W09 314 HW06 solution

W09 314 HW06 solution - Solution by Idong Ebong W09 EECS...

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Unformatted text preview: Solution by Idong Ebong W09 EECS 314 HW06 Problem 1 Solution Given – series RC circuit; V S is governed by the pulse function to the right. Assume DC steady state conditions for time less than 0 ‒ Part 1 Given R=150Ω, determine the range of capacitances such that the capacitor voltage reaches at least 3.5V within the pulse duration of 200ps. V C = V S 1 − e − t τ ¡ for t > 0 3.5V ≤ 5V 1 − e − 200ps 150 Ω ×C ¡ Solving for capacitance gives C ≤ 1.107pF Part 2 Given C=2.5pF, determine the range of resistances such that the capacitor voltage reaches at least 3.5V within the pulse duration of 200ps. 3.5V ≤ 5V ¢ 1 − e − 200ps 2.5pF×R £ Solving for resistance gives R ≤ 66.45Ω Part 3 V C From KVL , V is continuous . S = V R + V C V R = V S – V At t=0, V C C =0V but V R switches from 0V at t=0 – to 5V-0V=5V at t=0 + since V S At t= 2τ, V switches from 0V to 5V. V C = V S 1 − e − t τ ¡ for t > 0 during the pulse C =4.323V but V R switches from 5V-4.323V =0.677V at t= 2τ – to 0V-4.323V=-4.323V at t= 2τ + since V S switches from 5V to 0V. After Pulse At 4.5τ, V : The capacitor discharges from its initial voltage V C = 4.323V e − t − 2 τ τ for t > 2 ? C =0.355V so V R =-0.355V + V R- Solution by Idong Ebong V C,max V =4.323V R,max =5V Note to graders: Depending on how V R is defined on diagram , V R,max can be - 5V, and the graph of V R would be mirrored about the x- axis. Solution by Idong Ebong W09 EECS 314 HW06 Problem 2 Solution Given – series RL circuit; V IN is governed by the pulse function to the right. Assume DC steady state conditions for time less than 0 ‒ Part 1 Derive the algebraic equation for the inductor voltage and the resistor voltage during the pulse. Split up your time space into 3 regions – Before Pulse, During Pulse, and After Pulse. RL circuit so variable of interest is the inductor current – i L (t) – since it’s the variable whose graph must be continuous in time. Figure out initial conditions (i.c.) and final conditions (f.c.) on i L . Before Puls e , we have DC steady state so inductor is seen as a short (V L i L (t = 0 − ) = V IN R = 10V R =0V) (I.C.) During Pulse i L (t = ∞ ) = V IN R = 5V R , if we had a steady state, the current will want to settle to (F.C.) We know this is a first order circuit with τ=L/R so we can write a general form solution i L (t) = Ke − t τ + C i L (t = 0 − ) = i L (t = 0 + ) = 10V R = K + C i L (t = ∞ ) = 5V R = C yields ¡⎯⎯¢ K = 10V R − 5V R = 5V R ? £ ( ¤ ) = ¥¦ § ¨© − ¤ ª + «¬ where ª = £ § ­®¯ 0 < ° < 1 ±² ¦ § ( ¤ ) = ? £ ( ¤ ) × § = ¥¦ ¨© − ¤ ª + «¬ where ª = £ § ­®¯ 0 < ° < 1 ±² Note: this equation is only valid during the pulse because of the way it was derived. We all know the pulse is finite so we might not reach the steady state current of 5V/R depending on pulse width....
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This note was uploaded on 10/28/2009 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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W09 314 HW06 solution - Solution by Idong Ebong W09 EECS...

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