First Order RC and RL Transient Circuits

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Adrian Welch
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1 First Order R and RL Transient ircuits

2 Objectives To introduce the transients phenomena. To analyze step and natural responses of first order R circuits. To analyze step and natural responses of first order RL circuits.

3 Transients In circuits with inductors and capacitors voltage and current cannot change instantaneously. The application or removal of sources or circuit elements creates a transient behavior. Transient is the process of going from one steady state to another steady state following a sudden change in the circuit configuration. Sudden changes are mainly due to switching process or faults.

4 FIRST ORDER IRUITS ircuits that contain a single energy storing elements. Either a capacitor or an inductor. SEOND ORDER IRUITS ircuits with two independent energy storing elements in any combination

5 Transient Analysis The circuit is modeled in time domain using differential equations. The order of the differential equation equals the number of independent energy storing elements in the circuits. urrents and Voltages of circuits with just one or One L can be obtained using first order differential equations.

6 First Order ODF The first order ordinary differential equation in the form dy dt Ay B y(t ) 0 Y 0 Has a solution Steady state Transient part y(t) where, Y F ( Y 0 - Y F )e t - T Y F B A, T 1 A

7 Initial onditions Initial conditions are the values of the capacitors voltage or the inductor current at starting instant of the transient period. t = 0 - is the instant just before switching. t = 0 + is the instant just after switching. In apacitors V (0 - ) = V (0 + ). Where V is the capacitor voltage. In inductors i L (0 - ) = i L (0 + ). Where i L is the capacitor voltage.

8 First Order R circuit A. Step Response The response of the circuit to sudden application of an energy supply. V TH Then, R TH dv dt R TH v t=0 v TH + v c _ For t = 0 + to inf. V Since, TH i c i R R TH dv dt TH V i dv 1 v 1 dt R R R c TH 0 v TH

9 First Order R circuit o The following is the first order differential equation describing the hange in the capacitor voltage during the transient period. dv 1 v 1 dt R R TH o The solution of this equation requires the initial conditions of the capacitor voltage V(0). Note that V(0+) = V(0-) = Vo o The general solution of this equation is given by: TH v TH V (t) V F (V 0 - V F )e t - Where, - V F is the final value of V - is the circuit time constant

10 Final Value of V The final value of V occurs when the capacitor is fully charge i.e. the rate of change of V c = 0 R TH Using the differential equation, v F R Then, V F = V TH V F is usually refers to as V ( ) TH 1 R TH V F can be obtained by replacing the capacitor with an open circuit since i F = 0 v TH V TH t = v cf _

11 Time constant The time constant is a measure of how fast is the charging process of the apacitor. Mathematically it s the time required to reach 63% of the final value. R TH

12 Example alculate the capacitor voltage for t > 0 for the circuit shown Solution

13 Initial Value Vco Vco + 2*4=0 Vco = 4V Final Value V F V F = 12 V R TH = 3 Ohms T = R = 3*2=6 S V c (t) = 12 +(4-12) e -t/6

14 B. Natural Response of R circuits The response of the circuit due to the energy stored in the capacitor is known as natural response (no sources in the circuit). t=0 i For t >0 KVL equation around the circuit yields, ir v R dv dt 0 v 0 V i (t) 0 (V dv (t) dt 0-0)e V R o t - e t -

15 Natural Response apacitor Voltage Time constant : R

16 Example In the circuit shown in the figure, the switch opens at t = 0. Find the numerical expression for i(t).

17 Solution Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source. Initial onditions

18 Time onstant

19 Solution Final Value = 0 KL yields

20 First Order R-L circuits 0 ) (0 ) 0 ( i L i L The differential equation is Solving the equation yields At t = 0 the switch closes, The initial conditions t L R s L t L R S S t L R LF L LF e V dt t di L t v e R V R V t i I I I t i ) / ( ) / ( ) / ( ) ( ) ( ) (0 ) ( ) (0) ( ) ( The final conditions I LF = Vs/R

21 First Order R-L circuits step response L = R EQ R-L Step Response

22 Equivalent Resistance seen by an Inductor For the RL circuit in the previous example, it was determined that = L/R. As with the R circuit, the value of R should actually be the equivalent (or Thevenin) resistance seen by the inductor. In general, a first-order RL circuit has the following time constant: L = R EQ R = R EQ seen from the terminals of the inductor for t > 0 with independent sources killed ircuit L ircuit t > 0 independent R EQ sources killed

23 Natural Response of R-L ircuit When the switch is closed (ON) the inductor will store the energy. As a result, the inductor is said to be charged. When the switch is opened (OFF), this will result in the instantaneous change in the circuit. The inductor will supply the energy stored to the resistive.

24 Natural Response of R-L ircuit The circuit when the switch is opened Using the Kirchhoff s voltage law (KVL), to find the differential equation of the loop: L ir 0 By using differential and integral equations technique: di dt i( t) I(0) e t / v L ( t) RI (0) e t / where = L/R is a time constant.

25 Natural Response of R-L ircuit The current generated against time shows the inductor loses its energy exponentially i( t) I(0) e t /

26 The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0. Example Answer: i( t) 2 e 10t

First Order RC and RL Transient Circuits

First Order RC and RL Transient Circuits Firs Order R and RL Transien ircuis Objecives To inroduce he ransiens phenomena. To analyze sep and naural responses of firs order R circuis. To analyze sep and naural responses of firs order RL circuis.