anna university lab manual

EXP NO: 1

D A T E :

8086 PROGRAMMING

ADDITION & SUBTRACTION

AIM: To write an Assembly Language Program (ALP) for performing the Addition and subtraction operation of 16-bit numbers. APPARATUS REQUIRED:

SL.N O

ITEM SPECIFICATION QUANTITY

1. Microprocessor kit 2. Power Supply

8086 kit +5 V dc

1 1

ALGORITHM:

16-bit addition i. Initialize the MSBs of sum to 0ii) Get the first number. iii) Add the second number to the first number. iv) If there is any carry, increment MSBs of sum by 1. v. Store LSBs of sum. vi) Store MSBs of sum.

16-bit subtraction

i. Initialize the MSBs of difference to 0 ii) Get the first number iii) Subtract the second number from the first number. iv) If there is any borrow, increment MSBs of difference by 1. v. Store LSBs of difference vi) Store MSBs of difference.

1

FLOWCHART:

2

ADDITION PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg ADD AX,BX Add the contents of both the regs AX &

JNC L1 Check for carry

INC CX If carry exists, increment the CX L1 : MOV [1206],CX Store the carry

MOV [1204], AX Store the sum

HLT Stop the program SUBTRACTION

PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg

SUB AX,BX Subtract the contents of BX from AX

JNC L1 Check for borrow

INC CX If borrow exists, increment the CX L1 : MOV [1206],CX Store the borrow

MOV [1204], AX Store the difference

HLT Stop the program

3

3

OUTPUT:

MEMORY

DATA

MEMORY

DATA

RESULT:

ADDITION

SUBTRACTION

Thus addition & subtraction of two byte numbers are performed and the result is stored.

4

EXPT NO: 2

DATE:

MULTIPLICATION & DIVISION

AIM: To write an Assembly Language Program (ALP) for performing the multiplication and division operation of 16-bit numbers .

APPARATUS REQUIRED: SL.N

O ITEM SPECIFICATION QUANTITY


8086 +5 V dc

1 1

ALGORITHM:

Multiplication of 16-bit numbers:

i. Get the multiplier. ii) Get the multiplicand iii) Initialize the product to 0. iv) Product= product + multiplicand v. Decrement the multiplier by 1 vi) If multiplicand is not equal to 0,repeat from step (d) otherwise store the product.

Division of 16-bit numbers.

i. Get the dividend ii) Get the divisor iii) Initialize the quotient to 0. iv) Dividend = dividend divisor v. If the divisor is greater, store the quotient. Go to step g. vi) If dividend is greater, quotient = quotient + 1. Repeat from step (d) vii) Store the dividend value as remainder.

5

FLOWCHART:

6

MULTIPLICATION PROGRAM COMMENTS

MOV AX,[1200] Get the first data MOV BX, [1202]

MUL BX

MOV [1206],AX

MOV AX,DX

MOV [1208],AX

HLT

Get the second data

Multiply both

Store the lower order product

Copy the higher order product to AX

Store the higher order product

Stop the program DIVISION

PROGRAM COMMENTS MOV AX,[1200] Get the first data lower nibble MOV DX, [1202]

MOV BX, [1204]

DIV BX

MOV [1206],AX

MOV AX,DX

MOV [1208],AX

HLT

Get the first data higher nibble

Get the second data

Divide the dividend by divisor

Store the lower order product

Copy the higher order product to AX

Store the higher order product

Stop the program

7 7

OUTPUT:

MULTIPLICATION

MEMORY

DATA

DIVISON

MEMORY

DATA

RESULT:

Thus multiplication & division of two byte numbers are performed and the result is stored.

8

EXPT NO: 3

DATE:

ASCENDING & DESCENDING

AIM: To write an Assembly Language Program (ALP) to sort a given array in

Ascending and descending order. APPARATUS REQUIRED:

SL.N O

ITEM SPECIFICATION QUANTITY


8086 +5 V dc

1 1

ALGORITHM:

Sorting in ascending order:

i. Load the array count in two registers C ii) Get the first two numbers. 01 and C 2.iii) Compare the numbers and exchange if necessary so that the two numbers are in ascending order. iv) Decrement C 2. v. Get the third number from the array and repeat the process until C 2 is 0. vi) Decrement C1 and repeat the process until C 1 is 0.

