Chapter 1.1 Digital Computers : Hardware Organization

Chapter 1.1

Digital Computers :

Hardware Organization

The Sequential Computer

Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

Input-Output Devices

Printer

Process Control I/O


Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

Memories00000 00001 00010 00011

00100 00101 00110 00111

01000 01001 01010 01011

01100 01101 01110 01111

10000 10001 10010 00011

10100 10101 10110 10111

11000 11001 11010 11011

11100 11101 11110 11111


Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

Instructions Format

OPC OP1 OP2 RES NEXT

OPC OP1 OP2 NEXT1 NEXT2

Information handling instructions

Control instructions

Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Electronic Lock

4 5 6

1 2 3

* 0 #

7 8 9

KFL =

KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

Data Memory

Arithmetic Unit

Control Unit

Program Memory

ND =

SC =

p1 COPY #0 ND P2p2 COPY #0 SC P3p3 EQ? KFL #0 P3 P4p4 MUL SC #10 SC P5p5 ADD SC KDA SC P6

p7 ADD ND #1 ND P8p8 NE? ND #3 P3 P9p9 NE? SC #321 P1 P10p10 COPY #1 DDA P1

p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = SC =

COPY #0 ND P2



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC =

COPY #0 ND P2



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC =

COPY #0 SC P3



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

COPY #0 SC P3



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

EQ? KFL #0 P3 P4

(KFL = 0) = TRUE



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

EQ? KFL #0 P3 P4

(KFL = 0) = TRUE



p6 COPY #0 KFL P7

KFL = 1KDA = 3

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

EQ? KFL #0 P3 P4

(KFL = 0) = FALSE



p6 COPY #0 KFL P7

KFL = 1KDA = 3

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

(0 * 10) = 0

MUL SC #10 SC P5



p6 COPY #0 KFL P7

KFL = 1KDA = 3

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 0

(0 + 3) = 3

ADD SC KDA SC P 6



p6 COPY #0 KFL P7

KFL = 1KDA = 3

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 3

(0 + 3) = 3



p6 COPY #0 KFL P7

ADD SC KDA SC P 6

KFL = 1KDA = 3

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 3

COPY #0 KFL P7



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 3



p6 COPY #0 KFL P7

COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 3

(0 + 1) = 1

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 3

(0 + 1) = 1

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 3

(1 # 3) = TRUE

NE? ND #3 P3 P9



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 3

(0 = 0) = TRUE

EQ? KFL #0 P3 P4



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 3

(1 = 0) = FALSE

EQ? KFL #0 P3 P4



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 3

(3 * 10) = 30

MUL SC #10 SC P5



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 30

(3 * 10) = 30

MUL SC #10 SC P5



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 30

(30 + 2) = 32

ADD SC KDA SC P6



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 32

(30 + 2) = 32

ADD SC KDA SC P6



p6 COPY #0 KFL P7

KFL = 1KDA = 2

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 32

COPY #0 KFL P7



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 32



p6 COPY #0 KFL P7

COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 1SC = 32

(1 + 1) = 2

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 32

(1 + 1) = 2

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 32

(2 # 3) = TRUE

NE? ND #3 P3 P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 32

(0 = 0) = TRUE

EQ? KFL #0 P3 P4



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 32

(1 = 0) = FALSE

EQ? KFL #0 P3 P4



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 32

(32 * 10) = 320

MUL SC #10 SC P5



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 320

(32 * 10) = 320

MUL SC #10 SC P5



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 320

