microprocessor 1

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Microprocessor and Interface Systems

Kafrelsheikh Univerisity – Faculty of Engineering

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Dr. Hassanein Shaban AhmedDr. Hassanein Shaban Ahmed

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Course Content

ExitExit

Microprocessor ArchitectureMicroprocessor Architecture

Addressing ModesAddressing Modes

Data Movement, Arithmetic logic, and Program control InstructionsData Movement, Arithmetic logic, and Program control Instructions

Programming the MicroprocessorProgramming the Microprocessor

Memory InterfacingMemory Interfacing

InterruptsInterrupts

Sensor and TransducerSensor and Transducer

Digital and analog interfaceDigital and analog interface

D/A and A/D ConverterD/A and A/D Converter

Input / Output InterfacingInput / Output Interfacing

Microprocessor Based ControlMicroprocessor Based Control

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Microprocessor Architecture

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IntroductionWriting a program in assembly language requires knowledge of the

Microprocessor’s hardware (or Architecture) and the details of its instruction set. An explanation of the basic hardware : bit, bytes, registers, memory, microprocessor, and buses is provided in this lecture. The instruction set are listed in the rest lectures[2].

Components of a Microcomputer system. A microcomputer contains a microprocessor, but also contains other

circuits such as memory devices to store information, interface adapters to

connect it with the outside world, and a clock circuit to act as a master

timer for the system[3]

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Major components of Microcomputer

A Basic Microcomputer[3]

ClockClock MicroprocessorMicroprocessor

MemoryMemory

InterfaceInterface adapteradapter

Input/OutputInput/OutputDeviceDevice

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Memory

Information processed by the computer is stored in its memory [1]. A memory circuit element can store one bit of data. The memory circuits are usually organized into groups that can store eight bits which its called byte. Each memory byte is identified by a number that is called its address, like the street address of a house.

The first memory byte has address 0. The data stored in a memory byte are called its contents. The main difference between address and contents

1. The address of memory byte “location” is fixed but the contents of memory location can be changed

2. The contents of a memory location are always 8-bits, the number of bits in an address depends on the processor. For example, the Intel 8086 microprocessor assigns a 20-bit address, and the Intel 80286 microprocessor uses a 24-bit address

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Memory

Example: Suppose a processor uses 20-bits for address. How many memory locations can be accessed[1].

Solution: The number of bits used in the address determines the number of memory locations that can be accessed by the precursora bit can have two possible values, so in a 20-bit address there can be 220 =1,048,576 or 1 megabyte

MemoryMemory

AddressAddress Binary ContentsBinary Contents00110000110000001100001100000000

00001100000000000011

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The organization of memory locations

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MemoryDepending on model, the processor can access one or more bytes at a

time[2]. In the case two byte “word”, the processor allows any pair of successive memory locations to be treated as a single unit. The lower address of the two memory locations is used as the address of the memory word

Figure (1.3) shows the bit position in a microcomputer word and byte. The position are numbered from right to left starting from 0. in a word, the bits 0 to 7 form low-order “least significant” byte and bits 8 to 15 form high order “most significant” byte [1]

Example: Consider the hex number 0529H, requires 2 bytes or 1 word

Figure (1.3)

0011223344556677

00112233445566778899101011111212131314141515

Byte Bit Positions

Word Bit Positions

High Byte Low Byte

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Addressing data in Memory

Example : Consider the hex number 0529H, requires 2 bytes or 1 word. The most significant byte,05, and least significant , 29. The Microprocessor stores the data in memory in reverse-byte sequence

Low order byte in low memory address and the high order byte in the high memory address

Register

Memory

0505 2929

2929 0505

Memory Address 7612 7613

The processor reverses the bytes again to retrieve the word from memory

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Memory Operations

The processor perform two operations on memory : Read operation “Fetch”, the contents of a memory location “the

original contents of the location are unchanged”. Write operation “Store”, data at memory location “the data

written become the new contents of the location and the original contents of the location are lost” [1].

The processor communicates with memory and I/O circuits by using signals that travel along a set of wires called buses. There are three kinds of signals : address, data, and control. Also there are three buses: address bus, data bus, and control bus[1].

