microprocessors and microcontrollers module 1 - 5[1]

7/28/2019 Microprocessors and Microcontrollers Module 1 - 5[1]

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33 Days Just an analysis about

MICROPROCESSORS AND

MICROCONTROLLERS

(Designed as per Anna University Syllabus)

B.E. IV Semester CSE

MODULE I

By

Y. SYED SHA MUHAMED M.E., Ph.D*.

Associate Professor

Department of Computer Science and Engineering,

National College of Engineering, Maruthakulam

Tirunelveli.


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PREFACE

Dear Reader,

This book is designed for IV Sem B.E. CSE of Anna University Syllabus. It has 5

module. Every module is one unit. It is also helpful for MCA Students (Module 2, 3, 4) for

Unit 1, 2 & 3. It is also designed for dedication purpose to help the student community. If a

student (especially college first rank) who is not affordable to pay (or) a student who makes

the laudable comment for the improvement of this book. He will be given book for free of

cost.

Hope you might be one among them with regards.

Author.


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This book is dedicated especially for student

community who are my jewels

By

Y. SYED SHA MUHAMED


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SYLLABUS

UNITI

THE 8085 AND 8086 MICROPROCESSORS:

8085Microprocessors Architecture addressing modesinstruction set

programming it 8085.


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DAYI

Hello ! if you want to become tyro to adept in Microprocessor you

probe this book.

p

Day - I


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DAY 1

MICROPROCESSOR:

Single Chip CPU.

Processors has two giants

(i) Intel(8085,8086etc)(ii) Motorola (6800, 68000 etc)

Evolution:

1. Intel 4004 (4 bit p)2. Intel 8085 and 8080 (8 bit p)3. Intel 8086 (16 bit p)4. Intel (802686)

5.

Intel (80386)

6. Intel (80486)7. PentiumIII, IV etc8. Core Processor

Forfast computation numeric coprocessor are used

i) 8087, 808107 etc.About 8085:

8 bit p NMOS device.

6200 transistor,

IC has 40 pins.

8085 can accept process 8 bit data simultaneously. That is why it is called 8 bitprocessor.

5 V power supply needed.

SX

DX

SX

DX


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It can operates on 3MHz clockfree 8085A on 5MHz It has 16 Address lines.

150AA

kb12222610616

= 64 kb

8 bit I/O addresses2562

8 I/O Ports.

70

ADAD

It has the following addresses modesi) Immediateii) Registeriii) Directiv) Indirectv) Implied

It can perform Arithmetic and Logical Operations.Arithmetic:

i) 8 bit binary addition (with carry)ii) 16 bit binary additioniii) 8 bit binary subtraction (with borrow)iv) AND, OR, X-OR, Complement (NOT) and Shift operationShift leftMultiply by 2

Shift rightdivided by 2


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QUESTION ANSWERS

I) TWO MARKS QUESTIONS

1) What is Microprocessor? Give example.

Ans:

Single Chip CPU

8085 etc.

2) What is Numeric coprocessor?

Ans:

For CPUs fast computation it needs numeric coprocessor

Eg. 8087 etc.

3) Why 8085 is called 8bit processor?

Ans:

It can accept, process 8 bit data simultaneously.

4) What are the addressing modes in 8085?

Ans:

i)

Immediate ii) Register iii) Direct iv) Indirect v)Implied

II) ESSAY QUESTION

1. Explain evolution of Microprocessor.Ans:-

From Intel 4004 to core processor.


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III) OBJECTIVE QUESTIONS:

1. Who are the two giants in Microprocessor family?

a) Motorola, Apple b) Intel, Motorola2. Which is 4 bit processor?

a) 4004 b) 8085 A

3. What is bit number Motorola chip starts with?

a) 80. b) 68

4. Which is called Numeric coprocessor?

a) 6800 b)8087

5. 8085 is _________________ devices

a) NMOS b) CMOS

6. 8085 has _____________________ transistors

a) 6100 b) 6200

7. 8085 A needs _______________ clock frequency

a) 3 MHz b) 5 MHz

8. 8085 maps how many address lines

a) 8 address lines b) 16 address lines

9. 8085 has _____________ I/O ports.

a) 8 b) 256

10. What mechanism happens in shift left?

a) Multiply by 2 b) divided by 2.


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RESULT ANALYSIS

III) Objectives question carries / mark for every questions

810 MarksExcellent

< 8Poor

If your score less than 8 repeat day 1. Otherwise go to Day 2.

Answer:

1) b 2) a 3) b 4) b 5) a 6) b 7) b 8) a 9) b 10) a.


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DAYII

If you want to become meticulous with acumen in Microprocessor read DayII.

p

Day - II


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DAYII (INTROD)

8085 has

i) Accumulator (8 bit)ii) 6 GPR (8 bit each, BC, DE, HL)iii) Two 16 bit register

SPStack Pointer

PCProgram counter

8085 has software and hardware interrupts.

The hardware interrupts are

TRAP

RST 7.5 RST 6.5 RST 5.5 and INTR.

It has

Opcode Operand

1) Opcode Fetch cycle ( 4 T)(i.e. 4 time cycles)

2) Memory Read Cycle (3 T)3) Memory Write Cycle (3 T)

It has DMA (Direct Memory Access) facilities.Architecture of 8085:

8085 has following a functional blocks.

1) Registers.2) ALU3) Instruction decoded and machine cycle encoded4) Address buffer5) Address / Data buffer6) Increment / Decrement Address latch7) Interrupt control


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8) Serial I/O control9) Timing and Control circuitry.

QUESTIONS AND ANSWERS

I) Two marks questions:

1) What are 6 GPRegisters in 8085?

Ans:

BC, DE, HL

2) What are 2 16 bit register in 8085?

Ans:

i) PC ii) SP

3) What are the interrupts in 8085?

Ans:

*TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR

4) What is DMA?

Ans:

Direct Memory Access

OBJECTIVE QUESTIONS

1) Accumulator in 8085 has ___________?a) 8 bits b) 16 bits

2) GPR B&D register are ___________?a) 8 bits b) 16 bits

3) Who is the brightest priority interrupt?a)

TRAP b) RST 7.5

4) What is the address location of RST 6.5?a) 003 C x b) 0042 x

5) How many time cycles for OFC?a) 3 T b) 4 T

6) How many time cycle for memory read?a) 3 T b) 4 T

7) How many time cycle for memory write?


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a) 3 T b) 4 T8) DMA has _________________ ?

a) Cycle stealing b) Page replacements9) How many functional blocks in 8085?

a) 9 b) 1010)What is buffer?

a) Temporary storage b) RAM and ROM

Answers:

1) a 2) a 3) a 4) a 5) b 6) a 7) a 8) a 9) a 10) a

RESULT ANALYSIS

810 MarksExcellent

< 8Poor

If your score less than 8 repeat day 1. Otherwise go to Day 3


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DAYIII

If you want to bag with accolades in Microprocessor you better read dayIII.

p

Day - III


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Interrupt Control

INTR

INTA

RST

5.5

RST

6.5

RST

7.5

TRAP

Interrupt & High

priority interrupt

Serial I/O Control

STD SIO

8 Bit Internal Data Bus

AccumulatorTemporary

re ister Flag RegisterInstruction Reg.

ALU Instruction

decoder and m/c

cycle encoder

B C

D E

H L

W Z

SP

PC

Increment/

Decrement Counter

Timing and Control

Clk READY RD W ALE

OutSo S1 I/M HOLD HOLD A Reset In Reset Out

X1

Clk

InX2

Control Status DMA Reset

Address

Buffer

Address / Data

Buffer

A8A15

Hi her bit

AD0AD7

Lower bit

Data bus


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8085 Architecture has 9 functional block.

1) Registers:

a) General Purpose registers

B, C, D, E, H, L

b) Temporary Register

w, z

c) Special purpose registers

Accumulator, Flag, Instruction register.

Flag

D7 D6 D5 D4 D3 D2 D1 D0

S Z X AC X P X Cy

Zero Auxillary Carry Parity Carry

d) Sixteen Bit registers

SP, PC

II) ALU: (8 Bit)

i) Arithmetic Unit +, - (Add, Sub)

ii) Logical Unit AND, OR, XOR, ROTATE, CLEAR

III) Instruction Decoder:

Fetches by opcode fetch cycle.

IV) Address Buffer:

8 bit unidirectional buffer (A8A15)

V) Address / Data Buffer:

8 bit bidirectional buffer (ALE)


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QUESTION AND ANSWERS:

Two mark questions

1) What are the GPR?

Ans:

B, C, D, E, H, L registers

2) What are the special purpose registers?

Ans:

Accumulator, Flag, Instruction register.

3) Draw the flag register?