Sorting in descending order: i. Load the array count in two registers C ii) Get the first two numbers. 01 and C 2.iii) Compare the numbers and exchange if necessary so that the two numbers are in descending order. iv) Decrement C2. v. Get the third number from the array and repeat the process until C 2 is 0. vi) Decrement C 01 and repeat the process until C 1 is 0.

9

FLOWCHART

10

ASCENDING MOV SI,1200H Initialize memory location for array size MOV CL,[SI] Number of comparisons in CL

L4 : MOV SI,1200H Initialize memory location for array size MOV DL,[SI] Get the count in DL

INC SI Go to next memory location MOV AL,[SI] Get the first data in AL L3 : INC SI Go to next memory location

MOV BL,[SI] Get the second data in BL CMP AL,BL Compare two data s

JNB L1 If AL < BL go to L1 DEC SI Else, Decrement the memory location

MOV [SI],AL Store the smallest data MOV AL,BL Get the next data AL

JMP L2 Jump to L2 L1 : DEC SI Decrement the memory location

MOV [SI],BL Store the greatest data in memory location L2 : INC SI Go to next memory location

DEC DL Decrement the count JNZ L3 Jump to L3, if the count is not reached zero

MOV [SI],AL Store data in memory location DEC CL Decrement the count JNZ L4 Jump to L4, if the count is not reached zero

HLT Stop DESCENDING

PROGRAM COMMENTS

MOV SI,1200H Initialize memory location for array size MOV CL,[SI] Number of comparisons in CL

L4 : MOV SI,1200H Initialize memory location for array size MOV DL,[SI] Get the count in DL

INC SI Go to next memory location MOV AL,[SI] Get the first data in AL L3 : INC SI Go to next memory location

MOV BL,[SI] Get the second data in BL CMP AL,BL Compare two data s

11

JB L1 If AL > BL go to L1

DEC SI Else, Decrement the memory location

MOV [SI],AL Store the largest data MOV AL,BL Get the next data AL

JMP L2 Jump to L2 L1 : DEC SI Decrement the memory location

MOV [SI],BL Store the smallest data in memory location L2 : INC SI Go to next memory location

DEC DL Decrement the count JNZ L3 Jump to L3, if the count is not reached zero

MOV [SI],AL Store data in memory location DEC CL Decrement the count JNZ L4 Jump to L4, if the count is not reached zero

HLT Stop

ASCENDING

MEMORY

DATA

MEMORY

DATA

DESCENDING

RESULT:

Thus given array of numbers are sorted in ascending & descending order12

EX NO:4

AIM: To write an Assembly Language Program (ALP) to find the largest and smallest number in a given array.




8086 +5 V dc

1 1

ALGORITHM:

(i) Finding largest number: a. Load the array count in a register C bc. Compare the numbers and exchange if the number is small. d. Get the third number from the array and repeat the process until C

(ii) Finding smallest number: e. Load the array count in a register C1. f. Get the first two numbers.

g. Compare the

h. Get the third number from the array and repeat the process until C1 is 0.

13

FLOWCHART:

14

LARGEST

PROGRAM COMMENTS MOV SI,1200H Initialize array size MOV CL,[SI] Initialize the count

INC SI Go to next memory location MOV AL,[SI] Move the first data in AL

DEC CL Reduce the count L2 : INC SI Move the SI pointer to next data

CMP AL,[SI] Compare two data s JNB L1 If AL > [SI] then go to L1 ( no swap)

MOV AL,[SI] Else move the large number to AL L1 : DEC CL Decrement the count

JNZ L2 If count is not zero go to L2 MOV DI,1300H Initialize DI with 1300H MOV [DI],AL Else store the biggest number in 1300 location

HLT Stop SMALLEST

PROGRAM COMMENTS

MOV SI,1200H Initialize array size MOV CL,[SI] Initialize the count

INC SI Go to next memory location MOV AL,[SI] Move the first data in AL

DEC CL Reduce the count L2 : INC SI Move the SI pointer to next data

CMP AL,[SI] Compare two data s JB L1 If AL < [SI] then go to L1 ( no swap)

MOV AL,[SI] Else move the large number to AL L1 : DEC CL Decrement the count

JNZ L2 If count is not zero go to L2 MOV DI,1300H Initialize DI with 1300H MOV [DI],AL Else store the biggest number in 1300 location

HLT Stop

15

1

LARGEST

MEMORY DATA

MEMORY

DATA

SMALLEST

RESULT:

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

16

EXPT NO: 5 DATE: 8086 PROGRAMMING




8086 +5 V dc

1 1

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.