MicroprocessorMicroprocessor MemoryMemory I/O DevicesI/O Devices

data bus

control bus

address bus

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Instruction Execution

Any machine instruction has two parts:Opcode : to specify the type of operation. Operands : which are often given as memory address to the

data

The microprocessor goes through the following steps to execute a machine instruction (fetch and execute cycle)

Fetch cycleFetch an instruction from memoryDecode the instruction to determine the type of operation

Execute cycleRetrieve data from memory if necessaryPerform the operation on the dataStore the result in memory if needed

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Execution a Program

ExampleLDA 7ADD 10HLT

1000 0110 Opcode for LDA0000 0111 Operand (7)1000 1011 Opcode for ADD0000 1010 Operand (10)0011 1110 Opcode for HLT

1st instruction LDA 7

2st instruction ADD 10

3st instruction HLT

To understand how the microprocessor operates, consider the following sample of a program

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Execution a ProgramMicroprocessor Microprocessor

UNITUNIT Arithmetic LogicArithmetic Logic Unit (ALU)Unit (ALU)

ControllerController SequencerSequencer

InstructionInstruction DecoderDecoder

ProgramProgram CounterCounter

AddressAddress RegisterRegister

DataData RegisterRegister

AccumulatorAccumulator

MEMORY

AddressAddress Binary Binary ContentsContents1000 01101000 01100000 01110000 01111000 10111000 10110000 10100000 1010

0011 11100011 1110

Assembly Assembly CodeCode

0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0000000000000000

Before a Program can be run, it must be placed in memory

To begin executing the program, the program counter must be set to the address of the first instruction i.e 0000 0000 zero

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Fetch Cycle of LDA instructionMicroprocessor Microprocessor








MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0000000000000000

0000000000000000

1. The contents of the program counter (0000 0000) are transferred to the address register

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Microprocessor Microprocessor UNITUNIT Arithmetic LogicArithmetic Logic

Unit (ALU)Unit (ALU)







MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

0000000000000000

2. The program counter is incremented by one

Fetch Cycle of LDA instruction

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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

0000000000000000

3. The address is placed on the address bus

Address bus


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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

0000000000000000 0011110000000011

4. The contents of the selected memory location are transferred to the data register


Data bus

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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

0000000000000000 0011110000000011

5. The contents of the data register are decoded by the instruction decoder. The instruction decoder informs the controller sequencer to produce the necessary control signal to carry out the instruction

LDA


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Execute Cycle of LDA instructionMicroprocessor Microprocessor








MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

1100000000000000


0011110000000011

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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0011000000000000

1100000000000000 0011110000000011

2. The program counter is incremented by one for the next fetch cycle

Execute Cycle of LDA instruction

2020

Address bus

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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0011000000000000

1100000000000000 0011110000000011

3. The address of the operand is placed on the address bus


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MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1100000000000000

0000000000000000 1111110000000000

Data bus

4. The contents of the selected memory location are transferred to the data register and also transferred into the accumulator

1111110000000000


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3. The address is placed on the address bus4. The contents of the selected memory

location are transferred to the data register

5. The contents of the data register are decoded by the instruction decoder. The instruction decoder informs the controller sequencer to produce the necessary control signal to carry out the instruction

Fetch Cycle of ADD instruction

Address bus









MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0011000000000000

0011000000000000 1100111100000011

11

22

33 Data bus

44

55

ADDADD

1111110000000000

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2. The program counter is incremented by one for the next fetch cycle

3. The address of operand is placed on the address bus

4. The operand (1010) is transferred to the

data register

5a. The operand (1010) is transferred into one

input of the ALU

5b. Simultaneously, The other operand (710) is

transferred from the accumulator to the other input of the ALU

6. The ALU adds the two operands. The

sum (1710) is loaded into the accumulator

Execute Cycle of ADD instruction

Address bus









MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

1111000000000000

1111000000000000 0011001100000000

11

22

33 Data bus

44

1100000011000000

5a5a

5b5b

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Fetch cycle1. The contents of the program counter

(0000 0100) are transferred to the address register


3. The address is placed on the address bus4. The contents of the selected memory

location are transferred to the data register

5. The contents of the data register are decoded by the instruction decoder.

Execute cycle1. The execution of the HLT instruction is

very simple. The instruction sequencer stops producing control signals then all computer operations stop.

