Lecture Introduction

Lecture no. 18Microprocessors & Microcontrollers

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Stored Program Concept

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Stored Program Concept• There are three major parts

– The CPU (Central Processing Unit) which acts as the brain coordinating all activities within the computer

– The memory unit where the program instructions and data are temporarily stored– The I/O (Input/Output) devices which allow the computer to input information for

processing and then output the result• Today the CPU circuitry has been reduced to ICs called the microprocessor,

the entire computer with the three parts is called a microcomputer• Several registers (e.g., flip-flops wired in series with each other)

– Some are general purpose, the accumulator for example is reserved for performing complex mathematical operations like multiply and divide, and all I/O data has to go thru the accumulator

• The basic timing of the computer is controlled by a square wave oscillator or a clock generator circuit.

– Synchronization– Determines how fast the program can be fetched from memory and executed

• Memory Read or Fetch Cycle– IP: Instruction Pointer

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Stored Program Concept• Memory unit consists of a large number of storage locations each

with its own address– RAM (Random Access Memory) and its volatility

• Typically each memory location is 8 bits wide (byte accessible memory)– ROM (Read Only Memory)

• The memory unit’s address selector/decoder circuit examines the binary number on the address line and selects the proper memory location to be accessed.– In this example, CPU is reading from memory, it activates its MEMORY

READ control signal– This causes the selected data byte in memory to be placed onto the

data lines and routed to the instruction register in the CPU• Once in the CPU, the instruction is decoded and executed

– In this example, instruction has the decimal code 64 which for a 8086 microprocessor is decoded to be INC AX

– The ALU (Arithmetic Logic Unit) is instructed to add 1 to the contents of the AX

• The cycle repeats itself

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Internal Working Of Computers

1- the CPU program counter can have any value between 0000 FFFF H. This one is set to start with 1400

2- the CPU puts out 1400. The memory circuitry finds the location. Activates the read signal, indicating the memory location 1400. B0 is put on the bus and brought to the CPU

3- B0 is decoded internally it now knows it needs to fetch the next byte!. It brings 21h from 1401. The program counter automatically increments itself to the next location to fetch the next data/instruction.

MEM

CPUPC=1400

Read

B0Inst Decoder

B0 decode

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General Purpose Microprocessors

CPU General Purpose

MicroprocessorRAM

Serial COM Port

TimerI/OROM

These general microprocessors contain no RAM, ROM, or I/O ports on the chip itselfEx. Intel’s x86 family (8088, 8086, 80386, 80386, 80486, Pentium)Motorola’s 680x0 family (68000, 68010, 68020, etc)

Data bus

Address bus

Microprocessors lead to versatile products

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Microcontrollers

CPU RAM ROM

I/O TIMER Serial Com Port

Microcontroller

A microcontroller has a CPU in addition to a fixed amount of RAM, ROM, I/O ports on one single chip; this makes them ideal for applications in which cost and space are criticalExample: a TV remote control does not need the computing power of a 486

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Computer Operating Systems• What happens when the computer is first turned on?• MS-DOS

– A startup program in the BIOS is executed– This program in turn accesses the master boot record on the floppy or

hard disk drive– A loader then transfers the system files IO.SYS and MSDOS.SYS from

the disk drive to the main memory– Finally, the command interpreter COMMAND.COM is loaded into

memory which puts the DOS prompt on the screen that gives the user access to DOS’s built-in commands like DIR, COPY, VER.

• The 640 K Barrier– DOS was designed to run on the original IBM PC– 8088 microprocessor, 1Mbyte of main memory– IBM divided this 1Mb address space into specific blocks

• 640 K of RAM (user RAM)• 384 K reserved for ROM functions (control programs for the video system,

hard drive controller, and the basic input/output system)

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Memory Map

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MS-DOS Functions and BIOS Services

• Program Support• BIOS: usually stored in ROM these routines provide access to the

hardware of the PC• Access to the BIOS is done through the software interrupt instruction

Int n• For example, the BIOS keyboard services are accessed using the

instruction INT 16h• In addition to BIOS services DOS also provides higher level

functions– INT 21h– More details later

Revised on 2001/9/10 1:15 PM ELE3230A Microprocessor & Computer SystemsLecture on 10 September 2001

