Lect39LogicGates

# T2t z 50 ω z0 zl 50v t1 μ sec 200 ω v l 50 i l i l

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t=2T 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 + + V " + V " + I S = V S /50 50 V = 200 I S + V S I L V L + - V L = 50 I L I L I L = 2 V + " V L 50 Straight Line slope = +1/50 intercept = -2V - /50 I S = I + + I " + I " + I S = " 2 V " + 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 Solution for I and V at z=0 at t=2T is point “C” V L = 50 I L I L I S = " 2 V " + V S 50 C B A
I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S A Solution for I and V at z=l at t=3T is point “D” V L = 50 I L I L B C D

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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 A V L = 50 I L I L B C D Steady state: Intersection of source and load lines SS
I [A] V [V] 2 4 6 0.1 0.2 0.3 0.4 8 10 50 V = 200 I S + V S A Steady state: Intersection of source and load lines V L = 50 I L I L B C D SS

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Outline of load line method Determine “source line” from I-V relationship at the source Determine “load line” from I-V relationship at the load Plot source line and load line together on an I-V graph Start with V + wave at source. Plot a line with slope = +1/Z 0 and intercept at origin (line with zero init cond) Intersection with source line gives initial I and V at source side.
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