Fetch and Execute Cycle of HLT instruction

Address bus









MEMORY


0011 11100011 1110


0000 00000000 00000000 00010000 0001

0000 00100000 00100000 00110000 0011

0000 01000000 0100

LDALDA77

ADDADD1010

HLTHLT

0000110000000000

0000110000000000 0011001100000000

11

22

33 Data bus

44

1100000011000000

55

HLT

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Machine Cycle

TimeTime

VoltageVoltage

One machine cycleOne machine cycle

A clock circuit controls the microprocessor by generating a train of clock pulses as shown

A one machine cycle can be defined as the minimum time required to fetch a data from memory[1,12],[3,76]. If the microprocessor speed is one megahertz 1MHz, the machine cycle is expressed as:

T=1/f =1/1000000 =1 micro second

Thus, the fetch cycle of an instruction requires one machine cycle. In inherent and immediate addressing modes ,the execute cycle also requires one machine cycle. Therefore, the minimum time required to fetch and execute any instruction is two machine cycle. As will see later in a lecture of addressing modes

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Programming languages

We know that, the operations of the computer’s hardware are controlled by its software. The software program is written by one of the following programming languages:

Machine languageA microprocessor can only execute machine language instructions. As we’ve

seen, they are bit strings from 0 or 1.

Assembly languageA more convenient language to use is the assembly language. In assembly

language, we use symbolic names to represent operations, register, and memory location. Such as, mov ax,7

A program written in assembly language must be converted to machine language before the microprocessor can execute it. The assembler is used to translated from assembly language into machine language

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Programming languagesHigh-Level languageDifferent high-level languages are designed for different applications, but they

generally allow programmers to write programs that look more like natural language text than is possible in assembly language

A program called a compiler is needed to translate a high-level language program into machine code

Advantages of High-Level Language1. Because high-level languages are closer to natural languages, it easer to

convert a natural language algorithm to a high-level language program than to an assembly language program

2. Assembly language program generally contains more statements than an equivalent high-level language program. So more time is needed to code the assembly language program

3. Assembly instruction sets are unique for each type of the microprocessor, but high-level language program can be executed on any machine

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Programming languages

Advantages of Assembly Language1. Assembly language is so close to machine language, so the assembly

language program is efficiency because it produces a faster, shorter machine language program

2. Some operations, such as reading or written to specific memory locations I/O ports, can be done easily in assembly language than high level language

3. Assembly language helps to understand why the operations are executed inside the computer

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Representation of Numbers and Characters

Numbers can be expressed in any of the following number system Decimal number system Binary number system Hexadecimal number system

Decimal number systemNumber : 0 to b-1 where b is the base of the number systemNumbers : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (ten digit)Base = 10

Example : 3932=3x103+9x102+3x101+2x100

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Representation of Numbers and Characters

Binary number systemNumbers : 0, 1 (two digit)Base = 2Example : 1001

Hexadecimal number systemNumbers : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F (sixteen

digit)

Base = 16Example : 3EF1

Advantages of hexadecimal number system The conversion between binary and hex is easy but decimal

to binary is difficult The binary system is long and difficult to represent the

content of memory. For example, word needs 16 bits

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Conversion between Number Systems

Converting binary to decimal11101b = 1x24+1x23+1x22+0x21+1x20

= 16+8+4+0+1=29d

Converting Hex to decimal8A2Dh = 8x163+Ax162+2x161+Dx160

= 8x163+10x162+2x161+13x160 = 35373d

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Converting decimal to binary Example : Convert the decimal number 25 into its binary equivalent

25 2 = 12 with remainder 1 LSB12 2 = 6 with remainder 06 2 = 3 with remainder 03 2 = 1 with remainder 11 2 = 0 with remainder 1 MSB

Then,25d = 11001b

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Converting decimal to Hex Example : Convert the decimal number 11172 into its Hex equivalent

11172 16 = 698 with remainder 4LSB

698 16 = 43 with remainder 10 (A)43 16 = 2 with remainder 11 (B)2 16 = 0 with remainder 2

MSB

Then,11172d = 2BA4h

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Converting Hex to Binary Example : Convert 2B3Ch to binary

Solution :

2 B 3 C = 0010 1011 0011 1100 = 0010101100111100

Then,2B3Ch = 0010101100111100b

Converting Binary to Hex Example : Convert 1110101010b to Hex

Solution :

0011 1010 1010 = 3 A A

= 3AAh

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Addition in The Number SystemsDecimal addition

Example : 2546 + 1872

4418 The unit’s digit = 6+2 = 8The ten’s digit = 4+7 = 11 we write down 1 and carry 1 to hundred’s columnThe hundred’s digit = 5+8+1 = 14 we write down 1 and carry 4 to the last columnThe last digit = 2+1+1 = 4

