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CS 140L Lecture 1 Professor CK Cheng 3/31/02
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CMOS Logic (3.2 – 3.6) Complementary Metal-Oxide Semiconductor
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1. Transistors (FET) Field Effect Transistor G S D NMOS D WG S D L G =1 => D =S { D = 1 => S = 0.8 D = 0 => S = 0 G = 0 => D&S are separated.
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G C D S r L W, c WxL A C ’ C B G S D PMOS G = 0 => D =S { D = 1 => S = 1 D = 0 => S = 0.2 G = 1 => D&S are separated.
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2. Transmission Gate A C C BC = 0 => A&B are separated. C = 1 => A = B
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3. Inverter AB A = 1 => B = 0 A = 0 => B = 1 Thus B = A
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4. NAND gate B A A B A B Z 0 0 1 0 1 1 1 0 1 1 1 0 Z = D iff AB = 1. Z = (AB)’ Z
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5. NOR gate B A A B A B Z 0 0 1 0 1 0 1 0 0 1 1 0 Z = 1 iff A+B = 0. Z = (AB)’ = (A+B)’ Z
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6. XOR gate B A Z B’ A’ B A B’ A’ Z = ab’ + a’b = a + b
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7. Multiplexer d1 d2 S S S ‘ Z If S=1, Z = d1 Else (S=0), Z = d0 Z = S d1 + S’ d0
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Basic Boolean Algebra a + a = 1 a * a = 0 1 + a = 1 1*a = 1 a + a = a a * a = a a a
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DeMorgan’s Law (a + b)’ = a’ * b’ (a*b)’ = a’ + b’ ab
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