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EC6513 – MICROPROCESSOR AND MICROCONTROLLER LAB DEPT OF ECE

INDEX

EXPT.NO

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

NAME OF THE EXPERIMENT

Basic arithmetic and Logical operations

Move a Data Block Without Overlap

Code conversion, decimal arithmetic and Matrixoperations.

Floating point operations, string manipulations, sortingand searchingPassword Checking, Print Ram Size And System DateCounters and Time Delay

Traffic light control

Stepper motor control

Digital clock

Key board and Display

Printer status

Serial interface and Parallel interface

A/D and D/A interface and Waveform Generation

Basic arithmetic and Logical operations

Square and Cube program, Find 2’s complement of anumber

Unpacked BCD to ASCII

PAGENO 3

10

11

17

31

36

38

40

42

45

48

49

53

59

70

72

2


BASIC ARITHMETIC AND LOGICAL OPERATIONS USING 8086 PROGRAMMING

EXPT NO: 01 DATE:AIM: To write an Assembly Language Program (ALP) for performing the Arithmeticoperation of two byte numbers.

APPARATUS REQUIRED:

SL.N O 1. 2.

ITEM

Microprocessor kitPower Supply

SPECIFICATION

8086 kit+5 V dc

QUANTITY

11

PROBLEM STATEMENT: Write an ALP in 8086 to add and subtract two byte numbers stored in thememory location 1000H to 1003H and store the result in the memory location1004H to 1005H.Also provide an instruction in the above program to consider thecarry also and store the carry in the memory location 1006H.ALGORITHM: (iii) Multiplication of 16-bit(i) 16-bit additionInitialize the MSBs of sum to 0

Get the first number.Add the second number to the firstnumber.If there is any carry, incrementMSBs of sum by 1.Store LSBs of sum.Store MSBs of sum.

(ii) 16-bit subtraction Initialize the MSBs of difference to 0 Get the first number Subtract the second number from the first number. If there is any borrow, increment MSBs of difference by 1. Store LSBs of difference Store MSBs of difference.

numbers: Get the multiplier. Get the multiplicand Initialize the product to 0. Product=product+ multiplicand Decrement the multiplier by 1 If multiplicand is not equal to 0,repeat from step (d) otherwise store the product.(iv) Division of 16-bit numbers. Get the dividend Get the divisor Initialize the quotient to 0. Dividend = dividend – divisor If the divisor is greater, store the quotient. Go to step g.

If dividend is greater, quotient= quotient + 1. Repeat fromstep (d)Store the dividendvalue as remainder.

3

EC6513 – MICROPROCESSOR AND MICROCONTROLLER LAB

FLOWCHART

ADDITION

START

SET UP COUNTER (CY)

GET FIRST OPERAND

DEPT OF ECE

SUBTRACTION

START

SET UP COUNTER (CARRY)

GET FIRSTOPERAND TO A

GET SECOND OPERANDTOA

SUBTRACTSECOND OPERAND FROM MEMORY

A=A+BIS THEREANY CY

YESIS THEREANY CARRY

COUNTER =COUNTER + 1

NO

STORE THE SUM

NO

YES

COUNTER =COUNTER + 1

STORE THEDIFFERENCE

STORE THE CARRY

STORE THE CARRY STOP

STOP

4


ADDITION ADDRESS

DEPT OF ECE

Opcodes PROGRAM

MOV CX, 0000H

MOV AX,[1200]

MOV BX, [1202]

ADD AX,BX

JNC L1

INC CX

L1 : MOV [1206],CX

MOV [1204], AX

HLT

COMMENTS

Initialize counter CX

Get the first data in AX reg

Get the second data in BX reg

Add the contents of both theregs AX & BX

Check for carry

If carry exists, increment theCX

Store the carry

Store the sum

Stop the program

SUBTRACTION

ADDRESS OPCODES PROGRAM

MOV CX, 0000H

MOV AX,[1200]

MOV BX, [1202]

SUB AX,BX

JNC L1

INC CX

L1 : MOV [1206],CX

MOV [1204], AX

HLT

COMMENTS

Initialize counter CX

Get the first data in AX reg

Get the second data in BX reg

Subtract the contents of BX fromAX

Check for borrow

If borrow exists, increment theCX

Store the borrow

Store the difference

Stop the program

5


RESULT:.ADDITION

DEPT OF ECE

MEMORY

DATA

SUBTRACTION

MEMORY

DATA

MANUAL CALCULATION

.

6


FLOWCHART

MULTIPLICATION

StartStart

DIVISION

Start

Get Multiplier & MultiplicandGet Multiplier & Multiplicand MULTIPLICAND MULTIPLICAND

Load Divisor &Dividend

REGISTER=00 REGISTER=00

QUOTIENT = 0

REGISTER = REGISTER REGISTER = + MULTIPLICAND REGISTER + MULTIPLICAND

DIVIDEND =DIVIDEND-DIVISOR

QUOTIENT = QUOTIENT+1Multiplier=MULTIPLIER Multiplier=MU–1 LTIPLIER – 1

NO

NO IS IS MULTIPLIMULTIPLIERER=0?

=0?

YES

STORE THE RESULT

ISDIVIDEND<DIVISOR

?

YES

STORE QUOTIENTSTORE REMAINDER= DIVIDEND NOW

STOP STOP

7


MULTIPLICATION

ADDRESS Opcodes PROGRAM

MOV AX,[1200]

MOV BX, [1202]

MUL BX

MOV [1206],AX

MOV AX,DX

MOV [1208],AX

HLT

DIVISION

ADDRESS Opcodes PROGRAM

MOV AX,[1200]

MOV DX, [1202]

MOV BX, [1204]

DIV BX

MOV [1206],AX

MOV AX,DX

MOV [1208],AX

HLT

COMMENTS

Get the first data

Get the second data

Divide the dividend by divisor

Store the lower order product

Copy the higher order product toAXStore the higher order product

Stop the program

Get the first data

COMMENTS

Get the first data

Get the second data

Multiply both

Store the lower order product

Copy the higher order product toAXStore the higher order product

Stop the program

8


RESULT:.MULTIPLICATION

DEPT OF ECE

MEMORY

DATA

DIVISON

MEMORY

DATA

MANUAL CALCULATION

9


MOVE A DATA BLOCK WITHOUT OVERLAP

DEPT OF ECE

EXP.NO: 02AIM:To convert a given Move a data block without overlap .ALGORITHM:

1. Initialize the memory location to the data pointer.2. Increment B register.3. Increment accumulator by 1 and adjust it to decimal every time.4. Compare the given decimal number with accumulator value.5. When both matches, the equivalent hexadecimal value is in Bregister.6. Store the resultant in memory location.

PROGRAM:

DATE:

DATA SEGMENTX DB 01H,02H,03H,04H,05H ;Initialize Data Segments Memory LocationsY DB 05 DUP(0)DATA ENDSCODE SEGMENTASSUME CS:CODE,DS:DATASTART:MOV AX,DATA ; Initialize DS to point to start of the memoryMOV DS,AX ; set aside for string of dataMOV CX,05H ; Load counterLEA SI,X+04 ; SI pointer pointed to top of the memory blockLEA DI,X+04+03 ; 03 is displacement of over lapping, DI pointed to;the top of the destination blockCODE ENDSEND START

Output:

RESULT: Thus the output for the Move a data block without overlap was executed successfully

10


CODE CONVERSION, DECIMAL ARITHMETIC AND MATRIX OPERATIONS.