Hex addition Example : 5B39h

+ 7AF4hD62Dh

The unit’s column = 9h+4h = 13h=DhThe next column = 3h+Fh = 12h write down 2 and carry 1 to the next columnThe next column = Bh+Ah+1h = 16h write down 6 and carry 1 to the next columnThe last column = 5h+7h+1 = Dh

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Addition in The Number SystemsBinary addition

Example : 100101111 + 110110

101100101

The unit’s column = 1+0= 1The next column = 1+1= 10 write down 0 and carry 1 to the next columnThe next column = 1+1+1= 11 write down 1 and carry 1 to the next column And so on

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Subtraction in The Number SystemsDecimal subtraction

bbExample : 9145

- 72831862

The unit’s digit = 5 – 3 = 2The ten’s digit = 14 - 8 = 6 we first borrow 1 from the hundred’s columnThe hundred’s digit = 11 – 2 -1 = 8 we borrow 1 from the next columnThe last digit = 9 - 7 – 1 = 1

By the same way the Hex and Binary subtraction can be computed

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Integer Number Representation in the Computer

Unsigned Integers An unsigned integer is an integer that represents a

magnitude, so it is never negative no bit is needed to represent the sign, and so all 8 bits in a

byte, or 16 bits in a word, are available to represent the number

The largest unsigned integer in a byte is 1111 1111=FFh=255d while in the word is 1111 1111 1111 1111=FFFF=65535

If the least significant bit LSB of an integer is 1, the number is odd, and its even if the LSB is 0

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Signed Integers A signed integer can be positive or negative The most significant bit MSB is reserved for the sign : 1

means negative and 0 means positive Negative integers are stored in the computer in a special

way known as two’s complements

Example: Find the two’s complement of 55 = 0000 0000 0000 0101

One’s complement of 5 = 1111 1111 1111 1010 +1

Two’s complement of 5 = 1111 1111 1111 1011 =FFFBh

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Example: Find the two’s complement of the two’s complement of 5Two’s complement of 5 = 1111 1111 1111 1011one’s complement of 1111 1111 1111 1011 = 0000 0000 0000 0100

+1Two’s complement of 1111 1111 1111 1011 = 0000 0000 0000 0101 =5

Example: show how the decimal integer -97 would be represented a) in 8 bits and b) in 16 bits. Express the answers in Hex.

97d in Hex is 61h = 0110 0001b one’s complement of 61h = 1001 1110

+1Two’s complement of 61h = 1001 1111=9Fh in 8bitsTwo’s complement of 61h =FF9Fh in 16bits

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Decimal Interpretation

We show how signed and unsigned decimal integers may be represented in the computer. In reverse way. Now, we attempt to interpret the contents of a byte or word as signed or unsigned decimal integer

Unsigned decimal interpretationJust do a binary to decimal conversion (b to h and then h to d)

Signed decimal interpretation If the MSB is 0, the number is positive, the signed

decimal is the same as the signed decimal If the MSB is 1, the number is negative, so call it –N.

To find N, just take the two’s complement and the convert to decimal

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Decimal Interpretation

Example:Suppose AX contains FE0Ch. Give the unsigned and signed decimal interpretations

Solution:Unsigned decimal interpretation, convert FE0Ch to decimal yields 65036For the signed interpretation, FE0Ch=1111 1110 0000 1100. So the sign bit is 1 i.e negative number, call it –N. To find N, get the two’s complement

FE0Ch = 1111 1110 0000 1100one’s complement of FE0Ch = 0000 0001 1111 0011

+1Two’s complement of FE0Ch = 0000 0001 1111 0100Then, AX contains -500 =01F4h=500d

Product8088808580868088802868038680486

Year introduced1974197619781979198219851989

Clock Rate (MHz)2-33-85-105-86-1616-2325-50

No. Transistors4500650029000290001300002750001.2million

Physical Memory64K64K1M1M16M4G4G

Internal data bus881616163232

External data bus88168163232

Address bus16162020243232

Data type (bits)888, 168, 168, 168, 16, 328, 16, 32

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The 8086 family Microprocessors

The Intel 8086 family microprocessors includes the 8086, 8088, 80186, 80286, 80386, 80386sx, 80486, and 804 86sx [4]