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Microprocessors

An Intel Pentium II

Heart of a computer — “Central Processing Unit”Small integrated circuit with many transistorsProgrammableBe able to compute (arithmetic, logic, etc.)Work with I/O peripherals & memory


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Early history

1969 - 70 Intel 4004, the first microprocessor, 4-bitIntel 4040, faster then 4004

1971 Intel 8008, a 8-bit version of 4004

1973 Intel 8080, 10 times faster than 8008(Similar products: Motorola MC6800, Zilog Z80)

1974 MITS Altair 8800, the first microcomputerprogrammed with BASIC developed by Bill Gates and Paul Allen

1977 Apple II, the first popular home computerIntel 8085, the last 8-bit microprocessor

1978 Intel 8086, 16-bit microprocessor, much faster

1979 Intel 8088


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1980s1980 Motorola 68000

1981 IBM PC with Intel 8088, running at 4.77 MHz with a 160K floppydisk drive & MS-DOS 1.0/1.1

1982 Intel 80286

1984 Apple Macintosh, with Motorola 68000

1985 Intel 80386

1987 Macintosh II

1989 Intel 80486 at 25 MHz or higher speed


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More recently

1990 Microsoft Windows 3.0 releasedMotorola 68040 developed

1991 Apple and IBM formed a partnership to investigate RISC

1992 Microsoft Windows 3.1 became the standard for PCs.

1993 Intel Pentium (80586) released, MMX technology provided later

1995 Microsoft Windows 95

1995 Intel Pentium Pro (P6)

1997 Intel Pentium II

1998 Intel Pentium II Xeon

1999 Intel Pentium III

2001 Intel Pentium IV


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Growth of capability: the Intel family No. of

transistorsAddressable

MemoryClock Speed

(MHz)MIPs

4004 2,300 640 Bytes 0.108 0.06 8008 3,500 16 KBytes 0.2 0.06 8080 6,000 64 KBytes 2 – 3.1 0.64 – 1 8086 29 K 1 MBytes 5 – 10 0.33 – 0.75 8088 29 K 1 MBytes 5 – 10 0.33 – 0.68 80186 29 K 1 MBytes 8 – 12 1.2 – 1.66 80286 134 K 16 MBytes 8 – 12 1.2 – 1.66 80386SX 275 K 16 MBytes 16 – 33 2.5 – 5 80386DX 275 K 4 GBytes 16 – 33 6 – 12 80486DX 3,200 K 4 GBytes 25 – 100 20 – 80 Pentium 3,200 K 4 GBytes 60 – 133 100 – 200 Pentium Pro 5,500 K 64 GBytes 150 – 200 ~ 250 Pentium II (Klamatch) 7,500 K 64 GBytes 233 – 500 ??? Pentium III (Katmai) 9,500 K 64 GBytes up to 1,200 ??? Pentium IV ??? ??? ??? ???


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Other processors

Intel-compatible:AMD (e.g. K6, k7 series)CyrixIDT

Other general-purpose CPUs:

Motorola 68K series (Apple Macintosh)IBM PowerPCDEC AlphaSunMicro’s UltraSPARC, MAJC (for Java Computing)

Special-purpose processors

Microcontroller: e.g. Intel 8051,Digital Signal Processors: e.g. TMS320C seriesFor handheld products: Intel StrongArm


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Microprocessor-based computer


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Bus, memory & I/O

Bus: a set of connectionscarry the same type of information

binary Memory: ICs that store programs & data

I/O: Input/output devices via which themicroprocessor communicates with outside world


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Memory & addressComputer memory is organized on per byte basis: data are stored ingroups of 8 bits. The location of a data byte is represented by an address in thememory system. Each address points to a 8-bit data.

The address is also a binary number.

Address

11001111000

01110100001

11110001010

11001011011

00000000100

10000000101

01010010110

10000100111

Data in memory


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Address bits & address bus1-bit address can point to only two memory locations.

8-bit address can index 28 = 256 memory locations.

The number of address bits is referred to as the width of address bus.

Intel 8086/8088 uses 20-bit address bus. Thus the address range is0000 0000 0000 0000 0000 – 1111 1111 1111 1111 1111

or in hexadecimal00000H – FFFFFH

The number of memory locations is 220 = 1,048,576

210 = 1,024 = 1 K 220 = 1 K × 1 K = 1 M230 = 1 K × 1 M = 1 G

Question: How many address bits are required to fully access the RAMin your home computer ?


