MICROPROCESSORS LAB MANUAL ECE Dept. CONTENTS Page no. 1. List of Experiments (as per JNTU) -1, 2 2. Microprocessor 8086 Programming 2.1 Introduction to MASM and TASM - 5 2.2 Arithmetic operations a) Arithmetic Operations on 8 bit data - 8 b) Arithmetic Operations on 16 bit data - 14 c) Multibyte addition and subtraction - 20 d) Signed Operations on 8 bit data - 26 e) ASCII Arithmetic Operations - 31 2.3 Logic Operations a) BCD to ASCII Conversion - 37 b) ASCII to BCD Conversion - 41 c) No. of Positive no’s & Negative no’s. - 45 ADITYA ENGINEERING COLLEGE 1
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MICROPROCESSORS LAB MANUAL ECE Dept.
CONTENTS
Page no.
1. List of Experiments (as per JNTU) -1, 2
2. Microprocessor 8086 Programming
2.1 Introduction to MASM and TASM - 5
2.2 Arithmetic operationsa) Arithmetic Operations on 8 bit data - 8
b) Arithmetic Operations on 16 bit data -
14
c) Multibyte addition and subtraction -
20
d) Signed Operations on 8 bit data -
26
e) ASCII Arithmetic Operations -
31
2.3 Logic Operationsa) BCD to ASCII Conversion - 37
b) ASCII to BCD Conversion - 41
c) No. of Positive no’s & Negative no’s. - 45
d) No. of Odd no’s & Even no’s. - 51
e) Packed BCD to Unpacked BCD Conversion - 57
2.4 String Operationsa) Transfer Block of Data - 62
MICROPROCESSORS LAB MANUAL ECE Dept.LIST OF EXPERIMENTS
III Year B.Tech, ECE –II semester.II Year B.Tech, CSE & IT – II Semester
I MICROPROCESSOR 8086:
1. Introduction to MASM and TASM
2. Arithmetic Operations – Multi byte addition and subtraction, Multiplication and division-signed and unsigned Arithmetic operation, ASCII-arithmetic operation
3. Logic operations-Shift and rotate-Converting packed BCD to unpacked BCD, BCD to ASCII conversion.
4. By using string operation and Instruction prefix: Move Block, Reverse string, Sorting, Inserting. Deleting, length of the string, String comparison.
5. DOS/BIOS programming: Reading keyboard (with Buffered and without echo)-Display characters, Strings.
II. INTERFACING:
1. 8259-Interrupt controller : Generate an interrupt using 8259 timer.
2. 8279-Keyboard Display : Write a small program to display a string of Characters.
3. 8255-PPI : Write an ALP to generate sinusoidal wave Using PPI
4. 8251-USART : Write a program in ALP to establishcommunication between two processors
III. MICROCONTROLLER 8051:
1. Reading and Writing on a parallel port.
2. Timer in different modes.
3. Serial communication implementation.
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MICROPROCESSORS LAB MANUAL ECE Dept.LIST OF EXPERIMENTS
IV YEAR B.TECH EEE – I SEMESTERI MICROPROCESSOR 8086:
1. Introduction to MASM and TASM
2. Arithmetic Operations – Multi byte addition and subtraction, Multiplication and division-signed and unsigned Arithmetic operation, ASCII-arithmetic operation
3. Logic operations-Shift and rotate-Converting packed BCD to unpacked BCD, BCD to ASCII conversion.
4. By using string operation and Instruction prefix: Move Block, Reverse string, Sorting, Inserting. Deleting, length of the string, String comparison.
5. DOS/BIOS programming: Reading keyboard (with Buffered and without echo)-Display characters, Strings.
II. INTERFACING:
1. 8259-Interrupt controller : Generate an interrupt using 8259 timer.
2. 8279-Keyboard Display : Write a small program to display a string of Characters.
3. 8255-PPI : Write an ALP to generate sinusoidal wave Using PPI
4. 8251-USART : Write a program in ALP to establishcommunication between two processors
III. MICROCONTROLLER 8051:
1. Reading and Writing on a parallel port.
2. Timer in different modes.
3. Serial communication implementation.
4. Understanding three memory areas of 00-FF(Programs using above areas).
5. Using external Interrupts.
6. Programs using special instructions like swap, bit/byte, set/reset etc.
7. Programs based on short, page, absolute addressing.
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INTRODUCITION TO MASM/TASM
ASSEMBLY LANGUAGE PROGRAMMING USING MASM SOFTWARE:
This software used to write a program (8086, Pentium processors etc.)The programs are written using assembly language in editor then compile it. The complier converts assembly language statements into machine language statements/checks for errors. Then execute the compiled program. Programs for different processor instructions (Pentium, 8086) programming manner differ for each model.
There are different soft wares developed by different companies for assembly language programming are:
MASM - Microsoft Company. TASM - Bore Land Company.
MERIT OF MASM:
1. produces binary code2. Referring data items by their names rather than by their address.
HOW TO ENTER INTO MASM EDITOR:
Click “Start” on the desktop.
Then select Run
Then it Shows inbox
Then type Command (CMD) which enters You into DOS prompt
Path setting
Suppose it display path as C:\ DOCUME-\ADMIN>
Then type CD\
i.e.; C:\DOCUME\\ADMIN>CD\
Then the path is C :\>
Then type CD MASM
Then the path is C: MASM>
Then type edit i.e.; C: MASM>edit
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MICROPROCESSORS LAB MANUAL ECE Dept.Then you enter into MASM text editor.
Then enter to file and select New.
And name it and then write the ALP (Assembly Language Program) in this editor.After that save it as filename’sThen exit from the editor and go to prompt.Then type MASM filename.ASMI.e. C: MASM>MASM filename.ASM or C: MASM filename.ASM, , ;Then link this file using C: MASM>LINK filename.OBJ or C: MASM>LINK filename.OBJ , , ;i.e link the program in assembly with DOS then to debug to create exe fileC:MASM>debug filename. EXEThen it display “--” on the screenAfter that type ‘R’ displays the registers contents steps and starting step of the program.
‘T’ Tracing at contents of program step by step.Suppose you need to go for break point debugging. Then type that instruction no where you need to check your register. For example T10 it will display the contents of register after executing 10 instructions.