Evolution of Intel’s Microprocessors

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Notes

♣ Internally, the 8088 is essentially the same as the 8086

♣ Externally, the 8086 has a 16bit data bus while the 8088 has 8bit

♣ The 8086 also has a faster clock rate and better performance

♣ The 8088 is less expensive than 8086

♣ The 8086 and 8088 have the same instruction set[4]

8086 major improvement over the 8088/8085

♣ The 8086 has 1M memory capacity while 8088/8085 has 64K

♣ The 8086 is a 16 bit microprocessor which can access 16bit at a time

♣ The 8086 was a pipelining processor as opposed to the nonpipelined

8080/8085. In a system with pipelining, the data and address buses

are busy transferring data while the CPU is processing information

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The main features of 80286 microprocessor

♣ Two mode of operation, real mode or protected mode. In protected

mode, the 80286 supports multitasking which is the ability to

execute several programs a the same time

♣ More memory capacity (224 = 16megabytes)

♣ Virtual memory in protected mode. The processor 286 can treat

external storage (HD) as if it were physical memory up to 1 gigabyte

(230 byte) [1]

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The main features of 80386 microprocessorThe main features of 80386 microprocessor

♣ The 80386 also can operate in either real or protected mode.

♣ More memory capacity data bus 32bit (232 = 4gigabytes)

♣ Virtual memory in protected mode up to 64 terabyte (246 byte) [1]

♣ The 80386SX microprocessor has the same internal structure as

386 but it has only a 16 bit external data bus and 24 bit address

bus (16`MB)

The main features of 80486 microprocessorThe main features of 80486 microprocessor

♣ The 80486 also is 32 bit microprocessor.

♣ The 80486 has a numeric coprocessor 80487 and cash memory

♣ The 80486SX microprocessor has the same internal structure as

486DX but without coprocessor

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Inside The 8080/8086 Microprocessors

PipeliningPipelining

There are two ways to make the processor execute program faster♣ By increasing the working frequency, which depends on the

Integrated Circuit IC’s technology. This way is highly cost♣ By changing the internal working of the microprocessor

In 8085 microprocessor fetch an instruction from memory then execute it and so on. The idea of pipelining in its simplest form is to allow the processor to fetch and execute at the same time as shown

Fetch1Fetch1 Execute1Execute1 Fetch2Fetch2 Execute2Execute2 Fetch3Fetch3 Execute3Execute3

Fetch1Fetch1 Execute1Execute1



Nonpipelined8085

Pipelined8086

AX

BX

CX

DX

SP

BP

SI

DI

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Inside The 8080/8086 Microprocessors

Gen

era

l re

gis

ters

Internal BusInternal Bus

Temporary registersTemporary registers

ALU

FlagsFlags

CS

DS

SS

ES

IP

BusBusControllerController

LogicLogic Exte

rnal

Bu

s

Instruction queue

Execution Unit EU

Bus Interface Unit BIU

Inte

rnal B

lock D

iag

ram

of

Th

e 8

08

0/8

08

6

Mic

rop

rocessor

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Intel 8086 Microprocessors Organization

The 8086 microprocessor is organized into main components: the

execution unit EU and the bus interface unit BIU. ♣ The purpose of the execution unit is to execute instructions. It

contains ALU and a eight registers (AX, BX, CX, DX, SP, BP, SI,

and DI) for storing date. In addition, the EU contains temporary

registers for holding operands for the ALU, and the Flags

register♣ The bus interface unit BIU facilitates communication between the

EU and memory or I/O devices. It also contains a five registers

(CS, DS, ES, SS, and IP); they hold the address of memory

locations. ♣ The EU and BIU are connected by and internal bus. While the

EU is executing an instruction, the BIU fetches up to 6 bytes

(8086) or 4 bytes (8088) of the next instruction.This operation is

called instruction prefetch [1].

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Intel 8086 Microprocessors Organization

Notes♣ Intel implemented the concepts of pipelining in the 8088/8086

by splitting the internal structure of the microprocessor into

two sections the EU and BIU. These two units work

simultaneously[4]. ♣ The BIU of 8086 contains a buffer or instruction queue. ♣ A register is like a memory location except that we normally

refer to it by a name rather than a number

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8086 Registers Data Registers

AXAHAL

BXBHBL

CXCHCL

DXDHDL

Segment Registers

CS

DS

SS

ES

Pointer and Index Registers

SI

DI

SP

BP

IP

Flags Register

8086 R

eg

iste

rs

The 8086 has The 8086 has four general date

registers; the address registers are

divided into segment, pointer, and

index registers; and the status

register is called the Flags register.