EXP.NO: 03AIM: To convert a given decimal number to hexadecimal.ALGORITHM:

1. Initialize the memory location to the data pointer. 2. Increment B register. 3. Increment accumulator by 1 and adjust it to decimal every time. 4. Compare the given decimal number with accumulator value. 5. When both matches, the equivalent hexadecimal value is in Bregister. 6. Store the resultant in memory location.FLOWCHART :

CODE CONVERSIONS –DECIMAL TO HEX DATE:

11


PROGRAM:

DEPT OF ECE

INPUT

MEMORY

OUTPUT

DATA

12


CODE CONVERSION –HEXADECIMAL TO DECIMALAIM:To convert a given hexadecimal number to decimal.ALGORITHM: 1. Initialize the memory location to the data pointer. 2. Increment B register. 3. Increment accumulator by 1 and adjust it to decimal every time. 4. Compare the given hexadecimal number with B register value. 5. When both match, the equivalent decimal value is in A register. 6. Store the resultant in memory location.

13


14


DECIMAL ARITHMETIC AND MATRIX OPERATIONSAIM:

To write a program for addition of two matrix by using 8086.

APPARATUS REQUIRED:8086 Microprocessor Kit

ALGORITH:1. Initialize the pointer only for data and result2. Load AL with count3. Add two matrix by each element4. Process continues until CL is zero5. Store result.

FLOWCHART

15


PROGRAM

RESULT: Thus the output for the addition for two matrix was executed successfully.

16


FLOATING POINT OPERATIONS, STRING MANIPULATIONS,SORTING AND SEARCHING

EXPT NO:04COPYING A STRINGAIM: To move a string of length FF from source to destination.

ALGORITHM: a. Initialize the data segment .(DS) b. Initialize the extra data segment .(ES) c. Initialize the start of string in the DS. (SI) d. Initialize the start of string in the ES. (DI) e. Move the length of the string(FF) in CX register. f. Move the byte from DS TO ES, till CX=0.

START

DATE:

Initialize DS,ES,SI,DI

CX=length of string,DF=0.

Move a byte from source string (DS)to destination string (ES)

Decrement CX

NO

Check forZF=1

STOP

17


COPYING A STRINGADDRESS OPCODES

DEPT OF ECE

PROGRAM

MOV SI,1200H

MOV DI,1300H

MOV CX,0006H

CLD

REP MOVSB

HLT

COMMENTS

Initialize destination address

Initialize starting address

Initialize array size

Clear direction flag

Copy the contents of source intodestination until count reaches zeroStop

RESULT:

INPUT

MEMORY

DATA

OUTPUT

MEMORY

DATA

Thus a string of a particular length is moved from source segment to destination segment

18


SEARCHING A STRINGAIM: To scan for a given byte in the string and find the relative address of the byte from thestarting location of the string.ALGORITHM: a. Initialize the extra segment .(ES) b. Initialize the start of string in the ES. (DI) c. Move the number of elements in the string in CX register. d. Move the byte to be searched in the AL register. e. Scan for the byte in ES. If the byte is found ZF=0, move the address pointed by ES:DI to BX.

START

Initialize DS,ES ,SI,DI

CX=length of the string,DF=0.

Scan for a particular characterspecified in AL Register.

NO

Check forZF=1

Move DI to BX

STOP

19


SEARCHING FOR A CHARACTER IN THE STRINGADDRESS OPCODESPROGRAM

MOV DI,1300H

MOV SI, 1400H

MOV CX, 0006H

CLD

MOV AL, 08H

REPNE SCASB

DEC DI

MOV BL,[DI]

MOV [SI],BL

HLT

RESULT:

INPUT

MEMORY

DATA

DEPT OF ECE

COMMENTS





Store the string to be searched

Scan until the string is found

Decrement the destination address

Store the contents into BL reg

Store content of BL in source address

Stop

OUTPUT

MEMORY LOCATION

DATA

Thus a given byte or word in a string of a particular length in the extra segment(destination)is found .

20


FIND AND REPLACEAIM:

To find a character in the string and replace it with another character.

DEPT OF ECE

ALGORITHM: a. Initialize the extra segment .(E S) b. Initialize the start of string in the ES. (DI) c. Move the number of elements in the string in CX register. d. Move the byte to be searched in the AL register. e. Store the ASCII code of the character that has to replace the scanned byte in BL register. f. Scan for the byte in ES. If the byte is not found, ZF≠1 and repeat scanning. g. If the byte is found, ZF=1.Move the content of BL register to ES:DI.

START

Initialize DS, ES, SI, DI

CX=length of the string in ES,DF=0.DF=0.

Scan for a particular characterspecified in AL Register.

NO

Check for ZF=1

YES

Move the content of BLto ES:DI

STOP

21


FIND AND REPLACE A CHARACTER IN THE STRINGADDRESS OPCODESPROGRAMCOMMENTS

MOV DI,1300H

MOV SI,1400H

MOV CX, 0006H

CLD

MOV AL, 08H

MOV BH,30H

REPNE SCASB

DEC DI

MOV BL,[DI]

MOV [SI],BL

MOV [DI],BH

HLT

RESULT:

INPUT

MEMORY

DATA





Store the string to be searched

Store the string to be replaced

Scan until the string is found

Decrement the destination address

Store the contents into BL reg

Store content of BL in source address

Replace the string

Stop

OUTPUT

MEMORY

DATA

Thus a given byte or word in a string of a particular length in the extra segment(destination)is found and is replaced with another character.

22


ASCENDING & DESCENDING

AIM:

DEPT OF ECE

To write an Assembly Language Program (ALP) to sort a given array inascending and descending order.

APPARATUS REQUIRED:

SL.N O 1. 2.

ITEM


SPECIFICATION

8086+5 V dc

QUANTITY

11

PROBLEM STATEMENT:

An array of length 10 is given from the location. Sort it into descendingand ascending order and store the result.

ALGORITHM:(i)Sorting in ascending order:a. Load the array count in two registers C1 and C2.b. Get the first two numbers.c. Compare the numbers and exchange if necessary so that the two numbers are in ascending order.d. Decrement C2.e. Get the third number from the array and repeat the process until C2 is 0.f. Decrement C1 and repeat the process until C1 is 0.(ii) Sorting in descending order:a. Load the array count in two registers C1 and C2.b. Get the first two numbers.c. Compare the numbers and exchange if necessary so that the two numbers are in descending order.d. Decrement C2.e. Get the third number from the array and repeat the process until C2 is 0.f. Decrement C1 and repeat the process until C1 is 0.

