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Transcript
First Order Circuit • Capacitors and inductors • RC and RL circuits RC and RL circuits (first order circuits) Circuits containing no independent sources • Excitation from stored energy • ‘source-free’ circuits • Natural response Circuits containing independent sources • DC source (voltage or current source) • Sources are modeled by step functions • Step response • Forced response Complete response = Natural response + forced response RC circuit – natural response Assume that capacitor is initially charged at t = 0 + ic C iR vc vc(0) = Vo R Objective of analysis: to find expression for vc(t) for t >0 i.e. to get the voltage response of the circuit Taking KCL, C dv c v c 0 dt R t dv c dt vc (0) v c o RC v c (t ) v c (0)e dv c v c dt RC dv c dt vc RC vc (t ) 1 t RC ln OR v c (t) 1 t v c (0) RC v c (t ) Voe 1 t RC RC circuit – natural response t / • Can be written as v c ( t ) v c (0)e , = RC time constant • This response is known as the natural response vC(t) Voltage decays to zero exponentially Vo At t=, vc(t) decays to 37.68% of its initial value The smaller the time constant the faster the decay 0.3768Vo t t= v c (t ) v c (0)e 1 t RC RC circuit – natural response The capacitor current is given by: iC C dv c dt iC And the current through the resistor is given by iR Vo e R pR v RiR 2 t R The energy loss (as heat) in the resistor from 0 to t: t ER pR dt 0 t o Vo2 e 2 R t t Vo2 2 t dt e 2 R 0 t 2 1 2 ER CVo 1 e 2 t v C (t) Ve o R R The power absorbed by the resistor can be calculated as: Vo2e t RC circuit – natural response As t , ER 1 2 CVo 2 As t , energy initially stored in capacitor will be dissipated in the resistor in the form of heat t 2 1 2 ER CVo 1 e 2 RC circuit – natural response PSpice simulation 1 + ic C RC circuit c1 1 0 1e-6 IC=100 r1 1 0 1000 .tran 7e-6 7e-3 0 7e-6 UIC .probe .end iR vc R 0 100V 50V 0V 0s 1.0ms 2.0ms 3.0ms 4.0ms V(1) Time 5.0ms 6.0ms 7.0ms RC circuit – natural response PSpice simulation RC circuit .param c=1 c1 1 0 {c} IC=100 r1 1 0 1000 .step param c list 0.5e-6 1e-6 3e-6 .tran 7e-6 7e-3 0 7e-6 .probe .end 1 + ic C iR vc R 0 100V c1 = 1e-6 c1 = 3e-6 50V 0V 0s c1 = 0.5e-6 1.0ms 2.0ms 3.0ms 4.0ms V(1) Time 5.0ms 6.0ms 7.0ms RL circuit – natural response iL vL + L Assume initial magnetic energy stored in L at t = 0 + vR R iL(0) = Io Objective of analysis: to find expression for iL(t) for t >0 i.e. to get the current response of the circuit Taking KVL, L diL iLR 0 dt t diL R dt i L iL ( 0 ) L o iL (t ) iL diL iR L dt L diL R dt iL L iL ( t ) R t (0)e L OR ln iL ( t ) R t iL (0) L iL ( t ) Io R t e L RL circuit – natural response • Can be written as iL ( t ) iL (0)e t / , = L/R time constant • This response is known as the natural response iL(t) Current exponentially decays to zero Io At t=, iL(t) decays to 37.68% of its initial value The smaller the time constant the faster the decay 0.3768Io t t= iL (t ) iL R t (0)e L RL circuit – natural response di vL L L dt The inductor voltage is given by: v L Io Re And the voltage across the resistor is given by pR 2 t The energy loss (as heat) in the resistor from 0 to t: t ER pR dt 0 t o Io2 Re 2 t t 2 2 t dt Io Re 2 0 t 2 1 2 ER LIo 1 e 2 t v R iL ( t )R I Re o The power absorbed by the resistor can be calculated as: v RiR Io2 Re t RL circuit – natural response As t , ER 1 2 LIo 2 As t , energy initially stored in inductor will be dissipated in the resistor in the form of heat t 2 1 2 ER LIo 1 e 2 RL circuit – natural response PSpice simulation 1 vL + R L RL circuit L1 0 1 1 IC=10 r1 1 0 1000 .tran 7e-6 7e-3 0 7e-6 UIC .probe .end + vR 0 10A 5A 0A 0s 1.0ms 2.0ms 3.0ms 4.0ms I(L1) Time 5.0ms 6.0ms 7.0ms RL circuit – natural response PSpice simulation RL circuit .param L=1H L1 0 1 {L} IC=10 r1 1 0 1000 .step param L list 0.3 1 3 .tran 7e-6 7e-3 0 7e-6 UIC .probe .end 1 + vR vL R L + 0 10A L1 = 1H L1 = 3H 5A 0A 0s 1.0ms 2.0ms 3.0ms 4.0ms I(L1) L1 = 0.5H Time 5.0ms 6.0ms 7.0ms