DEBUG:This command utility enables to write and modify simple assembly language programs in
an easy fashion. It provides away to run and test any program in a controlled environment.We can change any part of the program and immediately execute the program with an
having to resemble it. We can also run machine language(Object files) directly by using DEBUG
DEBUG COMMANDS:
ASSEMBLE A [address] ; Assembly the instructions at a particular address
COMPARE C range address ; Compare two memory ranges
DUMP D [range] ; Display contents of memory
ENTER E address [list] ; Enter new or modifies memory contents beginning at specific
Location
FILL F range list ; Fill in a range of memory
GO G [=address] [addresses] ; Execute a program in memory
HEX H value1 value2 ; Add and subtract two Hex values
INPUT I port
LOAD L [address] [drive] [first sector] [number]
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MICROPROCESSORS LAB MANUAL ECE Dept.MOVE M range address
5. What are the different addressing modes available in 8086 P?
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2.2 (C) MULTIBYTE ADDITIONS AND SUBTRACTION
ABSTRACT: Assembly language program to perform multibyte addition and subtraction
PORT USED: None
REGISTERS USED: AL, BX, CX
ALGORITHM:
Step1: Start
Step2: Initialize data segment
Step3: Load the CX register with count
Step4: Load the BX register with no: of bytes
Step5: Copy the contents from the memory location n1 [bx] to AL
Step6: Perform addition with second number n2 [bx]
Step7: Move the result to the memory location sum [bx]
Step8: Decrement BX
Step9: Decrement CX, if CX not equal to Zero jump to step5
Step10: Load CX register with count
Step11: Load the BX register with no: of bytes
Step12: Move the contents from memory location n1 [bx] to AL
Step13: Perform subtraction with second number n2 [bx]
Step14: Move the result to the memory location sum [bx]
Step15: Decrement BX
Step16: Decrement CX, if CX not equal to Zero jump to step12
Step17: Stop
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Start
Copy the contents from the memory location n1 [bx] to AL
Perform addition with second number n2 [bx]
Move the result to the memory location sum [bx]
Decrement BX
1
Initialize data segment
Get count into CX register
Load the BX register with no: of bytes
Decrement CX, &CX = 0
NO
YES
FLOW CHART
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1
Move the contents from memory location n1 [bx] to AL
Perform subtraction with second number n2 [bx]
Move the result to the memory location sum [bx]
Stop
Decrement BX
Decrement CX, &If CX = 0
Get count into CX register &Load BX register with no of
bytes
NO
YES
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PROGRAM:ASSUME CS: CODE, DS: DATADATA SEGMENTN1 DB 33H, 33H, 33HN2 DB 11H, 11H, 11HCOUNT EQU 0003HSUM DB 03H DUP (00)DIFF DB 03H DUP (00)DATA ENDSCODE SEGMENTORG 1000HSTART: MOV AX, DATA
MOV DS, AX
MOV CX, COUNT
MOV BX, 0002H
CLCBACK: MOV AL, N1 [BX]
ADC AL, N2 [BX]
MOV SUM [BX], AL
DEC BX
LOOP BACK
MOV CX, COUNT
MOV BX, 0002H
BACK1: MOV AL, N1 [BX]
SBB AL, N2 [BX]
MOV DIFF [BX], AL
DEC BX
LOOP BACK1
MOV AH, 4CH
INT 21H
CODE ENDSEND START
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CODE TABLE:
Physical address Label
Hex code Mnemonic Operand commentsSegment address
Effective address
0D64 1000
1003
1005
1008
100B
100F
1013
1017
1018
101A
101D
1020
1024
1028
102C
102D
102F
1031
L1
L2
B8 63 0D
8E D8
B9 03 00
BB 02 00
8A 87 00 00
12 87 00 00
88 87 06 00
4B
E2 F1
B9 03 00
BB 02 00
8A 87 00 00
1A 87 03 00
88 87 09 00
4B
E2 F1
B4 4C
CD 21
MOV AX,0D63
MOV DS,AX
MOV CX,0003
MOV BX,0002
MOV AL,[BX+0000]
ADC AL,[BX+0003]
MOV [BX+0006],AL
DEC BX
LOOP L1
MOV CX,0003
MOV BX,0002
MOV AL,[BX+0000]
SBB AL,[BX+0003]
MOV [BX+0009],AL
DEC BX
LOOP 1020
MOV AH,4C
INT 21
;Initialize the data segment
;Initialize the count CX with 0003;Move the value 02 in BX register;Copy the contents of BX+0000
;Perform ADC with[BX+0000]
;Move the result in BX+0006 location;Decrement the contents of BX
;Decrement the counter and go to L1 until CX is zero;Initialize the counter CX with 0003;Move the number 02 in BX register ;Copy the contents of BX+0000
;Perform SBB with [BX+0003]
;Move the result in BX+0009 location;Decrement the contents of BX
;Decrement the counter and go to L2 until CX is zero
1003 8E D8 MOV DS, AX1005 B0 13 MOV AL,13 ;Move the13 into
AL register.1007 B4 00 MOV AH, 00 ;Initialize the AH
register with 00001009 CD 10 INT 10 ;Set the keyboard
display mode100B BE 00 00 MOV SI, 0000 ;Initialize SI with
0000 location100E L2 8A 80 40
00MOV AL,
[SI+00];Copy the contents of SI into AL
1012 3C 21 CMP AL, 21 ;Compare the value 21 with AL register
1014 74 09 JZ L1 ;If equal to zero go to L1
1016 B4 0E MOV AH, 0E ;Load AH register with 0Eh
1018 B3 05 MOV BL, 05 ;Load BL register with 05h
101A CD 10 INT 10 ;Set the keyboard display mode
101C 46 INC SI ;Increment value of SI
101D EB FF JMP L2 ;Jump go to L2 with out condition
101F L1 B4 4C MOV AH, 4Ch ;Terminate the program
1021 CD 21 INT 21 ;Dos command interrupt
RESULT:
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3.1. PROGRAMMABLE PERIPHERAL INTERFACING(8255)
AIM: To study the functioning of programmable peripheral interfacing 8255. With port A & port B as out put port With port A as input and port B as output.
REGISTERS USED: AX, CX, DX.
PORTS USED: port A and port B.
CONNECTIONS: Study card; J1 of the study card adapter to I5 of ESA 8086/88.
DESCRIPTION:The 8255 is a general purpose programmable I/O device with 24 I/O
lines. These I/OLines are grouped as shown below:
Group A …. Port A (8 bits – PA0 to PA7) and Port C (4 bits – PC7 toPC4).
16 Group B …. Port B (8 bits – PB0 to PB7) and Port C (4 bits – PC3 toPC0).These groups can be operated in 3 different modes: mode 0, mode 1, and mode 2. InMode 0 the three ports, A, B, and C, may be programmed as input or output. In mode 1Ports A and B may be programmed as input or output, but port C is used to generateHandshake and interrupt signals. In mode 2, port A becomes a bi-directional port and 5I/O lines of port C are used for handshaking and interrupt signals. The pin-out of the8255 is given on the picture shown below.