23


FLOWCHARTASCENDING ORDER

DEPT OF ECE

DESCENDING ORDER

STARTSTART

INITIALIZE POINTER

COUNT = COUNT – 1

FLAG = 0

IS POINTER POINTER +1

NO

TEMP = POINTERPOINTER = POINTER + 1 POINTER + 1 = TEMP

YES

INITIALIZE POINTER

COUNT = COUNT – 1

FLAG = 0

IS POINTER POINTER +1

NO

TEMP = POINTERPOINTER = POINTER + 1 POINTER + 1 = TEMP

YES

POINTER = POINTER +1COUNT = COUNT + 1

POINTER = POINTER +1COUNT = COUNT + 1

NO

IS

COUNT

=0

NOIS FLAG

= 0YES

STOP

YES

NOIS

COUNT

=0 YES

IS FLAG

= 0YES

STOP

24


ASCENDING

ADDRESS OPCODES PROGRAMMOV SI,1200H

MOV CL,[SI]L4 : MOV SI,1200HMOV DL,[SI]INC SIMOV AL,[SI]L3 : INC SIMOV BL,[SI]CMP AL,BLJNB L1DEC SIMOV [SI],ALMOV AL,BLJMP L2L1 : DEC SIMOV [SI],BL

L2 : INC SIDEC DLJNZ L3

MOV [SI],ALDEC CLJNZ L4

HLT

DESCENDING

ADDRESS OPCODES PROGRAMMOV SI,1200H

MOV CL,[SI]L4 : MOV SI,1200HMOV DL,[SI]INC SIMOV AL,[SI]L3 : INC SI

DEPT OF ECE

COMMENTSInitialize memory location for array sizeNumber of comparisons in CLInitialize memory location for array sizeGet the count in DLGo to next memory locationGet the first data in ALGo to next memory locationGet the second data in BLCompare two data’sIf AL < BL go to L1Else, Decrement the memory locationStore the smallest dataGet the next data ALJump to L2Decrement the memory locationStore the greatest data in memorylocationGo to next memory locationDecrement the countJump to L3, if the count is not reachedzeroStore data in memory locationDecrement the countJump to L4, if the count is not reachedzeroStop

COMMENTSInitialize memory location for array sizeNumber of comparisons in CLInitialize memory location for array sizeGet the count in DLGo to next memory locationGet the first data in ALGo to next memory location

25


MOV BL,[SI]CMP AL,BLJB L1DEC SIMOV [SI],ALMOV AL,BLJMP L2L1 : DEC SIMOV [SI],BL

L2 : INC SIDEC DLJNZ L3

MOV [SI],ALDEC CLJNZ L4

HLT

RESULT:.ASCENDING

DEPT OF ECE

Get the second data in BLCompare two data’sIf AL > BL go to L1Else, Decrement the memory locationStore the largest dataGet the next data ALJump to L2Decrement the memory locationStore the smallest data in memorylocationGo to next memory locationDecrement the countJump to L3, if the count is not reachedzeroStore data in memory locationDecrement the countJump to L4, if the count is not reachedzeroStop

MEMORY

DATA

DESCENDING

MEMORY

DATA

Thus given array of numbers are sorted in ascending & descending order.

26


LARGEST& SMALLEST

AIM:

DEPT OF ECE

To write an Assembly Language Program (ALP) to find the largest andsmallest number in a given array.

APPARATUS REQUIRED:

SL.N O 1. 2.

ITEM


SPECIFICATION

8086+5 V dc

QUANTITY

11

PROBLEM STATEMENT:

An array of length 10 is given from the location. Find the largest andsmallest number and store the result.

ALGORITHM:(i)Finding largest number:a.b.c.d.

(ii)e.f.g.h.

Load the array count in a register C1.Get the first two numbers.Compare the numbers and exchange if the number is small.Get the third number from the array and repeat the process until C1 is 0.

Finding smallest number:Load the array count in a register C1.Get the first two numbers.Compare the numbers and exchange if the number is large.Get the third number from the array and repeat the process until C1 is 0.

27


FLOWCHARTLARGEST NUMBER IN AN ARRAY

START

INITIALIZE COUNTPOINTER MAX = 0

PONITER =POINTER + 1

DEPT OF ECE

SMALLEST NUMBER IN AN ARRAYSTART

INITIALIZE COUNTPOINTER MIN = 0

PONITER =POINTER + 1

YESIS MAX

YES POINTER ? NO

IS MINPOINTER?

NO

MIN = POINTERMAX = POINTER

COUNT = COUNT-1

NO

IS COUNT = 0 ?

YES

STORE MAXIMUMYES

NO

COUNT = COUNT-1

IS COUNT = 0 ?

STORE MINIIMUM

STOP STOP

28


LARGEST


MOV SI,1200HMOV CL,[SI]INC SIMOV AL,[SI]DEC CLL2 : INC SICMP AL,[SI]JNB L1MOV AL,[SI]L1 : DEC CLJNZ L2MOV DI,1300HMOV [DI],AL

HLT

SMALLEST


MOV SI,1200HMOV CL,[SI]INC SIMOV AL,[SI]DEC CLL2 : INC SICMP AL,[SI]JB L1MOV AL,[SI]L1 : DEC CLJNZ L2MOV DI,1300HMOV [DI],AL

HLT

DEPT OF ECE

COMMENTS

Initialize array sizeInitialize the countGo to next memory locationMove the first data in ALReduce the countMove the SI pointer to next dataCompare two data’sIf AL > [SI] then go to L1 ( no swap)Else move the large number to ALDecrement the countIf count is not zero go to L2Initialize DI with 1300HElse store the biggest number in 1300locationStop

COMMENTS

Initialize array sizeInitialize the countGo to next memory locationMove the first data in ALReduce the countMove the SI pointer to next dataCompare two data’sIf AL < [SI] then go to L1 ( no swap)Else move the large number to ALDecrement the countIf count is not zero go to L2Initialize DI with 1300HElse store the biggest number in 1300locationStop

29


RESULT:.LARGEST

MEMORY

DATA

SMALLEST

MEMORY

DATA

Thus largest and smallest number is found in a given array.

30


PASSWORD CHECKING, PRINT RAM SIZE AND SYSTEM DATE

EXPT NO:05DATE:AIM: To write an Assembly Language Program (ALP) for performing the Arithmeticoperation of two byte numbers

APPARATUS REQUIRED:

SL.N O 1. 2.

