT 1:

ENTER THE ORDER OF FIRST MATRIX : 2 2

ENTER THE ORDER OF SECOND MATRIX : 2 2

ENTER THE ELEMENTS OF FIRST MATRIX:11 1111 11

ENTER THE ORDER OF SECOND MATRIX02 0202 02

SUM OF TWO MATRICES:013 013014 014

OUTPUT 2:ENTER THE ORDER OF FIRST MATRIX : 2 3

ENTER THE ORDER OF SECOND MATRIX : 2 2

MATRICES CANNOT BE ADDED

4.(A)TRAFFIC SIGNAL CONTROLLER USING 8255 PPI

AIM: To Implement Traffic signal controller using 8255 PPI

Apparatus:1.8086 trainer kit2. Traffic signal controller board3.power connectionProcedure:Step 1: Start the processStep 2: Connect the traffic controller interface kit along with the system using SMPSStep 3: Program the operation in system by program-> Assessories-> Communication-> Hyper Terminals-> Enter okStep 4: Enter the GND 1 serial in 8086 kitStep 5:Enter the 1205 for setup motorStep 6: Stop the program PROGRAMME:MEMORY ADDRESSLABELMNEMONICOPCODECOMMENTS

0100STRT:JMP SKIP DATAEB,01,90

0103SKIP DATA:MOV AX,11FFB8,FF

STRT1:MOV SP,AX8B,E0

PUSH CSBE

POP DS1F

NOP90

NOP90

NOP90

MOV AL,80B0,80

MOV DX,8006BA,80,06

OUT DX,ALEE

START:MOV AL,61HB0,61

MOV DX,8000HBA,80,00

OUT DX,ALEE

MOV AL,68HB0,68


OUT DX,ALEE

MOV AL,86HB0,86


OUT DX,ALEE

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,48HB0,48


OUT DX,ALEE

MOV AL,64HB0,64


OUT DX,ALEE

MOV AL,58HB0,58


OUT DX,ALEE

CALL DELAY1E8,01,27

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,48HB0,48


OUT DX,ALEE

MOV AL,62HB0,62


OUT DX,ALEE

CALL DELAY2E8,01,0B

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,61HB0,61


OUT DX,ALEE

MOV AL,68HB0,68


OUT DX,ALEE

CALL DELAY3E8,00,F6

MOV AL,41HB0,41


OUT DX,ALEE

MOV AL,06HB0,06


OUT DX,ALEE

MOV AL,49HB0,49


OUT DX,ALEE

MOV AL,26HB0,26

MOV DX,8004BA,80,04

OUT DX,ALEE

CALL DELAY1E8,00,E2

MOV AL,41HB0,41


OUT DX,ALEE

MOV AL,06HB0,06


OUT DX,ALEE

MOV AL,51HB0,51


OUT DX,ALEE

MOV AL,46HB0,46


OUT DX,ALEE

CALL DELAY2E8,00,C0

MOV AL,41HB0,41


OUT DX,ALEE

MOV AL,06HB0,06

MOV DX,8004BA,80,04

OUT DX,ALEE

MOV AL,61HB0,61


OUT DX,ALEE

MOV AL,86B0,86


OUT DX,ALEE

CALL DELAY2E8,00,AS

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,82HB0,82


OUT DX,ALEE

MOV AL,62HB0,62


OUT DX,ALEE

MOV AL,92HB0,92


OUT DX,ALEE

CALL DELAY1E8,00,91

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,82HB0,82


OUT DX,ALEE

MOV AL,64HB0,64


OUT DX,ALEE

MOV AL,8AHB0,8A


OUT DX,ALEE

CALL DELAY2E8,00,6F

MOV AL,60HB0,60


OUT DX,ALEE

MOV AL,82HB0,82


OUT DX,ALEE

MOV AL,68HB0,68


OUT DX,ALEE

MOV AL,86HB0,86


OUT DX,ALEE

CALL DELAY2E8,00,54

MOV AL,21HB0,21


OUT DX,ALEE

MOV AL,28HB0,28


OUT DX,ALEE

MOV AL,29HB0,29


OUT DX,ALEE

MOV AL,87HB0,87


OUT DX,ALEE

CALL DELAY1E8,00,40

MOV AL,21HB0,21


OUT DX,ALEE

MOV AL,86HB0,86


OUT DX,ALEE

MOV AL,0A1B0,A1

MOV DX,8000BA,80,00

OUT DX,ALEE

MOV AL,0A8B0,A8

MOV DX,8002BA,80,02

OUT DX,ALEE

CALL DELAY2E8,00,1E

MOV AL,21HB0,21


OUT DX,ALEE

MOV AL,28HB0,28

MOV DX,8002BA,80,02

OUT DX,ALEE

MOV AL,61HB0,61


OUT DX,ALEE

MOV AL,68HB0,68

MOV DX,8002BA,80,02

OUT DX,ALEE

JMP STARTE9,FE,B3

CALL DELAY2CD,AA

INT 0AACD,AA

INT OAACD,AA

RETC3

DELAY1:CALL DELAY2E8,FF,F6

CALL DELAY2E8,FF,F3

CALL DELAY2E8,FF,F0

CALL DELAY2E8,FF,ED

CALL DELAY2E8,FF,EA

RETC3

RESULT : Thus the interface was successfully executed and the output has been verified

4(B). STEPPER MOTOR SPEED CONTROLLER USING8255 PPI

AIM: To Implement the programme for Stepper motor controller using 8255 PPI to interface 8086

Apparatus:1.8086 trainer kit2. Stepper motor

PROCEDURE:

Step1: Start the programmeStep2: Connect the kit to the system.The kit connects one separate keyboardStep3: After connecting the kit ,in kit press R then enter,it shows the kit name in monitorStep4: Press I write then E038 start & ending address givenStep5: Then menu has transfer->send dataStep6: See the output in kit through the monitorStep7: Stop the programme