D7 D6 D5 D4 D3 D2 D1 D0

S Z X AC X P X Cy

Auxillary Carry

4) What are the functions performed by ALU?

i) Add ii) Sub iii) AND iv) OR v) X-OR vi) ROTATE vii) CLEAR

5) What is ALE?Address Latch Enable


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OBJECTIVES QUESTIONS

1) HL is what type register?a) GPR b) Special purpose register

2) w, z is what type register?a) GPR b) Temporary register

3) Following is Special purpose register?a) Accumulator b) w, z

4) SP, PC are what type register?a) 8 bit b) 16 bit

5)

Following are the logical operations?a) ADD b) AND, X-OR

6) What are the address lines for address buffer?a) AD0AD7 (bidirectional) ii) A8A15 (Unidirectional)

7) What is the Flip Flop used in ALE?a) DEF b) JK

Answer:1) a 2) b 3) c 4) b 5) b 6) b 7) a

If your score is 6 (or) more excellent. You can go to Day 4. Otherwise repeat Day 3


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DAYIV

To become unconquerable in Microprocessor read dayIV

p

Day - IV


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DAY4

vi) Incrementer / Decrementer Address Latch 16 bit register

vii) Interrupt Control:

RST 5.5, 6.5, 7.5, TRAP & INTA

viii) Serial I/O control:

STD, SOD

ix) Timing and Control:

i) STATUSii) DMAiii)

RESET

The pin diagram (8085A):

1

2

3

4

5

6

7

8

9

10

11

12

19

20

X1X2

Reset

SOD

SID

TRAP

RST 7.5

RST 6.5

RST 5.5

INR

INTA

AD0

AD7VSS

40

39

38

37

36

35

34

33

32

31

30

29

28

21

Vcc

Hold

Hold A

Clk

Reset INRec 4

IO/M

S

RP

WR

ALE

S0A15

A0

8085 A


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QUESTION AND ANSWERS

Two Marks question:

1) What is the mechanism of incrementer / Decrementer Address Latch?

Increment (or) Decrement the contains of program counter or stack pointer.

2) What is Interrupt? Say priority?

Temporary stop Tray

Essay Questions:

Explain the architecture of 8085?

Answer:

Say the functional blocks with diagram.

RST 7.5

RST 6.5

RST 5.5


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DAYV

To become primitive to sophisticated in microprocessor read dayV.

p

Day - V


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DAY5

Instruction formats:

i) One Byte:

ii) Two Byte:

iii)Three byte

i) Immediate:

OPCODE Operand ADD B

Operand 2OPCODE Operand 1 MOV A, B

Immediate

Addressing Modes

Register Direct Indirect Implied

M V I A , 10 H

IMust Coon Destination

data

OPCODE PUSH


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ii) Register Addressing Mode:

- (GPR is used)

iii)Direct addresses mode:L D A 1000

A (1000)

iv) Indirect Addresses Mode:Contend of conforms.

L D A X B

A ( (BC) )

v) Implied Addressing Mode:CMA

A Complement A

M O V A , B

Destination

Source


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1) What is one address instruction?Answer:

Opcode - Instruction

Operand - address

Ex: 7

ADD B

SUB B2) What is two address instructions?

Answer:

MOV A, B

3) What is zero address instruction?Answer:

PUSH

POP

ESSAY QUESTIONS:

1) Discuss the different addressing modes in 8085?Answer:

i) Immediate ii) Register iii) Direct iv) Indirect v) Implied

OPCODE Operand

OPCODE Operand1 Operand2

OPCODE


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OBJECTIVES QUESTIONS

i) PUSH is ___________ Address instruction

a) Zero b) One

ii) SUB A is ___________ Address instruction

a) Zero b) Oneiii) ADD A, B is _______ Address instruction

a) One b) Twoiv) In immediate address mode contains register and ___________ (source)

a) Register b) Datav) MOV B, C is ___________ addressing mode

a) RegisterRegister b) RegisterMemoryvi) The difference between direct and indirect is ___________

a) Direct is content of indirect is content of contentb) Indirect is content of direct is content of content

RESULT ANALYSIS

Answer:

1) a 2) b 3) b 4) b 5) a 6) aIf your score is greater than 5 go to day 6 otherwise repeat day5.


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If you want to bombard in microprocessor read this dayVI.

p

Day - VI


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DAY - 6

Data Transfer Arithmetic

Increment /

Decrement Branch Group Logic Rotate Stack

1

2

3

4

5

6

7

8

9

10

11

12

13

MVI r, data

MVI M, data

MOV rd, rs

MOV M, rs

MOV rd, M

LXI rp, data

STA addr

LDA addr

SHLD addr

LHLD addr

STAX rp

LDAX rp

XCHL

1

2

3

4

5

6

7

8

9

10

11

12

13

ADD r

ADD M

ADI data

ADC r

ADC M

ACI data

DAD rp

SUB r

SUB M

SUI data

SOB r

SOB M

SOI data

1

2

3

4

5

6

INR r

INR M

INX rp

DCR r

DCR M

DCX rp

1

2

3

JMP addr

J condn addr

PCHL

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

ANA r

ANA M

ANI data

XRA r

XRA M

XRI data

ORA r

ORA M

ORI data

CMP rCMP M

CPI data

STC

CMC

CMA

1

2

3

4

RLC

RRC

RAC

RAR

1

2

PUS

POF

Instruction Set


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IData Transfer:

1) MVI A, 08- 2 byte instruction

Note: (Within GPR no need of MR, MW 3 time cycles)

H or OF (opcode fetch) - 4T

MR (data) - 3T

7 Time cycles.

Immediate addressing mode.

2) MVI M, 4OH(H L Pair) H F 20 L = 10

2010

OF - 4T

MR - 3T (data read)

MW - 3T (write)

10 T

3) MOV A,B

Register (B) Register (A)

40

Stack of I/O and M/C control group

PUSH

POP

CALL

RET

IN addr

OUT addr

EI

DT

PORT (I/O) Interrupt


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(within GPR so no need of MR, MV)

OFC - 4T

TOTAL - 4T

4) MOV M, B

(H = 20) (L) = 50

2050

OF + MW =

4T + 3T = 7 T

5) MOV C,MH = 20 L = 50

2050

OF + MR

4T + 3T = 7 T

6) L X I B, 1020 HX - for register pair

Ifor I monodial

B C

10

1515

10 20


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OF - 4 T Data read 3 T (10) & 4 T (10)

= 10 T

7) STA 2000 H

A (10) 2000

OF 4T MW 30 (3T) MW 00 (3T)

8) LDA 2000 H

2000 A

OF 4T + MR 20 (2T) + MR 00 (3T) + (2000) MR3T

= 13 T

9) SHLD 2500 H(L) (2500)

(H) (2501)

H = 30 L = 20

10

15 15

20

30

2500

2501


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MR

+

MR

+

MW

+

MW

OF (H) (L) 2500 2501 = 16 T

(4T) 3T 3T 3T 3T

10)LHLD 2500 H

+(2500) = 20 (2501) = 30

(H) (H)

OF + MW + MW + MR + MR= 16 T

(4T) 3T 3T 3T 3T

11)STAX B A = 10

BC = 1020

OF 4T Register pair so 3T

= 7T

12)LDAX B

BC = 1030

= 7T

13)XCHGH D

L E

101020

151030 A


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4T

Small Program:

1) Store 16 bin in location 3000 isAnswer:

MVI A, 16 H,

STA 3000 H

HLT

(example for direct address)

Another method:

LXI H, 3000 H

MVI M, 16 H

HLT

(example for indirect addressing)


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Two Mark Questions:

1) How many time cycles for OF, MR, MW?Answer:

OF 4T MR 3T MW 3T

2) Write a program to store 20 in location 5000 HAnswer:

MVI A, 20 H

STA 5000 H

HLT

Essay Questions:

1) Write all data transfer instruction?(1 to 13) instructions.

2) Swap the content of location 2000 and 5000?Answer:

LDA 2000 H

MOV B,A

LDA 5000 H

STA 2000 H

MOV A,B

STA 5000 H

HLT

3) Swap the content of location 2000 and 5000 in indirect address?


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Now introduction anchors.

p

Day - VII


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DAYVII

Arithmetic Group:

In 8085 it has only ADD & SUBTRACT.