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Following is the description of the inputs and outputs used for interfacing to the 8255:CS’ – A low on this input pin enables communication between the 8255 and the deviceControlling it.47RD’ – A low on this input pin enables the 8255 to send data or status information to theDevice controlling the 8255.WR’ – A low on this input pin enables the device controlling the 8255 to write data orControl words to the 8255.A0 and A1 – These input address lines, in conjunction with the RD’ and WR’ signals,allow the selection of one of three ports or the control register, as defined by the tablegiven below:A1 A0 Location0 0 Port A0 1 Port B1 0 Port C1 1 Control RegisterRESET – A high on this input pin clears the control register and all ports are set to inputMode.To program the 8255 PPI, one must write a control word to the control register. ThisControl word will define how the 8255 is going to behave. The table shown below definesthe value of each bit of the control word:Control word bit Function
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MICROPROCESSORS LAB MANUAL ECE Dept.D0 1 = Port C (lower) is input, 0 = Port C (lower) is outputD1 1 = Port A is input, 0 = Port A is outputD2 1 = Port B mode 1, 0 = Port B mode 0D3 1 = Port C (upper) is input, 0 = Port C (upper) is outputD4 1 = Port A is input, 0 = Port A is outputD5
D6
Port A00 = Mode 0, 01 = Mode 1, 1X = Mode 2D7 1 = Mode set flag active48Procedure
Write a program to:
1. Reset and initialize the 8255 to mode 0 with ports A, B, and C defined as outputs.
2. Create a menu of options to allow for the selection of which port is going to biased to simulate a three bits up-counter, and to quit the program.
3. When a port is selected from the menu, the screen should be cleared and message indicating which port is in use should be displayed.
The counting sequence should be displayed both on the screen and on the LED s associated withThe selected port until a key is pressed on the keyboard, at which time you start back at step 2.
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1. PORT A AND PORT B AS OUT PUT PORT:
ALGORITHM:
Step 1: start
Step 2: move the control word address 0FF46 to register DX
Step 3: move 80 to AL register.
Step 4: locate the contents in AL register to DX register using port out.
Step 5: move 55 to AL register.
Step 6: Initialize port A address.
Step 7: locate the contents in AL register to DX register using port out.
5016 EF OUT DX , AX Move the contents of AX into DX.
501A EB F1 JMP RPT Go to 500A location.
RESULT:
VIVA-VOCE QUESTIONS
72. How data can be transferred in programmed I/O?
73. What is the difference between machine language and assembly language?
74. Expand USART?
75. What are the signals required for memory read operation?
76. What are the signals required for memory write operation?
77. Why 8086 uses hexadecimal number system?
78. On which data stepper motor works?
79. How the stepper motor can be interfaced to 8086 microprocessor?
80. Write the syntax for 32-bit division operation?
81. Give an example for relative based indexed addressing mode?
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3.1 a) DIGITAL TO ANALOG CONVERTER
GENERATION OF WAVE FORMS:
AIM: program to generate the following wave forms: Triangular wave forms Saw tooth wave forms Square wave Sine wave
REGISTERS USED: general purpose registers: AL, DX, and CX
PORTS USED: out (port-B)
CONNECTION:J4 of ESA 86/88 to J1 DAC interface.
DESCRIPTIONS: As can be from the circuit only 17 lines from the connector are used totally. The port A and port B of 8255 programmable peripheral interface are used as output ports. The digital inputs to the DAC’s are provided through the port A and port B of 8255.the analog outputs of the DAC’s are connected to the inverting inputs of op-amps µA741 which acts as current to voltage converters. The out puts from the op- amps are connected to points marked Xout and Yout at which the wave forms are observed on a CRO. (port A is used to control Xout port B is used to control Yout).the difference voltage for the DAC’s is derived from an on-board voltage regulator µA 723 .it generates a voltage of about 8V.the offset balancing of the op-amps is done by making use of the two 10k pots provided. The output wave forms are observed at Xout and Yout on an oscillator.THEORY:BASIC DAC TECHNIQUE:
Vo = K VFS (d1 .2-1 + d2 . 2-2 + . . . . . . . .+dn . 2-n )Where d1 = MSB, d2 = LSBVFS = Full scale reading / out put voltageK --- Conversion factor is adjusted to ‘unity’.D/A converters consist of ‘n’ bit binary word ‘D’and is combined with a reference voltage VR to give an analog output. The out put can be either voltage or currentOut put voltage Vo = K VFS (d1 .2-1 + d2. 2-2 + . . . . . . . . +dn. 2-n)MSB weight = ½ VFS if d1 = 1 and all are zero’s, K = 1.LSB weight = VFS/2n if dn = 1 and all are zero’s, K = 1
DUAL DAC INTERFACE: This program generates a square wave or a Triangular wave at points Xout or Yout of
interface. The waveforms may be observed on an oscilloscope. This program can be executed in STAND-ALONE MODE or SERIAL MODE of operation. The program starts at memory location 3000H
ALGORITHM: (FOR TRIANGULAR WAVE):
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Step 1: Start
Step 2: Initialize the control word register
Step 3: Move 80 to AL register.
Step 4: Locate the contents in AL register to DX register using port out.
Step 5: Move 00FF to CX register.
Step 6: Move 00 to AL register.
Step 7: Initialize the port A address.
Step 8 : Locate the contents of AL to DX register.
Step 9: Increment the value in AL by one.
Step 10: Locate AL contents to DX register.
Step 11: Decrement the value of CX register by one and move to step 6 if CX not equal to zero.
Step 11: Other wise move 00FF to CX register.
Step 12: Decrement the value of AL by one.
Step 13: Locate the contents in AL register to DX register.
Step 14: Decrement the value of CX by one and move to step 11 if CX not equal to zero.
Step 15: Otherwise move to step 5.
Step 16: Stop.