PROGRAMME:MEMORYADDRESSLABELMNEMONICOPCODECOMMENTS

0100MOV AX,0000B8,00,00

0103MOV ES,AX8E,C0

0105MOV SS,AX8E,D0

0107MOV AX,11F0B8,11,F0

010AMOV SP,AX89,C4

010CPUSH CSBE

010DPOP DS1F

010EMOV AL,80B0,80

0110MOV DX,CR 55BA,80,06

0113OUT DX,ALEE

0114MOV AL,00B0,00

0116MOV DX,OFFSET CODEBA,01,C0

0119OUT DX,ALEE

011AMOV DX,OFFSET CODEBA,01,C0

011DMOV CL,40B1,40

011FMOV CH,80B5,80

0121MOV BH,10B7,10

0123SPEED: MOV BL,08B3,08

0125BACK: CALL STPRE8,00,58

0128INT 0A AHCD,AA

012ADEC BLFE,CB

012CJNZ BACK75,F7

012EDEC CHFE,CD

0130DEC BHFE,CF

0132JNZ SPEED 75,EF

0134MOV BL,11B3,11

0136BACK1: CALL STPRE8,00,47

0139INT 0A AHCD,AA

013BDEC BLFE,CB

013DJNZ BACK175,F7

013FMOV BH,10B7,10

0141MOV BL,08B3,08

0143CALL STPRE8,00,3A,BACK2

0146INT OS 5HFD,55

0148DEC BLFE,CB

014AJNZ BACK275,E7

014CINC CHFE,C5

014EDEC BHFE,CF

0150JNZ SPDCONT75,EF

0152MOV AL,00HB0,00

0154MOV DX,PORTABA,80,00

0157OUT DX,ALEE

0158MOV AH,08B4,08

015AINT OA 1HCD,A1

015CCMP AL,1B3C,1B

015EJNZ STRT75,A0

0160INT OA 3HCD,A3

0180ORG 01 80----

0180STPR:PUSH BX53

0181NOP90

0182MOV BX,BX89,D3

0184MOV AL,[BX]8A,07

0186MOV DX,PORTABA,80,00

0189OUT DX,ALEE

018ANOP90

018BMOV AL,[0175]A0,01,75

018ECMP AL,013C,01

0190JNZ ANTICK0F

0192INC BX43


0195CMP AL,003C,00

0197JNZ NEXT75,03

0199SUB BX,08EB,08

019CNEXT: MOV DX,BX89,DA

019ENOP90

019FPOP BX5B

01A0RETC3

01A1ANTICK:DEC BX4B

01A2MOV AL,[BX]8A,07

01A4CMP AL,00BC,00

01A6JNZ NEXT75,FH

01A8ADD BX,0883,C3,08

01ABJMP NEXTEB,FF

01BFDUM: DB 0000

01C0CODE:DB 00H 02H 06H 04H 0CH 00,02,06,04,0C

01C5DB08H 09H 01H 03H 00H08,09,01,03,00


4.(C)1. ADC INTERFACE

AIM: To interface an ADC to 8086

Apparatus: 8086 trainer kitADC kit

PROCEDURE:

Step1: Start the processStep2: Connect 8086 with ADC connectionStep3: Give the required opcode for the programme in the memory address 1000Step4: Give the input voltage and press reset key in the convertersStep5: compile the ProgrammeStep6: Execute the programme and view the result is the output memoryStep7: Stop the process

