Gates – Part 1

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04 GATES1 Page 1 Sistem Komputer NAROTAMA Gates – Part 1

description

Gates – Part 1. drain. drain. drain. no current flows. gate. 3 volts. 0 volts. current flows. source. source. source. Gates are Built With Transistors. nFet. nFetOn. nFetOff. N-type field-effect transistor = nFet. source. source. source. no current flows. gate. 0 volts. - PowerPoint PPT Presentation

Transcript of Gates – Part 1

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Gates – Part 1

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Gates are Built With Transistors

nFet

gate

drain

source

3 volts

drain

source

0 volts

drain

source

currentflows

nocurrentflows

nFetOn nFetOff

N-type field-effect transistor = nFet

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Gates are Built With Transistors

pFet pFetOn pFetOff

gate

source

drain

0 volts 3 voltscurrentflows

nocurrentflows

source

drain

source

drain

P-type field-effect transistor = pFet

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Complement

Also known as invert or not.

x x'0 11 0

x’x

This is a schematic symbol.It is a graphical representationof a circuit which implementsthe operation.

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FET-Based Inverter

Vin Vout

Vcc=3V

GND=0V

3V0V

0V 3V

off

on

on

off

Vcc=3V

GND=0V

Vcc=3V

GND=0V

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AND and OR Gates

A B Q=A•B

0 0 00 1 01 0 01 1 1

A

BQ

A

BQ

A B Q=A+B

0 0 00 1 11 0 11 1 1

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Boolean Expressions and GatesEach Boolean expression has a corresponding realization with logic gates.

A’

BC

F

F = A’ + B C

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NAND Gates

A

BQ

A B Q=(A•B)'

0 0 10 1 11 0 11 1 0

NAND

A

BQ

A

BQ

Q is true iff A AND B are true

Q is false iff A AND B are true

AND

NAND

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FET-Based NAND Gate

A

B

BA

Vcc

GND

F

Vcc

GND

1

1

11 off

on

off

on

0

Vcc

GND

1

0

01 off

on

on

off

1

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NOR Gates

A B Q=(A+B)'

0 0 10 1 01 0 01 1 0

A

BQ

NOR

A

BQ

A

BQ

Q is true if A OR B is true

Q is false if A OR B is true

OR

NOR

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FET-Based NOR Gate

A

B

BA

F

Can you complete the truth table?

A B F

0v 0v ?0v 5v ?5v 0v ?5v 5v ?

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FET-Based Gates

• P-type FETs must be on top of gate• N-type FETs must be on bottom of gate

– Due to electrical characteristics of the two FET types

• Output is driven to either ‘1’ or ‘0’– never both– never neither

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ECEn/CS 224

Exclusive-OR (XOR)

Output is true iff inputs are different.

A

B

Q = A B = A’B + AB’

Another definition: C is true if A /= B

A B Q = A B

0 0 00 1 11 0 11 1 0

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Exclusive-OR Theorems

X 0 = X

X 1 = X'

X X = 0

X X' = 1

X Y = Y X Commutative law

( X Y) Z = X ( Y Z ) = X Y Z Associative law

( X Y)' = X Y' = X' Y

The first 4 are important,

the others are used less infrequently

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Equivalence Operation

denotes equivalence (also written as X==Y)

Output is true iff inputs are equal

X

Y

Q = (X==Y)= X’Y’ + XY

X Y X==Y

0 0 10 1 01 0 01 1 1

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XOR and EQUIV are Complements !!

X Y XY X==Y

0 0 0 10 1 1 01 0 1 01 1 0 1

Gate often called exclusive NOR or XNOR

Bubble means NOT

Equivalence – alternate symbol

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Multi-Input Gates

A

B

BA

Vcc

GND

F

C

C

A

B

BA

Vcc

GND

F

Which one will be slower/faster?

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Alternative Gate Symbols

The symbolic meaning of the circuit should be clear from what

you draw…

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Alternative Gate Symbols

Q is true if A is false OR B is false

A

BQ

Q is true iff A is false AND B is false

A B Q

0 0 10 1 11 0 11 1 0

A B Q

0 0 10 1 01 0 01 1 0

A

BQ A

BQ

A

BQ

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Alternative Gate Symbols

Turn on sprinklers if it is not a holiday and it is not a weekend

or?

The problem statement uses AND, so use the AND symbol

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Alternative Gate Symbols

Turn off the sprinklers if it is a holiday or it is a weekend

or?