ITEM


SPECIFICATION

8086 kit+5 V dc

QUANTITY

11

PROGRAM:;PASSWORD IS MASM1234DATA SEGMENTPASSWORD DB 'MASM1234'LEN EQU ($-PASSWORD)MSG1 DB 10,13,'ENTER YOUR PASSWORD: $'MSG2 DB 10,13,'WELCOME TO ELECTRONICS WORLD!!$'MSG3 DB 10,13,'INCORRECT PASSWORD!$'NEW DB 10,13,'$'INST DB 10 DUP(0)DATA ENDSCODE SEGMENTASSUME CS:CODE,DS:DATASTART:MOV AX,DATAMOV DS,AXLEA DX,MSG1MOV AH,09HINT 21HMOV SI,00UP1:MOV AH,08HINT 21HCMP AL,0DHJE DOWNMOV [INST+SI],ALMOV DL,'*'MOV AH,02HINT 21HINC SIJMP UP1DOWN:

31


MOV BX,00MOV CX,LENCHECK:MOV AL,[INST+BX]MOV DL,[PASSWORD+BX]CMP AL,DLJNE FAILINC BXLOOP CHECKLEA DX,MSG2MOV AH,09HINT 21HJMP FINISHFAIL:LEA DX,MSG3MOV AH,009HINT 21HFINISH:INT 3CODE ENDSEND STARTEND

;Today.asm Display month/day/year.; Feb 1st, 2012;CIS 206 Ken Howard .MODEL small .STACK 100h .DATA mess1 DB 10, 13, 'Today is $' ; 10=LF, 13=CR .CODE

Today PROC MOV AX, @data MOV DS, AX MOV DX, OFFSET mess1 ; Move string to DX MOV AH, 09h; 09h call to display string (DX > AH > DOS) INT 21H; Send to DOS ; CX year, DH month, DL day MOV AH, 2AH; Get the date (appendix D) INT 21H; Send to DOS PUSH CX; Move year to the stack MOV CX, 0; Clear CX

DEPT OF ECE

MOV CL, DL PUSH CX; Move day to stack MOV CL, DH; Move month > CL PUSH CX; Move month to stackMOV DH, 0; Clear DH; ************************** DISPLAY MONTH ************************

32


; Set up for division ; Dividend will be in DX/AX pair (4 bytes) ; Quotient will be in AX ; Remainder will be in DX MOV DX, 0; Clear DX POP AX; Remove month from stack into AX MOV CX, 0; Initialize the counter MOV BX, 10; Set up the divisordividem: DIV BX; Divide (will be word sized) PUSH DX; Save remainder to stack ADD CX, 1; Add one to counter MOV DX, 0; Clear the remainder CMP AX, 0; Compare quotient to zero JNE dividem; If quoient is not zero, go to "dividem:"divdispm:

DEPT OF ECE

POP DX; Remove top of stack into DX ADD DL, 30h; ADD 30h (2) to DL MOV AH, 02h; 02h to display AH (DL) INT 21H; Send to DOS LOOP divdispm; If more to do, divdispm again ; LOOP subtracts 1 from CX. If non-zero, loop. MOV DL, '/'; Character to display goes in DL MOV AH, 02h; 02h to display AH (DL) INT 21H; Send to DOS; ************************** DISPLAY DAY ************************ ; Set up for division ; Dividend will be in DX/AX pair (4 bytes) ; Quotient will be in AX ; Remainder will be in DX MOV DX, 0; Clear DX POP AX; Remove day from stack into AX MOV CX, 0; Initialize the counter MOV BX, 10; Set up the divisordivided: DIV BX; Divide (will be word sized) PUSH DX; Save remainder to stack

ADD CX, 1; Add one to counter MOV DX, 0; Clear the remainder CMP AX, 0; Compare quotient to zero JNE divided; If quoient is not zero, go to "divided:"divdispd: POP DX; Remove top of stack ADD DL, 30h; ADD 30h (2) to DL MOV AH, 02h; 02h to display AH (DL) INT 21H; Send to DOS LOOP divdispd; If more to do, divdispd again ; LOOP subtracts 1 from CX. If non-zero, loop.

33


MOV DL, '/'; Character to display goes in DL MOV AH, 02h; 02h to display AH (DL) INT 21H; Send to DOS ; ************************** DISPLAY YEAR ************************ ; Set up for division ; Dividend will be in DX/AX pair (4 bytes) ; Quotient will be in AX ; Remainder will be in DX MOV DX, 0; Clear DX POP AX; Remove month from stack into AX MOV CX, 0; Initialize the counter MOV BX, 10; Set up the divisor dividey: DIV BX; Divide (will be word sized) PUSH DX; Save remainder to stack ADD CX, 1; Add one to counter MOV DX, 0; Clear the remainder CMP AX, 0; Compare quotient to zero JNE dividey; If quoient is not zero, go to "dividey:"divdispy: POP DX; Remove top of stack into DX ADD DL, 30h; ADD 30h (2) to DL MOV AH, 02h; 02h to display AH (DL) INT 21H; Send to DOS LOOP divdispy; If more to do, divdisp again ; LOOP subtracts 1 from CX. If non-zero, loop. MOV al, 0; Use 0 as return code MOV AH, 4ch; Send return code to AH INT 21H; Send return code to DOS to exit.

Today ENDP; End procedureEND Today; End code. Start using "Today" procedure.MVI A, 80H: Initialize 8255, port A and port BOUT 83H (CR): in output modeSTART: MVI A, 09HOUT 80H (PA): Send data on PA to glow R1 and R2MVI A, 24HOUT 81H (PB): Send data on PB to glow G3 and G4MVI C, 28H: Load multiplier count (40ıο) for delayCALL DELAY: Call delay subroutineMVI A, 12HOUT (81H) PA: Send data on Port A to glow Y1 and Y2OUT (81H) PB: Send data on port B to glow Y3 and Y4MVI C, 0AH: Load multiplier count (10ıο) for delayCALL: DELAY: Call delay subroutineMVI A, 24HOUT (80H) PA: Send data on port A to glow G1 and G2MVI A, 09HOUT (81H) PB: Send data on port B to glow R3 and R4MVI C, 28H: Load multiplier count (40ıο) for delay

34


CALL DELAY: Call delay subroutineMVI A, 12HOUT PA: Send data on port A to glow Y1 and Y2OUT PB: Send data on port B to glow Y3 and Y4MVI C, 0AH: Load multiplier count (10ıο) for delayCALL DELAY: Call delay subroutineJMP STARTDelay Subroutine:DELAY: LXI D, Count: Load count to give 0.5 sec delayBACK: DCX D: Decrement counterMOV A, DORA E: Check whether count is 0JNZ BACK: If not zero, repeatDCR C: Check if multiplier zero, otherwise repeatJNZ DELAYRET: Return to main program

Ram size:

ORG 0000H CLR PSW3 CLR PSW4 CPL A ADD A, #01H MOV A,R3AGAIN: SJMP AGAIN

DEPT OF ECE

RESULT: Thus the output for the Password checking, Print RAM size and system date wasexecuted successfully

35


Counters and Time Delay

EXP.NO: 06

AIM:

DATE:

To write an assembly language program in 8086 to Counters and Time Delay

SPECIFICATION8086+5 V, dc,+12 V dc--

QUANTITY 1 1 1 1

APPARATUS REQUIRED:SL.NOITEM 1.Microprocessor kit 2.Power Supply 3.Stepper Motor Interface board 4.Stepper Motor

PROGRAM: .MODEL SMALL .DATA

MSGIN DB 'Enter delay duration (0-50): $'MSG1 DB 'This is Microprocessor!$'DELAYTIME DW 0000H

.CODE

MOV DX,@DATAMOV DS,DXLEA DX,MSGINMOV AH,09HINT 21H

IN1: MOV AH,01H INT 21H CMP AL,0DH ; JE NXT SUB AL,30H MOV DL,AL MOV AX,BX MOV CL,0AH MUL CL MOV BX,AX AND DX,00FFH ADD BX,DX MOV DELAYTIME,BX LOOP IN1

