Logic Gates

CS 202, Spring 2007

Epp, setions 1.4 and 1.5
Aaron Bloomfield

1
 Review of Boolean algebra
• Just like Boolean logic
• Variables can only be 1 or 0
– Instead of true / false

2
 Review of Boolean algebra
• Not_ is a horizontal bar above the number
–0_ =1
– 1=0
• Or is a plus
– 0+0 = 0
– 0+1 = 1
– 1+0 = 1
– 1+1 = 1
• And is multiplication
– 0*0 = 0
– 0*1 = 0
– 1*0 = 0
– 1*1 = 1 3
 Review of Boolean algebra
_ __
• Example: translate (x+y+z)(xyz) to a Boolean
logic expression
– (xyz)(xyz)
• We can define a Boolean function:
– F(x,y) = (xy)(xy)
• And then write a “truth table” for it:
x y F(x,y)
1 1 0
1 0 0
0 1 0
0 0 0 4
 Basic logic gates
x
• Not x
x xy x xyz
• And y y
z
x+y x x+y+z
x y
• Or y z
x xy
• Nand y
x x+y
• Nor y
x xÅy
• Xor y
5
 Converting between circuits and
equations
• Find the output of the following circuit

x x+y
y (x+y)y

y y

__
• Answer: (x+y)y
– Or (xy)y 6
 Converting between circuits and
equations
• Find the output of the following circuit

x
x xy xy
y
y
___
__
• Answer: xy
– Or (xy) ≡ xy 7
 Converting between circuits and
equations
• Write the circuits for the following
Boolean algebraic expressions
__
a) x+y

x
x x+y

8
 Converting between circuits and
equations
• Write the circuits for the following
Boolean
_______
algebraic expressions
b) (x+y)x

x x+y
x+y (x+y)x
y

9
 Writing xor using and/or/not
• p  q  (p  q)  ¬(p  q) x y xy
____ 1 1 0
• x  y  (x + y)(xy) 1 0 1
0 1 1
0 0 0

x x+y (x+y)(xy)
y
xy xy
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 Converting decimal numbers to
binary
• 53 = 32 + 16 + 4 + 1
= 25 + 24 + 22 + 20
= 1*25 + 1*24 + 0*23 + 1*22 + 0*21 + 1*20
= 110101 in binary
= 00110101 as a full byte in binary

• 211= 128 + 64 + 16 + 2 + 1
= 27 + 26 + 24 + 21 + 20
= 1*27 + 1*26 + 0*25 + 1*24 + 0*23 + 0*22 +
1*21 + 1*20
= 11010011 in binary
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 Converting binary numbers to
decimal
• What is 10011010 in decimal?
10011010 = 1*27 + 0*26 + 0*25 + 1*24 + 1*23 +
0*22 + 1*21 + 0*20
= 27 + 24 + 23 + 21
= 128 + 16 + 8 + 2
= 154

• What is 00101001 in decimal?

00101001 = 0*27 + 0*26 + 1*25 + 0*24 + 1*23 +
0*22 + 0*21 + 1*20
= 25 + 23 + 20
= 32 + 8 + 1
= 41 12
 A note on binary numbers
• In this slide set we are only dealing with
non-negative numbers
• The book (section 1.5) talks about two’s-
complement binary numbers
– Positive (and zero) two’s-complement binary
numbers is what was presented here
– We won’t be getting into negative two’s-
complmeent numbers

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End of lecture on 24
January 2007

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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 x y Carry Sum
– 1+0 = 1
– 1+1 = 10 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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 The half-adder
x y Carry Sum
• Sum = x XOR y 0 0 0 0
0 1 0 1
• Carry = x AND y
1 0 0 1
1 1 1 0

x x
y y Sum
Sum
Carry
Carry
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 Using half adders
• We can then use a half-adder to compute
the sum of two Boolean numbers

1 0 0
1 1 0 0
+1 1 1 0
? 0 1 0

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 How to fix this
• We need to create an adder that can take a
carry bit as an additional input
x y c carry sum
– Inputs: x, y, carry in
1 1 1 1 1
– Outputs: sum, carry out
1 1 0 1 0
• This is called a full adder
– Will add x and y with a half-adder 1 0 1 1 0
– Will add the sum of that to the 1 0 0 0 1
carry in 0 1 1 1 0
• What about the carry out? 0 1 0 0 1
– It’s 1 if either (or both): 0 0 1 0 1
– x+y = 10
0 0 0 0 0
– x+y = 01 and carry in = 1 18
 x y c s1 c1 carry sum
The full adder 1 1 1 0 1 1 1
1 1 0 0 1 1 0
• The “HA” boxes are 1 0 1 1 0 1 0
1 0 0 1 0 0 1
half-adders
0 1 1 1 0 1 0
0 1 0 1 0 0 1
0 0 1 0 0 0 1
0 0 0 0 0 0 0
c X HA S
s
s1 Y C

x X HA S

c
y Y C

c1
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 The full adder
• The full circuitry of the full adder

c
s

x
y
c

20
 Adding bigger binary numbers
• Just chain full adders together

x0 X HA S
s0
y0 Y C

x1
C

X
FA S
s1
y1 Y C

x2
C

X
FA S

s2
y2 Y C

x3
C

X
FA S
s3
y3 Y C
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? 24

 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
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Hexadecimal
• A numerical range
from 0-15
– Where A is 10, B is 11,
… and F is 15
• Often written with a
‘0x’ prefix
• So 0x10 is 10 hex, or
16
– 0x100 is 100 hex, or
256
• Binary numbers easily
translate: 26
 From
ThinkGeek
(http://www.thinkgeek.com)

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 Also from
ThinkGeek
(http://www.thinkgeek.com)

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 DEADBEEF
• Many IBM machines would fill allocated
(but uninitialized) memory with the hexa-
decimal pattern 0xDEADBEEF
– Decimal -21524111
– See http://www.jargon.net/jargonfile/d/DEADBEEF.html

• Makes it easier to spot in a debugger

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 Also also from ThinkGeek
(http://www.thinkgeek.com)

• 0xDEAD = 57005
• Now add one to that...

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