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 AB = 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

10. Inversion (NOT)
A Q
0 1
1 0
AQ =Logic:

11. 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

12. Getting the XOR
A B S
0 0 0
1 0 1
0 1 1
1 1 0
BAorBA ⋅⋅
Two ways of getting S = 1

13. Circuit for XOR
BABABA ⋅+⋅=⊕
Accumulating our results: Binary addition is the
result of XOR plus AND

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)

15. 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

16. Full Adder Circuit

17. Chaining the Full Adder
Possible to use the same
scheme for subtraction by
noting that
A – B = A + (-B)

18. Binary Counting
Use 1 for ON
Use 0 for OFF
= 00101011
Binary Counter
So our example has 25
+ 23
+ 21
+ 20
= 32 + 8 + 2 + 1 = 43

19. 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

20. NAND (NOT AND)
A B Q
0 0 1
0 1 1
1 0 1
1 1 0
BAQ ⋅=

21. NOR (NOT OR)
A B Q
0 0 1
0 1 0
1 0 0
1 1 0
BAQ +=

22. Exclusive NOR
A B Q
0 0 1
0 1 0
1 0 0
1 1 1
BAQ ⊕=
Equality Detector

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

25. 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

26. 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

27. Example
1 1 0 1 1 0 0 1
27
26
25
20
21
22
24 23

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

30. Another Example
 144
 27
= 128 144 – 128 = 16
 24
= 16 16 – 16 = 0
 Result 10010000

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

32. 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

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