Download Logic Gates

Chapter 10.3:
Logic Gates
Based on Slides from
Discrete Mathematical Structures:
Theory and Applications
and by Aaron Bloomfield
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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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?
F(x,y) = xy + x’y’
Draw a circuit for F,
i.e., a circuit to control two light switches.
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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
x
y
Carry Sum
0
0
0
0
0
1
0
1
1
0
0
1
1
1
1
0
 Carry is x AND y
 Sum is x XOR y
 The circuit to compute this is called a half-adder
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= s (sum)
c (carry)
x
1
1
0
0
y
1
0
1
0
s
0
1
1
0
c
1
0
0
0
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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.
c
HA
X
Y
x
y
x
y
c
s (sum)
c (carry)
X
HA
Y
s
S
S
C
C
S
c
C
1
1
1
1
1
0
1
0
0
1
0
1
0
0
0
0
1
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
0
0
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The full adder
The full circuitry of the full adder
c
s
x
y
c
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Adding bigger binary numbers
We can use a half-adder and full adders to
compute the sum of two Boolean numbers
1
1
+1
?
0
1
1
0
0
0 0
1 0
1 0
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Adding bigger binary numbers
 Just chain one half adder and full adders together,
e.g., to add x=x2x1x0 and y=y2y1y0 we need:
x0
y0
x1
y1
x2
y2
x3
y3
X
Y
HA
s0
S
C
C
FA
s1
S
X
Y
C
C
FA
s2
S
X
Y
C
C
FA
S
X
Y
C
s3
c
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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?
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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 flipflop
 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
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