PROGRAMME:

MEMORYADDRESSLABELMNEMONICOPCODECOMMENTS

1000MOV AL,03B0,03

1002OUT C8,ALE6,C8

1004MOV AL,23B0,23

1006OUT C8,ALE6,C8

1008MOV AL,03B0,03

100AOUT C8,ALE6,C8

100CMOV AL,01B0,01

100EOUT 00,ALE6,00

1010MOV AL,00B0,00

1012OUT D0,ALE6,00

1014IN AL,E6E4,E6

1016AND AL,013C,01

1018CMP AL,013C,01

101AJNZ LOOP75,F8

101CIN AL,C0E4,C0

101EMOV BX,1100BB,00,11

1021MOV [BX],AL88,07,F4

1023INTCD,A5


4(C) 2.DAC INTERFACE(TRIANGULAR WAVEFORM)

AIM: To Perform DAC using 8086 and verify the output waveform

Apparatus: 8086 trainer kitDAC kit

PROCEDURE:Step1: Start the processStep2: Connect 8086 with DAC connectionStep3: Give the required opcode for the programme in the memory address 1000Step4: Give the input voltage and press reset key in the convertersStep5: compile the ProgrammeStep6: Execute the programme and view the result is the output memoryStep7: Stop the process

Program:MEMORYADDRESSLABELMNEMONICOPCODECOMMENTS

1000L1MOV BL,0DB3,0D

1002MOV AL,BL88,DB

1004OUT DAC ALE6,C0

1006INC BLFE,C3

1008JNZ L175,FA

100AL2MOV AL,BL88,DB

100COUT DAC,ALE6,C0

100EDEC BLFE,CB

1010JNZ L275,F8

1012JMP STARTEB,FA

RESULT: Thus the above program has been successfully executed and the output is verified.4(C) 3.DAC INTERFACE(SAWTOOTH WAVEFORM)

AIM: To Perform DAC using 8086 and verify the output waveform

Apparatus: 8086 trainer kitDAC kit

PROCEDURE:Step1: Start the processStep2: Connect 8086 with DAC connectionStep3: Give the required opcode for the programme in the memory address 1000Step4: Give the input voltage and press reset key in the convertersStep5: compile the ProgrammeStep6: Execute the programme and view the result is the output memoryStep7: Stop the process

Program:MEMORY ADDRESSLABELMNEMONICOPCODECOMMENTS

1000L1MOV AL,00B0,00

1002OUT DAC ALE6,C0

1004INC ALFE,C0

1006JNZ L175,FA

1008JMP STARTEB,F6


4.(D) WAVEFORM GENERATION USING 8253 TIMERS

AIM: To implement programme for Waveform Generation using 8253 Timer.

