Logic Gates

Transistors as Switches

VBB voltage controls whether the transistor conducts in a common base configuration.

Logic circuits can be built

AND

In order for current to flow, both switches must be closed¤ Logic notation AB = C

(Sometimes AB = C)

A B C

0 0 0

0 1 0

1 0 0

1 1 1

OR

Current flows if either switch is closed¤ Logic notation A + B = C

A B C

0 0 0

0 1 1

1 0 1

1 1 1

Properties of AND and OR

Commutation¤ A + B = B + A¤ A B = B A

Same as

Same as

Properties of AND and OR

Associative Property¤ A + (B + C) = (A + B) + C

¤ A (B C) = (A B) C

=

Properties of AND and OR

Distributive Property¤ A + B C = (A + B) (A + C)¤ A + B C

A B C Q

0 0 0 0

0 0 1 0

0 1 0 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

Distributive Property

(A + B) (A + C)

A B C Q

0 0 0 0

0 0 1 0

0 1 0 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

Binary Addition

A B S C(arry)

0 0 0 0

1 0 1 0

0 1 1 0

1 1 0 1

Notice that the carry results are the same as AND

C = A B

Inversion (NOT)

A Q

0 1

1 0Logic:

AQ

Exclusive OR (XOR)

Either A or B, but not both

This is sometimes called the inequality detector, because the result will be 0 when the inputs are the same and 1 when they are different.

The truth table is the same as for S on Binary Addition. S = A B

A B S

0 0 0

1 0 1

0 1 1

1 1 0

Getting the XOR

A B S

0 0 0

1 0 1

0 1 1

1 1 0

Two ways of getting S = 1

BAor BA

Circuit for XOR

Accumulating our results: Binary addition is the result of XOR plus AND

BA BABA

Half Adder

Called a half adder because we haven’t allowed for any carry bit on input. In elementary addition of numbers, we always need to allow for a carry from one column to the next.

18

25

4

3 (plus a carry)

Full Adder

INPUTS OUTPUTS

A B CIN COUT S

0 0 0 0 0

0 0 1 0 1

0 1 0 0 1

0 1 1 1 0

1 0 0 0 1

1 0 1 1 0

1 1 0 1 0

1 1 1 1 1

Full Adder Circuit

Chaining the Full Adder

Possible to use the same scheme for subtraction by

noting that

A – B = A + (-B)

Binary CountingUse 1 for ON

Use 0 for OFF

= 00101011

Binary Counter

So our example has 25 + 23 + 21 + 20 = 32 + 8 + 2 + 1 = 43

Counting in Binary

1 1 11 1011 21 10101

2 10 12 1100 22 10110

3 11 13 1101 23 10111

4 100 14 1110 24 11000

5 101 15 1111 25 11001

6 110 16 10000 26 11010

7 111 17 10001 27 11011

8 1000 18 10010 28 11100

9 1001 19 10011 29 11101

10 1010 20 10100 30 11110

NAND (NOT AND)

A B Q

0 0 1

0 1 1

1 0 1

1 1 0

BAQ

NOR (NOT OR)

A B Q

0 0 1

0 1 0

1 0 0

1 1 0

BAQ

Exclusive NOR

A B Q

0 0 1

0 1 0

1 0 0

1 1 1

Equality Detector

BAQ

Summary

Summary for all 2-input gates

Inputs Output of each gate

 A   B  AND NAND  OR  NOR XOR XNOR

0 0 0 1 0 1 0 1

0 1 0 1 1 0 1 0

1 0 0 1 1 0 1 0

1 1 1 0 1 0 0 1

Number Systems

Decimal (base 10) {0 1 2 3 4 5 6 7 8 9}¤ Place value gives a logarithmic representation

of the number¤ Ex. 4378 means

۞ 4 X 103 = 4000۞ 3 X 102 = 300۞ 7 X 101 = 70۞ 8 X 100 = 8

¤ The place also gives the exponent of the base

Example

432,600

4 3 2 6 0 0

105

104

103

100

101

102

Powers of ten:

100 = 1 102 = 100 104 = 10000

101 = 10 103 = 1000 105 = 100000

Binary (base 2) {0 1}Binary Decimal

0 0

1 1

10 2

11 3

100 4

101 5

110 6

111 7

1000 8

1001 9

1010 10

Example

1 1 0 1 1 0 0 1

27

26

25

20

21

22

24 23

Decimal Equivalent

1101 1001 1 X 27 = 128 + 1 X 26 = 64 + 0 X 25 = 0 + 1 X 24 = 16 + 1 X 23 = 8 + 0 X 22 = 0 + 0 X 21 = 0 + 1 X 20 = 1 217

Notice how powers of two stand out:

20 = 1

21 = 10

22 = 100

23 = 1000

Decimal to Binary Conversion

Ex. 575¤ Find the largest power of two less than the number

۞ 29 = 512

¤ Subtract that power of two from the number۞ 575 – 512 = 63

¤ Repeat steps 1 and 2 for the new result until you reach zero.۞ 25 = 32 63 – 32 = 31۞ 24 = 16 31 – 16 = 15۞ 23 = 8 15 – 8 = 7۞ 22 = 4 7 – 4 = 3۞ 21 = 2 3 – 2 = 1۞ 20 = 1 1 – 1 = 0

¤ Construct the number۞ 1000111111

Another Example

144¤ 27 = 128 144 – 128 = 16¤ 24 = 16 16 – 16 = 0

Result 10010000

Hexadecimal (base 16)

{0 1 2 3 4 5 6 7 8 9 A B C D E F} Assignments Dec Hex Dec Hex

0 0 8 8

1 1 9 9

2 2 10 A

3 3 11 B

4 4 12 C

5 5 13 D

6 6 14 E

7 7 15 F

Example

163

162

160

161

3 B 6 E

3 X 163 = 12288

11 X 162 = 2816

6 X 161 = 96

14 X 160 = 14

15214

Hexadecimal is Convenient for Binary Conversion

Binary Hex Binary Hex

0 0 1001 29

1 1 1010 A

10 2 1011 B

11 3 1100 C

100 4 1101 D

101 5 1110 E

110 6 1111 F

111 7 1 0000 10

1000 8 Nibble

Binary to Hex Conversion

Group binary number by fours (nibbles)¤ 1101 1001 0110

Convert each nibble into hex equivalent¤ 1101 1001 0110

D 9 6

Decimal to Hex Conversion

Ex. 284¤ 162 = 256 284 – 256 = 28¤ 161 = 16 28 - 16 = 12 (Hex C)

¤ Result 1 1 C

Another Example with an Extension

1054¤ 162 = 256

۞But we have several multiples of 256 in 1054– 1054/256 = 4.12 take integer part– This eliminates 4*256 = 1024

۞ 1054 – 1024 = 30

¤ 161 = 16 30 – 16 = 14 (Hex E)

¤ Result 4 1 E