Chapter 10.3: Logic Gates

Based on Slides from

Discrete Mathematical Structures:

Theory and Applications

and by Aaron Bloomfield

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Learning Objectives

Explore the application of Boolean algebra in the design of electronic circuits. The basic elements of circuits are gates. Each type of gate implements a Boolean operation.

We will study combinational circuits - the circuits whose output depends only on the input and not on the current state of the circuit (no memory).

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Logical Gates and Combinatorial Circuits

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Logical Gates and Combinatorial Circuits

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Logical Gates and Combinatorial Circuits

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Logical Gates and Combinatorial Circuits

In circuitry theory, NOT, AND, and OR gates are the basic gates. Any circuit can be designed using these gates. The circuits designed depend only on the inputs, not on the output. In other words, these circuits have no memory. Also these circuits are called combinatorial circuits.

The symbols NOT gate, AND gate, and OR gate are also considered as basic circuit symbols, which are used to build general circuits.

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Logical Gates and Combinatorial Circuits

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Draw a circuit diagram for = (xy' + x'y)z.

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A circuit for two light switches

EXAMPLE 3, p. 714

F(x,y)=1 when the light is on

F(x,y)=0 when the light is off

When both switches are closed, the light is on:F(1,1)=1, this implies

When we open one switch, the light is off:F(1,0)=F(0,1)=0

When the other switch is also open, the light is on:F(0,0)=1

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Thus, we get:

x y F(x,y)

1 1 1

1 0 0

0 1 0

0 0 1

Which Boolean expression is given by F?

Draw a circuit for F,i.e., a circuit to control two light switches.

F(x,y) = xy + x'y'

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Logical Gates and Combinatorial Circuits

A NOT gate can be implemented using a NAND gate (a).

An AND gate can be implemented using NAND gates (b).

An OR gate can be implemented using NAND gates (c).

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Logical Gates and Combinatorial Circuits

Any circuit which is designed by using NOT, AND, and OR gates can also be designed using only NAND gates.

Any circuit which is designed by using NOT, AND, and OR gates can also be designed using only NOR gates.

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Adders: Logical gates to add two numbers

We need to use a circuit with more than one output, which clearly more powerful than a Boolean expression.

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How to add binary numbers

Consider adding two 1-bit binary numbers x and y

0+0 = 0

0+1 = 1

1+0 = 1

1+1 = 10

Carry is x AND y

Sum is x XOR y

The circuit to compute this is called a half-adder

x y Carry Sum

0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 0

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x y s c

1 1 0 1

1 0 1 0

0 1 1 0

0 0 0 0

= s (sum)

c (carry)

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x 1 1 1 1 0 0 0 0

y 1 1 0 0 1 1 0 0

c 1 0 1 0 1 0 1 0

s (sum) 1 0 0 1 0 1 1 0

c (carry) 1 1 1 0 1 0 0 0

HAX

Y

S

C

HAX

Y

S

C

xy

c

c

s

HAX

Y

S

C

HAX

Y

S

C

xy

c

A full adder is a circuit that accepts as input thee bits x, y, and c, and produces as output the binary sum cs of a, b, and c.

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The full adder

The full circuitry of the full adder

xy

s

c

c

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We can use a half-adder and full adders to compute the sum of two Boolean numbers

1 1 0 0

+ 1 1 1 0010?

001

Adding bigger binary numbers

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Adding bigger binary numbers

Just chain one half adder and full adders together, e.g., to add x=x3x2x1x0 and y=y3y2y1y0 we need:

HAX

Y

S

C

FAC

Y

X

S

C

FAC

Y

X

S

C

FAC

Y

X

S

C

x1y1

x2y2

x3y3

x0y0

s0

s1

s2

s3c

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Adding bigger binary numbers

A half adder has 4 logic gates

A full adder has two half adders plus a OR gate Total of 9 logic gates

To add n bit binary numbers, you need 1 HA and n-1 FAs

To add 32 bit binary numbers, you need 1 HA and 31 FAs Total of 4+9*31 = 283 logic gates

To add 64 bit binary numbers, you need 1 HA and 63 FAs Total of 4+9*63 = 571 logic gates

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More about logic gates

To implement a logic gate in hardware, you use a transistor

Transistors are all enclosed in an “IC”, or integrated circuit

The current Intel Pentium IV processors have 55 million transistors!

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Flip-flops

Consider the following circuit:

What does it do?

R

S

Q

Q’

R S Function

1 0 Reset

0 1 Set

1 1 Hold

0 0 1/1

It holds a single bit of memory.

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Memory

A flip-flop holds a single bit of memory

The bit “flip-flops” between the two NAND gates

In reality, flip-flops are a bit more complicated

Have 5 (or so) logic gates (transistors) per flip-flop

Consider a 1 Gb memory chip

1 Gb = 8,589,934,592 bits of memory

That’s about 43 million transistors!

In reality, those transistors are split into 9 ICs of about 5 million transistors each