ECE 231 -7

ECE 231 -7 - ECE-231 Circuits and Systems I Spring 2011...

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ECE-231 Circuits and Systems I Spring 2011 Session 7 Professor Stewart Personick Office: ECEC Room 321 [email protected]
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Review of the last part of Session 6
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Network component: Capacitor + - Vs= V Rs = 100 a b Let’s assume that, for t<0 seconds (when the switch is open), there is no charge, Q, on the capacitor . i.e. no extra electrons on the plate on the right, and no missing electrons on the plate on the left. Vab = Q/C = 0 Coulombs /C (Farads) = 0 Volts; for t<0 seconds t<0 seconds C (Farads)
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Network component: Capacitor + - Rs = 100 a b (KVL): V = Rs x i(t) + Vab(t) Vab(t) = Q(t)/C Therefore: V= Rs x i(t) + Q(t)/C; for t>0 seconds i(t) - + Vab Vs= V t>0 seconds C (Farads)
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Network component: Capacitor + - Vs= V Rs = 100 i(t) - e - e - e - e e e e e electrons i(t) = d[Q(t)]/dt = -1.6 x 10-19 x the number of electrons flowing (counterclockwise) per second through any node in the mesh Q(t)= C x Vab(t) Therefore: i(t) = C d[Vab(t)]/dt a b electrons C (Farads)
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Network component: Capacitor + - Rs = 100 a b V= Rs x i(t) + Q(t)/C, Vab(t) =Q(t)/C i(t) = d[Q(t)]/dt = C x d[Vab(t)]/dt Therefore: Rs x d[Q(t)]/dt + Q(t)/C - V = 0; for t>0 seconds RsC x d[Q(t)/C]/dt + Q(t)/C - V = 0; for t>0 seconds RsC x d[Vab(t)]/dt + Vab(t) - V = 0 ; for t>0 seconds i(t) Vs= V t>0 seconds - + Vab C (Farads)
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Network component: Capacitor + - Rs = 100 a b At t=0, Vab(t)= 0V (there is no charge on either plate of the capacitor) When the switch is closed (at t=0), current starts to flow Vab(t) + RsC x d[Vab(t)]/dt - V = 0 ; for t>0 seconds i(t) Vab(t ) = V [1 - e -(t/ R s C ) ] RsC = τ , where τ is called the “time constant of the circuit” t V Vs= V C
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Network component: Capacitor + - a b i(t) Vab(t) = V [1 - e -(t/ R s C) ] i(t) = C x d[Vab(t)]dt =[V/Rs ] e -(t/ R s C ) t Vs /Rs Vs= V C Rs At t=0, Vab(t)= 0V (there is no charge on either plate of the capacitor) When the switch is closed (at t=0), current starts to flow Vab(t) + RsC x d[Vab(t)]/dt - V = 0 ; for t>0 seconds
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Simulation Vs = 18V, Rs=20 Ohms, and C= 0.015F Rs x C = 0.3 seconds n003
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18V x [1- exp(-3)] = 17.1V
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18V/20 Ohms = 900mA
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Network component: Capacitor + - Vs= V - e - e - e - e e e e e - + The energy stored in the capacitor, in the steady state = the integral ( from t=0 to t>> RsC ) of Vab(t) x i(t) dt = the integral ( from t=0 to t>> RsC ) of Vab(t) x {C d[Vab(t)]/dt} dt = the integral ( from Vab =0 to Vab=V ) of Vab(t) x C d[Vab(t)] = ½ x C x V2 (Joules) b i(t) = 0 Vab= V t>> RsC Rs
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Network component: Capacitor + - Vs= V - e - e - e - e e e e e - + More generally, the energy stored in the capacitor, at any time t , = ½ x C x [Vab(t)]2 (Joules) b i(t) = 0 Vab= V t>> RsC Rs
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New materials
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It’s time for: the  Professor’s Basement Kids: you can do this at home if you have the parts in your basement
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Demonstration: Pre-charging phase + - Rs = 50 a b For t<0s, the switch is open Vab(t)=0 (no charge on either plate of the capacitor) i(t)=0 i(t) t Vs= 18V C=15,000 μ F Lamp: RL = 50 The switch is open for: t<0s
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Demonstration: Charging phase + - Rs = 50 a b At t=0+, the switch is closed. Vab=0 When the switch is closed (at t=0s), current starts to flow Vab(t) + 1.5s x d[Vab(t)]/dt - 18V = 0; for t>0s The maximum current is 18V/100 = 0.18 Amperes i(t) t Vs= 18V 4.5s C=15,000 μ F Lamp: RL = 50 i(t) = [18V /100 Ω] e - (t/ 1.5 s ) ] 0.18A RC = 100 x 0.015F = 1.5 seconds The switch is closed at t=0s
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Demonstration: Charged phase Rs = 50 a b
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ECE 231 -7 - ECE-231 Circuits and Systems I Spring 2011...

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