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Start
Initialize the control word register
FLOW CHART
Copy the contents from AX to 80 &Locate the contents in AL to DX register using port
out Move 0FF to CX register and
1
Locate the contents of AX to DX register&
Increment the value in AX by one
Locate AX contents to DX register
Move 00 to AL register
Initialize port A address
Decrement CX & CX =0
NO
YES
2
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1
NO
YES
Move 0FF to CX register
Decrement the value of AL
Locate the contents in AL register to DX register
Stop
Decrement CX &CX=0
Jump
2
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PROGRAM (FOR TRIANGULAR WAVE):
MOV DX, 0FFE6
MOV AL, 80 OUT DX, AL
MOV DX, 0FFE0
MOV AX, 00
RPT: MOV CX, 0FF
L1: OUT DX, AX
INC AX
LOOP L1
MOV CX, 0FF
L2 : OUT DX,AX
DEC AX
LOOP L2
JMP RPT
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CODE TABLE (FOR TRIANGULAR WAVE):
Physical address Label Hex code Mnemonic
Op code operands
Comments
Segment address
Effective address
0000 2000 BA E6 FF MOV DX , FFE6
Move FFE6 to DX register
2003 B8 80 00 MOV AX, 0080 Load AX with 80
2006 EF OUT DX , AX Move DX content to AX
2007 BA E0 FF MOV DX , FFE0 Move FFE0 to DX register
5007 BA E0 FF MOV DX , FFE0 Move FFE0 to DX register
500A Rpt B8 FF 00 MOV AX , 00FF Load AX with 00FF
500D EF OUT DX,AX Move AX content to DX
500E E8 FF EF CALL 6000
5011 B8 00 00 MOV AX , 0000 Load AX with 0000
5014 EF OUT DX , AX Move the contents of AX into DX
5015 E8 E8 FF CALL 6000
5018 E8 F0 JMP RPT Jump to 500A location
DELAY PROGRAM:
LOCATION 0000:6000
Physical addressBase adder offset adder
Label Hex codes
Mnemonics Op code Operand
comment
0000 6000 B9 1E 00 MOV CX,001E Move 001E into CX register
6003 NOP
6004 NOP
6005 C3 RET Return to main program
RESULT:
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ALGORITHM (FOR SINE WAVE):
Step 1: Start
Step 2: Move the control word address 0FFE6 to register DX
Step 3: Move 80 to AL register.
Step 4: Locate the contents in AL register to DX register using port out.
Step 5: Initialize the SI with 4000h location.
Step 6: Move the number 46 into the counter CX register.
Step 7: Initialize the Port A address ie. 0FFE0
Step 8: Move the contents SI into AL register.
Step 9: Locate the contents in AL register to DX register using port out..
Step 10: Increment the value of SI.
Step 11: Decrement the counter and go to Step 8 until CX =0
Step 12: Jump to location step 5.
Step 13: Stop.
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PROGRAM FOR SINE WAVE:
MOV DX, 0FFE6
MOV AL, 80
OUT DX, AL
RPT: MOV SI, 4000
MOV CL, 46
MOV DX, 0FFE0
L1: MOV AL, [SI]
OUT DX, AL
INC SI
LOOP L1
JMP RPT
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CODE TABLE:
Base address
Effective address
Label Hex code Mnemonics Operands Comments
0000 2000 BA E6 FF MOV DX, 0FFE6 ;Initialize control word register
2003 B8 80 00 MOV AL, 80 ;Load AL with 802006 EF OUT DX, AL ;Read the data from
AL register using port out
2007 RPT BE 0040 MOV SI, 4000 ;Initialize the SI with 4000 ie,LOOK-UP TABLE
200A 8B 46 00 MOV CX , 46 ;Initialize the counter with 46h
200D BA E0 FF MOV DX, 0FFE0 Initialize port A
2010 L1 B8 C6 MOV AL, [SI] ;move the number in [SI] into al register
2012 EF OUT DX, AL ;read the data from Al register using port out
2013 46 INC SI ;Increment the value of SI
2014 E2 F5 LOOP L1 ;Decrement counter and go to L1 until CX=0
2016 EB EA JMP RPT ;Jump go to rpt with out condition
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LOOK – UP TABLE:
EFFECTIVE ADDRESS
HEX CODES
4000 7F 8A 95 A0
4004AA B5 BF C8
4008D1 D9 E0 E7
400CED F2 F7 FA
4010FC FE FF FE
4014FC FA F7 F2
4018ED E7 E0 D9
401CD1 C8 BF B5
4020AA A0 95 8A
40247F 74 69 5F
402853 49 3F 36
402C2D 25 1D 17
403010 0B 07 04
403401 00 01 04
403807 0B 10 17
403C
1D 25 2D 364040
3F 49 53 5F4044
69 74
RESULT:
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VIVA-VOCE QUESTIONS
82. How does the 8086 microprocessor instruction set is classified?
83. Which operation is performed by the Negate instruction?
84. What is the difference between RET and IRET instruction?
85. If AH=00, AL=0F then what is the content of AH and AL after AAA instruction?
86. What is the function of SAHF instruction?
87. What is the difference between IN and OUT instructions?
88. Which flags are affected after CMP (compare) instruction?
89. If AL=71, AH=81 then what is the content of AL after DAA instruction?
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3.2. KEYBOARD INTERFACING
AIM: To display a string through interfacing 8279.
REGISTERS USED: AX, BX, CX, DX, SI
PORTS USED: command port, data port
CONNECTIONS: 8279 study card P1 to J2 of ESA 86/88 study card adapter.
DESCRIPTION:8279 study card provides keyboard as well as display section. The display section features
size 8 digit seven segment displays while the keyboard connections comprises a 4x 4 matrix hex key pad and associated circuitry. The options for using shift and controls keys during key scanning are also provided. The interface has 3 connectors, P1, J1, & J4 to interface the card with ESA 86/88E trainers. Connect the 50 pin FRC connectors P1 to connector J2 to ESA 86/88E study card adapter.
INTRODUCTION:In many microprocessors-based systems, calculator keypad is used as an input device. A
calculator keypad can be interfaced to a microprocessor using a dedicated peripheral controller like INTEL 8279Akeyboard/display controller. In this case, the controller can handle the interface problems like key debounce, 2-key lock-out, N-key roll-over etc,. Further such an alternative approach, the calculator keypad interface is passive and software is used for encoding the key positions and for handling problems like key debounce, roll-over etc.
The present interface module provides a calculator style calculator keypad consisting of the key 0 to 9 , + ,- , ×,= ,% , . , C, CE and two spare keys. These 20 keys are arranged in a 3×8 matrix (the third row has only four keys). The row lines can be driven through port C and the status of column lines can be read through port A. this interface allows the user to study a number of techniques generally used in calculator keypad interfacing. User can write programs for software debouncing of key closures, two key understanding of keyboard interface. Further , user can become familiar with the arithmetic group of processor instructions by implementing the calculator functions like addition, subtraction, multiplication , diversion, percentage etc..
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THEORY: A programmable keyboard and display interfacing chip.
o Scans and encodes up to a 64-key keyboard. o Controls up to a 16-digit numerical display.