1) ADD BA (A) + (B)

OF 4T

(B) source destination A is implied

2) ADD M

A (A) + (M) M = H L Pair

OF 4T MR 3T H (20) L (10)

= 7T

3) ADI 08H

A (A) + 08

OF MR = 7T

4T 3T

4) ADC BA

(A) + (cy) + (B)OF 4T

5) ADC MA (A) + (cy) + (M)

152010 A 25A+

10


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HL

OF4T MR 3T = 7T

6) ACI 20 H

A (A) + (cy) + 20

HL

= OF + data read

= 4T + 3T = 7 T

7) DAD BDouble Add

(HL) (H) (L)

+ +

(B) (C)

Result

H L

Similarly Subtract:

1) SUB BA (A)(B)

OF 4 T

2) SUB MA (A)(M)

OF 4T MR 3T = 7T

3) SUI 08HA (A)08

OF / 4T + Data Read (3T) = 7T

4) SBB BA (A)(B)cy

OF = 4T


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5) SBB MA (A)(M)cy

OF data read

4T 3T = 7T

6) SBI 08A (A)08cy

OF + data read

4T + 3T = 7T

7) DAA (Decimal Adjust Accumulator)A = 0100 1010 = 4 A

DAA

0100 1010

0000 0110

0101 0000

AD = 50 (BCD)

Increment (or) Decrement

1) INR B(B) (B) + 1

OF = 4T

2) INR M(H L)

20 50

2050 2

OF + MR + MW

10 11


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4T + 3T + 3T = 10 T

3) INX rpINX H

rp register pair

if H = 20 L = 10

= 6T

4) DCR B(B) (B)1

4T

5) DCR M

H L 2050 2

10 T

6) DCX BB = 20 C = 10

6 T

10 9

20 9

20 11H L


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Two Marks questions:

1) B = FF C = FF What is the value of INRB & INX B?INR B =

FF

01 B = 00 C = FF

00

INX B =

FF FF

01 01

00 00

B C


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Essay Questions

1) Explain Arithmetic operations is 8085?Answer:

ADD and Subtract discuss 6 + 6 instructions.

2) Discuss the different increment and decrement operator?Answer:

3 Increment 3 Decrement

3) Write a program Add two 8 Bit numbers? 5000 Location 5001 Location Add shot inlocation 5002.

Answer:

LXI H, 5000 H

MOV A, M

INX H

ADD M

INX H

MOV M, A

HLT

4) Write a program to subtract two 8 bit numbers?Answer:LXI OH 5000

MOV A, M

INX H

SUB M

INX H

MOV M, A

HLT

5) Add two 16 bit numbers?Answer:

LHLD 5000

XCHGLHLD 5002


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DAD D

SHLD 5004

HLT

6) Sub two 16 bit numbers?Answer:

LHLD 5000

XCHG

LHLD 5002

MOV A, E

SUB L

MOV L, A

MOV A, D

SBB H

MOV H,A

LHLD 5004

HLT


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To become no to know in microprocessors read dayVIII

p

Day - VIII


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DAY8

IV Branch Group

(i) JMP 2000

PC 2000 is assigned to PC (Program Counter)

OF + MR + MR

4T + 3T + 3T = 10 T

ii) a) JC 2000 jump if carry = 1 at loc 2000b) JNC 2000 jump if carry = 0 at loc 2000

c) JP 2000 jump on possible at loc 2000

d)JM 2000 jump on Minus at loc 2000

e) JPE 2000 jump on parity odd at loc 2000

f) JZ 2000 jump on zero at loc 2000

g) JNZ 2000 jump on not zero at loc 2000

iii) PCHLPC HL

6 T

A B

A N D Clamp


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V Logical Group:

(i) ANA BA A ^ B

4 T

Application: Maskid

(ii)ANA MO A X (M)

7T

(iii)ANI 60

A A X 60

7 T

(iv)XRA AA A X A

4T

Clear Accumulator

Or gate:

1 0 1 0 1 1 1 1A

1 1 1 1 0 0 0 0B

1 0 1 0 0 0 0 0A

A

B

C


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(v) XRA MA A + M

7T

(vi)XRI 40A A + 40

7 T

(vii)ORA AA A V A

4T

(viii) ORA MA A V M

7T

(ix)ORI 107 T

A A V 10 V

(x)CMP B (Compare B)A AB

4 T

(xi)CMP M7T

A A(HL)

(xii) CPI 10 (Compare immedial with data 10)A A10

7 T

(xiii) STCSet carry = 1

Cy 1 (4T)

(xiv) CMC Complement Carry (4 T)(xv) CMA Complement Accumulator (4 T)


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QUESTIONS

Essay Questions:

1) Discuss Branch group instructions?Answer:

i) JMP ii) Conditional Group iii) PCHL

2) Discuss logical group instruction?Answer:

AND, OR, XOR, Compare, Complement etc.

OBJECTIVE TYPE QUESTIONS

1) What is the mechanism of JMP 2000?i) PC 2000 ii) HL 2000

2) JC _________i) Jump on carry ii) jump on carry = 1

3) JNC ________i) Jump no carry ii) Jump on carry = 0

4) JP _____________i) Jump on positive ii) Jump zero

5) JM ______________i) Jump on HL ii) Jump on negative

6) JZ _______________i) Jump on zero ii) Jump on HL

7) JNZ _____________Jump no zero ii) Jump on HL

8) What is XRA A?i) Clears accumulators ii) Adds accumulator

9) CMP A ___________i) Clear Accumulator ii) Adds accumulator

10) CMA _____________i) 1s Complement Accumulator ii) 2s Complement


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RESULT ANALYSIS

1) i 2) ii 3) ii 4) i 5) ii 6) i 7) i 8) i 9) i 10) i

If your score is 8 (or) more you are excellent. If less than 8 please repeat Day 8.


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Now it is dayIX.

p

Day - IX


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DAY9

V) Rotate group

i) R L C

Rotate Accumulator left where B7 goes to carry and B0

ii) R R C:

Rotate Accumulator right. B0 goes to cy and B7

Reverse of RLC

B7 B0

cy

4 T

B7 B0

cy


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iii) R A L:

Rotate Accumulator left through carry.

iv) R A R:

Programs:

i) Computer 1s complement of 54 in 1000 left loc?

Answer:

LDA 1000

CMA

STA 1000

HLT

B7 B0

cy

4 TB7 B0

cy

4 T


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VI) Stack Operations:

i) L X I sp, data

(or) sp data

S P H L

SP HL

6 T

i) PUSH rp

eg. PUSH D

ii) P O P rp

P O P D

E 20

D 10

iii) C A L L 2000 H

D 10 C 20

Sp - 2 20

Sp - 1 10

Sp

SP 1000 1000 10

20


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pc 2000 H

R E T

iv) C C 2000

Call on carry = 1 to 2000

v) C N C 2000

call on no carry

similarly return

v) R C return on carry

vi ) R N C return on no carry

vii) R S T r (restart o to n)

R S T 0 = 0 x 8 = 0000

R S T 5.5 = 5 x 8 + 5 x 8 = 2 C H

R S T 6.5 = 3 C H (Address)

R S T 7.5 = 4 C H (Address)

viii) I/O:

a) I N 80 H

80 H (device no) to Accumulator

b) O U T 50 H

Acc 50 H (device)

ix) X T H L:

Change top of stack with H L

L

H

x) i) E I: (Enable interrupt)

F F: (flip flop enabled)

F F 1

ii) D I (Disable)

F F 0

iii) N O P (No operation)


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iv) H L T

Hales

v) S I M:

Set interrupt mask interrupts R S T 7.5, 6.5, 5.5 are masked

Read Interrupt Mask: (R I M)

It display left status of interrupt

1Mask

O - Unmask

1 1

7.5 6.5 R S T 5.5


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QUESTION ANSWERS

Two marks:

1) What is R L C?

Answer:

2) What R A L?

3) STACK definition?

Stores left data and retriew left data by PUSH and POP operation.

4) Explain Restart n ?

Answer:

R S T nR S T 5.5 = 5.5 x 8 = 2 CH

R S T 6.5 = 6.5 x 8 = 3 CH

R S T 7.5 = 7.5 x 8 = 4 CH

B7 B0

cy

B7 B0

cy


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5) What are the port operation?

Answer:

Port mean carry

a) IN 50H

Acc 50 H

b) OUT 50 H

50 H Acc

6) Define E I and DI?

Answer:

i) EI: Enable Interrupt

FF 1

ii) DI: Disable Interrupt

FF 0

2) How initialize SP?

i) LXI SP ii) LDAX

3) What are the operation in stack?

i) CALL ii) PUSH, POP

4) What is CNC 2000?

i) Call on carry ii) Call no carry

5) What is the vector address of RST 5.5?

a) 2 CH b) 3 CH

6) What is in port?

a) (device no to accumulator) b) (Accumulator to device number)

7) What is EI?

i) FF 1 ii) FF - 1

8) What is DI?

i) FF 0 ii) ALE 0

9) What is S I M?

i) Set interrupt mask ii) Reset interrupt mask


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Essay Questions:

I. Explain rotate group instruction?

Answer:R L C, R R C, R A L, R A R

II. Explain Stack, call sub routine, I/O, Interrupt masking instruction?

Answer:

Condition

PUSH, POP, Call , EI, DT, STM, etc.

Uncondition

OBJECTIVE QUESTIONS

i) R A L is _________________

a) Rotate accumulator through carry

b) Without carry rotation.