In total, there are fourteen 16 bit

registers

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Data Registers : AX, BX, CX, DX

These four registers are available to the programmer for general data

manipulation, also they perform special functions such as: ♣ AX ( accumulator register )

AX is used for the accumulator♣ BX ( Base register )

BX is used as a base addressing register♣ CX ( Counter register )

CX is used as a counter in loop operation♣ DX ( Counter register )

DX is used to point data in I/O operation

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Flags Register

The Flags register is an individual 16 bits reflect the result of

a computation. There are two kinds of flags:

♣ Status flags : many instructions involving comparison

and arithmetic change the status flags (such as ZE zero

flag)

♣ Control flags : which enable or disable certain operations

of the processor ( such as IF interrupt flag )

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Segment Registers CS, DS, SS, ES

The address (segment) registers store addresses of instructions

and data in memory

CS : Code Segment DS : Data Segment

SS : Stack Segment ES : Extra Segment

A typical assembly language program consists of at least three

segments: a code segment, a data segment, and a stack segment.

The code segment contains the assembly language instructions that

perform the tasks. The data segment is used to store information

(data) that needs to be processed by the instructions in the code

segment. The stack is used to store information temporarily

If a program needs to access a second data segment, it can use

the ES register

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Memory Segment

A segment is an area of consecutive memory bytes up to 216

(64kb). Each segment is identified by a segment number,

starting with 0. A segment number is 16 bit, so the highest segment

number is FFFFh. A memory location may be specified by

providing a segment number and an offset, written in the form

segment:offset .

For example Segment 0 start at address 0000:0000=00000h and ends at

0000:FFFF=0FFFFh. Segment 1 starts at address 0001:0000=00010h and

ends at 0001:FFFF==1000Fh

In the 8085 there are only 64 kb of memory for all code, data,

and stack information, but in the 8086 there can be up to 64kb of

memory assigned to each category

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Logical and Physical address

In the Intel 8086, there are three type of addresses: the physical

address, the offset address, and the logical address. The physical

address is the 20 bit address that is actually put on the address

pins of the 8086 microprocessor and decoded by address decoder.

This address range from 00000h to FFFFFh for the 8086. The

offset address is the location within a 64KB segment range. So

the offset range from 0000h to FFFFh. The logical address

consists of a segment value and an offset address.

To obtain a 20 bit physical address, the 8086 microprocessor first

shifts the segment address 4 bit to the left, and then adds the offset.

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Example

If CS=24F6 and IP=634Ah, show

1.The logical address 24F6:634A

2.The offset address 634A

And calculate :

3.The physical address 24F60+634A=2B2AA

4.The lower and upper rang of that segment

Lower range=24F60+0000=24F60

Upper range=24F60+FFFF=34F5F

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Pointer and Index Registers: SP, BP, SI, DIThe registers SP, BP, SI, and DI normally point (contain the offset

address of) memory locations. Unlike segment registers, the pointer and index register can be used in arithmetic and other operations.

SP: Stack Pointer The stack pointer register is used in connection with SS for accessing the stack segment.

BP: Base Pointer The base pointer register is used primarily to access data on the stack. However, unlike SP, we can also use BP to access data in the other segments

SI: Source IndexThe SI register is used to point the memory locations in the data segment addressed by DS.

DI: Destination IndexThe DI register is used to point the memory locations in the extra segment addressed by ES

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Instruction Pointer: IP

To access instructions in the code segment, the 8086 uses the registers CS and IP. The CS register contains the segment number and the IP contains the offset

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8086 Addressing Modes

The microprocessor can access operands (data) in various ways called

addressing modes. The 8086 provides a total of seven distinct addressing

modes:

1. Register

2. Immediate

3. Direct

4. Register indirect

5. Based relative

6. Indexed relative

7. Based index relative

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Register Addressing Mode

The register addressing mode involves the use of registers to hold

the data to be manipulated. Memory is not accessed when this

addressing mode is executed; therefore, it is relatively fast. As

example:

MOV BX,DX ;copy the content of DX into BX

MOV ES,AX ;copy the content of AX into ES

MOV AL,BH ;add the content of BH to the content of AL

It should be noted that the source and destination registers must

match in size. In other words coding “ MOV CL,AX “ will give an

error. Since the source is a 16-bit register and the destination is an 8-

bit

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Immediate Addressing Mode

In the immediate addressing mode, the source operand is a constant.

