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First Order Circuit

First Order Circuit. Capacitors and inductors. RC and RL circuits. Excitation from stored energy. ‘source-free’ circuits. DC source (voltage or current source). Natural response. Sources are modeled by step functions. Step response. Forced response.

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First Order Circuit

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  1. First Order Circuit • Capacitors and inductors • RC and RL circuits

  2. Excitation from stored energy • ‘source-free’ circuits • DC source (voltage or current source) • Natural response • Sources are modeled by step functions • Step response • Forced response RC and RL circuits (first order circuits) Circuits containing no independent sources Circuits containing independent sources Complete response = Natural response + forced response

  3. + vc  ic iR R C Taking KCL,    OR RC circuit – natural response Assume that capacitor is initially charged at t = 0  vc(0) = Vo Objective of analysis: to find expression for vc(t) for t >0 i.e. to get the voltage response of the circuit

  4. ,  = RC  time constant • Can be written as vC(t) Vo 0.3768Vo t t= RC circuit – natural response • This response is known as the natural response  Voltage decays to zero exponentially  At t=, vc(t) decays to 37.68% of its initial value  The smaller the time constant the faster the decay

  5. The capacitor current is given by:  And the current through the resistor is given by RC circuit – natural response The power absorbed by the resistor can be calculated as: The energy loss (as heat) in the resistor from 0 to t:

  6. As t  , ER  RC circuit – natural response As t  , energy initially stored in capacitor will be dissipated in the resistor in the form of heat

  7. + vc  ic iR R C RC circuit – natural response PSpice simulation 1 RC circuit c1 1 0 1e-6 IC=100 r1 1 0 1000 .tran 7e-6 7e-3 0 7e-6 UIC .probe .end 0

  8. c1 = 1e-6 c1 = 3e-6 c1 = 0.5e-6 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 + vc  ic iR R C 0

  9. iL  vL + + vR  R L Taking KVL,    OR RL circuit – natural response Assume initial magnetic energy stored in L at t = 0  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

  10. ,  = L/R  time constant • Can be written as iL(t) Io 0.3768Io t t= RL circuit – natural response • This response is known as the natural response  Current exponentiallydecays to zero  At t=, iL(t) decays to 37.68% of its initial value  The smaller the time constant the faster the decay

  11. The inductor voltage is given by:  And the voltage across the resistor is given by RL circuit – natural response The power absorbed by the resistor can be calculated as: The energy loss (as heat) in the resistor from 0 to t:

  12. As t  , ER  RL circuit – natural response As t  , energy initially stored in inductor will be dissipated in the resistor in the form of heat

  13. vL + + vR  R L RL circuit – natural response PSpice simulation RL circuit L1 0 1 1 IC=10 r1 1 0 1000 .tran 7e-6 7e-3 0 7e-6 UIC .probe .end 1 0

  14. L1 = 1H L1 = 3H L1 = 0.5H 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  vL + + vR  R L 0

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