RESULT ANALYSIS

1) a 2) a 3) ii 4) ii 5) a 6) a 7) 1 8) 1 9) i

If your score is 8 or more you can go to Day 10. Otherwise repeat day 9.


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The laudable module finishes.

p

Day - X


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DAY10

Programs:

1) Shift right 8 bit data?

Answer:

MOV A,C

RAR

RAR

RAR

RAR

MOV C, A

H L T2) Shift left 16 bit data?

Answer:

MOV A,B

RAL

MOV B,A

MOV A,C

RAL

MOV C,A

HLT

3) Sum of series of number?

LDA 3000 H

MOV C, A

XRA A

LXI H, 3000 H

Loop ADD MINX HDCR C

JNZ Loop

STA 3004

HLT


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4) Data Transfer (Block)

MVI C, 05H

LXI H 2200

LXI D 2300

Loop MOV A, M

STA X, D

INX H

INX D

DCR C

JNZ Loop

HLT

4) Find number of negative numbers in block?

Answer:

MSB is 1 then negative

LDA 3000

MOV C, A

MVI B, 00

LXI H 3001

Loop MOV A, M

ANI 80 H

JZ Ship

INR B

Ship INX H

DCR C

JNZ Loop

MOV A,B

STA 4000

HLT


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QUESTION AND ANSWERS (ESSAY)

1) Find the largest of given number?

Answer:

CMP M2) Calculate the sum of odd number

Answer:

ANI 01, H

JZ


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Annexure


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PROGRAMS FOR 8085 MICROPROCESSOR LEARNERS

1. Store 8-bit data in memory of 8085 microprocessor2. Exchange the contents of memory locations in a 8085 microprocessor3.

Add two 8-bit numbers in a 8085 microprocessor

4. Subtract two 8-bit numbers in a 8085 microprocessor5. Add two 16-bit numbers in a 8085 microprocessor6. Add contents of two memory locations in a 8085 microprocessor7. Subtract two 16-bit numbers in a 8085 microprocessor.8. Finding one's complement of a number9. Finding Two's complement of a number10.Pack the unpacked BCD numbers11.Unpack a BCD number12.Execution format of instructions13.Right shift bit of data14.Left Shifting of a 16-bit data15.Alter the contents of flag register in 8085

1.Store 8-bit data in memory of 8085 microprocessor

MVI A, 52H : Store 32H in the accumulator

STA 4000H : Copy accumulator contents at address 4000H

HLT : Terminate program execution

2.Exchange the contents of memory locations in a 8085 microprocessor

LDA 2000H : Get the contents of memory location 2000H into accumulator

MOV B, A : Save the contents into B register

LDA 4000H : Get the contents of memory location 4000Hinto accumulator

STA 2000H : Store the contents of accumulator at address 2000H

MOV A, B : Get the saved contents back into A register

STA 4000H : Store the contents of accumulator at address 4000H

3.Add two 8-bit numbers in a 8085 microprocessor

Sample problem
http://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Subtract-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Subtract-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Add-contents-of-two-memory-locations.aspxhttp://www.8085projects.info/post/Add-contents-of-two-memory-locations.aspxhttp://www.8085projects.info/post/Subtract-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Subtract-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Finding-ones-complement-of-a-number.aspxhttp://www.8085projects.info/post/Finding-ones-complement-of-a-number.aspxhttp://www.8085projects.info/post/Finding-Twos-complement-of-a-number.aspxhttp://www.8085projects.info/post/Finding-Twos-complement-of-a-number.aspxhttp://www.8085projects.info/post/Pack-the-unpacked-BCD-numbers.aspxhttp://www.8085projects.info/post/Pack-the-unpacked-BCD-numbers.aspxhttp://www.8085projects.info/post/Pack-the-unpacked-BCD-numbers.aspxhttp://www.8085projects.info/post/Unpack-a-BCD-number.aspxhttp://www.8085projects.info/post/Unpack-a-BCD-number.aspxhttp://www.8085projects.info/post/Unpack-a-BCD-number.aspxhttp://www.8085projects.info/post/Execution-format-of-instructions.aspxhttp://www.8085projects.info/post/Execution-format-of-instructions.aspxhttp://www.8085projects.info/post/Execution-format-of-instructions.aspxhttp://www.8085projects.info/post/Right-shift-bit-of-data%28-8-bit-and-16-bit%29.aspxhttp://www.8085projects.info/post/Right-shift-bit-of-data%28-8-bit-and-16-bit%29.aspxhttp://www.8085projects.info/post/Right-shift-bit-of-data%28-8-bit-and-16-bit%29.aspxhttp://www.8085projects.info/post/Left-shifting-of-a-16-bit-data.aspxhttp://www.8085projects.info/post/Left-shifting-of-a-16-bit-data.aspxhttp://www.8085projects.info/post/Left-shifting-of-a-16-bit-data.aspxhttp://www.8085projects.info/post/Alter-the-contents-of-flag-register-in-8085.aspxhttp://www.8085projects.info/post/Alter-the-contents-of-flag-register-in-8085.aspxhttp://www.8085projects.info/post/Alter-the-contents-of-flag-register-in-8085.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspxhttp://www.8085projects.info/post/Alter-the-contents-of-flag-register-in-8085.aspxhttp://www.8085projects.info/post/Left-shifting-of-a-16-bit-data.aspxhttp://www.8085projects.info/post/Right-shift-bit-of-data%28-8-bit-and-16-bit%29.aspxhttp://www.8085projects.info/post/Execution-format-of-instructions.aspxhttp://www.8085projects.info/post/Unpack-a-BCD-number.aspxhttp://www.8085projects.info/post/Pack-the-unpacked-BCD-numbers.aspxhttp://www.8085projects.info/post/Finding-Twos-complement-of-a-number.aspxhttp://www.8085projects.info/post/Finding-ones-complement-of-a-number.aspxhttp://www.8085projects.info/post/Subtract-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Add-contents-of-two-memory-locations.aspxhttp://www.8085projects.info/post/Add-two-16-bit-numbers.aspxhttp://www.8085projects.info/post/Subtract-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Add-two-8-bit-numbers.aspxhttp://www.8085projects.info/post/Exchange-the-contents-of-memory-locations.aspxhttp://www.8085projects.info/post/Store-8-bit-data-in-memory.aspx


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(4000H) = 14H

(4001H) = 89H

Result = 14H + 89H = 9DH

Source program

LXI H 4000H : HL points 4000H

MOV A, M : Get first operand

INX H : HL points 4001H

ADD M : Add second operand


MOV M, A : Store result at 4002H


Flowchart


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Program - 4: Subtract two 8-bit numbers

Sample problem:

(4000H) = 51H

(4001H) = 19H

Result = 51H - 19H = 38H

Source program:

LXI H, 4000H : HL points 4000H

MOV A, M : Get first operand


SUB M : Subtract second operand


MOV M, A : Store result at 4002H.


Flowchart


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Program - 5.a: Add two 16-bit numbers - Source Program 1

Sample problem:

(4000H) = 15H

(4001H) = 1CH

(4002H) = B7H

(4003H) = 5AH

Result = 1C15 + 5AB7H = 76CCH

(4004H) = CCH

(4005H) = 76H

Source Program 1:

LHLD 4000H : Get first I6-bit number in HL

XCHG : Save first I6-bit number in DE

LHLD 4002H : Get second I6-bit number in HL

MOV A, E : Get lower byte of the first number

ADD L : Add lower byte of the second number

MOV L, A : Store result in L register

MOV A, D : Get higher byte of the first number

ADC H : Add higher byte of the second number with CARRY

MOV H, A : Store result in H register

SHLD 4004H : Store I6-bit result in memory locations 4004H and 4005H.



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Flowchart

6.Sample problem:

(4000H) = 7FH

(400lH) = 89H

Result = 7FH + 89H = lO8H

(4002H) = 08H

(4003H) = 0lH

Source program:

LXI H, 4000H :HL Points 4000H

MOV A, M :Get first operand

INX H :HL Points 4001H


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ADD M :Add second operand


MOV M, A :Store the lower byte of result at 4002H

MVIA, 00 :Initialize higher byte result with 00H

ADC A :Add carry in the high byte result


MOV M, A :Store the higher byte of result at 4003H

HLT :Terminate program execution

Flowchart


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7. Sample problem:

(4000H) = 19H

(400IH) = 6AH

(4004H) = I5H (4003H) = 5CH

Result = 6A19H - 5C15H = OE04H

(4004H) = 04H

(4005H) = OEH


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Source program:

LHLD 4000H : Get first 16-bit number in HL

XCHG : Save first 16-bit number in DE

LHLD 4002H : Get second 16-bit number in HL

MOV A, E : Get lower byte of the first number

SUB L : Subtract lower byte of the second number

MOV L, A : Store the result in L register

MOV A, D : Get higher byte of the first number

SBB H : Subtract higher byte of second number with borrow

MOV H, A : Store l6-bit result in memory locations 4004H and 4005H.