Apparatus :8086 trainer kit8253 counterPower supply

Procedure:Step 1: Start the programStep 2: Connect the 8086 trainer kit and generate the waveformStep 3: Enter OPCODE in the memory address 0100Step 4: Memory segment for this 12 DCStep 5: Compile the programStep 6: Check out the add and even counterStep 7: Stop the program

THEORY: The programmable timer device 8253 contains 3 independent 16 bit counters, each with a maximum count rate of 2.6mhz

It is thus possible to generate three totally independent delays or maximum three independent counters simultaneously all the 3 counters may be independently controlled by programming the 3 word registersPROGRAM:


0100STARTEB,01,90

0103MOV AX,10FFB8,FF,10

0106MOV SP,AX8B,E0

0108PUSH CS0E

0109POP DS1F

010ANOP90

010BNOP90

010CNOP90

010DMOV AL,96B0,96

010FMOV DX,01E6BA,E6,01

0112OUT DX,ALEE

0113MOV AL,08B0,08

0115MOV DX,01E4BA,E4,01

0118OUT DX,ALEF

0119MOV DX,01E6BA,E6,01

011CIN AL,DXEC

011DIN AL,DXEC

011EUP: MOV DX,01E4BA,E4,01

0121IN AL,DXEC

0122JMP UPEB,FA

OUTPUT: ODD COUNT

STEPSDATAOUT2OUT1OUT0A1A0RDWRCS

STEP196--------------11-----LL

STEP2051---------10----LL

STEP3FF1---------11L---L

STEP4FF1--------11L---L

STEP5051--------10L---L

STEP6041--------10L---L

STEP7021--------10L---L

STEP8050--------10L---L

STEP9020--------10L---L

STEP10051--------10L---L

CONNECTIONS: Make a connection to pin D to out2, then make a connection from clk/src to clk2

Memory segment -------- 12DCMemory segment -------- 0100

FOR EVEN COUNT:

STEPSDATA BUSOUT2OUT1OUT0A1A0WRRDCS

START96---------11L---L

STEP1081------10L---L

STEP2FF1------11---LL

STEP3FF1------11---LL

STEP4081------10---LL

STEP5061------10---LL


4.(E) DC MOTOR SPEED CONTROLLER

AIM: To Implement the program for DC motor speed controller using 8253

Apparatus:1.8086 trainer kit2. DC motor

PROCEDURE:

Step1: Start the programStep2: Connect the kit to the system.The kit connects one separate keyboardStep3: After connecting the kit ,in kit press R then enter,it shows the kit name in monitorStep4: Press I write then E038 start & ending address givenStep5: Then menu has transfer->send dataStep6: See the output in kit through the monitorStep7: Stop the program

PROGRAMME:


