The RLC circuits have a wide range of applications, including oscillators and frequency filters

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1 9. The RL ircuit The RL circuits have a wide range of applications, including oscillators and frequency filters This chapter considers the responses of RL circuits The result is a second-order differential equation for any voltage or current of interest We consider the following analysis The Natural Response of a Parallel RL ircuit The Natural Response of a Series RL ircuit The omplete (Natural and Step) Response of RL ircuits

2 9. The Source-Free Parallel ircuit Obtaining the differential equation for a parallel RL circuit Apply KL v i t i t =0 R L v t dv t v d t il t0 =0 R L dt il(t) 0 Differentiate both sides with respect to time d v d v v=0 R d t L dt Two initial conditions The capacitor voltage cannot change abruptly v 0 =v 0 =v 0 =V 0 The inductor current cannot change abruptly il 0 il 0 il 0 =I 0 I 0 V 0 / R d v t = dt t=0 i 0 = il 0 ir 0 V = I 0 0 R d v t i 0 = dt t=0 i(t)

3 3 Solution of the differential equation We assume the exponential form of the natural response is st v t =Ae Substitute into the ordinary differential equation, we got the characteristic equation of s determined by the circuit parameters d v d v s s =0 v=0 R L R d t L dt The characteristic equation has two roots s s v=0 R L s, = ± L R R Thus, the natural response has the following form s t s t v t =A e A e where the constants A and A are determined using the initial conditions Definition of frequency terms The resonant frequency 0 = L The damping coefficient = R

4 4 The roots of the characteristic equation can be expressed as s = 0 s= 0 Three types of natural response Response riteria Solutions Overdamped α>ω0 real, distinct roots s, s Underdamped α<ω0 complex, conjugate roots s, s* ritically damped α=ω0 real, equal roots s, s

5 5 9. The Overdamped Parallel RL ircuit The condition of overdamped response ( 0 ) implies that The roots of the characteristic equation s and s are distinct negative real numbers The response, v(t), can be seen as a sum of two decreasing exponential terms st s t t v t =A e A e 0 as Finding values for A and A For the shown circuit, we determine 0 = = 6 = =3.5 R L s = s = 6 The general form of the natural response v t =A e t A e 6 t From the initial conditions v(0)=0 and il(0)=-0 A v 0 =A A =0 () A 6 A =40 I 0 V 0 /R d v t = dt t=0 =A s A s

6 The final numerical solution is v t =84 e t e 6 t V Graphing the response The maximum point can be determined as d v t =0 dt t 6 t =84 e 6 e =0 max We determine the time t max =0.358 s Then v tmax =48.9 V max

7 Find an expression for v(t) valid for t > 0 in the circuit ompute the initial conditions (t < 0) The capacitor acts as open circuits 00 v 0 =50 =60 V The inductor acts as short circuits 50 il 0 = = 300 ma After the switch is thrown (t > 0) The capacitor is left in parallel with a 00 Ω resistor and a 5 mh inductor 0 = =00000 = =5000 R L s = s = 00000

8 Solve the capacitor voltage Since α > ω0, the circuit is overdamped and so we expect a capacitor voltage of the form v t =A e t A e t Finding values for A and A From the initial conditions v(0)=60 V and il(0)=-0.3 A v 0 =A A = A 00000A =0 Solving, A = 80 V and A = 0 V, so that v t =80 e t 0 e t t 0 I 0 V 0 / R d v t = dt t=0 =A s A s

9 9 (a) Sketch the voltage vr(t)= e t 4e 3t V in the range 0<t<5 s (b) Estimate the settling time (c) alculate the maximum positive value and the time at which it occurs Graphing the response The maximum point can be determined as d vr t =0 dt tmax e 3 tmax 4 3 e =0 ln 6 t max = =0.895 s vr tmax =544 mv We compute the settling time as follows v R tmax vr tsettling = 00 tsettling e 3 tsettling 4e =5.44 mv tsettling e =5.44 mv t settling=5.9 s

10 0 9.3 ritical Damping The condition of a critical damping ( = 0 ) implies that The roots of the characteristic equation s and s are equal and negative real numbers s=s = For repeated roots, the response, v(t), can be seen as v t =A te t A e t Finding values for A and A For the shown circuit, we determine 0 = = 6 s s =s = 6 s The general form of the natural response v t =A te 6 t A e 6 t From the initial conditions v(0)=0 and il(0)=-0 A v 0 =A =0 A A = 0 =40 I 0 V 0 / R d v t = dt t=0 =A A

11 The solution is v t =40 te 6 t V Graphing the response The maximum point can be determined as d v t =0 dt =40e t max 40 tmax e t =0 We determine the time tmax 40e t max tmax =0 t max = =0.408 s Then v t max =63. V The settling time v tmax 6t =40tsettling e 00 settling

12 Find R such that the circuit is critically damped for t>0 and R so that v(0)= V For t < 0 The capacitor acts as open circuits The inductor acts as short circuits v 0 =5 R R = V R R After the switch is thrown (t > 0) The current source has turned itself off and R is shorted The capacitor is left in parallel with R and a 4 H inductor = = 0 = =580 R R 0 9 L Since the critically damping implies that 0 =, we have R = R = R R =0.4

