2. Transistors as Switches
VBB voltage controls whether the transistor
conducts in a common base configuration.
Logic circuits can be built
3. 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
4. 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
5. Properties of AND and OR
Commutation
¤ A + B = B + A
¤ A B = B A
Same as
Same as
6. Properties of AND and OR
Associative Property
¤ A + (B + C) = (A + B) + C
¤ A (B C) = (A B) C
=
7. 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
8. 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
9. 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
11. 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
12. Getting the XOR
A B S
0 0 0
1 0 1
0 1 1
1 1 0
Two ways of getting S = 1
BAorBA ⋅⋅
14. 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)
22. Exclusive NOR
A B Q
0 0 1
0 1 0
1 0 0
1 1 1
Equality Detector
BAQ ⊕=
23. 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
24. 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
28. 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
29. 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
31. 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
33. 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
34. 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
35. Decimal to Hex Conversion
Ex. 284
¤ 162
= 256 284 – 256 = 28
¤ 161
= 16 28 - 16 = 12 (Hex C)
¤ Result 1 1 C
36. 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