SHLD 4004H : Store l6-bit result in memory locations 4004H and 4005H.

HLT : Terminate program execution.

Flowchart


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8.Sample problem:

(4400H) = 55H

Result = (4300B) = AAB

Source program:LDA 4400B : Get the number

CMA : Complement number

STA 4300H : Store the result


Flowchart

9. Sample problem:

(4200H) = 55H

Result = (4300H) = AAH + 1 = ABH

Source program:

LDA 4200H : Get the number

CMA : Complement the number

ADI, 01 H : Add one in the number


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Flowchart

10. Sample problem:

(4200H) = 04

(4201H) = 09

Result = (4300H) = 94

Source program:

LDA 4201H : Get the Most significant BCD digit RLC RLC RLC RLC : Adjust the position of the second digit (09 is changed to 90) ANI FOH : Make least significant BCD digit zero MOV C, A : store the partial result LDA 4200H : Get the lower BCD digit ADD C : Add lower BCD digit


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STA 4300H : Store the result HLT : Terminate program execution

Flowchart

11. Sample problem:

(4200H) = 58

Result = (4300H) = 08 and

(4301H) = 05

Source program:

LDA 4200H : Get the packed BCD number

ANI FOH : Mask lower nibble

RRC

RRC

RRC


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RRC : Adjust higher BCD digit as a lower digit

STA 4301H : Store the partial result

LDA 4200H : .Get the original BCD number

ANI OFH : Mask higher nibble



Flowchart

12. Main program:

4000H LXI SP, 27FFH

4003H LXI H, 2000H

4006H LXI B, 1020H

4009H CALL SUB

400CH HLT


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Subroutine program:

4100H SUB: PUSH B

4101H PUSH H

4102H LXI B, 4080H

4105H LXI H, 4090H

4108H SHLD 2200H

4109H DAD B

410CH POP H

410DH POP B

410EH RET

13. Statement: Write a program to shift an eight bit data four bits right. Assume data is in register

C.

Sample problem:

(4200H) = 58

Result = (4300H) = 08 and

(4301H) = 05

Source program 1:

MOV A, C

RAR RAR RAR RAR MOV C, A HLT

Flowchart for Source program1


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14. Statement: Write a program to shift a 16 bit data, 1 bit right. Assume that data is in BC

register pair.

Source program 2

MOV A, B RAR MOV B, A MOV A, C RAR MOV C, A HLT

Flowchart for Source program1

15. PUSH PSW: Save flags on stack

POP H: Retrieve flags in 'L' MOV A, L :Flags in accumulator CMA:Complement accumulator MOV L, A:Accumulator in 'L' PUSH H:Save on stack POP PSW:Back to flag register HLT:Terminate program execution


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16.8085 Program to calculate GCD of two numbers.

OBJECTIVE :-

To Determine GCD of two numbers. The Two Numbers are stored at consecutive memory

locations :8000H and 8001H. The GCD of two numbers is stored at memory location 8003H.

LOGIC:-

STEP 1:-Compare number 1(no1) with number 2(no2).If no1==no2,go to step 3.

Else if no1


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MODULE II


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SYLLABUS

UNITII

8086 SOFTWARE ASPECT:

Intel 8086 Microprocessor - Architecture


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Slowly you are mincing towards exemplary in microprocessor

Day XI in

Microprocessor


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Day11

Introduction to 8086 Microprocessor

Year 1978, HMOS

Flavors:ALU, internal registers all are 16 bit. So it is called 16 bit microprocessor.

Reads or writes data to port in 16 bit

It has 20 address lines, so it can map 220

1 mb

216

I/O devices (port)

i) 10 MHz 80861ii) 8 MHz 80862iii) 8 MHz 8086

It functions in MIN & MAX mode.

In micro computer only one 8086 microprocessor is available then it is min mode.

If multiprocessor max mode.


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3

6

5

4

2

1

ES

CS

SS

DS

IPControl

systems

AL

BL

CL

DL

SP

BP

SI

DI

AX

BX

CX

DX

AH

CH

BH

DHOperands

Flags

A - BusEu

BIU

B - bus

Memory Interface

C - Bus

Instructionstream byte

queue

Architecture of 8086

ALU


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Register organization:

(i) GPR

All are 16 bit register.

Accumulator AX is 16 bit. AL can be 8 bit.

(ii) Segment Registers:

(iii) Pointers and Index Registers:IP points code segment

BP points data segment

SP Points stock segment

SI, DI used in addressing.

SISource Index

DIDestination Index

ES

SS

DS

CS

Base address(or) segment

base

64 K

64 K

64 K

64 K

1 Mb Physical

Memory

AX AH AL

BX BH BL

CX CH CL

DX DH DL


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iv) Flag register:

OOverflow flag SSign

DDirection ZZero

IInterrupt ACAuxillary carry

TTrap PParity

CyCarry

Questions and Answers

Two Marks:

1) How many address and data line in 8086?Answer:

Address - 20

Data - 16

2) What is the clock frequency of 8086 -1 , 80862?Answer:

80861 10 MHz

80862 8 MHz

3) What is MIN Mode?Answer:

In micro computer if only one 8086 microprocessor is there

than min mode.

4) How many GPR define in 8086?Answer:

4 GPR

AX, BX, CX, DX

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

X X X X O D I T S Z X AC X P X cy


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5) IP, BP and SP points to which segment?Answer:

IP points to code segment

BP points to data segment

SP points to stock segment

Objective Question

1) Which MOS technology in 8086?

i)

HMOS ii) CMOS

2) 8086 Maps ____________i) 1 Mb ii) 64 K

3) The clock frequency of 8086 _________________i) 8 MHz ii) 5 MHz

4) Min mode 8086 microprocessor is _______________i) Many ii) One

5) Multiprocessor functions in __________________i) Min mode ii) Max mode

6) What is AX and AL?i) AX16 bit

AL8 bit

ii)AX16 bitAL16 bit

7) BP points __________ segmenti) Code segment ii) Data segment


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8) SI, DI used in _________________i) Interrupt ii) Addressing

Result analysis

Answers:

1) i 2) i 3) ii 4) ii 5) ii 6) i 7) ii 8) iiIf your score is more than 7 go to day 12 otherwise repeat day 11.


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It is ineffable to say about 8086 microprocessorplease go to Day XII

Day XII


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DAY12

Architecture continuation:

In 8086 it has two units i) Bus interface unit ii) Execution unit

i) BIU:Interface with outer world

The jobs of BIU is

1) Sends address of I/O and memory2) Fetches instruction from memory3) Reads data4) Writes data5) Executes instruction queue6) Address relocation

Hence when you pass the light on BIU it has i) segment register ii) IP iii) summer

iv) instruction queue

Instruction queue:

BIU fetches six instruction bytes. Feature of fetching the next instruction.

While the current instruction execution is called pipelines.

BU F1 F2 F3

U1 U2 etc

F + U + X + ST


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Describe about segmentation

i) Direct Addressing: IP 2000

MOV AL, [ 2000 H]

ii) Register Indirect: (IP)MOV BX, (CX)

Content of content

10

DS 1000 X 10 10,000

+

3000 HCX

16

15

BX

13,000

ADDRESSING MODES (8086)

Direct

Addressing

mode

Register Indirect

Addressing mode

Based

Addressing

Mode

Indexed

Addressing

Mode

Based Indexed

Addressing

Mode

String

Addressing

Mode

x 101000 + 2000

1012,000

DS/Ppr

10,000


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iii)Base plusindex ( BP )MOV CX, (BX + DI)

Question and Answers

Essay Question:

1) Explain the architecture of 8086?2) Discuss the functioning mechanism of BIU and EU?

BX 2000

12,000

+

3000 HCX

20

10

CX13,500

1000DSX 10

+


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It is ineffable to say microprocessor now please read day XIII.

Microprocessor

Day XIII


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Day13

Addressing Mode8086

4) Register relative addressing:

MOV CX, [BX + 003 H]

5) Base relative plus index address

MOV AL, [BX + SI + 10H]

DS 5000

+1000BX

10

20

SP / CX

50,000

51,003

3

51,004

51,003

10 20

BX 1000

SI

10

2000

AL 10

22,010

+

DS

+ 22,010

1000

2000


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6) String Addresses mode:

MOV S BYTE

7) I/O Ports accessing by address modes:

i) Direct

IN AX, 80 H

ii) Indirect

IN AL, DX

8) STACK Memory addressing mode:

PUSH & POP

Instruction set

Here it is like 8086 by multiplication and division also instruction are available.


