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 0AQ Logic:
Exclusive OR (XOR)
A B S
0 0 0
1 0 1
0 1 1
1 1 0
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
Getting the XOR
A B S
0 0 0
1 0 1
0 1 1
1 1 0
BAor BA
Two ways of getting S = 1
Circuit for XOR
BA BABA
Accumulating our results: Binary addition is the result of XOR plus AND
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
BAQ
Equality Detector
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
432,6004 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
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