36


NXT:MOV CX,DELAYTIME MOV DL,10 MOV AH,02H INT 21H

LEA SI,MSG1

LP: PUSH DX MOV DL,[SI] CMP DL,'$' JE NXT2 MOV AH,02HINT 21HADD SI,1POP DXMOV DI,DELAYTIMEMOV AH, 0INT 1AhMOV BX, DX

Delay:MOV AH, 0INT 1AhSUB DX, BXCMP DI, DXJA Delay

LOOP LP

NXT2: MOV AH,4CH INT 21H

END

DEPT OF ECE

RESULT: Thus the output for the Counters and Time Delay was executed successfully

37


TRAFFIC LIGHT CONTROL

EXP.NO: 07

AIM:

DATE:

To write an assembly language program in 8086 to Traffic light controlAPPARATUS REQUIRED:SL.NOSPECIFICATIONQUANTITYITEM 1.80861Microprocessor kit 2.+5 V, dc,+12 V dc1Power Supply 3.-1Traffic light control Interface board

PROGRAM:

38


39


STEPPER MOTOR INTERFACING

DEPT OF ECE

EXP.NO: 08 DATE:

AIM: To write an assembly language program in 8086 to rotate the motor at different speeds.

APPARATUS REQUIRED:

SL.NO 1. 2. 3. 4.

ITEMMicroprocessor kitPower SupplyStepper Motor Interface boardStepper Motor

SPECIFICATION8086+5 V, dc,+12 V dc--

QUANTITY 1 1 1 1

PROBLEM STATEMENT:

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

THEORY:

A motor in which the rotor is able to assume only discrete stationary angularposition is a stepper motor. The rotary motion occurs in a stepwise manner from oneequilibrium position to the next.Two-phase scheme: Any two adjacent stator windings areenergized. There are two magnetic fields active in quadrature and none of the rotor pole facescan be in direct alignment with the stator poles. A partial but symmetric alignment of therotor 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:

MEMORYLOCATION4500450145024503

A1

1001

A2

0110

B1

0011

B2

0100

HEXCODE09 H05 H06 H0A H

40


FLOWCHART:

DEPT OF ECE

START

INTIALIZE COUNTER FOR LOOK UP TABLE

GET THE FIRST DATA FROM THE ACCUMULATOR

MOVE DATA INTO THE ACCUMULATOR

DRIVE THE MOTOR CIRCUITARY

DELAY

DECREMENT COUNTER

IS B = 0 ?

GET THE DATA FROM LOOK UP TABLE

PROGRAM TABLEADDRESS OPCODE PROGRAM

START : MOV DI, 1200H

MOV CX, 0004HLOOP 1 : MOV AL,[DI]OUT 0C0,ALMOV DX, 1010HL1 : DEC DXJNZ L1INC DILOOP LOOP1JMP START

1200 : 09,05,06,0A

Go to next memory locationLoop until all the data’s have been sentGo to start location for continuousrotationArray of data’s

Introduce delay

COMMENTSInitialize memory location to store thearray of numberInitialize array sizeCopy the first data in ALSend it through port address

RESULT: Thus the assembly language program for rotating stepper motor in both clockwiseand anticlockwise directions is written and verified.

41


DIGITAL CLOCK

EXP.NO: 09

Aim

DATE:

To display the digital clock specifically by displaying the hours, minutes and secondsusing 8086 kits

Apparatus required

S.No12

ItemMicroprocessor kit Power Supply

Specification80865V +5 V, dc, +12 V dcPreliminary Settings

Org 1000hStore time value in memory location 1500- Seconds 1501- Minutes 1502- HoursDigital clock program

42


43


Result Thus the digital clock program has been written and executed using 8086microprocessor kit and the output of digital clock was displayed as [hours: minutes: seconds]successfully.

44


INTERFACING PRGRAMMABLE KEYBOARD AND DISPLAY CONTROLLER- 8279

EXP.NO:10

AIM :To display the rolling message “ HELP US “ in the display.

DATE:

APPARATUS REQUIRED: 8086 Microprocessor kit, Power supply, Interfacing board.

ALGORITHM : Display of rolling message “ HELP US “ 1. Initialize the counter 2. Set 8279 for 8 digit character display, right entry 3. Set 8279 for clearing the display 4. Write the command to display 5. Load the character into accumulator and display it 6. Introduce the delay 7. Repeat from step 1.

1. Display Mode Setup: Control word-10 H

00

DD 00- 8Bit character display left entry 01- 16Bit character display left entry 10- 8Bit character display right entry 11- 16Bit character display right entryKKK- Key Board Mode 000-2Key lockout.2.Clear Display: Control word-DC H

11

11

00

1CD

1CD

1CD

0CF

0CA

00

00

1 D

0D

0K

0K

0K

11 A0-3; B0-3 =FF

1-Enables Clear display0-Contents of RAM will be displayed

1-FIFO Status is cleared

451-Clear all bits(Combined effect of CD)


3. Write Display: Control word-90H

11

00

00

1 0 0 0 0

DEPT OF ECE

AI A A A A

Selects one of the 16 rows of display.

Auto increment = 1, the row address selected will be incremented after each of read andwrite operation of the display RAM.FLOWCHART

SEGMENT DEFINITION

DATA BUS D7 D6

c

D5

b

D4

a

D3

dp

D2

g

D1

f

D0

eSEGMENTS d

46


PROGRAM TABLE

DEPT OF ECE

PROGRAMSTART : MOV SI,1200HMOV CX,000FHMOV AL,10OUT C2,ALMOV AL,CCOUT C2,ALMOV AL,90OUT C2,ALL1 : MOV AL,[SI]OUT C0,ALCALL DELAYINC SILOOP L1JMP STARTDELAY : MOV DX,0A0FFHLOOP1 : DEC DXJNZ LOOP1RET

LOOK-UP TABLE:

12001204

98FF

681C

7C29

C8FF

Initialize arrayCOMMENTS

Initialize array sizeStore the control word for display modeSend through output portStore the control word to clear displaySend through output portStore the control word to write displaySend through output portGet the first dataSend through output portGive delayGo & get next dataLoop until all the data’s have been takenGo to starting locationStore 16bit count valueDecrement count valueLoop until count values becomes zeroReturn to main program

RESULT:

MEMORYLOCATION1200H1201H1202H1203H1204H1205H1206H1207H

7-SEGMENT LED FORMATcbadp eg001100110100111110100100111111000110010100111111

HEX DATAf00001011

98687CC8FF1C29FF

d10011001

Thus the rolling message “HELP US” is displayed using 8279 interface kit.

47


PRINTER STATUS

EXP.NO:11AIM:To display the Printer Status in the display

APPARATUS REQUIRED: 8086 Microprocessor kit, Power supply, interfacing board.