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Instruction Set

DAA

DAS

Data Movement

Instruction

Add Sub

CompareMVL

DIVBCD &ASCII

Arithmetic

Logic

Instruction JumpM/C Control

MOV

PUSH

POP

XCHG

XLOT

ADD

ADC

INC

SUB

SBB

DEC

NEG

CMP

MVL

IMUL

DIV

CALL

RET

JMP

JCON

INT

INTO

RET

NOT

AND

OR

XORTEST

Shift and

Rotate

SAL/SHL

SHR

SAR

String

Compare

REP

MOVS

CMPS

STC

CLC

CMC

STDCLD

Interrupt

instruction


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Data Transfer (8086) Day 17

i) MOV BX, AX ( 5 addresses modes) implement itii) PUSH BX ( Copy BX to Stack)iii) POP CX (top of stock to CX)iv) XCHG BX, CX

I) Arithmetic instruction:

i) ADD AX, BX [ AX AX + BX ]ii) ADC DX, BX, DX DX + BX + Cyiii) INC BXiv) ADD AL, BLv) DAA

II) Subtract Instruction:

i) SUB AX, BX [ AX AXBX]ii) SBB DX, BX [ DX DXBXCy]iii) DEC BXiv) NEG AL ( zs complements)v) CMP BX (AXBX)vi) SUB AL, BL

DAS (Decimal After Sub)

AL = 0010 0011

CL = 0101 1000


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1100 1011

Decimal After Sub 1100 0110

1100 01 01

Sub 0110 0001

0110 01 01

- (65)

III) Multiplication:

i) MUL BXAX AX, BX

ii) IMUL (signed byte)

IMUL CX

IV) DIVISION:

i) DIV BXAX AX / BX

Quotient in AX remainder in BX

IV ) Logical Instruction: (Prefer 8 bit data)

i) NOT CXNOT CL

ii) AND BXAND AL, BL

iii)OR CX

OR AL CL

iv)XOR:XOR AL, BL


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v)i) SAL / SHL:

Cy B7 B1 B0

0 1 0 . 1 0

1 0 1 ..... 1 0

ii) SHR:SHR destination, count

Cy

1 0 . 1 0

0 1 0 ..... 0 1

iii)RCL:RCL destination, count

Cy0 0

iv)RCR:B Cy


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Question Answers

Essay:

i) Explain the addressing modes in 8086?ii) Discuss Data Transfer instruction in 8086?iii) Discuss Arithmetic and Logic instruction in 8086?iv) Discuss about shift and rotate in 8086?


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The laudable man is reading the laudable microprocessor

Microprocessor

Day XIV


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Day14

Near Intra segment

i) CALL :Far Inter segment

CALL PRO Name of the procedure

ii) Res returniii)JMP: (Unconditional)

JMP NEXT

iv)J (Condition)VI) Processor control instruction:

i) STC - Set carryii) CLC - Reset the carry flag to zeroiii) CMC - Complements the carry flagiv) STD - Set direction flagv) CLD - Reset the direction flag to zero

VII) Interrupt instruction

1)

2) INT OOver Flow flag is set

3) RET

INT Type

Far

(0255 type)


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VIII) String instructions:

1) REP C X = 02) MOV S BX, CX3) CMP BX, CX

Essay questions

1) Discuss about Jump, processor instruction, interrupt instruction and stringinstruction?


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This book makes you as Man with meticulous awards (accolades) in

microprocessor

DAY

XV


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DAY15

Assembler Directives (Pseudo operations)

1) ALIGN 8 (numbers)2) ASSUME

ASSUME CS: Code Segment;

3) CODE4) DATA

DB - Define byte

BW - Define word

DD - Define double word

DQ - Define qued word

DT - Define Ten Bytes

Eg: COST DB 10,20,30

5) DUP: InitializeTABLE BW 10 DUP (0)

6) END7) EQU8) EXTRN

10

20Cost

50


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9) Groups: All segment of same data byte10)Label11)MACRO

END M

12)Name: Name of each module13)OFF set: Displacement14)ORG:

ORG 1000

Start with 1000

15)PTR16)PAGE [ length] [ width ]17)Proc and End P18)Public19)Segment and End S20)Short: Short address21)STACK STACK [ SIZE]22)TITLE:

TITLE rent

ALP : (Assembly Language Programs)

Addition of two numbers

NAME addition

PAGE 52, 80

TITLE Add of two numbers

Model Small


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Stack 100

Data

N1 DB 42H

N2 DB 43H

Result DW ?

Code

START MOV AX, @data (Initialize data segment)

MOV DS, AX

MOV AL, N1

ADD AL, N2

ADC AH, 00

MOV Result, AX

END Stack


Essay question:

1) Discuss the different assembler directives?

Objective Type questioni) Assembler directives are also called _____________________

a) Pseudo operations b) conditional operationii) Four physical segments are

a) Code b) data c) stack iv) ?iii)DB is in ________________in assembler directive

a) Data buffer b) Define byteiv)DT ___________

a) Define Terminals ii) Define Ten bytes


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v) DUP ___________a) Initialize b) duplicate

vi)EQU is __________a) Equate numeric b) Equate numeric and string

vii)MACRO means ____________ subroutinea) Null sub routine b) Open subroutine

Result Analysis

1) a 2) b 3) b 4) b 5) a 6) b 7) bIf your score is greater than 6 go to day 16 otherwise repeat day 15.


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The erudite lured by this microprocessor

DAY XVI


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DAY 16

Open (Macro)

1. Procedures:

Closed

Near

2. Procedure

Far

But for far procedures CALL instruction pushes lot IP and CS on the stack.

Parameter passing mechanism in procedure

i) Using registersii) Using general memoryiii) Using pointersiv) Using stack

i) Parameter passing using register:Main Prog

CS

MOV AL, d1;

:

CALL PRO 1

:

; procedure

PRO 1 PROC NEAR

MOV INPUT , AL;

:


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RET

PRO 1 END P

CODE END

ii) Parameter passing using memory:

Main Program

DS

BCD _ INPUT DB 42; storage

HEX _ VALUE DB 2; storage

DATA END,

CS

CALL PRO 1

Procedure

PRO 1 PROC NEAR

MOV AL, BCD, INPUT

iii) Passing parameter using pointers:

BCD_INPUT DB 42;

:

MOV SI, BCD _ INPUT

MOV AL, [SI]iv) Passing parameters using array:

BCD _ INPUT DW 4209;

MOV AX, OFFSET BCD_INPUT

--

-

MOV AX, [BTR]


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ii) Reentrant Procedure

iii) Recursive procedure

Macros:

INI T MACRO

MOV AX, @ data

MOV DS

:

END M

MAIN LINE PROC 1 PROC 2

RET main

prog

RET

Cal PL

PROC 2 re enters PROC 1

Call pro

Call Call Cc

RET

Important


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i) Passing parameters in Macro:

INT T MACRO HEY

LEA HEY

ii) Local

iii) Global


Two mark question:

i) What are the two types of procedure?

Answer:

i) Open ii) Closed

ii) What are the parameter passing mechanism in procedures?

Answer:

i) Using registersii) Using general memoryiii)Using pointersiv)Using stack

Essay questions:

1) Describe about procedure and parameter passing mechanism?2) Describe about Macros and Parameter passing in macros?


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You are reaching to the apex of 8086 microprocessor

Day

XVII


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Day 17

Please probe about 8086 expeditely


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Day 17

Assembly language prog:

i) . IF _ ELSE END IF

. IF (Condn)

.

. ELSE (Optional)

:

. END IF

ii) WHILE _ END W

. WHILE

. END Wiii) REPEAT E

iv) WHILE

v) FOR

Interrupts

Normal program can be interrupted by three ways

An interrupt caused by an external signal is referred as a hardware interrupts.

Conditional interrupts (or) interrupts cased by special instructions are called software

interrupt.

i) By external signalii) By a special instruction in the programiii)By the occurrence of some condition

i) External Signal: (Hardware signals)

i) NMI input pin

ii) INTR input pin

ii) Special instruction:

It (8086) supports a special instruction, INT to execute special program.


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iii) Condition produced by instruction:

DIV by zero

PUSH FLAG

PUSH CS

PUSH IP

POP IPPOP CS

POP FLAG

ISR

Push register

Pop register

INT

MAIN prog

Type 255 points

.

.

.

Type 2

NMI

Type 1

Single - step

DIV by zero

Type 0


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Interrupt (Vector) Pointer tabe.

Type 0 : Divide by zero interrupt

Type 1: Single step interrupt

The single step mode, systems will execute one instruction and

wait for further direction from user.

Type 2 : NMI

Type 3 : Break point

Type 4 : Over flow

Software interrupts:

TYPE 0255 ;

INT 2 instruction from NMI ISR

With the software interrupts you can call the desired routines from many different

programs in a system by BIOS.