17

FLOWCHART:

18

COPYING A STRING PROGRAM COMMENTS

MOV SI,1200H

MOV DI,1300H

MOV CX,0006H

CLD

REP MOVSB

HLT

MEMORY

Initialize starting address

Initialize destination address

Initialize array size

Clear direction flag

Copy the contents of source into destination until count reaches zero Stop

INPUT

DATA

MEMORY

DATA

RESULT:

OUTPUT

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

19

EXPT NO: 6 SEARCHING A STRING

DATE:

AIM:To scan for a given byte in the string and find the relative address of the bytefrom the starting 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.

19

20

FLOWCHART :

21

MOV DI,1300H

MOV SI, 1400H

MOV CX, 0006H

CLD

MOV AL, 08H

REPNE SCASB

DEC DI

MOV BL,[DI]

MOV [SI],BL

HLT

OUTPUT:


Initialize starting address



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 INPUT

MEMORY

DATA

OUTPUT

MEMORY LOCATION DATA

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

22 EXNO: 7 8086 PROGRAMMING FIND AND REPLACE

DATE:

AIM:

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

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 scannedbyte in BL register. f. Scan for the byte in ES. If the byte is not found, ZF?1 and repeatscanning. g. If the byte is found, ZF=1.Move the content of BL register to

ES:DI.

21

FLOWCHART:

24

FIND AND REPLACE A CHARACTER IN THE STRING PROGRAM COMMENTS

MOV DI,1300H

MOV SI,1400H


Initialize starting address MOV CX, 0006H

CLD

MOV AL, 08H

MOV BH,30H

REPNE SCASB

DEC DI

MOV BL,[DI]

MOV [SI],BL

MOV [DI],BH

HLT



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 INPUT

MEMORY DATA

MEMORY

DATA

RESULT:

OUTPUT

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.

25 EXNO: 8 8086 INTERFACING

INTERFACING ANALOG TO DIGITAL CONVERTER DATE: AIM:

To write an assembly language program to convert an analog signal into a digital signal using an ADC interfacing.

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY1. Microprocessor kit 2. Power Supply 3. ADC Interface board

8086 +5 V dc,+12 V dc -

1 1 1

THEORY: An ADC usually has two additional control lines: the SOC input to tell theADC when to start the conversion and the EOC output to announce when theconversion is complete. The following program initiates the conversionprocess, checks the EOC pin of ADC 0809 as to whether the conversion is overand then inputs the data to the processor. It also instructs the processor to storethe 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.

26

FLOW CHART:

27

PROGRAM TABLE PROGRAM

MOV AL,00

COMMENTS

Load accumulator with value for ALE high OUT 0D0H,AL

MOV AL,00 MOV AL,00 MOV AL,00 MOV AL,00

OUT 0D0H,AL

Send through output port

Introduce delay Store the value to make SOC low the accumulator

Send through output port

L1 : IN AL, 0D8H AND AL,01 CMP AL,01

JNZ L1

Read the EOC signal from port & check for end of conversion

If the conversion is not yet completed, read EOC signal

from port again

IN AL,0C0H Read data from port ANALOG DIGITAL DATA ON LED HEX CODE IN MEMORY

VOLTAGE DISPLAY LOCATION

RESULT:

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

28 EXNO: 9 8086 INTERFACING

INTERFACING DIGITAL TO ANALOG CONVERTER DATE: AIM:

1. To write an assembly language program for digital to analog conversion 2. To convert digital inputs into analog outputs & To generatedifferent waveforms.

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY1. Microprocessor kit 2. Power Supply 3.

DAC Interface board

8086 Vi Microsystems +5 V, dc,+12 V dc -

1 1 1

THEORY:

Since DAC 0800 is an 8 bit DAC and the output voltage variation is between 5v and +5v. The output voltage varies in steps of 10/256 = 0.04 (approximately). The digital data input and the corresponding output voltages are presented in the table. Outputting digital data 00 and FF at regular intervals, to DAC2, results in a square wave of amplitude 5v.Output digital data from 00 to FF in constant steps of 01 to DAC2. Repeat this sequence again and again. As a result a saw-tooth wave will be generated at DAC2 output. Output digital data from 00 to FF in constant steps of 01 to DAC2. Output digital data from FF to00 in constant steps of 01 to DAC2. Repeat this sequence again and again. As a result 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.

