04 Logic Gates1

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Logic GatesLogic GatesCS/APMA 202, Spring 2005CS/APMA 202, Spring 2005

Rosen, section 10.3Rosen, section 10.3 Aaron Bloomfield Aaron Bloomfield





33

Review of Boolean algebraReview of Boolean algebra

Not is a horizontal bar above the number Not is a horizontal bar above the number 0 = 10 = 1 1 = 01 = 0

Or is a plusOr is a plus 0+0 = 00+0 = 0 0+1 = 10+1 = 1 1+0 = 11+0 = 1 1+1 = 11+1 = 1

And is multiplication And is multiplication 0*0 = 00*0 = 0 0*1 = 00*1 = 0 1*0 = 01*0 = 0

1*1 = 11*1 = 1

_ _

_ _





5

Quick surveyQuick survey

I understand the basics of BooleanI understand the basics of Boolean

algebraalgebra

a)a)

Absolutely!Absolutely!

b)b) More or lessMore or less

c)c) Not reallyNot really

d)d) Boolean what?Boolean what?



66

Today’s demotivatorsToday’s demotivators



77

Basic logic gatesBasic logic gates

NotNot

And And

Or Or

NandNand

Nor Nor

Xor Xor

x x

x y

xy x y

xyz

z x +y x y

x y

x +y +z

z

x y

xy

x +y x y

x ⊕y x y



88

Rosen, §10.3 question 1Rosen, §10.3 question 1

Find the output of the following circuitFind the output of the following circuit

Answer: ( Answer: ( x+y x+y )y)y Or (Or ( x x ∨∨y y ))∧¬∧¬yy

x y x +y

y

( x +y )y

__ __

x y

y



99

x

y


Find the output of the following circuitFind the output of the following circuit

Answer: xy Answer: xy Or Or ¬¬((¬¬ x x ∧¬∧¬y y ) ≡) ≡ x x ∨∨y y

x

y

x y x y

_ _ _ _ ___ ___



10


I understand how to figure out whatI understand how to figure out what

a logic gate doesa logic gate does

a)a)

Absolutely!Absolutely!



d)d) Not at allNot at all



1111


Write the circuits for the followingWrite the circuits for the following

Boolean algebraic expressionsBoolean algebraic expressions

a)a)

x x

++

y y

x

y

x

y

x

y

__ __

x x +y



1212

x

y

x

y

x

y


Write the circuits for the followingWrite the circuits for the following

Boolean algebraic expressionsBoolean algebraic expressions

b)b)

(( x x

++

y y )) x x

_______ _______

x

y

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



1313

Writing xor using and/or/notWriting xor using and/or/not

p p ⊕⊕ qq ≡≡ (( p p ∨∨ qq)) ∧∧¬(¬( p p ∧∧ qq))

x x ⊕⊕ yy ≡≡ (x + y)((x + y)( xy xy ))

x y x ⊕y

1 1 0

1 0 1

0 1 1

0 0 0

x

y

x +y

xy xy

( x +y )(xy)

____ ____



14


I understand how to write a logicI understand how to write a logic

circuit for simple Boolean formulacircuit for simple Boolean formula

a)a) Absolutely!Absolutely!






1515

Converting decimal numbers toConverting decimal numbers to

binarybinary

5353 = 32 + 16 + 4 + 1= 32 + 16 + 4 + 1

= 2= 255 + 2+ 244 + 2+ 222 + 2+ 200

= 1*2= 1*255 + 1*2+ 1*244 + 0*2+ 0*233 + 1*2+ 1*222 + 0*2+ 0*211 + 1*2+ 1*200

= 110101 in binary= 110101 in binary= 00110101 as a full byte in binary= 00110101 as a full byte in binary

211= 128 + 64 + 16 + 2 + 1211= 128 + 64 + 16 + 2 + 1

= 2= 277 + 2+ 266 + 2+ 244 + 2+ 211 + 2+ 200

= 1*2= 1*277 + 1*2+ 1*266 + 0*2+ 0*255 + 1*2+ 1*244 + 0*2+ 0*233 + 0*2+ 0*222 ++

1*21*211 + 1*2+ 1*200

= 11010011 in binary= 11010011 in binary



1616

Converting binary numbers toConverting binary numbers to

decimaldecimalWhat is 10011010 in decimal?What is 10011010 in decimal?1001101010011010 = 1*2= 1*277 + 0*2+ 0*266 + 0*2+ 0*255 + 1*2+ 1*244 + 1*2+ 1*233 ++

0*20*222 + 1*2+ 1*211 + 0*2+ 0*200

= 2= 277 + 2+ 244 + 2+ 233 + 2+ 211

= 128 + 16 + 8 + 2= 128 + 16 + 8 + 2= 154= 154

What is 00101001 in decimal?What is 00101001 in decimal?

