Lect39LogicGates - Interconnection between logic gates L e...

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Interconnection between logic gates Lecture 39
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Nonlinear circuit elements What if the TL is attached to a “device” with a complex I-V characteristic? – Diode – Transistor – Amplifier “Load Line” method can be used as a substitute for bounce diagrams
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Example Z 0 =50 Ω z=0 z=l 50V t=0, close T=1 μ sec 200 Ω V L = 50 I L I L Source Side Load Side
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Example, t=0+ Z 0 =50 Ω z=0 z=l 50V T=1 μ sec 200 Ω V L = 50 I L I L Source Side Load Side V s I s + - V S = V + I S = I + = V + / Z 0 = V S /50 50 V = 200 I S + V S
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Source Line I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S
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I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S I S = V S /50
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I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S I S = V S /50 A Point “A” is solution to V + and I +
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t=T Z 0 =50 Ω z=0 z=l 50V T=1 μ sec 200 Ω V L = 50 I L I L Source Side Load Side V s I s + - V S = V + I S = V S /50 50 V = 200 I S + V S I L V L + - V L = 50 I L I L V L = V + + V " I L = I + + I " I L = 2 V + " V L 50 Straight Line slope = -1/50 intercept = 2V + /50
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I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S Load Line V L = 50 I L I L A
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I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S Solution for I and V at z=l at t=T is point “B” V L = 50 I L I L I L = 2 V + " V L 50 B A
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