29

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. (vi) 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)

30

FLOWCHART:

31

PROGRAM COMMENTS

MOV AL,7FH OUT C0,AL HLT

Load digital value 00 in accumulator Send through output port Stop

DIGITAL DATA ANALOG VOLTAGE

PROGRAM TABLE: Square Wave PROGRAM COMMENTS

PROGRAM TABLE: Saw tooth Wave PROGRAM COMMENTS

L2 : MOV AL,00H L1 : OUT C0,AL INC AL JNZ L1 JMP L2

Load 00 in accumulator Send through output port Increment contents of accumulator Send through output port until it reaches FF Go to starting location

27

32

PROGRAM TABLE: Triangular Wave PROGRAM COMMENTS

L3 : MOV AL,00H L1 : OUT C0,AL INC AL JNZ L1 MOV AL,0FFH L2 : OUT C0,AL DEC AL JNZ L2 JMP L3

WAVEFORM GENERATION:

Load 00 in accumulator Send through output port Increment contents of accumulator Send through output port until it reaches FF Load FF in accumulator Send through output port Decrement contents of accumulator Send through output port until it reaches 00 Go to starting location

WAVEFORMS AMPLITUDE TIMEPERIOD Square Waveform Saw-tooth waveform Triangular waveform

MODEL GRAPH: Square Waveform

33

Saw-tooth waveform

Triangular waveform

RESULT:

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

34 EXNO: 10 STEPPER MOTOR INTERFACING

DATE:

AIM:

To write an assembly language program in 8086 to rotate the motor at differentspeeds.

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY1. Microprocessor kit808612. Power Supply+5 V, dc,+12 V dc 13. Stepper Motor Interface board- 14. Stepper Motor - 1

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 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 theabove procedure both for backward and forward directions.

SWITCHING SEQUENCE OF STEPPER MOTOR: MEMORY LOCATION 4500 4501 4502 4503

A1 A2 B1 B2 HEX CODE

01 0 0 0 09 H 00 1 0 1 05 H 10 1 1 0 06 H 01 0 1 0 0A H

35

FLOWCHART:

36

PROGRAM TABLE PROGRAM START : MOV DI, 1200H MOV CX, 0004H

COMMENTS Initialize memory location to store the array ofnumber Initialize array size

LOOP 1 : MOV AL,[DI] Copy the first data in AL OUT 0C0,AL Send it through port address MOV DX, 1010H L1 : DEC DX JNZ L1 INC DI

Introduce delay Go to next memory location

LOOP LOOP1 Loop until all the data s have been sent JMP START Go to start location for continuous rotation 1200 : 09,05,06,0A Array of data s

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

37 EXNO: 11 INTERFACING PROGRAMMABLE KEYBOARD AND

DISPLAY CONTROLLER- 8279 DATE:

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

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 00 00 01 00 00 00 0 0

DD

0 0 D D

K K

K

00- 8Bit character display left entry 1. - 16Bit character display left entry 2. 8Bit character display right entry 3. 16Bit character display right entry KKK- Key Board Mode 000-2Key lockout.

38

2.Clear Display: Control word-DC H 1 1

1 1

0 0

1 CD

1 CD

1 CD

0

CF

0 CA

11 A0-3; B0-3 =FF 1-Enables Clear display 0-Contents of RAM will be displayed

1-FIFO Status is cleared 1-Clear all bits (Combined effect of CD)

39

3. Write Display: Control word-90H 1 0 0 1 0 0 0 0

1 0 0 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:

32

40

PROGRAM TABLE: PROGRAM COMMENTS

START : MOV SI,1200H MOV CX,000FH MOV AL,10 OUT C2,AL MOV AL,CC OUT C2,AL MOV AL,90 OUT C2,AL L1 : MOV AL,[SI] OUT C0,AL CALL DELAY INC SI LOOP L1 JMP START DELAY : MOV DX,0A0FFH LOOP1 : DEC DX JNZ LOOP1 RET

LOOK-UP TABLE:

1200 98 68 7C C8

1204 FF 1C 29 FF

Initialize array Initialize array size Store the control word for display mode Send through output port Store the control word to clear display Send through output port Store the control word to write display Send through output port Get the first data Send through output port Give delay Go & get next data Loop until all the data s have been taken Go to starting location Store 16bit count value Decrement count value Loop until count values becomes zero Return to main program