00101001 = 0*200101001 = 0*277 + 0*2+ 0*266 + 1*2+ 1*255 + 0*2+ 0*244 + 1*2+ 1*233 ++0*20*222 + 0*2+ 0*211 + 1*2+ 1*200

= 2= 255 + 2+ 233 + 2+ 200

= 32 + 8 + 1= 32 + 8 + 1

= 41= 41



1717

A bit of binary humor A bit of binary humor

Available for $15 at Available for $15 athttp://www.thinkgeek.com/http://www.thinkgeek.com/tshirts/frustrations/5aa9/tshirts/frustrations/5aa9/



18


I understand the basics of I understand the basics of

converting numbers betweenconverting numbers between

decimal and binarydecimal and binary



c)c) Not reallyNot reallyd)d) Not at allNot at all



1919

How to add binary numbersHow to add binary numbers

Consider adding two 1-bit binary numbersConsider adding two 1-bit binary numbers x x andand y y 0+0 = 00+0 = 0

0+1 = 10+1 = 1

1+0 = 11+0 = 1

1+1 = 101+1 = 10

Carry isCarry is x x AND AND y y

Sum isSum is x x XORXOR y y

The circuit to compute this is called a half-adder The circuit to compute this is called a half-adder

x y Carry Sum

0 0 0 00 1 0 1

1 0 0 1

1 1 1 0



2020

The half-adder The half-adder

Sum =Sum = x x XORXOR y y

Carry =Carry = x x AND AND y y

x y Sum

Carry

x y Sum

Carry



2121

Using half addersUsing half adders

We can then use a half-adder to computeWe can then use a half-adder to compute

the sum of two Boolean numbersthe sum of two Boolean numbers

1 1 0 0

+ 1 1 1 0

010?

001



22


I understand half addersI understand half adders


b)b) More or lessMore or lessc)c) Not reallyNot really






2424

The full adder The full adder

The “HA” boxes are half-addersThe “HA” boxes are half-adders

HAX

Y

S

C

HAX

Y

S

C

x

y

c

c

s

HAX

Y

S

C

HAX

Y

S

C

x

y

c



2525

The full adder The full adder

The full circuitry of the full adder The full circuitry of the full adder

x

y

s

c

c



2626

Adding bigger binary numbers Adding bigger binary numbers

Just chain full adders together Just chain full adders together

HAX

Y

S

C

FAC

Y

X

S

C

FAC

Y

X

S

C

FAC

Y

X

S

C

x 1y 1

x 2 y 2 x 3y 3

x 0

y 0 s0

s1

s2

s3c



2727

Adding bigger binary numbers Adding bigger binary numbers

A half adder has 4 logic gates A half adder has 4 logic gates

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

To addTo add nn bit binary numbers, you need 1 HA andbit binary numbers, you need 1 HA andnn-1 FAs-1 FAs

To add 32 bit binary numbers, you need 1 HATo add 32 bit binary numbers, you need 1 HAand 31 FAsand 31 FAs Total of 4+9*31 = 283 logic gatesTotal of 4+9*31 = 283 logic gatesTo add 64 bit binary numbers, you need 1 HATo add 64 bit binary numbers, you need 1 HAand 63 FAsand 63 FAs Total of 4+9*63 = 571 logic gatesTotal of 4+9*63 = 571 logic gates



28


I understand (more or less) aboutI understand (more or less) about

adding binary numbers using logicadding binary numbers using logic

gatesgates



c)c)

Not reallyNot reallyd)d) Not at allNot at all



2929

More about logic gatesMore about logic gates

To implement a logic gate in hardware,To implement a logic gate in hardware,

you use a transistor you use a transistor

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

integrated circuitintegrated circuit

The current Intel Pentium IV processorsThe current Intel Pentium IV processors

have 55 million transistors!have 55 million transistors!



3030

Pentium math error 1Pentium math error 1

Intel’s PentiumsIntel’s Pentiums(60Mhz – 100 Mhz)(60Mhz – 100 Mhz)had a floating pointhad a floating pointerrorerror

Graph of z = y/xGraph of z = y/x

Intel reluctantlyIntel reluctantly

agreed to replaceagreed to replacethem in 1994them in 1994

Graph from http://kuhttp.cc.ukans.edu/cwis/units/IPPBR/pentium_fdiv/pentgrph.htmlGraph from http://kuhttp.cc.ukans.edu/cwis/units/IPPBR/pentium_fdiv/pentgrph.html





3232

Flip-flopsFlip-flops

Consider the following circuit:Consider the following circuit:

What does it do?What does it do?



3333

MemoryMemory

A flip-flop holds a single bit of memory A flip-flop holds a single bit of memory The bit “flip-flops” between the two NANDThe bit “flip-flops” between the two NAND

gatesgates

In reality, flip-flops are a bit moreIn reality, flip-flops are a bit more

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

flopflop

Consider a 1 Gb memory chipConsider a 1 Gb memory chip

1 Gb = 8,589,934,592 bits of memory1 Gb = 8,589,934,592 bits of memory That’s about 43 million transistors!That’s about 43 million transistors!

In reality, those transistors are split into 9In reality, those transistors are split into 9

ICs of about 5 million transistors eachICs of about 5 million transistors each



34


I felt I understood the material in thisI felt I understood the material in this

slide set…slide set…

a)a) Very wellVery well

b)b) With some review, I’ll be goodWith some review, I’ll be good





35


The pace of the lecture for this slide The pace of the lecture for this slide

set was…set was…

a)a) FastFast

b)b) About rightAbout right

c)c) A little slowA little slow

d)d) Too slow Too slow



3636

End of lecture on 27 January 2005End of lecture on 27 January 2005

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