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I/O spaceI/O space is organized and addressed in the same way as the memoryspace. Each I/O address corresponds to an 8-bit I/O port. For Intel 8088, 64K (65,536) I/O ports are addressed. This requires only16 address lines.

For memory space addressing, all 20 address lines are used. The I/O and memory share the same address bus. A separate controlsignal is needed to indicate whether the bus is carrying I/O address ormemory address.


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8088 based microcomputer


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Microprocessor operation

There is little memory inside a microprocessor (µµµµP). This internalmemory is referred as register.The µµµµP has to load data and program from outside into itsinternal registersBoth data and program are binaryProgram is often stored in the memoryProgram is divided into many tiny steps. Each step is performedby an instructionTypically an instruction is several bytes longData are either loaded from the memory or supplied by the I/O


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Fetch

Decode

Execution

load an instruction from memoryinto an internal register

interpret the fetched instruction

execute the instructione.g. read/write data, arithmeticcomputation on loaded data

Fetch-decode-execution cycle

During the fetch-decode-execution cycle, the µµµµP may receive aninterrupt signal via the control bus

Upon the interrupt request, the µµµµP will finish the execution of currentinstruction and then responds to the interrupt


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Programming the microprocessor

An instruction is a simple command telling the µµµµP what to doEach model of microprocessor has its own instruction setIntel 8088 has 90 basic instructions.The µµµµP understands binary numbers only. Any instruction is nothingbut a string of 0’s and 1’s, which is called machine language.Each instruction may occupy multiple bytes:

Op code Operand ... ... Operand

indicatingwhat to do

data (or the addresseswhere data are stored)


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Machine language

B8 23 01 [Move the value 0123H to AX]05 25 00 [Add the value 0025H to AX]8B D8 [Move the content of AX to BX]03 D8 [Add the content of AX to BX]8B CB [Move the content of BX to CX]2B C8 [Subtract the content of AX from CX]2B C0 [Subtract the content of AX from AX]C3 [Return to system]

AX, BX, CX, etc. are the internalregisters in the µµµµP.

0123H means hexadecimal value 0123,i.e. 0000 0001 0010 0011

Binary code Action

1st instruction2nd instruction


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Assembly languageProgramming directly with machine codes is neither desirable norfeasible. It is very difficult to debug.

Assembly language is represented by word-like terms that aremuch more comprehensible than binary number.

B8 23 01 MOV AX,0123H [Move the value 0123H to AX]

05 25 00 ADD AX,25H [Add the value 0025H to AX]

8B D8 MOV BX,AX [Move the content of AX to BX]

03 D8 ADD BX,AX [Add the content of AX to BX]

8B CB MOV CX,BX [Move the content of BX to CX]

2B C8 SUB CX,AX [Subtract the content of AX from CX]

2B C0 SUB AX,AX [Subtract the content of AX from AX]

C3 RET [Return to system]


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Arithmetic& logic unit

EU registerset

Segmentregisters

Instructionqueue

Execution unit (EU)

Bus interface unit (EU)

Buscontrol

unit

databus

Databus

Addressbus

Controlbus

Registers: small, fastmemory that store data oraddresses temporarily.

In a 8086/8088, there areten 16-bit EU registers andanother four 16-bit segmentregisters.

16 bit

(16 bit in 8086,8 bit in 8088)

(20 bit)

8088/8086 internal architecture


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Internal registers

Flag register

8 bit

8 bit 8 bit

16 bit

16 bit

16 bit


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Segment and offset8086/8088 uses a 20-bit address bus

However, all of its internal registers are 16 bit. How are these registersused to store the 20 bit address ?

16-bit segmentaddress

16-bit offset address

20-bit physicaladdress

0371115

+)

shift by 4 bit

0371115

037111519

The offset address can specify up to 216 = 64 KBytes of memory.

The segment address gives the beginning of such a 64 K block.


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CS is a segment register, holding thevalue A000H.

IP holds the offset address 5F00H.

The physical address is A5F00H

Exercises:

CS = 2300H and IP = 1000H

CS = 1A00H and IP = B000H

Segment and offset (continued)

⇒ 23000H + 01000H = 24000H

⇒ 1A000H + 0B000H = 25000H

Lecture Introduction

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