41

OUTPUT: E 7-SEGMENT

LED FORMAT

HEX DATA

1200H 1201H 1202H 1203H 1204H 1205H 1206H 1207H

01 0 0 1 1 0 0 0 98 00 1 1 0 1 0 0 0 68 10 1 1 1 1 1 0 0 7C 01 1 0 0 1 0 0 0 C8 11 1 1 1 1 1 1 1 FF 00 0 0 0 1 1 0 0 1C 10 0 1 0 1 0 0 1 29 01 1 1 1 1 1 1 1 FF

RESULT:

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

42 EXNO: 12 INTERFACING PROGRAMMABLE TIMER-8253

DATE: A IM:

To study different modes of operation of programmable timer 8253.

THEORY: The main features of the timer are,

i. Three independent 16-bit counters, ii. Input clock from DC to 2 MHz,

iii. Programmable counter modes, iv. Count binary or BCD. The control signals with which the 8253 interfaces with the CPU are CS, RD, WR, A1, A2.The basic operations performed by 8253 are determined bythese control signals. It has six different modes of operation, viz, mode 0 to mode 5.

MODE 2 RATE GENERATOR It is a simple divide - by N counter. The output will be low for one input clock period. The period from one output pulse to the next equals the numberof input counts in the count register. If the count register is reloaded between output pulses, the present period will not be affected, but the subsequent period will reflect the new value.

MODE 3 SQUARE WAVE GENERATOR It is similar to mode 2, except that the output will remain high until one half for even number count, If the count is odd, the output will be high for (count+1)/2 counts and low for (count-1)/2 counts

ALGORITHM:

Mode 2- 1. Initialize channel 0 in mode 2 2. Initialize the LSB of the count. 3. Initialize the MSB of the count. 4. Trigger the count. 5. Read the corresponding output in CRO.

43

34 Mode 3- 1. Initialize channel 0 in mode 3 2. Initialize the LSB of the count. 3. Initialize the MSB of the count. 4. Trigger the count 5. Read the corresponding output in CRO.

PORT ADDRESS: 1. CONTROL REGISTER2. COUNTER OF CHANNEL 0 - 3. COUNTER OF CHANNEL 1 - 4. COUNTER OF CHANNEL 2 - 5. O/P PORT OF CHANNEL 0 - 6. O/P PORT OF CHANNEL 1 - 7. O/P PORT OF CHANNEL 2 -

CONTROL WORD FORMAT:

D7 D6 D5 D4 D3 D2 D1 D0 SC1 SC0 RL1 RL0 M2 M1 M0 BCD

0 0 1 1 00 1 0 0 Mode 2 = 34 H

00 01 1 0 1 1 0 Mode 3 = 36 H

SC1 SC0 CHANNEL SELECT RL1 RL0 READ/LOAD 0 0 1 1

0 1 0 1

CHANNEL 0 CHANNEL 1 CHANNEL 2 -----

0 0 1 1

0 1 0 1

LATCH LSB MSB LSB FIRST, MSB NEXT

BCD --0 BINARY COUNTER

M2 M1 M0 MODE

01 --BCD COUNTER

0 0 0 0 1 1

0 0 1 1 0 0

0 1 0 1 0 1

MODE 0

MODE 1

MODE 2

MODE 3

MODE 4

MODE 5

PORT PIN ARRANGEMENT DEBOUNCE CIRCUIT CONNECTION

1 CLK 0 2 CLK 1 3 CLK 2

44

FLOW CHART:

45

04 OUT 0

05 OUT 1

06 OUT 2

07 GATE 0

08 GATE 1

09 GATE 2

10 GND

MODE 2 RATE GENERATOR: PROGRAM COMMENTS

MOV AL, 34H OUT 0BH MOV AL, 0AH OUT 08H MOV AL, 00H OUT 08H HLT

Store the control word in accumulator Send through output port Copy lower order count value in accumulator Send through output port Copy higher order count value in accumulator Send through output port Stop

MODE 3 SQUARE WAVE GENERATOR: PROGRAM COMMENTS

MOV AL, 36H OUT 0BH MOV AL, 0AH OUT 08H MOV AL, 00H OUT 08H HLT

Store the control word in accumulator Send through output port Copy lower order count value in accumulator Send through output port Copy higher order count value in accumulator Send through output port Stop

6 46

MODEL GRAPH: RATE GENERATOR SQUARE WAVE GENERATOR

RESULT: Thus an ALP for rate generator and square wave generator are written and executed.