Keyboard has a built-in FIFO 8 character buffer
The display is controlled from an internal 16x8 RAM that stores the coded display information
PIN OUT DEFINITION 8279
A0: Selects data (0) or control/status (1) for reads and writes between micro and 8279. BD: Output that blanks the displays. CLK: Used internally for timing. Max is 3 MHz. CN/ST: Control/strobe, connected to the control key on the keyboard. CS: Chip select that enables programming, reading the keyboard, etc. DB7-DB0: Consists of bidirectional pins that connect to data bus on micro. IRQ: Interrupt request, becomes 1 when a key is pressed, data is available.
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MICROPROCESSORS LAB MANUAL ECE Dept. OUT A3-A0/B3-B0: Outputs that sends data to the most significant/least significant nibble
of display. RD (WR): Connects to micro’s IORC or RD signal, reads data/status registers. RESET: Connects to system RESET. RL7-RL0: Return lines are inputs used to sense key depression in the keyboard matrix. Shift: Shift connects to Shift key on keyboard. SL3-SL0: Scan line outputs scan both the keyboard and displays.
8279 Interfaced to the 8088
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Keyboard Interface of 8279
KEYBOARD INTERFACE OF 8279:
The keyboard matrix can be any size from 2x2 to 8x8.
Pins SL2-SL0 sequentially scan each column through a counting operation. o The 74LS138 drives 0’s on one line at a time. o The 8279 scans RL pins synchronously with the scan. o RL pins incorporate internal pull-ups, no need for external resistor pull-ups.
Unlike the 82C55, the 8279 must be programmed first.
D7 D6 D5 Function Purpose
0 0 0 Mode set Selects the number of display positions, type of key scan…
0 1 0 Read FIFO Selects type of FIFO read and address of the read.
0 1 1 Read Display Selects type of display read and address of the read.
1 0 0 Write Display Selects type of write and the address of the write.
1 0 1 Display write inhibit Allows half-bytes to be blanked.
1 1 0 Clear Clears the display or FIFO
1 1 1 End interrupt Clears the IRQ signal to the microprocessor.
o The first 3 bits of # sent to control port selects one of 8 control words.
Keyboard Interface of 8279
First three bits given below select one of 8 control registers (opcode).
000DDMMM o Mode set: Opcode 000.
DD sets displays mode. MMM sets keyboard mode.
o DD field selects either: 8- or 16-digit display Whether new data are entered to the rightmost or leftmost display position.
DD Function
00 8-digit display with left entry
01 16-digit display with left entry
10 8-digit display with right entry
11 16-digit display with right entry
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Keyboard Interface of 8279
o MMM field:
DD Function
000 Encoded keyboard with 2-key lockout
001 Decoded keyboard with 2-key lockout
010 Encoded keyboard with N-key rollover
011 Decoded keyboard with N-key rollover
100 Encoded sensor matrix
101 Decoded sensor matrix
110 Strobed keyboard, encoded display scan
111 Strobed keyboard, decoded display scan
o Encoded: SL outputs are active-high, follow binary bit pattern 0-7 or 0-15. o Decoded: SL outputs are active-low (only one low at any time).
Pattern output: 1110, 1101, 1011, 0111. o Strobed: An active high pulse on the CN/ST input pin strobes data from the RL pins
into an internal FIFO for reading by micro later.
o 2-key lockout/N-key rollover: Prevents 2 keys from being recognized if pressed simultaneously/Accepts all keys pressed from 1st to last.
INTERFACE OF 8279
001PPPPP o The clock command word programs the internal clock driver. o The code PPPPP divides the clock input pin (CLK) to achieve the desired operating
frequency, e.g. 100KHz requires 01010 for a 1 MHz CLK input.
010Z0AAA o The read FIFO control word selects the address (AAA) of a keystroke from the
FIFO buffer (000 to 111). o Z selects auto-increment for the address.
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MICROPROCESSORS LAB MANUAL ECE Dept. 011ZAAAA
o The display read control word selects the read address of one of the display RAM positions for reading through the data port.
100ZAAAA o Selects write address – Z selects auto-increment so subsequent writes go to
subsequent display positions.
INTERFACE OF 8279
1010WWBB o The display write inhibit control word inhibits writing to either the leftmost 4 bits of
the display (left W) or rightmost 4 bits. o BB works similarly except that they blank (turn off) half of the output pins.
1100CCFA o The clear control word clears the display, FIFO or both o Bit F clears FIFO and the display RAM status, and sets address pointer to 000.
If CC are 00 or 01, all display RAM locations become 00000000. If CC is 10, 00100000, if CC is 11, 11111111.
1110E000 o End of Interrupt control word is issued to clear IRQ pin in sensor matrix mode.
1) Clock must be programmed first. If 3.0 MHz drives CLK input, PPPPP is programmed to 30 or 11110.
INTERFACE OF 8279
2) Keyboard type is programmed next. o The previous example illustrates an encoded keyboard, external decoder used to
drive matrix.
3) Program the FIFO.
Once done, a procedure is needed to read data from the keyboard. o To determine if a character has been typed, the FIFO status register is checked.
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MICROPROCESSORS LAB MANUAL ECE Dept.o When this control port is addressed by the IN instruction, the contents of the FIFO
status word is copied into register AL:
INTERFACE OF 8279
Code given in text for reading keyboard.
Data returned from 8279 contains raw data that need to be translated to ASCII:
o Row and column number are given the rightmost 6 bits (scan/return).
o This can be converted to ASCII using the XLAT instruction with an ASCII code lookup table.
o The CT and SH indicate whether the control or shift keys were pressed.
o The Strobed Keyboard code is just the state of the RLx bits at the time a 1 was `strobed’ on the strobe input pin.
ALGORITHM:
Step1: Move the value 00 to BX register.
Step2: Move the value 00 to AX register.
Step3: Move the 0FF42 to DX & out AX to DX.
Step4: Move the 90 to AL , and out value AL to DX.
Step5: Initialize counters CX with the value 08h
Step6: Move the 0ff40 to DX register and Load AL with 00
Step7: Write the value in AL register into DX register using port OUT.
Step8: Go to step 6 until equal to zero.
Step9: Move the address 2050 into SI register.
Step10: Move the value 0FF42 to DX register.
Step11: Read the value in DX register into AL register using port IN.
Step12: Add the contents of AL register with the value 07h.
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MICROPROCESSORS LAB MANUAL ECE Dept.Step 13: Go to step 8 if the result not equal to zero.
Step14: Get the value 40 to AL register.
Step15: Write the value in AL register into DX register using port OUT.
Step 16: move the 0ff40 to DX register.
Step 17: read the value in DX register into AL register using port IN.
Step 18: Logical AND ed the contents of AL with 01F.
Step19: Move the value of AL into BL register.
Step 20: Add the contents of SI with BX register.
Step21: Initialize data port and move the value 94 into AL register.
Step22: Write the data in AL register into DX register using port OUT.