The problem statement uses OR, so use the OR symbol

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Another Example

• Design a circuit to determine whether the bits of a 4-bit wire are all zero

This is the appropriate symbol to use…

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Mixed Symbols

• Such a gate doesn’t likely exist• Build from AND gate and inverter• Simplifies schematics

– enhanced readability

Q is true iff A is false AND B is true

A

BQ

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Single Gate Conversion Rules

• Change symbol– AND to OR– OR to AND

• Invert all inputs and outputs• No change in behavior – merely a symbol change

Q is true iff A is false AND B is true Q is false if A is true OR B is false

A

BQ

A

BQ

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Alternative Gate Symbols - Summary

• Use the symbol that matches the problem statement– Clarity– Documentation– Maintennance

• If function is correct but symbol is wrong– your schematic is wrong

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Positive vs. Negative Logic

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Positive Logic and Negative Logic

v1LogicGate

v1 v2 v3 vout

0v 0v 0v 0v0v 0v 5v 0v0v 5v 0v 0v0v 5v 5v 0v5v 0v 0v 0v5v 0v 5v 0v5v 5v 0v 0v5v 5v 5v 5v

v2

v3vout

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Positive LogicLet:

0 volts => 05 volts => 1

The circuit is a logical AND gate

v1 v2 v3 vout

0 0 0 00 0 1 00 1 0 00 1 1 01 0 0 01 0 1 01 1 0 01 1 1 1

v1 v2 v3 vout

0v 0v 0v 0v0v 0v 5v 0v0v 5v 0v 0v0v 5v 5v 0v5v 0v 0v 0v5v 0v 5v 0v5v 5v 0v 0v5v 5v 5v 5v

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Negative LogicLet:

0 volts => 15 volts => 0

The same circuit is a logical OR gate

v1 v2 v3 vout

1 1 1 11 1 0 11 0 1 11 0 0 10 1 1 10 1 0 10 0 1 10 0 0 0

v1 v2 v3 vout

0v 0v 0v 0v0v 0v 5v 0v0v 5v 0v 0v0v 5v 5v 0v5v 0v 0v 0v5v 0v 5v 0v5v 5v 0v 0v5v 5v 5v 5v

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Positive/Negative Logic

• The most common mapping is:

+V 10V 0

– Different systems have used different mappings in the past

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Multi-Level Logic

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Levels of a Network

Maximum number of gates between an input and the output

5 Levels

3 LevelsIn general:

- the more levels - the slower the circuit

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Number of Levels

• Number of levels can be increased by factoring

• Number of levels can be decreased by multiplying out

G = AB + ACDE + ACF = A(B+CDE+CF)

G = A(B+CDE+CF) = AB + ACDE + ACF

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Example

G = AB + ACDE + ACFLevels = 2#Gates = 4Delay = tAND4 + tOR3

#Inputs = 12#transistors = 24Largest gate = 4 inputs

Area Calculations: - each input to a gate costs ~2 transistors - area # of transistorsDelay Calculations: - find slowest path from inputs to output

tdelay = tAND4 + tOR3

- the 4-input AND is likely slower than the other AND gates

A

B

ACDE

ACF

G

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Change the number of levels by factoring

G = ACDE + ACF + AB = A(CDE + CF + B)

CDE

C

F

BA

G

Levels = 3#Gates = 4Delay = tAND3 + tOR3 + tAND2

#Inputs = 10#transistors = 20Largest gate = 3 inputs

factor

This is a 3-level circuit…

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Factor Again

G = A(CDE + CF + B) = A[B+C(F+DE)]

A

BC

F

D

E

G

Levels = 5#Gates = 5Delay = 3 x tAND2 + 2 x tOR2

#Inputs = 10#transistors = 20Largest gate = 2 inputs

factor

This is a 5-level circuit…

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Changing the number of levels

Three alternative solutions for same function…

2-level 3-level 5-level

Logic Levels 2 3 5Delay tAND4 + tOR3 tAND3 + tOR3 + tAND2 3 x tAND2 + 2 x tOR2

Gate Count 4 4 5Gate Inputs 12 10 10Transistors 24 20 20

Largest Gate 4 3 2

Each has different area/speed characteristics

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Two-Level vs. Multi-Level

• In general:– two-level is fastest– multi-level can be smaller

• Exploring by hand to find just the right solution can be difficult

• We will focus on two-level– easy to get from truth table– minimization techniques in later chapters focus on

it