47

EXNO: 13 DATE:

AIM:

INTERFACING USART 8251

To study interfacing technique of 8251 (USART) with microprocessor 8086 and an 8086 ALP to transmit and receive data between two serial ports with RS232cable.

APPARATUS REQUIRED:

8086 kit (2 Nos), RS232 cable.

THEORY: The 8251 is used as a peripheral device for serial communication and is programmed 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 them into a continuous serial data stream for transmission. Simultaneously, it can receive serial data streams and convert them into parallel data characters for the CPU. The CPU can read the status of the USART at any time. These include data transmission errors and control signals. The control signals define the complete functional definition of the 8251. Control words should be written into the control register of 8251.These control words are split into two formats: 1) Mode instruction word & 2) Command instruction word. Status word format is used to examine the error during functional operation.

1...transmit enable 1...data terminal ready 1... receive enable1... send break character

1.... reset error flags (pe,oe,fe) 1..... request to send (rts)

1...... internal reset 1....... enter hunt mode (enable search for sync characters)

48

01 ransmitter ready1. receiver ready 1.. transmitter empty 1... parity error (pe)

1.... overrun error (oe) 1..... framing error (fe), async only

1...... sync detect, sync only 1....... data set ready (dsr)

ALGORITHM:

1. Initialize 8253 and 8251 to check the transmission and reception of a character 2. Initialize8253 to give an output of 150Khz at channel 0 which will give a9600 baud rate of 8251. 3. The command word and mode word is written to the 8251 to set up forsubsequent operations 4. The status word is read from the 8251 on completion of a serial I/O

49

operation, or when the host CPU is checking the status of the device beforestarting the next I/O operation.

FLOW CHART:

50

PROGRAM: TRANSMITTER END PROGRAM COMMENTS

MOV AL,36 OUT CE,AL MOV AL,10 OUT C8,AL MOV AL,00 OUT C8,AL MOV AL,4E OUT C2,AL MOV AL,37 OUT C2,AL

Initialize 8253 in mode 3 square wave generator Send through port address Initialize AL with lower value of count (clock frequency 150KHz) Send through port address Initialize AL with higher value of count Send through port address Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit) Send through port address Set command instruction(enables transmit enable & receive enable bits) Send through port address

L1:IN AL,C2 Read status word AND AL,04 JZ L1 MOV AL,41 OUT C0,AL INT 2

Check whether transmitter ready If not wait until transmitter becomes ready Set the data as 41 Send through port address Restart the system

RECEIVER END PROGRAM COMMENTS

MOV AL,36 OUT CE,AL MOV AL,10 OUT C8,AL MOV AL,00 OUT C8,AL MOV AL,4E OUT C2,AL MOV AL,37 OUT C2,AL

Initialize 8253 in mode 3 square wave generator Send through port address Initialize AL with lower value of count (clock frequency 150KHz) Send through port address Initialize AL with higher value of count Send through port address Set mode for 8251(8bit data, No parity, baud rate factor 16x & 1 stop bit) Send through port address Set command instruction(enables transmit enable & receive enable bits) Send through port address

L1:IN AL,C2 Read status word MOV BX,1500 Initialize BX register with memory location to store the data MOV [BX],AL Store the data in the memory location INT 2 Restart the system

RESULT:

Thus ALP for serial data communication using USART 8251 is written and the equivalent ASCII 41 for character A is been transmitted & received.

51 EXNO: 14 INTERFACING PPI 8255

DATE:

AIM:

To write ALP by interfacing 8255 with 8086 in mode 0, mode 1 and mode 2.

APPARATUS REQUIRED:

8086 kit, 8255 interface kit.

ALGORITHM:

Mode 0

1. Initialize accumulator to hold control word 2. store control word in control word register 3. Read data port A. 4. Store data from port A in memory 5. Place contents in port B.

Mode 1 & Mode 2

1. Initialize accumulator to hold control word (for port A) 2. Store control word in control word register 3. Initialize accumulator to hold control word (for port B) 4. Place contents in control word register. 5. Disable all maskable interrupts, enable RST 5.5 6. send interrupt mask for RST 6.5 & 7.5 7. Enable interrupt flag 8. Read data from port A, place contents in port B.