PROGRAM:

XOR AX, AXXOR BX, BX

;this divides my 3digit number by 100 giving me my, hundredth digitMOV AX, RESMOV BX, 100DIV BX

;prints the hundredth digitADD AL, '0'MOV DL, ALPUSH AX ; save AX on the stackMOV AH, 02hINT 21hPOP AX ; restore ax

;divides the remainder by 10 giving me my tens digitMOV BX, 10DIV BX

;prints my tens digitADD AL, '0'MOV DL, ALPUSH AX ; save AX on the stackMOV AH, 02hINT 21hPOP AX ; restore ax

;print my last remainder which is my onesADD AH, '0'MOV DL, AHMOV AH, 02hINT 21h

RESULT: Thus the output for the Move a data block without overlap was executed successfully

DATE:

48


SERIAL INTERFACE AND PARALLEL INTERFACE

EXP.NO: 12 DATE:

To connect two 8086 microprocessor kits and to serially communicate witheach other by considering transmitter and receiver kits.Apparatus required

Aim

Procedure1. Take two no of 8086 microprocessor kits.2. Enter the transmitter program in transmitter kit.3. Enter the receiver program in receiver kit.4. Interface the two kits with 9-9 serial cable in the serial port of the microprocessor kits.(LCD kit means PC-PC cable. LED kit means kit-kit cable)5. Enter the data in transmitter kit use the memory location 1500.6. Execute the receiver kit.7. Execute the transmitter kit.8. Result will be available in receiver kit memory location 1500.

Transmitter Program

49


Receiver Program

ResultThus the serial communication between two 8086 microprocessor kits has beenestablished and the data is transmitted in one kit and received in the other kitsuccessfully

50


PARALLEL COMMUNICATION BETWEEN TWO 8086 MICROPROCESSORS KITS AimTo connect two 8086 microprocessor kits and to establish parallel communication with eachother by considering transmitter and receiver kits.Apparatus required

Procedure1. Take two 8086 microprocessor kits.2. Enter the transmitter program in transmitter kit.3. Enter the receiver program in receiver kit.4. Interface the two kits with 26-core cable on PPI-1.5. Execute the receiver kit.6. Execute the transmitter kit.7. Go and see the memory location 1200 in receiver to get same eight data.8. Data is available in transmitter kit in the memory location.9. Change the data & execute the following procedure & get the result in receiver kit.

Transmitter program

51


Receiver Program

DEPT OF ECE

ResultThus the serial communication between two 8086 microprocessor kits has been establishedand the data is transmitted in one kit and received in the other kit successfully.

52


A/D AND D/A INTERFACE AND WAVEFORM GENERATION

EXPT NO:13DATE:AIM: To write an assembly language program to convert an analog signal into a digital signalusing an ADC interfacing.APPARATUS REQUIRED: ITEMSL.NOSPECIFICATIONQUANTITY 1.80861Microprocessor kit 2.+5 V dc,+12 V dc1Power Supply 3.-1ADC Interface boardTHEORY: An ADC usually has two additional control lines: the SOC input to tell the ADC when tostart the conversion and the EOC output to announce when the conversion is complete. Thefollowing program initiates the conversion process, checks the EOC pin of ADC 0809 as towhether the conversion is over and then inputs the data to the processor. It also instructs theprocessor to store the converted digital data at RAM location.ALGORITHM: (i)Select the channel and latch the address. (ii)Send the start conversion pulse. (iii) Read EOC signal. (iv)If EOC = 1 continue else go to step (iii) (v)Read the digital output. (vi)Store it in a memory location.FLOW CHART: START

SELECT THE CHANNEL AND LATCH ADDRESS

SEND THE START CONVERSION PULSE

NOIS EOC = 1?

YES

READ THE DIGITALOUTPUT

STORE THE DIGITAL VALUE IN THE MEMORY LOCATION SPECIFIED

STOP

53


PROGRAM TABLE

DEPT OF ECE

PROGRAM

MOV AL,00OUT 0C8H,ALMOV AL,08OUT 0C8H,ALMOV AL,01OUT 0D0H,ALMOV AL,00MOV AL,00MOV AL,00MOV AL,00OUT 0D0H,ALL1 : IN AL, 0D8HAND AL,01CMP AL,01JNZ L1

IN AL,0C0HMOV BX,1100MOV [BX],ALHLTRESULT:

ANALOGVOLTAGE

COMMENTS

Load accumulator with value for ALE highSend through output portLoad accumulator with value for ALE lowSend through output portStore the value to make SOC high in the accumulatorSend through output port

Introduce delay

Store the value to make SOC low the accumulatorSend through output port

Read the EOC signal from port & check for end ofconversion

If the conversion is not yet completed, read EOC signalfrom port againRead data from portInitialize the memory location to store dataStore the dataStop

DIGITAL DATA ON LEDDISPLAY

HEX CODE IN MEMORYLOCATION

Thus the ADC was interfaced with 8086 and the given analog inputs were converted into itsdigital equivalent.

54


INTERFACING DIGITAL – TO – ANALOG CONVERTERAIM : 1. To write an assembly language program for digital to analog conversion 2. To convert digital inputs into analog outputs & To generate different waveformsAPPARATUS REQUIRED: SL.NOSPECIFICATIONQUANTITYITEM 1.8086 Vi Microsystems1Microprocessor kit 2.+5 V, dc,+12 V dc1Power Supply 3.-1DAC Interface boardPROBLEM STATEMENT: The program is executed for various digital values and equivalent analog voltages aremeasured and also the waveforms are measured at the output ports using CRO.THEORY: Since DAC 0800 is an 8 bit DAC and the output voltage variation is between –5vand +5v. The output voltage varies in steps of 10/256 = 0.04 (approximately). The digital datainput and the corresponding output voltages are presented in the table. The basic idea behindthe generation of waveforms is the continuous generation of analog output of DAC. With 00(Hex) as input to DAC2 the analog output is –5v. Similarly with FF H as input, the output is+5v. Outputting digital data 00 and FF at regular intervals, to DAC2, results in a square waveof amplitude 5v.Output digital data from 00 to FF in constant steps of 01 to DAC2. Repeatthis sequence again and again. As a result a saw-tooth wave will be generated at DAC2output. Output digital data from 00 to FF in constant steps of 01 to DAC2. Output digital datafrom FF to 00 in constant steps of 01 to DAC2. Repeat this sequence again and again. As aresult a triangular wave will be generated at DAC2 output.ALGORITHM: Measurement of analog voltage: (i)Send the digital value of DAC. (ii)Read the corresponding analog value of its output. Waveform generation: Square Waveform: (i)Send low value (00) to the DAC. (ii)Introduce suitable delay. (iii) Send high value to DAC. (iv)Introduce delay. (v)Repeat the above procedure. Saw-tooth waveform: (i)Load low value (00) to accumulator. (ii)Send this value to DAC. (iii) Increment the accumulator. (iv)Repeat step (ii) and (iii) until accumulator value reaches FF. (v)Repeat the above procedure from step 1. Triangular waveform: (i)Load the low value (00) in accumulator. (ii)Send this accumulator content to DAC. (iii) Increment the accumulator. (iv)Repeat step 2 and 3 until the accumulator reaches FF, decrement the accumulator and send the accumulator contents to DAC. (v)Decrementing and sending the accumulator contents to DAC. (vi)The above procedure is repeated from step (i)