Step 23: Initialize Command port and move the value F3 into AL register.
Step24: Write the value in AL register into DX register using port OUT.
Step25: Initialize the data port and move the value 95 into AL register.
Step 26: Write the value in AL register into DX register using port OUT.
Step 27: Initialize the Command port.
Step 28: Move the contents of SI into AL register.
Step29: Write the value in AL register into DX register using port OUT.
2003 BA 42 FF MOV DX,0FF42 Initialize control port
2006 B0 00 MOV AL,00 Load 00 with Al register
2008 EE OUT DX,AL Routine to clear
2009 B0 90 MOV AL, 90 All display LEDs
200B EE OUT DX,AL
200C B9 08 00 MOV CX,08 Set the count
200F B0 00 MOV AL,00 Load 00 with AL register
2011 BA 40 FF MOV DX,0FF40 Initialize data port
2014 RPT EE OUT DX,AL Read the data from port
2015 E2 FD LOOP RPT Continuous loop
2017 BACK
BE 50 20 MOV SI, 2050 Initialize SI with 2050
201A BA 42 FF MOV DX,0FF42 Read 8279 status to check if any character is available
201D KEY EC IN AL,DX Write the data from port IN
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201E 24 07 AND AL,07 Logical AND between [AL] and 07h
2020 74 FB JZ KEY Jump if condition
2022 B0 40 MOV AL,40 Load AL register with 40
2024 EE OUT DX,AL Read the data from port
2025 BA 40 FF MOV DX,0FF40 Read FIFO RAM
2028 EC IN AL,DX Mask SIFT
2029 24 1F AND AL,1F AND AL with 1F
202B 8A D8 MOV BL,AL Move the contents of AL into BL
202D 03 F3 ADD SI,BX Add SI with BX
202F BA 42 FF MOV DX,0FF42 Address to display 0
2032 B0 94 MOV AL,94 Load AL register with 94
2034 EE OUT DX,AL Read the data from port
2035 BA 40 FF
MOV DX,0FF40 Addressed control port
2038 B0 F3 MOV AL,F3 Move the value 0F3 into AL register
203A EE OUT DX,AL Read the data from port
203B BA 42 FF MOV DX,0FF42 Addressed data port
203E B0 95 MOV AL,95 Set 95 into AL register
2040 EE OUT DX,AL Read the data from port
2041 BA 40 FF MOV DX,0FF40 Initialize control port
2044 8A 04 MOV AL,[SI] Move the contents of SI into the AL register
2046 EE OUT DX,AL Read the data from port
2047 E9 CD FF JMP BACK Unconditional jump
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INPUT DATA:
Segment address
Offset address
DATA INPUT
2050 F3 60 B5 F4
2054 66 D6 D7 70
2058 F7 76 77 C7
205C 93 E5 97 17
RESULT:
VIVA-VOCE QUESTIONS
90. What is the difference between opcode and operand?
91. What is the function of instruction pointer (IP) register?
92. What does assembler mean?
93. What is the difference between RCL and ROL instructions?
94. Which signal is used to restart the 8086 microprocessor?
95. What is the power supply required for 8086 microprocessor?
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ADDITIONAL PROGARMS
DUAL SLOPE ANALOG TO DIGITAL CONVERTERS
AIM: To Convert digital data into Analog voltage.
INTRODUCTION:The analog to digital conversion can be done in many ways. One of the methods is the Dual
Slope method, used to achieve high noise immunity.The input voltage is integrated for a fixed time T1. Also a known reference voltage VR is
integrated for a time TX. Now the input voltage VX is given by
But =K (a constant)
Hence VX = K . TX I.e. the input voltage is proportional to the measured time TX. If integrated time TC is chosen as 20 m sec.the microprocessor measures time as the number of counts, namelyTX = NX. TC where
NX is number of counts. TX is counter period.By proper choice of the counter period TC scaling can be incorporated into the counter such that a software multiplication can be avoided.
THEORY: Voltage,current,temperature,pressure,time,etc, are available in analog form.ot is difficult to
process ,store or transmit the analog signal with out introducing considerable error because of the superimposition of noise as in the case of amplitude modulation. Therefore, for processing, transmission and storage purpose. It is often convenient to express these variables in digital form, it gives better accuracy system is based.
The operation of any digital communication system is based upon analog to digital and digital to analog conversion.
A/D converter requires sample and hold(S/H) circuit. The ADC output is a sequence of binary digit.D/A converter is to convert digital signal to analog signal and function of DAC is exactly opposite to that of ADC. The D/A converter is usually operated at the same frequency as the ADC.the output of D/A converter is commonly a stair case. This stair case like digital output is passed through a smoothing filter to reduce the effect of quantization noise.
ADC’s are classified broadly into two groups according to their conversion technique:1. Direct type ADC’s and2. Integrating type ADC’s.
Direct type ADC’s compare a given analog signal with the internally generated equivalent signal. This group includes:1. Flash (comparator) type converter2. Counter type converter.
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MICROPROCESSORS LAB MANUAL ECE Dept.3. Tracking or servo converter4. Successive approximation type converter.
Integral type ADC’s perform conversion in an indirect manner by first changing the analog input signal to a linear function of time or frequency and then to a digital code .the two most widely used integrating type converters are:1. Charge balancing ADC.2. Dual slope ADC
The flash (comparator) type is expensive for high of accuracy the integrated type converter is used in application such as digital meter, panel meter and so on.
PARALLEL COMPARATOR (FLASH) A/D CONVERTER:It is simplest possible converter and the fastest and most expensive technique.
3-BIT A/D CONVERTER:Circuit consists of resistive divider network, 8 op-amp comparators and 8-5line encoder. At each node of the resistive divider, a comparison voltage is available .since all the resistors are of equal value, the voltage levels available at the nodes are equally divided between the reference voltage VR and the ground.
The purpose of circuit is to compare the analog input voltage VA with each of the voltages.The circuit has the advantage of high speed as the conversion take place
simultaneously rather than sequentially. Typical conversion time is loons or less. Conversion time is limited only by the speed of the comparator and of the priority encoder by using advanced micro devices AMA 686A comparator and a T1147 priority encoder ,conversion delays of the order of 20 no’s can be obtained.
This type of ADC has the disadvantage that the number of comparators required almost doubles for each added bit. A 2-bit ADC requires 3 comparators .in general, the number of comparators required are 2n-1 where n is the desired number of bits. Hence the number of comparators approximately doubles for each added bit. Also the larger the value of n the more complex is the priority encoder.
INTEGRATING TYPE OF ADC’S: the generating type of ADC’s do not require a S/H circuit at the input. If the input changes during conversion, the ADC’s output code will be proportional to the value of the input averaged over the integration period.
