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Transcript

First Order Circuits Objective of Lecture Explain the operation of a RC circuit in dc circuits As the capacitor releases energy when there is: a transition in a unit step function voltage or current source or a voltage or current source is switched out of the circuit. Explain the operation of a RL circuit in dc circuit As the inductor releases energy when there is: a transition in a unit step function voltage or current source or a voltage or current source is switched out of the circuit. Natural Response The behavior of the circuit with no external sources of excitation. There was a transition in the source in the circuit where the unit step function changed from 1 to 0 at t ≤ 0s. There is stored energy in the capacitor or inductor at time t = 0 s. For t > 0 s, the stored energy is released Current flows through the circuit and voltages exist across components in the circuit as the stored energy is released. The stored energy will decays to zero as time approaches infinite, at which point the currents and voltages in the circuit become zero. RC Circuit Suppose there is some charge on a capacitor at time t = 0s. This charge could have been stored because a voltage or current source had been in the circuit at t<0s, but was switched off at t = 0s. We can use the equations relating voltage and current to determine how the charge on the capacitor is removed as a function of time. The charge flows from one plate of the capacitor through the resistor R to the other plate to neutralize the charge on the opposite plate of the capacitor. Equations for RC Circuit vC vR 0 iC iR dvC iC C dt vR iR R C dVC VR 0 dt R VR VC dVC VC 0 dt RC 1 dVC 1 0 VC dt RC dVC 1 dt VC RC ln VC t ln VC RC t to 2 If Vo VC VC (t ) Vo e t 0 s and RC t t when t 0s Vo I R (t ) I C (t ) e R t Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor. 2t Vo p R (t ) VR I R e R CVo w(t ) p R (t )dt 2 0s 2 2t 1 e RL Circuits VL VR 0 IL IR dI L VL L dt I R VR R dI L L RI R 0 dt dI L RI L 0 dt L 1 dI L R 0 I L dt L dI L R dt IL L R ln I L t ln I L t 0 s L If I o I L t 0 s L and R 2t 2 p R (t ) VR I R RI o e t I L (t ) I o e t when t 0 s VR (t ) VL (t ) RI o e t Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor. LI o w(t ) p R (t )dt 2 0s 2 2t 1 e Initial Condition Can be obtained by inserting a d.c. source to the circuit for a time much longer than at least t = -5. Capacitor Vo is the voltage calculated by replacing the capacitor with a resistor with infinite resistance (an open circuit) after the voltage across the capacitor has reached a constant value (steady state). Inductor Io is the current flowing through the inductor calculated by replacing the inductor with a resistor with zero resistance (a short circuit) after the current flowing through the inductor has reached a constant value (steady-state). PSpice You can set the initial condition on a capacitor or inductor by doubling clicking on the part symbol. Then, enter a value for IC in the pop-up window that opens. Time constant, The time required for the voltage across the capacitor or current in the inductor to decay by a factor of 1/e or 36.8% of its initial value. Example #1 IL(t) V(t) = 6V [1 - u(t)] Example #1 (con’t) IL(t) Example #1 (con’t) Find the initial condition. Io t < 0s VL = 0V VR = 6V IL = IR = 2mA Therefore, Io = 2mA Example #1 (con’t) IL(t) Example #1 (con’t) IL(t) t > 0s t = L/R = 10mH/3kW/ = 3.33ms IL = IR =Ioe-t/ = 2mA e-(t/3.33ms) VR = 3kW IR = 6V e-(t/3.33ms) VL = L dIL/dt = -6V e-(t/3.33ms) Note VR + VL = 0 V Example #2 + VC _ Example #2 (con’t) + VC _ Example #2 (con’t) Calculate the initial condition - the voltage on the capacitor. Replace the capacitor with an open circuit and find the voltage across the two terminals. Note that in this circuit, current will flow through R3 so there will be a voltage across C, but it will not be equal to the magnitude of the voltage source in the circuit. Example #2 (con’t) The voltage across the capacitor is equal to the voltage across the 12kW resistor. VC = Vo = [12kW /15kW] 5V = 4V Example #2 (con’t) + VC _ + VC _ Example #2 (con’t) Further simplification of the circuit + VC _ where Req 1kW 3kW 12kW 3kW Example #2 (con’t) Req C 3kW(2 mF ) 6ms when t 2ms VC VC (t ) t 2 ms e ( t 2 ms ) / IR(t) VC (t ) 4Ve(t 2 ms ) / 6 ms VC (t ) VReq (t ) dVC (t ) 2 mF (4V / 6ms)e (t 2 ms ) / 6 ms dt I C (t ) 1.33mA e (t 2 ms ) / 6 ms I C (t ) C I Req (t ) I C (t ) 1.33mA e (t 2 ms ) / 6 ms I Req (t ) I C (t ) 0 IC(t) Summary The initial condition for: the capacitor voltage (Vo) is determined by replacing the capacitor with an open circuit and then calculating the voltage across the terminals. The inductor current (Io) is determined by replacing the inductor with a short circuit and then calculating the current flowing through the short. The time constant for: an RC circuit is RC an RL circuit is L/R The general equations for the natural response of: the voltage across a capacitor is VC (t ) Vo e (t t ) / the current through an inductor is I (t ) I e (t t ) / o o L o