When the instruction is assembled, the operand comes immediately

after the opcode. For this reason, this addressing mode executes

quickly. As an example:

MOV AX,2550H ;move 2550h into AX

MOV CX,625 ;load the decimal value 625 into CX

MOV BL,40H ;load 625 into BL

In the first two addressing modes, the operands are either inside

the microprocessor or tagged along with the instruction which that

are not referred to the memory. There are many ways of accessing

the data in the data segment as the following of addressing modes.

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Direct Addressing Mode

In the direct addressing mode the data is in some memory

locations and the address of the data in memory comes immediately

after the instruction. Note that in immediate addressing, the operand

itself is provided with the instruction, whereas in direct addressing

mode, the address of the operand is provided with the instruction. As

an example

MOV DL,[2400] ;move contents of DS:2400H into DL

In this case the physical address is calculated by combining the

contents of offset location 2400 with DS, the data segment register.

Note, if the absence of the bracket it will give an error since it is

interpreted to move 16bit into 8bit

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Example

Find the physical address of the memory location and its contents

after the execution of the following, assuming that DS=1512H.

MOV AL,99

MOV [3518],AL

Solution

First AL is initialized to 99H, then in line two, the contents of AL

are stored to logical address DS:3515 which is 1512:3518. shifting DS

left and adding it to the offset gives the physical address of

(15120H+3518H=18638H). That means after execution of the second

instruction, the memory location with address 18638H will contain

the value 99H

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Register indirect Addressing Mode

In the register indirect addressing mode, the address of memory

location where the operand resides is held by a register. The registers

used for this purpose are SI, DI, and BX as a pointer. For example

MOV AL,[BX] ;moves into AL the contents of the

;memory location pointed to by

DS:BXExample : assume that DS=1120, SI=2498, and AX=17FE. Show

the contents of memory locations after the execution of

MOV [SI],AX

Solution

The contents of AX moves into memory locations with The logical

address DS:SI and DS:SI+1; therefore the physical address starts at

DS (shifted left) +SI=13698. Low address 13698Hcontains FE, low

byte, and high address 13698H will contain 17, the high byte

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Based relative Addressing Mode

In this mode, base registers BX and BP, as well as a displacement

value, are used to calculate the offset address. The physical address

(PA) are DS for BX and SS for BP, For example

MOV CX,[BX]+10 ;move DS:BX+10 and DS:BX+10+1 into CX

;physical address=DS(shifted left)

+BX+10

MOV AL,[BP]+5 ;physical address=SS(shifted left)+BP+5

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indexed relative Addressing Mode

The indexed relative addressing mode works as the same as the

based relative addressing mode, except that registers DI and SI hold

offset address. For examples

MOV DX,[SI]+5 ;PA=DS(shifted left)+SI+5

MOV CL,[DI]+20 ;PA=DS(shifted left)+DI+20

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Based indexed Addressing Mode

By combining based and indexed addressing modes, a new

addressing mode is derived called the based indexed addressing mode.

In this mode, one base register and one index register are used. For

examples

MOV CL,[BX][DI]+5 ;PA=DS(shifted left)+BX+DI+5

MOV CH,[BX][SI]+20 ;PA=DS(shifted left)+BX+SI+20

MOV AH,[BP][DI]+12 ;PA=SS(shifted left)+BP+DI+12

MOV AH,[BP][SI]+29 ;PA=SS(shifted left)+BP+SI+29

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Example

Assume that DS=4500,SS=2000,BX=2100,SI=1486,DI=8500,BP=7814,

and AX=2512. Show the exact physical memory location where AX is

stored in each of the following. All value are in hex

a) MOV [BX]+20,AX b) MOV [SI]+10,AX

c) MOV [DI]+4,AX d) MOV [BP]+12,AX

Solution

In each case PA = segment register (shifted left)+ offset

register + displacement

a) DS:BX+20 location 47120=(12) and 47121=(25)

b) DS:SI+10 location 46496=(12) and 46497=(25)

c) DS:DI+4 location 4D504=(12) and 4D505=(25)

a) SS:BP+12 location 27826=(12) and 27827

=(25)

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References

1. Assembly Language Programming and Organization of the

IBM PC, Ytha Ya and charles marut

2. Fundamentals of PC Hardware and Software

3. Microprocessors a step by step introduction

4. 8086 Microprocessor assembly language and experimental October

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Thank You

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