Priorities:

Div error High

Int A, Int 0

NMI

INTRSINGLESTEP Lower


Two mark questions:

1) What are the ways the normal program was interrupted?Answer:

3 ways

2) What is hardware interrupt?Answer:

By external signal


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3) What is software interrupt?Answer:

Special instruction

4) What are the hardware signal interrupt pin?Answer:

i) NMIii) INTR

5) Write the priority of interrupt?Essay question:

1) Discuss about the interrupts and its priorites?


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The second unit concludes

Day - 18


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Day18

I. 8086 Minimum mode: ((i) to (ii) Common Signals for Min and Max)

i) AD0AD15

ii) A16 / S3A19 / S4

S3 S4 Register

0 0 ES

0 1 SS

1 0 CS

1 1 DS

iii)B HE / S7 : (Bus High Enable)BHE A0 Data accesses

0 0 Word

0 1 Upper byte from odd address

1 0 Lower byte from even address

1 1 NON

iv)NMIv)INTN

MODES (8086)

Minimum Maximum


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vi)Clock8086 5Mhz

vii)Resetviii)Readix)TEST: wait signalx)RD (o/p)xi)MN / MX (input) For min mode onlyxii)INTA (Min mode only)xiii)ALE:

AD0AD15

xiv) DEN : This signal informs the transceivers that the CPU is ready tosend or receive data.

xv)DR / R (Data transmit / receive)xvi) M / IO xvii)WR xviii)HOLD:


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MAXIMUM MODE ONLY:

1)

QS1 QS0 Status

0 0 No operation

0 1 First byte of an opcode

1 0 Queue is empty

2 2Subsequent byte of an

opcode

2)

2S

1S

0S M/c Cycle

0 0 0 Interrupt Ack

0 0 1 I/O read

0 1 0 I/O unit

0 1 1 Halt

1 0 0 Instruction fetch

1 0 1 Memory read

1 1 0 Memory unit

1 1 1 inactive

3) LOCK : Bus not for another processor

4)1TG

RQand

0TG

RQ


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Essay question

1) Explain common in Min and Max mode?2) Explain pin only in Min mode?3) Explain pin only in Max mode?


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ANNEXURE FOR 8086 DIAGRAM


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ANNEXURE FOR 8086

MICROPROCESSOR

PROGRAMMING


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Ex. No: 1 32 BIT ADDITION AND SUBTRACTON

AIM:

To write an assembly language program to add and subtract two 32-bitnumbers using 8086 microprocessor kit.

APPARATUS REQUIRED:

8086 Microprocessor Kit

Power Chord

Key Board

33- BIT ADDITION:

ALGORITHM:

Step1: Start the program.

Step2: Move immediately the number 0000H to CX

register. Step3: Copy the contents of the memory 3000 to

AX register.

Step4: Add the content of the memory 3004 with the content of AX

register. Step5: Copy the content to AX register to two memories from

2000.

Step6: Copy the contents of the memory 3002 to AX register.

Step7: Add the content of the memory 3006 with the content of AX

register. Step8: Jump to specified memory location if there is no carry i.e.

CF=0. Step9: Increment the content of CX register once.

Step10: Copy the content to AX register to two memories from

2002. Step11: Copy the content to CX register to two memories

from 2004. Step12: End.


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MNEMONICS:

MOV CX, 0000

MOV AX, [3000]

ADD AX, [3004]

MOV [2000], AX

MOV AX, [3002]

ADC AX, [3006]

JNC loop1

INC CX

Loop1 MOV [2002], AX

MOV [2004], CX

HLTTABLE: 1

Memory LabelMnemonics

DescriptionInstruction Operand

1000 MOV CX,0000Move immediately 0000H to CXregister

1004 MOV AX, [3000] Copy contents of 3000 to AX register

1008 ADD AX, [3004]Add content of memory 3004 with

content of AX register

100C MOV [2000], AXCopy content to AX register to two

memories from 2000

1010 MOV AX, [3002]Copy contents of memory 3002 to

AX register

1014 ADC AX, [3006]Add content of memory 3006 with


1018 JNC loop1 Jump to specified memory CF=0

101A INC CX Increment content of CX registeronce

101B Loop1 MOV [2002], AXCopy content to AX register to two

memories from 2002


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101F MOV [2004], CXCopy content to CX register to two

memories from 2004

1023 HLT Halt

OUTPUT:

INPUT DATA: OUTPUT DATA:

3000: 9999 2000: 3332

3002: 9999 2002: 3333

3004: 9999 2004: 1

3006: 9999


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32 - BIT SUBTRACTION:

ALGORITHM:


Step2: Move immediately the number 0000H to CX register.


Step4: Add the content of the memory 3004 with the content of AX register.

Step5: Copy the content to AX register to two memories from 2000.


Step7: Subtract the content of the memory 3006 from AX register.

Step8: Jump to specified memory location if there is no carry i.e. CF=0.

Step9: Increment the content of CX register once.

Step10: Copy the content to AX register to two memories from 2002.

Step11: Copy the content to CX register to two memories from 2004.

Step12: End.

MNEMONICS:

MOV CX, 0000

MOV AX, [3000]ADD AX, [3004]

MOV [2000], AX

MOV AX, [3002]

SBB AX, [3006]

JNC loop1

INC CX

Loop1 MOV [2002], AX

MOV [2004], CX

HLT


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TABLE: 2



1000 MOV CX,0000

Move immediately 0000H to CX

register




100C MOV [2000], AXCopy content to AX register to two

memories from 2000

1010 MOV AX, [3002]Copy contents of memory 3002 to

AX register

1014 SBB AX, [3006]Subtract content of memory 3006

from content of AX register

1018 JNC loop1 Jump to specified memory CF=0

101A INC CXIncrement content of CX register

once

101B Loop1 MOV [2002], AXCopy content to AX register to two

memories from 2002

101F MOV [2004], CXCopy content to CX register to two

memories from 2004

1023 HLT Halt

OUTPUT:


3000: 9999 2000: 0000

3002: 9799 2002: FE00

3004: 9999

3006: 9999


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RESULT:

Thus an assembly language program to add and subtract two 32-bit numbers

as written and executed using 8086 microprocessor kit.


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Ex. No: 2 16 BIT MULTIPLICATION AND DIVISION

AIM:

To write an assembly language program to multiply and divide twounsigned 16-bit numbers using 8086 microprocessor kit.

APPARATUS REQUIRED:


Power Chord

Key Board

MULTIPLICATION:

ALGORITHM:

Step 1: Start the program.

Step2: Copy the contents of the memory 3000 to AX

register. Step3: Copy the contents of the memory 3002

to CX register.

Step4: Multiply the content of the CX register with the content of

accumulator. Step5: Copy the content to AX register to the memory

2000.

Step6: Copy the contents of DX register to the memory

2002. Step7: End.

MNEMONICS:

MOV AX, [3000]

MOV CX, [3002]

MUL CX

MOV [2000], AX

MOV [2002], DX


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HLT

TABLE: 1




1004 MOV CX, [3002] Copy contents of 3002 to CX register

1008 MUL CX

Multiply the content of the CX

register

with the content of

accumulator

100A MOV [2000], AXCopy content to AX registerto the

memory 2000

100E MOV [2004], DX

Copy content to DX registerto the

memory 2002

1012 HLT Halt

OUTPUT:


3000: 1234 2000: 0060

3002: 5678 2002: 0626


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DIVISION:

ALGORITHM:


Step2: Copy the contents of the memory 3000 to AXregister. Step3: Copy the contents of the memory

3002 to CX register.

Step4: Divide the content of the CX register from the content of

accumulator. Step5: Copy the content to AX register to the memory

2000.

Step6: Copy the contents of DX register to the

memory 2002. Step7: End.

MNEMONICS:

MOV AX, [3000]

MOV CX, [3002]

DIV CX

MOV [2000], AX

MOV [2002], DX

HLT

TABLE: 2




1004 MOV CX, [3002] Copy contents of 3002 to CX register

1008 DIV CX

Divide thecontent of the CXregisterwith the content ofaccumulator

100A MOV [2000], AXCopy content to AX register to the

memory 2000

100E MOV [2004], DXCopy content to DX register to the

memory 2002

1012 HLT Halt


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OUTPUT:


3000: 1234 2000: 0000

3002: 5678 2002: 4444

RESULT:

Thus an assembly language program to multiply and divide two

unsigned 16-bit numbers was written and executed using 8086

microprocessor kit.


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Ex. No: 3 FACTORIAL

AIM:

To write an assembly language program to calculate factorial of n-numbers using 8086 microprocessor kit.

APPARATUS REQUIRED:


Power Chord

Key Board

ALGORITHM:


Step2: Move immediately the number 0000H to AX

register. Step3: Copy the contents of the memory 3000

to CX register. Step4: Move immediately the number

0001H to AX register.