55


FLOWCHART:MEASUREMENT OF ANALOG VOLTAGE

START

DEPT OF ECE

SQUARE WAVE FORMSTART

INTIALISE THEACCUMULATOR SEND ACC CONTENT TO DAC

DELAY

LOAD THE ACC WITH MAXVALUE SEND ACC CONTENT TO DAC

DELAY

SEND THEDIGITALVALUE TO ACCUMULATOR

TRANSFER THE ACCUMULATORCONTENTS TO DAC

READ THE CORRESPONDING ANALOG VALUE

STOP

SAWTOOTH WAVEFORMSTART

TRIANGULAR WAVEFORMSTART

INITIALIZEACCUMULATOR

SEND ACCUMULATOR CONTENT TO DAC

INCREMENTACCUMULATOR CONTENT

YESIS ACC FF

NOYES

DECREMENTACCUMULATOR CONTENT

INITIALIZEACCUMULATOR

SEND ACCUMULATOR

CONTENT TO DAC

INCREMENTACCUMULATOR CONTENT

NOIS ACC FF

SENDACCUMULATORCONTENT TO DAC

IS ACC 00YES NO

56


MEASUREMENT OF ANALOG VOLTAGE:

PROGRAM

MOV AL,7FHOUT C0,ALHLT

COMMENTS

DEPT OF ECE

Load digital value 00 in accumulatorSend through output portStop

DIGITAL DATA ANALOG VOLTAGE

PROGRAM TABLE: Square Wave

PROGRAM

L2 : MOV AL,00HOUT C0,ALCALL L1MOV AL,FFHOUT C0,ALCALL L1JMP L2L1 : MOV CX,05FFHL3 : LOOP L3RET

COMMENTS

Load 00 in accumulatorSend through output portGive a delayLoad FF in accumulatorSend through output portGive a delayGo to starting locationLoad count value in CX registerDecrement until it reaches zeroReturn to main program

PROGRAM TABLE: Saw tooth Wave

PROGRAM

L2 : MOV AL,00HL1 : OUT C0,ALINC ALJNZ L1JMP L2

COMMENTS

Load 00 in accumulatorSend through output portIncrement contents of accumulatorSend through output port until it reaches FFGo to starting location

57


PROGRAM TABLE: Triangular Wave

PROGRAM

L3 : MOV AL,00HL1 : OUT C0,ALINC ALJNZ L1MOV AL,0FFHL2 : OUT C0,ALDEC ALJNZ L2JMP L3

COMMENTS

Load 00 in accumulatorSend through output portIncrement contents of accumulatorSend through output port until it reaches FFLoad FF in accumulatorSend through output portDecrement contents of accumulatorSend through output port until it reaches 00Go to starting location

RESULT:WAVEFORM GENERATION:

WAVEFORMS AMPLITUDE TIMEPERIOD

Square WaveformSaw-tooth waveform

Triangular waveform

MODEL GRAPH:

Square Waveform Saw-tooth waveform

Triangular waveform

Thus the DAC was interfaced with 8085 and different waveforms have been generated.

58


BASIC ARITHMETIC AND LOGICAL OPERATIONS

EXPT NO:14

8 BIT ADDITIONAIM:

DATE:

To write a program to add two 8-bit numbers using 8051 microcontroller.

ALGORITHM:

1. Clear Program Status Word.2. Select Register bank by giving proper values to RS1 & RS0 of PSW.3. Load accumulator A with any desired 8-bit data.4. Load the register R 0 with the second 8- bit data.5. Add these two 8-bit numbers.6. Store the result.7. Stop the program.

FLOW CHART:

START

Clear PSW

Select RegisterBank

Load A and R 0with 8- bit datas

Add A & R 0

Store the sum

STOP

59


8 Bit Addition (Immediate Addressing)ADDRESS LABEL MNEMONIC OPERAND

4100

4101

4103

4105

4108

4109 L1

CLR

MOV

ADDC

MOV

MOVX

SJMP

C

A, data1

A, # data 2

DPTR, #4500H@ DPTR, A

L1

DEPT OF ECE

HEXCODEC3

74,data1

24,data2

90,45,00

F0

80,FE

COMMENTS

Clear CY Flag

Get the data1 inAccumulatorAdd the data1 withdata2Initialize the memorylocationStore the result inmemory locationStop the program

RESULT:

OUTPUTMEMORY LOCATION

4500

DATA

Thus the 8051 ALP for addition of two 8 bit numbers is executed.

60


8 BIT SUBTRACTIONAIM: To perform subtraction of two 8 bit data and store the result in memory.

ALGORITHM: a. Clear the carry flag. b. Initialize the register for borrow. c. Get the first operand into the accumulator. d. Subtract the second operand from the accumulator. e. If a borrow results increment the carry register. f. Store the result in memory.

FLOWCHART:START

DEPT OF ECE

CLEAR CARRYFLAG

GET I’STOPERAND INACCR

SUBTRACT THE2’ND OPERANDFROM ACCR

NIS CF=1

Y

INCREMENTTHE BORROWREGISTER

STORERESULT INMEMORY

STOP

61


8 Bit Subtraction (Immediate Addressing)ADDRESS LABELMNEMONIC OPERAND

4100

4101

4103

4105

4108

4109 L1

CLR

MOV

SUBB

MOV

MOVX

SJMP

C

A, # data1

A, # data2

DPTR, # 4500

@ DPTR, A

L1

DEPT OF ECE

HEXCODEC3

74, data1

94,data2

90,45,00

F0

80,FE

COMMENTS

Clear CY flag

Store data1 inaccumulatorSubtract data2 fromdata1Initialize memorylocationStore the differencein memory locationStop

RESULT:

OUTPUTMEMORY LOCATION

4500

DATA

Thus the 8051 ALP for subtraction of two 8 bit numbers is executed.

62


8 BIT MULTIPLICATION

DEPT OF ECE

AIM: To perform multiplication of two 8 bit data and store the result in memory.

ALGORITHM: a. Get the multiplier in the accumulator. b. Get the multiplicand in the B register. c. Multiply A with B. d. Store the product in memory.

FLOWCHART:

START

GETMULTIPLIERIN ACCR

GETMULTIPLICANDIN B REG

MULTIPLY AWITH B

STORERESULT INMEMORY

STOP

63


8 Bit MultiplicationADDRESS LABEL

4100

4102

4104

4106

4109

401A

410B

410D

410E STOP

DEPT OF ECE

MNEMONIC

MOV

MOV

MUL

MOV

MOVX

INC

MOV

MOV

SJMP

OPERAND

A ,#data1

B, #data2

A,B

DPTR, #4500H@ DPTR, A

DPTR

A,B

@ DPTR, A

STOP

HEXCODE74, data1

75,data2

F5,F0

90,45,00

F0

A3

E5,F0

F0

80,FE

COMMENTS

Store data1 inaccumulatorStore data2 in B reg

Multiply both

Initialize memorylocationStore lower orderresultGo to next memorylocation

Store higher orderresult

Stop

RESULT:

INPUTMEMORY LOCATION

4500

4501

DATA OUTPUTMEMORY LOCATION

4502

4503

DATA

Thus the 8051 ALP for multiplication of two 8 bit numbers is executed.

64


8 BIT DIVISION

AIM: To perform division of two 8 bit data and store the result in memory.