0100STRT:JMP SKIP-DATAEB,01,90

SKIP-DATA:MOV AX,11FFEB,11,FF

STRT1:MOV SP,AX3B,E0

PUSH CSBE

POP DS1F

NOP90

NOP90

NOP90

MOV AL,80B0,80

MOV DX,8006BA,80,06

OUT DX,ALEE

START:INT 0ACCD,AC

MOV BX,OFFSET DIRBB,01,C9,R

INT 0AFCD,AF

MOV AH,08B4,08

INT 0A1CD,A1

MOV BL,AL8A,D8

MOV CL,46B1,46

CMP AL,CL3A,C1

JNE RVD75,0A

MOV AL,01B0,01

MOV DI,OFFSET DIRFRBF,02,30,R

MOV [DI[,AL88,05

JMP DISPLAYEB,1C,90

RVD:MOV AL,BL8A,C3

MOV CL,52B1,52

CMP AL,CL3A,C1

JE L174,0C

CMP AL,1B3C,1B

JNE START75,D7

MOV AL,80B0,80

MOV DX,8006BA,80,06

OUT DX,ALEF

INT 0A3CD,A3

L1:MOV AL,02B0,02


MOV [DI],AL88,05

DISPLAY: INT 0ACCD,AC

MOV BX,OFFSET SPEEDBB,01,D1,R

INT 0AFCD,AF

MOV AL,02B0,02

MOV AH,0AB4,0A

INT 0ADCD,AD

AND AX,400025,40,00

CMP AX,40003D,40,00

JNE DISPLAY75,EB

PUSH DX52

MOV AL,DL8A,C2

MOV CL,50B1,50

CMP AL,CL3A,C1

JC INVERT72,11


MOV BX,[DI]8B,1D

OR BL,0880,CB,08

MOV AL,BL8A,C3

MOV DX,8004BA,80,04

OUT DX,ALEE

JMP WAVEE8,0C,90

INVERT:MOV DI,OFFSET DIRFRBF,02,30,R

MOV BX,[DI]8B,1D

MOV AL,BL8A,C3

MOV DX,8004BA,80,04

OUT DX,ALEE

WAVE:MOV AL,36B0,3

MOV DX,8C07BA,8C,07

OUT DX,ALEE

POP DX5A

MOV AL,00B0,00

CMP AL,DL3A,C2

JNE 0D275,05

MOV AL,90B0,90

JMP MOD1EB,03,90

MOD2:MOV AL,94B0,94

MOD1:PUSH DX52

MOV DX,8C07BA,8C,07

OUT DX,ALEE

MOV BX,OFFSET TABLEBB,01,D8,R

POP DX5A

BACK:MOV AL,CS[BX]2E,8A,07

CMP AL,DL3A,C2

JC NEXT72,08

JZ NEXT74,06

INC BX43

INC BX43

INC BX43

INC BX43

JMP BACKEB,F1

NEXT:INC BX43

MOV AL,CS[BX]2E,8A,07

MOV DX,8C01BA,8C,01

OUT DX,ALEE

INC BX43

MOV AL,CS[BX]2E,8A,07

OUT DX,ALEE

INC BX43

MOV AL.CS[BX]2E,8A,07

MOV DX,8C05BA,8C,05

OUT DX,ALEE

JMP STARTE9,01,13,R

DIR: DB 44H,49H,52H,20H,46H,2FH,52H,03H44H,49H,52H,20H,46H,2FH,52H,03H

SPEED:DB 53H,50H,45H,45H,44H,20H,03H53H,50H,45H,45H,44H,20H,03H

TABLE:DB 98H,9BH,00H,30H,96H98H,9BH,00H,30H,96H

DB 0FAH,00H,1CH,94H,77H,01H0FAH,00H,1CH,94H,77H,01H

DB:13H,92H,0F4H,01H,0EH,89H,80H,02H,0BH13H,92H,0F4H,01H,0EH,89H,80H,02H,0BH

DB 86H,0E8H,03H,07H,83H,65H,04H,06H,80H86H,0E8H,03H,07H,83H,65H,04H,06H,80H

DB 0E2H,04H,05H,75H,0D0H,06H,04H,66H,0D0H0E2H,04H,05H,75H,0D0H,06H,04H,66H,0D0H

DB 08H,03H,50H,0B8H,0BH,02H,34H,0D0H,08H,03H08H,03H,50H,0B8H,0BH,02H,34H,0D0H,08H,03H

DB 25H,0D0H,06H,04H,20H,0E2H,04H,05H,17H,65H,04H25H,0D0H,06H,04H,20H,0E2H,04H,05H,17H,65H,04H

DB 06H,14H,0E8H,03H,07H,11H,80H,02H,0BH,08H06H,14H,0E8H,03H,07H,11H,80H,02H,0BH,08H

DB 0F4H,01H,0EH,06H,77H,01H,13H,04H,0FAH,00H,1CH0F4H,01H,0EH,06H,77H,01H,13H,04H,0FAH,00H,1CH

DB 02H,9BH,00H,30H,00H,9BH,00H,50H02H,9BH,00H,30H,00H,9BH,00H,50H


4.(F) KEYBOARD DISPLAY INTERFACE

AIM: To perform the program for keyboard display interfaces and drives the output