(320 + 1) = 321

ADD SCKDA SC P6



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 321

(320 + 1) = 321

ADD SCKDA SC P6



p6 COPY #0 KFL P7

KFL = 1KDA = 1

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 321

COPY #0 KFL P7



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 321



p6 COPY #0 KFL P7

COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 2SC = 321

(2 + 1) = 3

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 3SC = 321

(2 + 1) = 3

ADD ND #1 ND P8



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 3SC = 321

(3 # 3) = FALSE

NE? ND #3 P3 P9



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 3SC = 321

(321 # 321) = FALSE

NE? SC #321 P1 P10



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA = 1

4 5 6

1 2 3

* 0 #

7 8 9

ND = 3SC = 321

COPY #1 DDA P1



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 3SC = 321

COPY #0 ND P2



p6 COPY #0 KFL P7

KFL = 0KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

ND = 0SC = 321

COPY #0 ND P2



p6 COPY #0 KFL P7

Instructions Format

with P-Register

OPC OP1 OP2 RES

OPC OP1 OP2 NEXT

Information handling instructions

Control instructions

KFL =

KDA =

DDA =

4 5 6

1 2 3

* 0 #

7 8 9

Data Memory

Arithmetic Unit

Control Unit

Program Memory

ND =

SC =

p1 COPY #0 NDp2 COPY #0 SCp3 EQ? KFL #0 P3p4 MUL SC #10 SCp5 ADD SC KDA SC

p7 ADD ND #1 NDp8 NE? ND #3 P3p9 NE? SC #321 P1p10 COPY #1 DDA

p6 COPY #0 KFL

p11 JUMP P1

The

sequential Computer

Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

The

“Von Neumann” Compute

r

Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

Memories

00000 00001 00010 00011

00100 00101 00110 00111

01000 01001 01010 01011

01100 01101 01110 01111

10000 10001 10010 00011

10100 10101 10110 10111

11000 11001 11010 11011

11100 11101 11110 11111

Cost of Memory

Accesstime

10-710-8 10-6 10-5 10-4 10-3 10-2 10-1 100 S

Relative cost per bit

Semiconductor memories

Magnetic memoriesOptical memories

1

1000

Semiconductor Memories

(RAM, ROM, PROM)

083

Semiconductor Memories

• Read access time < 100 nS.

• Cost strongly influenced by access time

• RAM (“Random Access Memory “/ “Read And Modify”):

– volatile !

– Read and write access times equal

• ROM (“Read Only Memory”):

– non volatile

– Can only be written in factory

• PROM (“Programmable Read Only Memory”):

– non volatile

– Can be written by the user

– Write access time >> read access time

Peripheral Memories

14

Writing on magnetic memories

i

0 0 0 0 011 11

Reading from a magnetic memory

e

0 0 0 0 011 11 0 0 0 0 011 11

Manchester Code

0 00 0 1 11 11i

t

Data blocks

Header Data Block Check

0101010101...010101XXXXXXXXX

Check = f(data block)

Synchronization sequence

Disk Organization

Sector

Track

Cylinder

Tracks/cylinder

Cylinders

Sectors/track

Bytes/sector

Total Capacity(in bytes)

Double Density

2

80

9

512

737 280

High density

2

80

18

512

1 474 560

Format of 3.5” diskettes for PC’s.

Tracks/cylinder

Cylinders

Sectors/track

Bytes/sector

Total Capacity(in bytes)

4

16 383

Variable

512

12 072 517 632

Format of 12 GBytes Hard Disk.

Total # sectors 23 579 136

Hard-

disk drive (2)

17

Compact Disk Technology

Laser

Photodetector

Rewritable

CD Technology

Laser

Photodetector

DVD Technology

Laser

Photodetector

Laser

Photodetector

The

“Von Neumann” Compute

r

Data Memory

Arithmetic Unit

ControlUnit

Program Memory

Program Interface

Inp

ut

Inte

rfac

e

Ou

tpu

t In

terf

ace

Minimal Memory Hierarchy

Registers

Central Memory

Disks

CD-ROM

Size (log scale)

SpeedMostlyVolatile

Non-Volatile

RAMIn CPU

RAM(+small ROM)

Traditional “Von Neumann” Computer

Input-Output

Equipment

Central

Memory

Central Processing

Unit

PeripheralMemories

Personal Computer

047

Central Processor

76

Processor Board

79

Documents

Chapter 1.1 Digital Computers : Hardware Organization