52

FLOWCHART:

53

MODE 0 PROGRAM COMMENTS

MOV AL,90H OUT C6,AL IN AL,C0 OUT C2,AL HLT

MODE 1 PROGRAM

MOV AL,0B0H OUT C6,AL MOV AL,09H OUT C6,AL MOV AL,13H OUT 30,AL MOV AL,0AH OUT 32,AL MOV AL,0FH OUT 32,AL MOV AL,00H OUT 32,AL STI HLT ISR: IN AL,C0 OUT C2,AL HLT MODE 2

PROGRAM MOV AL,0C0H OUT C6,AL MOV AL,09H OUT C6,AL MOV AL,13H OUT 30,AL MOV AL,0AH OUT 32,AL

Set the control word Send it to control port Get the contents of port A in AL Send the contents of port B to port address Stop

COMMENTS Set the control word for mode 1 Send it to control port Control for BSR mode Send it to control port Interrupt generation Through 8259 Using IR2 interrupt(lower order count) Higher order count Set trap flag Stop Subroutine Read from Port A Send it to Port B Stop

COMMENTS Set the control word for mode 2 Send it to control port Control for BSR mode Send it to control port Interrupt generation Through 8259

54

MOV AL,0FH OUT 32,AL MOV AL,00H OUT 32,AL STI HLT ISR: IN AL,C0 OUT C2,AL HLT BSR mode

Using IR2 interrupt(lower order count) Higher order count Set trap flag Stop Subroutine Read from Port A Send it to Port B Stop

Bit set/reset, applicable to PC only. One bit is S/R at a time. Control word: D7 00 (0=BSR)

D6 D5 D4 D3 D2 D1 D0 X X X B2 B1 B0 S/R (1=S,0=R)

Bit select: (Taking Don't care's as 0) B2 B1 B0 PC bit Control word (Set) Control word (reset) 00 0 0 0 0000 0001 = 01h 10 0 1 1 0000 0011 = 03h 00 1 0 2 0000 0101 = 05h 00 1 1 3 0000 0111 = 07h 01 0 0 4 0000 1001 = 09h 01 0 1 5 0000 1011 = 0Bh 11 1 0 6 0000 1101 = 0Dh 01 1 1 7 0000 1111 = 0Fh

0000 0000 = 00h 0000 0010 = 02h 0000 0100 = 04h 0000 0110 = 06h 0000 1000 = 08h 0000 1010 = 0Ah 0000 1100 = 0Ch 0000 1110 = 0Eh

I/O mode D7 D6 D5 D4 D3 D2 D1 D0 01 (1=I/O)

GA mode select PA PCU GB mode

select PB PCL

0 D6, D5: GA mode select: o 00 = mode0 o

01 = mode1 o1X = mode2

0 D4(PA), D3(PCU): 1=input 0=output1 D2: GB mode select: 0=mode0, 1=mode12 D1(PB), D0(PCL): 1=input 0=output

Result:Mode 0 Mode 1 Mode 2

Input OutputInput OutputInput Output

The programs for interfacing 8255 with 8085 are executed & the output isobtained for modes 0,1 & 2.

55 EXNO: 15 8051 PROGRAMMING

8 BIT ADDITION DATE:

AIM:

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.

with

t5. Add these two 8-bit numbers. 6. Store the result. 7. Stop the program.

56

FLOW CHART:

57

08 Bit Addition (Immediate Addressing)ODE C

OMMENTS

4100 CLR C C3 Clear CY Flag 4101

4103

4105

MOV A,# data1

ADDC A. # data 2

MOV DPTR, # 4500H

74,data1 Get the data1 in Accumulator

24,data2 Add the data1 with

data2 90,45,00 Initialize the memory

location 4108

Store the result in

MOVX 0@ DPTR, A F0 memory location 4109 L1 SJMP L1 80,FE Stop the program

OUTPUT:

OUTPUT MEMORY LOCATION DATA

4500 RESULT:

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

58 EXNO: 16 8051 PROGRAMMING 8 BIT SUBTRACTION

DATE: AIM:

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.

59

FLOWCHART: -

60

08 Bit Subtraction (Immediate Addressing)ADDRESS LABEL MNEMONIC

OPERAND H COMMENTS

4100 CLR C

C3 Clear CY flag

4101

4103

MOV

SUBB

A. # data1

A. # data2

74. data1 Store data1 in accumulator

94,data2 Subtract data2 from

data1

4105

4108

MOV

MOVX

DPTR, # 4500 90,45,00 Initialize memory

location Store the difference

0@ DPTR, A

F0 in memory location

4109 L1 SJMP L1 80,FE Stop OUTPUT:

OUTPUT RESULT:

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

61

EXNO: 17 AIM:

To perform division 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.