Apparatus required:1.8086 microprocessor trainer kit2.8279 keyboard display interface

Procedure:Step1: Connect the microprocessor 8086 trainer kit with SMPSStep2: First reset and press S (substitute) to type the starting address Step3: Then type the opcode and press ESCStep4: Then press S followed by the loading addressStep5: Then press enter and get the input dataStep6: Then press ESC and then GO and execute restore and finally SubstituteStep7: Store it in the output address and get the output data

Theory: 1. The INTEL 8279 is a general purpose keyboard/display controller that simultaneously drives the display of a system and interfaces a keyboard with the CPU leaving it free for its routine task2. The keyboard display interface scans the keyboard to identify if any key has been pressed and serves the code of the pressed key to the CPU3. It also transmits the data received from the CPU to the display device. Both of these functions are performed by the controller bin respective fashion without involving the CPU4. The keyboard is interfaced either in the interrupt or parallel mode. In Interrupt mode the processor is requested service only if any key is pressed, otherwise the CPU periodically reads an internal flag of 8279 to check for a key press

Procedure:Step1: Connect the microprocessor 8086 trainer kit with SMPSStep2: First reset and press S (substitute) to type the starting address Step3: Then type the opcode and press ESCStep4: Then press S followed by the loading addressStep5: Then press enter and get the input dataStep6: Then press ESC and then GO and execute restore and finally SubstituteStep7: Store it in the output address and get the output data

PROGRAMME:

MEMORYADDRESSLABELMNEMONICSOPCODECOMMENTS

0100STARTEB,01,90

0103MOV AX,10FFB8,10,FF

0106MOV SP,AX8B,E0

0108PUSH CS0E

0109POP AX58

010AMOV DS,AX8E,D8

010CNOP90

010DNOP90

010ENOP90

010FSTART:

010FMOV AL,18HB0,18

0111MOV DX,01E2HBA,01,E2

0114OUT DX,ALEE

0115MOV AL,DFB0,DF

0117OUT DX,ALEE

0118MOV AL,8FB0,8F

011AOUT DX,ALEE

011BMOV CL,0FB1,0F

011DCALL DISPLAYE8,02,00

SUBPROGRAMME:

0120JMP STARTEB,ED

0122DISPLAY PROC NEARFB,ED

0122MOV BX,OFFSET DATABB,01,31 R

0125NEXT


0127MOV DX,01E0BA,O1,E0

012AOUT DX,ALEE

012BINC BX43

012CDEC CLFE,C9

012EJNS NEXT79,F5

0130RETC3

0131DISPLAY ENDPDATA:DB 0C0,0F9,0A4,0B0,99,92,82,0F8,80,900C0,0F9,0A4,0B0,99,92,82,0F8,80,90

013BDB 88,83,0C6,0A1,86,8E88,83,0C6,0A1,86,8E

OUTPUT:

STEPDATA BUSDISPLAYWRRDCSA0INT

START18------L---L1---

STEP 1DF------L---L1---

STEP 28FBLOCKL---L1---

STEP 3C0BLOCKL---L------

STEP 4F90L---L------

STEP 5A41L---L------

STEP 6B02---------------

STEP 7993L---L------

STEP 8924L---L------

STEP 9825L---L------

STEP 10E26L---L------

STEP 11807L---L------

STEP 12908L---L------

STEP 13880L---L------

STEP 1483AL---L------

STEP 15C6BL---L------

STEP 16A1CL---L------

STEP 1786DL---L------

STEP 188EEL---L------

STEP 1918FL---L------

RESULT : The keyboard interface was successfully executed and the output has been verified

EXP.NO: 7 DATE:(a)Interfacing Programmable Keyboard and Display Controller- 8279 ( With 8086)AIM: To display the rolling message in the display.