Step5: Multiply the content of the CX register with the content of

accumulator. Step6: Decrement the content of CX register once.

Step7: Jump to specified memory location if there is no zero in CX

register. Step8: Copy the content to AX register to two memories

from 2000. Step10: End.


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MNEMONICS:

MOV AX, 0001

MOV CX, [3000]

MOV AX, 0001

Loop1 MUL CX

DEC CX

JNZ loop1

MOV [2000], AX

HLT

TABLE: 1



1000 MOV AX, 0001Move immediately the number

0001H to AX register

1004 MOV CX, [3000]Copy the contents of memory 3000 to

CX register

1006 MOV AX, 0001Move immediately the number

0000H to AX register

100A loop1 MUL CX

Multiply content of CX registerwithcontent ofaccumulator

100B DEC CXDecrement content of CX register

once

100C JNZ loop1Jump to specified memory location if

there is no zero in CX register

100E MOV [2000], AXCopy content to AX register to

memory 2000

1012 HLT Halt


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OUTPUT:


3000: 0008 2000: 9d80

RESULT:

Thus an assembly language program to calculate factorial of n-

numbers was written and executed using 8086 microprocessor kit.


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Ex. No: 4 SORTING IN ASCENDING ORDER

AIM:

To write an assembly language program to sort n-numbers inascending order using 8086 microprocessor kit.

APPARATUS REQUIRED:


Power Chord

Key Board

ALGORITHM:


Step2: Load datas into the memory.

Step3: Set the conditions to sort n-numbers in ascending order.

Step4: Sort the n given numbers in ascending order.

Step5: Store the result in the memory.

Step6: Display the sorted result from memory.

Step7: End.


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MNEMONICS:

MOV BX, 2000

MOV CX, [BX]

MOV CH, CL

Loop2 INC BX

INC BX

MOV AX, [BX]

INC BX

INC BX

CMP AX, [BX]

JC loop1

MOV DX, [BX]

MOV [BX], AX

DEC BX

DEC BX

MOV [BX], DX

INC BX

INC BX

Loop1 DEC BX

DEC BX

DEC CL

JNZ loop2

MOV BX, 2000

MOV CH, CL

DEC CH

JNZ loop2

HLT


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TABLE: 1

Memory LabelMnemonics Description

Instruction Operand

1000 MOV BX, 2000 Move2000 to BX register

1004 MOV CX, [BX] Move BX memory data to CX register1006 MOV CH, CL Move data from CL to CH

1008 Loop2 INC BX Increment BX register content once

1009 INC BX Increment BX register content once

100A MOV AX, [BX]Move BX memory data to AX

register

100C INC BX Increment BX register content once

100D INC BX Increment BX register content once

100E CMP AX, [BX]Compare AX register content and

BX memory

1011 JC loop1Jump to specified memory location

if carry is 1

1013 MOV DX, [BX]Move BX memory data to DX

register

1015 MOV [BX], AXMove data from AX register to BX

memory data

1017 DEC BX Decrement BX register content once

1018 DEC BX Decrement BX register content once

1019 MOV [BX], DXMove data from DX register to BX

memory data

101B INC BX Increment BX register content once

101C INC BX Increment BX register content once

101D Loop1 DEC BX Decrement BX register content once

101E DEC BX Decrement BX register content once

101F DEC CL Decrement CL register content once

1020 JNZ loop2Jump to specified memory location

if there is no zero in CX register


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1022 MOV BX, 2000 Move2000 to BX register

1026 MOV CH, CL

Copy CL register data to CH

register

1028 DEC CH Decrement CH register content once

1029 JNZ loop2 Jump to specified memory location

if there is no zero in CX register

102B HLT Halt

OUTPUT:


2000: 0004 2002: 0001

2002: 0003 2004: 0002

2004: 0005 2006: 0003

2006: 0004 2008: 0004

2008: 0002 200A: 0005

200A: 0001

RESULT:

Thus an assembly language program to sort n-numbers in ascending

order was written and executed using 8086 microprocessor kit.


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Ex. No: 5 SOLVING AN EXPRESSION

AIM:

To write an assembly language program for solving an expressionusing 8086 microprocessor kit.

APPARATUS REQUIRED:


Power Chord

Key Board

ALGORITHM:


Step2: Load datas from memory to AX

register. Step3: Set the conditions to solve

an expression.

Step4: Solve the expression given below using the conditions

assumed. Step5: Store the result in the memory.

Step6: Display the sorted result from

memory. Step7: End.


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MNEMONICS:

MOV BX, [2000]

MUL AX

MOV BX, [2002]

MUL BX

MOV [3000], AX

MOV AX, [2000]

MOV BX, [2004]

MUL BX

ADD AX, [3000]

ADD AX, 0001

MOV [2006], AX

HLT


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TABLE: 1



1000 MOV AX, [2000]Move data from memory 2000 to

AX register

1004 MUL AX

Multiply content of AX register

with


1005 MOV BX, [2002]Move data from memory 2002 to

BX register

1009 MUL BX

Multiply content of BX registerwith


100A MOV [3000], AXCopy content to AX register to

memory 3000

100E MOV AX, [2000]Move data from memory 2000 to

AX register

1012 MOV BX, [2004]Move data from memory 2004 to

BX register

1016 MUL BX

Multiply content of BX register

with




101B ADD AX, 0001Add the number 0001 to AX

register

101F MOV [2006], AXCopy content to AX register to

memory 2006

1023 HLT Halt


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OUTPUT:


2000: 0002 2006: 1F

2002: 0004

2004: 0007

RESULT:

Thus an assembly language program for solving an expression was

written and executed using 8086 microprocessor kit.


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Ex No: 6 SUM OF N NUMBERS IN AN ARRAY

AIM:

To write a program to find sum of n numbers in an array.

APPARATUS REQUIRED:


Power Chord

ALGORITHM:


Step2: Initialize the counter.

Step3: Get the first number.

Step4: Decrement the

counter.

Step5: Load the base address of an array in to BX

Step6: By using the loop get the next number in to DX and add it with

AX. Step7: Increment the pointer and decrement the counter.

Step8: If the counter value is not equal to zero then go to

step6 Step9: Else store the result.

Step10:Stop the program.


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MNEMONICS:

MOV CL,[2000]

MOV AX,[2002]

DEC CL

XOR D1,D1

LEA BX,[2004]

LOOP1 MOV DX,[BX+D1]

ADD AX,BX

INC D1

INC D1

DEC CL

JNZ LOOP1

MOV [3000],AX

HLT


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TABLE:

LABEL OPCODE OPERAND DESCRIPTION

MOV CL,[2000] Move the memory content to CL.

MOV AX,[2002] Move the memory content to AX

DEC CL Decrement the CL register.

XOR D1,D1 XOR,D1 registers

LEA BX,[2004] Move the content of 2004 to BX

MOV DX,[BX+DI] Move the content of BX+D1 to DX

ADD AX,BX Add AX with DX content.

INC DI Increment D1

INC DI Increment D1

DEC CL Decrement CL

LOOP 1 JNZ LOOP 1 If zero flag is reseted go to loop1

MOV [3000],AX Move the content to memory

location

HLT Halt


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OUTPUT:


2000:0003 3000:0006

2002:0002

2004:0003

2006:0001

RESULT:

Thus the sum of n numbers in an array has been done using 8086 microprocessor and the

output is verified.


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MODULE III


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Unit III

Multiprocessor configurations

Coprocessor configuration closely coupled configuration Loosely coupled

configuration8087 Numeric data processorData typesArchitecture8089 I/O processor

ArchitectureCommunication between CPU and IOP.


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Leaping as well as engrossing microprocessor

Day 19th

Microprocessor


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Day 19

a) Bus contension:More than one processor shares the system memory and I/O devices through acommon system bus.

b) IPC:Interprocessor communication

c) Resource sharing:Deadlock may occur. If two processors unknowingly waiting for other processors.

Coprocessor configuration

Both the CPU and external (Coprocessor) Share entire memory and the I/O subsystem

ESCAPEMonitor

8086

Deactivate

TEST exec

Activate

TESTWait

Execute the

8086

COP

Call the

CPU COP

8086/888087


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Check S6 if S6 = 0 then it is main processor. S6 = 1 then it is 8089 (I/O processor)

BHC/ S7 is zero then COP

i) Closely coupled configuration

ii) Loosely coupled configuration:It contains different modules.

Each mode may consists of an 8086.

8086

8087

Bus controllog.c

System

M

I/O

Clock


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Local

I/O

device

s

Local

Memory

Local Bus

Localbus

control

System bus

control log

8086 +

8087Clock

S

stem

bus

MODULE 1

Similarly Module 2

System

Memory

SystemI/O

7/28/2019 Microprocessors and Microcontroller

microprocessors and microcontrollers module 1 - 5[1]

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