DUAL SLOPE ADC:
DUAL SLOPE OR DUAL RAMP CONVERTERS:
The analog part of the circuit consists of a high input impedance buffer A1, precision integrator A2 and a voltage comparator. The converter fort integrates the analog input signal VA for a fixed duration of 2n clock periods. Then to integrate an internal reference voltage VR of opposite polarity until the integrator output is zero. The number N of clock cycles required to return the integrated period. Hence N represents the desired output code. Since VR and n are constant, the analog voltage VA is proportional to the count reading N and is independent of R,C and T.
The dual-slope ADC integrates the input signal for a fixed time, hence to provides excellent noise rejection of ac signals whose periods are integral multiples of the integration time T1.
The main disadvantage of the dual-slope ADC is the long conversion time. For instance, if 2n-T =1 / 50 is used to reject line pick-up , the conversion time will be 20 m sec.
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MICROPROCESSORS LAB MANUAL ECE Dept.Dual-slope converters are particularly suitable for accurate measurement of slowly varying
signal, such as thermo couples and weighing scales. Dual slope ADC’s also form the basics of digital panel meters and millimeters.
Dual slope converters are available in monolithic form and are available both in microprocessors compatible and in display oriented versions.
REGISTERS USED: AX, BX, CX, DX.
PORTS USED: Port A and Port B.
ALGORITHM:
Step1: Initialize 8255 Control Word Register and Port A as O/P port and Port B as I/P port.
Step2: Load the number 02 in AL register.
Step3: Initialize Port A address and read the data from port out.
Step4: Perform NOP five times.
Step5: Move the number 01h into AL register.
Step6: Read the data from port out.
Step7: Load CX register with 1000h.
Step8: Continue the loop if CX is not equal to zero.
Step9: Load AL with 04 and read the data from port out.
Step10: Initialized port A with 0FFE0 and initialized BX register with 0000h.
Step11: Initialized port B with 0FFE2 and write the data from register AL using port IN.
Step12: Logical AND operation AL register with 01h value.
Step13: Jump the corresponding condition to specify in instruction.
Step14: After the increment the BL register jump step 11.
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MICROPROCESSORS LAB MANUAL ECE Dept.Step15: Move the contents of BL register into the AL register.
Step16: Initialize SI with 3000h and move the value in AL register into the SI location.
Step 17: Load AL register with 02h
Step 18: Initialize port A and read the data from port out.
Step19: Get the value 03h into BX register.
Step20: Initialize CX with 0FFFFh and continue the loop until CX is equal to zero.
Step21: Decrement the value of BX register and jump step 20 if BX equal to zero.
Step22: Jump step 2 Unconditional
Step23: Stop.
PROGRAM:MOV DX, 0FFE6
MOV AL, 82
OUT DX, AL
START: MOV AL, 02H
MOV DX, 0FFE0
OUT DX, AL
NOP
NOP
NOP
NOP
NOP
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MICROPROCESSORS LAB MANUAL ECE Dept.MOV AX, 01
OUT DX, AL
MOV CX, 1000H
L2: LOOP L1
MOV AL, 04H
OUT DX, AL
MOV DX, 0FFE0
MOV BL, 00H
L1: MOV DX, 0FFE2
IN AX, DX
AND AL, 01
JE DISP
INC BL
JMP L1
DISP: MOV AL, BL
MOV SI, 3000H
MOV [SI], AL
MOV AL, 02H
MOV DX, 0FFE0
OUT DX, AL
MOV BX, 20
DELAY: MOV CX, 0FFFF
$: LOOP $
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JMP START
CODE TABLE:
Physical address Label Hex code
Mnemonic
Op code operands
Comments
Segment address
Effective address
0000 4000 BA E6 FF MOV DX,0FFE6 Initialize the Control word
register4003 B0 82 MOV AL,82 Port A as output &
port B as input4004 EE OUT DX,AL
4006 START B0 02 MOV AL,02 Reset integrator
4008 BA E0 FF MOV DX,0FFE0 Addressed port A ie initializes.
400B EE OUT DX,AL Read the data from port out
400C 90 NOPNo operation
4011 90 NOP
4014 90 NOP
4015 90 NOP
4018 90 NOP
401A B8 01 00 MOV AX,01 Load AX with 01h
401C EE OUT DX,AL Read the data
401D B9 00 10 MOV CX, 1000H Set the count with 1000h
4020 L1 E2 FE LOOP L1 Continuous loop
4022 B0 04 MOV AL,04H Load AL with 04number
4025 EE OUT DX,AL Read the data
4026 BA E0 FF MOV DX,0FFE0 Initialized port A addressed
4028 B3 00 MOV BL,00 Clear BL register
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402A L2 BA E2 FF MOV DX,0FFE2 Initialize port B
402C EC IN AL,DX Write the data from port IN
402E 24 01 AND AL, 01H AND ed operation between AL and 01
4030 74 04 JE DISP Jump conditionally4033 FE C3 INC BL Increment value of
BL4035 EB F4 JMP L2 Jump
Unconditionally4037 DSP 8A C3 MOV AL,BL Move the contents
from AL into BL403A B8 00 30 MOV SI,3000 Initialize SI with
3000h403B MOV [SI],AL Move the value in
AL register into contents of SI
403E B0 02 MOV AL,02 Load AL with the value 02h
4041 BA E0 FF MOV DX,0FFE0 Initialize port A
4043 EE OUT DX,AL Read the data from port out
4044 BB 20 00 MOV BX,20 Set the 20h in BX register
4046 DY B9 FF FF MOV CX,0FFFH Set the count with 0FFFFh value into
CX register4049 L3 F2 EE LOOP L3 Continuous loop
404B 4B DEC BX Decrement the value of BX
register404C 75 F8 JNZ DY Jump
Conditionally404E EB AD JMP START Jump
Unconditionally.
RESULT:
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STEPPER MOTOR INTERFACING
AIM: program to design a stepper motor to rotate shaft of a 4 phase stepper motor in clockwise 15 rotations.
REGISTERS USED: General purpose registers: AL , DX , CX
PORTS USED: Port B, port C (out)
CONNECTIONS: J4 of ESA 86/88E to J1 of stepper motor.OPERATING PRINCIPLE OF PERMANENT MAGNET STEPPER MOTOR:
It consists of two stator windings A,B and a motor having two magnetic poles N and S. when a voltage +v is applied to stator winding A, a magnetic field Fa is generated. The rotor positions itself such that its poles lock with corresponding stator poles.