ALGORITHM: 1. Get the Dividend in the accumulator. 2. Get the Divisor in the B register. 3. Divide A by B. 4. Store the Quotient and Remainder in memory.

FLOWCHART:

START

DEPT OF ECE

GET DIVIDENDIN ACCR

GET DIVISOR INB REG

DIVIDE A BY B

STOREQUOTIENT &REMAINDERIN MEMORY

STOP

65


8 Bit DivisionADDRESS LABEL MNEMONIC

4100

4102

4104

4015

4018

4109

410A

410C

410D STOP

MOV

MOV

DIV

MOV

MOVX

INC

MOV

MOV

SJMP

DEPT OF ECE

OPERAND

A, # data1

B, # data2

A,B

DPTR, # 4500H

@ DPTR, A

DPTR

A,B

@ DPTR, A

STOP

HEXCODE74,data1

75,data2

84

90,45,00

F0

A3

E5,F0

COMMENTS

Store data1 inaccumulatorStore data2 in B reg

Divide

Initialize memorylocationStore remainder

Go to next memorylocation

Store quotientF0

80,FE Stop

RESULT:

INPUTMEMORY LOCATION4500(dividend)

4501 (divisor)

OUTPUTMEMORY LOCATION4502(remainder)

4503 (quotient)

DATA DATA

Thus the 8051 ALP for division of two 8 bit numbers is executed.

66


LOGICAL AND BIT MANIPULATIONAIM:

DEPT OF ECE

To write an ALP to perform logical and bit manipulation operations using 8051microcontroller.

APPARATUS REQUIRED: 8051 microcontroller kit

ALGORITHM:

1.2.3.4.5.6.7.8.

Initialize content of accumulator as FFHSet carry flag (cy = 1).AND bit 7 of accumulator with cy and store PSW format.OR bit 6 of PSW and store the PSW format.Set bit 5 of SCON.Clear bit 1 of SCON.Move SCON.1 to carry register.Stop the execution of program.

FLOWCHART:

START

Set CY flag, AND CY with MSB ofACC

Store the PSW format, OR CY with bit 2 IE reg

Clear bit 6 of PSW, Store PSW

Set bit 5 of SCON , clear bit 1 and storeSCON

Move bit 1 of SCON to CY and store PSW

STOP

67


PROGRAM TABLE

ADDRESS HEX CODE410090,45,00

4103

4105

4016

4018410A

410B

410C

410E

41104112

4113

41144116

4118411A

411B

411C

411E

4120

4122

74,FF

D3

82,EF

E5,D0F0

A3

72,AA

C2,D6

E5,D0F0

A3

D2,90C2,99

E5,98F0

A3

A2,99

E5,D0

F0

80,FE L2

LABEL MNEMONICS

MOV

MOV

SETB

ANL

MOVMOVX

INC

ORL

CLR

MOVMOVX

INC

SETBCLR

MOVMOVX

INC

MOV

MOV

MOVX

SJMP

OPERAND

DPTR,#4500

A,#FF

C

C, ACC.7

A,DOH@DPTR,A

DPTR

C, IE.2

PSW.6

A,DOH@DPTR,A

DPTR

SCON.5SCON.1

A,98H@DPTR,A

DPTR

C,SCON.1

A,DOH

DEPT OF ECE

COMMENT

InitializememorylocationGet the data inaccumulatorSet CY bit

Perform AND with 7thbit of accumulator

Store the result

Go to next location

OR CY bit with 2nd bitif IE regClear 6th bit of PSW

Store the result

Go to next location

Set 5th of SCON regClear 1st bit of SCONreg

Store the result

Go to next location

Copy 1st bit of SCONreg to CY flag

Store the result@DPTR,A

L2 Stop

68


RESULT:

DEPT OF ECE

MEMORYLOCATION

4500H (PSW)

SPECIAL FUNCTION REGISTER FORMAT

CY AC FO RS1 RS0 OV - P

BEFORE

00H

AFTER

88HEXECUTION EXECUTION

4501H (PSW) CY AC FO RS1 RS0 OV - P 40H 88H

4502H (SCON) SM0 SM1 SM2 REN TB8 RB8 TI RI 0FH 20H

4503H (PSW) CY AC FO RS1 RS0 OV - P FFH 09H

Thus the bit manipulation operation is done in 8051 microcontroller.

69


Square and Cube program, Find 2’s complement of a number

EXPT NO: 15

AIM:

DATE:

To convert Square and Cube program, Find 2’s complement of a number using 8051micro controller

RESOURCES REQUIERED: 8051 microcontroller kit Keyboard Power supply

PROGRAM:org 0000h; sets the program counter to 0000hmov a,#n;assign value 'n' in decimal to A which is converted to it'sequivalent hexadecimal valuemov b,#n;assign value 'n' in decimal to B which is converted to it'sequivalent hexadecimal valuemov r0,#n;assign value 'n' in decimal to R0 which is converted to it'sequivalent hexadecimal valuemul ab; multiplying 'A' with 'B'mov 40h,a; lower byte is stored in address 40hmov 41h,b; higher byte is stored in address 41hmov r1,a; move value of 'A' to R1mov a,b; move value of 'B' to 'A'mov b,r0; move value of R0 to bmul ab; multiply 'A' and 'B'mov b,a; lower byte obtained is moved from 'A' to 'B'mov r2,b; move value of 'B' to R2mov a,r1; move value of R1 to 'A'mov b,r0; move value of R0 to 'B'mul ab; multiplying 'A' and 'B'mov 50h,a; Lower byte obtained is stored in address 50h

70


mov r3,b; higher byte obtained is stored in R3mov a,r2; move value from R2 to 'A'add a,r3; add value of 'A' with R3 and store the value in 'A'mov b,a; move value from 'A' to 'B'mov 51h,b; store value obtained in address 51hend

SQUARE PGM USING 8051ORG 00h02LJMP MAIN

03 DELAY:04 ;MOV R0,#2

05 MOV TMOD, #01H06 MOV TH0, #HIGH (-50000)

07 MOV TL0, #LOW (-50000)08 SETB TR0

09 JNB TF0,10 CLR TF0

12 ;DJNZ R0,DELAY13 RET14 MAIN:

15 MOV DPTR,#300H16 MOV A,#0FFH

17 MOV P1,A18 BACK:

19 LCALL DELAY20 MOV A,P1

21 MOVC A,@A+DPTR22 ;MOV P2,#00H

23 ;LCALL DELAY24 MOV P2,A

25 SJMP BACK26 ORG 300H

27 XSQR_TABLE:28 DB 0,1,4,9,16,25,36,49,64,81

29 END Thus the Square and Cube program, Find 2’s complement of a number is done in8051 microcontroller

71


Unpacked BCD to ASCII

EXPT NO:16

AIM:To convert BCD number into ASCII by using 8051 micro controller

DATE:

RESOURCES REQUIERED: 8051 microcontroller kit Keyboard Power supply

72


FLOWCHART:

DEPT OF ECE

RESULT: The given number is converted into ASCII using 8051 microcontroller kit.

73

stannescet.ac.in · web view2019. 6. 13. · thus a given byte or word in a string of a...

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