APPARATUS REQUIRED:

(i)8086 Microprocessor kit (ii)Power supply (iii) interfacing board.

ALGORITHM: ( Display of rolling message HELP US)

1. Initialize the counter2. Set 8279 for 8 digit character display, right entry3. Set 8279 for clearing the display4. Write the command to display5. Load the character into accumulator and display it6. Introduce the delay7. Repeat from step 1.

INTERFACE - BLOCK DIAGRAM:

K/DISPLAY 8279INTERFACE BOARDMICROPROCESSORKIT 8086

CABLE

ROLLING DISPLAY (DISPLAY MESSAGE IS ECE- b) USINGH 8051ADDRESSLABELMNEMONICSOP-CODECOMMANDS

4000MOV DPTR,#FFC290 FF C2

4003MOV R0,#0078 00

4005MOV R1,#4479 44

4007MOV A,#1074 10

4009MOVX@DPTR,AF0

400AMOV A,#CC74 CC

400CMOVX@DPTR,AF0

400DMOV A,#9074 90

401FMOVX@DPTR,AF0

4010LOOPMOV DPH,R189 83

4012MOV DPL,R088 82

4014MOVX@DPTRE0

4015MOV DPTR,FFC090 FF C0

4018MOVX@DPTR,AF0

4019LCALL DELAY12 45 00

401CINC R008

401DCJNE R0,#0F,LOOPB8 0F F0

401ALJMP START02 81 00

ORG 4500ADDRESSLABELMNEMONICSOP-CODECOMMANDS

4500MOV R4,#A07C A04500

4502LOOP2MOV R5,#FF7D FF4502

4504LOOP1NOP004504

4505DJNZ R5,LOOP1DD FD4505

4507DJNZ R4,LOOP2DC F94507

4509RET224509

LOOK-UP TABLEADDRESSDATA

4400FF,FF,FF,FF

4404FF,FF,FF,FF

440868,6C,68,FF

440CFF,38,FF,FF

The above program initializes 8279 in scanned keyboard 2 key lock-out. Press two keys simultaneously and verify that only one key is accepted by 8279

RESULT: Thus the rolling message is displayed using 8279 interface kit.

EXP. NO: 8 DATE:

SERIAL COMMUNICATION INTERFACING USART 8251

AIM: To study interfacing technique of 8251 (USART) with microprocessor 8086 and write an 8086 ALP to transmit and receive data between two serial ports with RS-232 cable.

APPARATUS REQUIRED:(i)8086 kit with keyboard (2 Nos),(ii) RS232 cable.

THEORY:

The 8251 is used as a peripheral device for serial communication and isprogrammed by the CPU to operate using virtually any serial data transmission technique.The USART accepts data characters from the CPU in parallel format and then converts theminto a continuous serial data stream for transmission. Simultaneously, it can receive serialdata streams and convert them into parallel data characters for the CPU. The CPU canreadthe status of the USART at any time. These include data transmission errors and controlsignals. The control signals define the complete functional definition of the 8251. Controlwords should be written into the control register of 8251.

These control words are split intotwo formats: 1) Mode instruction word &2) Command instruction word. Status word formats used to examine the error during functional operation.