62

FLOWCHART:

63

08 Bit MultiplicationADDRESS LABEL MNEMONIC

OPERAND H COMMENTS

4100 MOV A ,#data1

74. data1 Store data1 in accumulator

4102 MOV B, #data2 75,data2 Store data2 in B reg4104 MUL A,B F5,F0 Multiply both 4106

4109

401A

410B

410D

MOV

MOVX

INC

MOV

MOV

DPTR, # 4500H 0@ DPTR, A

DPTR

A,B

0@ DPTR, A

90,45,00 Initialize memory location Store lower order

F0 result

A3 Go to next memory location

E5,F0

Store higherorder result F0

410E STOP SJMP STOP 80,FE Stop OUTPUT:

INPUT OUTPUTMEMORY LOCATION DATA

4500

MEMORY LOCATION

4502

DATA

RESULT:

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

64 EXNO: 18 8051 PROGRAMMING 8 BIT DIVISION

DATE:

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.

65

FLOWCHART:

66

08 Bit DivisionADDRESS LABEL MNEMONIC

OPERAND HEX CODE

COMMENTS

4100

MOV A. # data1 74,data1 Store data1 in

accumulator

4102 MOV B, # data2 75,data2 Store data2 in B reg4104 DIV A,B 84 Divide 4015

MOV DPTR, # 4500H 90,45,00

Initialize memory location

4018

4109

410A

410C

410D

MOVX

INC

MOV

MOV

STOP SJMP

0@ DPTR, A

DPTR

A,B

0@ DPTR, A

STOP

F0

A3

E5,F0

F0

80,FE

Store remainder Go to

next memory location

Store quotient Stop

OUTPUT:

INPUT OUTPUTMEMORY LOCATION DATA MEMORY LOCATION DATA4500 (dividend) 4502 (remainder)

4501 (divisor) 4503 (quotient) RESULT:

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

67 EXNO: 19 8051 PROGRAMMING

LOGICAL AND BIT MANIPULATION DATE:

AIM:

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

microcontroller.

APPARATUS REQUIRED:

8051 microcontroller kit

ALGORITHM:

1. Initialize content of accumulator as FFH 2. Set carry flag (cy = 1). 3. AND bit 7 of accumulator with cy and store PSW format. 4. OR bit 6 of PSW and store the PSW format. 5. Set bit 5 of SCON. 6. Clear bit 1 of SCON. 7. Move SCON.1 to carry register. 8. Stop the execution of program.

68

FLOWCHART:

69

PROGRAM TABLE: A LABEL MNEMONICS OPERAND

COMMENT 4100

90,45,0

0 Initialize memory

MOV DPTR,#4500 location 4103

74,FF

Get the data in

MOV A,#FF accumulator4105

4016

D3

82,EF

SETB

ANL

C

C. ACC.7

Set CY bit Perform AND with 7th

bit of accumulator

4018E5,D0 MOV A,DOH410A

F0

Store the result

MOVX @DPTR,A 410B

410C

A3

72,AA

INC

ORL

DPTR

C. IE.2

Go to next location OR CY bit with 2nd bit

if IE reg

410E C2,D6 CLR PSW.6 Clear 6th bit of PSW4110 E5,D0 MOV A,DOH

MOVX @DPTR,A 4113 4114 4116

A3D2,90C2,99

INC SETB CLR

DPTR SCON.5

SCON.1

Go to next location Set 5th of SCON reg

Clear 1st bit of SCON reg

4118E5,98 MOV A,98H411A F0 MOVX @DPTR,A Store the result411B

411C

411E

4120

4122

A3

A2,99

E5,D0

F0

80,FE

L2

INC

MOV

MOV

MOVX

SJMP

70 DPTR

C,SCON.1

A,DOH

@DPTR,A

L2

Go tonextlocation Co

py 1st bit of SCON

regto

CY

flag

Store the result

Stop

OUTPUT: MEMORY BEFORE AFTER

LOCATION SPECIAL FUNCTION REGISTER FORMAT

EXECUTION EXECUTION

4500H (PSW) CY AC FO RS1 RS0 OV - P 00H

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

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

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

88H 88H 20H 09H

RESULT:

Thus the bit manipulation operation is done in 8051 microcontroller.

71

anna university lab manual

Documents