With the winding ‘A’ excited as before ,winding ‘b’ is now to Fa. the resulting magnetic field F makes an angle of 450. the rotor consequently moves through 450 in anti clockwise direction,again to cause locking of rotor poles with corresponding stator poles.
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MICROPROCESSORS LAB MANUAL ECE Dept.While winding ‘B’ has voltage +V applied to it, winding ‘A’ is switched off. The rotor then
moves through a further 450 in anti-clockwise direction to aligne itself with stator field Fb. with voltage +V on winding B, a voltage –V is applied to winding A. then the stator magnetic field has two components Fa , Fb and their resultant F makes an angle of 1350 position.
In this way it can be seen that ,as the pattern of excitation of the state of winding is changed, the rotor moves successively through450 steps. And completes one full revolution in anti clock-wise direction. A practical PM stepper motor will have 1.80 step angle and 50 tooth on it’s rotor;there are 8 main poles on the stator, each having five tooth in the pole face. The step angle is given by
A = 360 / (N * K) degrees
Where N = number of rotor tooth.
K = execution sequence factor.
PM stepper motors have three modes of excitation i,e..
Single phase mode
Two phase mode
Hybrid mode
Single phase mode: in this mode only one of the motor winding is excited at a time. There are four steps in the sequence, the excitation sequence factor K=2 ,so that step angle is 900.
Two phase mode: Here both stators phase are excited at a time. There are four steps in the excitation sequence, K = 2 and the step angle is 900. However, the rotor positions in the two phase mode are 450 way from those in single phase mode.
Hybrid mode: this is a combination of single and two phase modes. There are 8 steps in excitation sequence=2 and step angle = 450. a voltage +V is applied to a stator winding during some steps, which voltage V is applied during certain other steps. This requires a bipolar regulated power supply capable of yielding +V,-V and zero outputs and a air of SPDT switches, which is quite cumbersome. Consequently each of the two stator windings is split into two sections A1-A2 , B1-B2. these sections are wound differentially. These winding sections can now be excited from a univocal regulated power supply through switcher S1 to S4. this type of construction is called bipolar winding construction. Bipolar windingesults in reduced winding inductance and consequently improved torque stepping rate.Description: the stepper motor interfaces uses four transistor pairs (SL 100 and 2N 3055) in a Darlington pair configuration. Each Darlington pair is used to excite the particular winding of the motor connected to 4 pin connector on the interface. The inputs to these transistors are from the 8255 PPI I/O lines of the microprocessor kit or from digital I/O card plugged in the PC. “port A” lower nibble PA0 , PA1, PA2 , PA3 are the four lines brought out to the 26 pin FRC male connector(J1) on the interface module. The freewheeling diodes across each winding protect transistors from switching transients.Theory:
A motor used for moving things in small increments is known as stepper motor. Stepper motor rotate from one fixed position to next position rather than continuous rotation as in case of other ac or dc motor stepper motors are used in printers to advance the paper from one position to advance the paper from one position to another in steps. They are also used to position the read/write head on the desired track of a floppy disk. To rotate the shaft the stepper motor a
ADITYA ENGINEERING COLLEGE 152
MICROPROCESSORS LAB MANUAL ECE Dept.sequence of pulses are applied to the windings in a predefined sequence. The number of pulses required for one complete rotation per pulse is given by 3600/NT. where “NT” is the number of teeth on rotot. Generally the stepper motor is available with 10 to 300 rotation. They are available with two phase and four phase common field connections.
Instead of rotating smoothly around and around as most motors, stepper motors rotate or step one fixed position to next. Common step size range from 0.90 to 300. it is stepped from one position to next by changing the currents through the fields in the motor.
The two common field connections are referred to as two phase and four phase. The drive circuitry is simpler in 4 phase stepper. The figure shows a circuitry that can interface a small 4 stepper motor to four microcomputer port lines.
The 7406 buffers are inverting , so. A high on ah output port pin turns on current to a winding. The purpose of clamp diodes across each winging is to save transistors from inductive kick. Resistors R1 and R2 are current limiting resistors.Typical parameters of stepper motor:1. Operating voltage - 12 volts2. Current rating - 1.2 Amp3. Step angle - 1.80
4. Step for revolution - 200(No. of teeth on rotor)5. Torque - 3 kg/cmWorking of stepper motor:
Suppose that SW1 and SW2 are turned ON. Turning OFF SW2 and turning ON SW4 cause the motor to rotate one step of 1.80 clockwise. Changing to SW4 and SW3 ON will cause the motor to rotate 1.80 clockwise another. Changing SW3 and SW2 ON will cause another step. To step the motor in counter clock wise direction simply work through the switch sequence in the reverse direction.
The switch pattern for changing from one step to another step in clockwise direction is simply rotated right one position. For counter clockwise direction rotated left one position.
ALGORITHM:
Step 1: Start
Step 2: move the control word address 0FFE6 to register DX
Step 3: move 80 to AL register.
Step 4: locate the contents in AL register to DX register using port out.
Step 5: move port A address ie.,,0FFE0 to DX register.
Step 6: move 11 to AL register.
Step 7: locate the contents in AL register to DX register using port out.
Step 8: move 300 to CX register.
Step 9: repeat step 8 until the content in CX register becomes equal to zero.
Step 10: Rotate carry left through bit.
Step 11: jump to location / step 7.
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Step 12: stop.
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Start
Initialize the control word register
FLOW CHART
Copy the contents from AX to 80 &Locate the contents in AL to DX register using port
out
Move the data 11h into AL register &Locate the contents in AL to DX register using port
out
Initialize port A address
1
Load CX register with 300
2
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1
NO YES
Stop
Decrement CX &CX=0
Jump
2
Rotate carry left through bit
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PROGRAM:
MOV DX, 0FFE6
MOV AX, 80
RPT: OUT DX, AX
MOV DX, 0FFE0
MOV AX, 0011
RPT: OUT DX, AX
MOV CX, 0300
L1: LOOP L1
RCL AL, 01H
JMP RPT
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CODE TABLE:
Physical address Label Hex code Mnemonic
Op code operands
Comments
Segment address
Effective address
0000 5000 BA E6 FF Mov DX , FFE6 Move FFE6 to DX register
5003 B8 80 00 Mov AX, 0080 Load AX with 80
5006 EF Out DX , AX Move AX content to DX
5007 BA E0 FF Mov DX , FFE0 Move FFE0 to DX register
500A Rpt B8 FF 00 Mov AX , 00FF Load AX with 00FF
500D EF Out DX,AX Move AX content to DX
500E E8 FF EF Call 6000
5011 B8 00 00 Mov AX , 0000 Load AX with 0000
5014 EF Out DX , AX Move the contents of AX into DX