Command instruction:

TRANSMITTER PROGRAMADDRESSMNEMONICSOP-CODECOMMANDS

1000MOV AL,36BO 36MOVE THE VALUE TOACCUMULATOR

1002OUT CE,ALE6 CESEND DATA TO OUTPUT PORT

1004MOV AL,10B0 10MOVE THE VALUE TOACCUMULATOR

1006OUT C8,ALE6 C8SEND DATA TO OUTPUT PORT


100AOUT C8,ALE6 C8 SEND DATA TO OUTPUT PORT

100CMOV AL,4EB0 4EMOVE THE VALUE TOACCUMULATOR

100EOUT C2,ALE6 C2SEND DATA TO OUTPUT PORT


1012OUT C2,ALE6 C2SEND DATA TO OUTPUT PORT

1014MOV AL,AABO AAMOVE THE VALUE TOACCUMULATOR

1016OUT CO,ALE6 COSEND DATA TO OUTPUT PORT

1018INT 02CD 02INTERUPT

RECEIVER PROGRAMADDRESSMNEMONICSOP-CODECOMMANDS

1200IN AL,C0E4 C0INTERUPT

1202MOV BX[1250]BB 50 12MOVE THE VALUE TO ADDRESS

1205MOV BX,AL88 07MOVE ACCUMULATORDATA TO BX

1207INT 02CD 02INTERUPT

RESULT:Thus ALP for serial data communication using USART 8251 is written and the equivalent ASCII 41 for character A is been tx&rx ed

EXP.NO:9 DATE: (a) STEPPER MOTOR INTERFACING

AIM: To write an assembly language program in 8086 to rotate the motor at different speeds in clock wise and anti-clock directions .

APPARATUS REQUIRED:1. Microprocessor kit 8086 -1No2. Power Supply +5 V, dc,+12 V dc-1No 3. Stepper Motor Interface board- 1No 4. Stepper Motor - 1No

PROBLEM STATEMENT:

Write a code for achieving a specific angle of rotation in a given time and particular number of rotations in a specific time.

THEORY:A motor in which the rotor is able to assume only discrete stationary angular position is a stepper motor. The rotary motion occurs in a stepwise manner from one equilibrium position to the next. Two-phase scheme: Any two adjacent stator windings are energized. There are two magnetic fields active in quadrature and none of the rotor pole faces can be in direct alignment with the stator poles. A partial but symmetric alignment of the rotor poles is of course possible.

ALGORITHM:For running stepper motor clockwise and anticlockwise directions(i) Get the first data from the lookup table.(ii) Initialize the counter and move data into accumulator.(iii) Drive the stepper motor circuitry and introduce delay(iv) Decrement the counter is not zero repeat from step(iii)(v) Repeat the above procedure both for backward and forward directions.

SWITCHING SEQUENCE OF STEPPER MOTOR:MEMORY LOCATION A1 A2 B1 B2 HEX CODE4500 1 0 0 0 09 H4501 0 1 0 105 H4502 0 1 1 0 06 H4503 1 0 1 0 0A H

INTERFACE - BLOCK DIAGRAM:

STEPPER MOTOR INTERFACE BOARDMICROPROCESSORKIT 8086

CABLE

PROGRAM:

A-TO RUN STEPPER MOTOR AT DIFFERENT SPEED

MEMORY ADDRESSLABELMNEMONICSOPCODECOMMENTS

1000STARTMOV D1 offset table (1014)BF,14,10

1003MOV CL, 04

B1,04

1005LOOP1MOV AL,[D1]

8A,05

1007OUT PORT1,AL

E6,C0

1009MOV DX,1010

BA,10,10

100CDELAYDEC DX

4A

100DJNZ DELAY

75,FD

100FINC D1

47

1010LOOP LOOP1

E2,F3

1012JMP START

EB,EC

1014TABLEDB

09,05,06,0A

note: for anti-clock direction only change the order of the input data.

RESULT: Thus the assembly language program for rotating stepper motor in Clockwise and anti-clock directions were written and verified.(b)INTERFACING AND PROGRAMMING OF DC MOTOR SPEED CONTROL

AIM: To interface and control the speed of a DC motor using 8253 and 8086 microprocessor kit.

APPARATUS REQUIRED:ApparatusSpecification/modelQuantity

Microcontrolle

mp manual programs

Documents

Transcript of mp manual programs