13 3 9.4 The Underdamped Parallel RL ircuit The condition of a critical damping ( 0 ) implies that The roots of the characteristic equation s and s are complex conjugate numbers s = j s = s = d 0 s = j d 0 where d= 0 is the natural resonant frequency For complex conjugate roots, the response, v(t), can be seen as v t =e t A ej t A e j t d d =e t [A cos d t j sin d t A cos d t j sin d t ] =e t [ A A cos d t j A A sin d t] =e t [B cos d t B sin d t] The derivative of v(t) is dv t = e t [B cos d t B sin d t] dt t e [B d sin d t B d cos d t ] =e t [ B d B cos d t B d B sin d t]

14 4 Finding values for B and B For the shown circuit, we determine = = s 0 = = 6 s R L d= 0 = s The natural response is v t =e t [B cos t B sin t] From the initial conditions v(0)=0 and il(0)=-0 A v 0 =B =0 B =40 The final numerical solution is v t =0 e t sin t I 0 V 0 / R d v t = dt t=0 = B d B

15 Graphing the response v t =0 e t sin t The voltage oscillates (~ωd) and approaches to the final value (~α) The voltage response has two extreme points (minimum and maximum points) 5

16 Find an expression for v(t) valid for t > 0 in the circuit ompute the initial conditions (t < 0) The capacitor acts as open circuits v 0 =3 =97.3 V The inductor acts as short circuits 00 il 0 =3 =.07 A After the switch is thrown (t > 0) The current source is off The capacitor is left in parallel with a 48 Ω resistor and a 0 H inductor 0 = =4.99 s = =. s L R Since 0, the circuit is underdamped v t =e t [B cos d t B sin d t] where d= 0 =4.75 s 6

17 7 Finding values for B and B The natural response is v t =e. t [B cos 4.75 t B sin 4.75 t] From the initial conditions v(0)=97.3 and il(0)=.07 A v 0 =B = B = /00 The final numerical solution is I 0 V 0 / R d v t = dt t=0 = B d B v t =e. t [97.3 cos 4.75 t 5.57 sin 4.75 t] V

18 8 9.5 The Source-Free Series ircuit Obtaining the differential equation of a series RL circuit Apply KVL (V0, I0, i(t) must satisfy the passive sign convention) R i vl t v t =0 di t R i L t i d t v t0 =0 dt 0 Differentiate both sides with respect to time d i di L R i=0 d t dt The two initial conditions The inductor current cannot change abruptly i 0 =I 0 The capacitor voltage cannot change abruptly v 0 =V 0 V 0 I 0 R d il t = dt t=0 L vl 0 = v 0 vr 0 = V 0 I 0 R vl 0 =L d il t dt t=0

19 9 Solution of the differential equation We assume the exponential form of the natural response is v t =Aest Substitute into the ordinary differential equation, we got the characteristic equation of s determined by the circuit parameters Ls R s =0 The characteristic equation has two roots R s, = ± L R L L = ± 0 where The resonant frequency The damping coefficient L R = L 0 =

20 0 Summary of Relevant Equations for Series Source-Free RL ircuits ondition Overdamped Underdamped ritically damped riteria α ω0 α>ω0 R L L α<ω0 R L L α=ω0 R L L Response s t i t =A e A e where s t s, = ± 0 i t =e t [B cos d t B sin d t] where d= 0 i t =A te t A e t

21 9.6 The omplete Response of the RL ircuit The response of RL circuits with dc sources and switches consists of the natural response and the forced response: Forced Response Natural Response v t = v f t Forced Response i t = i t f v t n Natural Response i t n The general solution is obtained by the same procedure that was followed for RL and R circuits

22 The Solution Steps of RL ircuits Determine the initial conditions ompute the circuit current, il(t), ir(t), and i(t), and voltages, vl(t), vr(t), and v(t), at t=0- and t=0+ Note that the inductor current the capacitor voltage cannot change abruptly, il(0-)=il(0)=il(0+) and v(0-)=v(0)=v(0+) hf(t) Upon we are confronted with a series or a parallel circuit R = (series RL) = (parallel RL) L R 0= L t hn t =A te t = 0 0 s t hn t =A e A e A e d= 0 t hn t =e 0 s, = ± 0 The forced response s t The complete response hf(t)+hn(t) [B cos d t B sin d t] Find unknown constants given the initial conditions

23 3 Find an expression for v(t) and il(t) valid for t > 0 in the circuit - +. Determine the forced response ( t > 0 ) ilf t = 9 A t 0 vf t =50 V t 0. Determine the natural response -. Write the differential equation R i vl t v t =0 + The differential equation in terms of i reduces to d i di L R i=0 d t dt Note: Independent current sources open circuits Independent voltage sources short circuits The differential equation in terms of v(t) is given as L d v dt R d v v =0 dt. ompute αand ω0 0= =3 s L s = R = =5 s L s = 9 3. Since α>ω0, the response is overdamped, we have and t 9 t t 9 t iln t =A e A e v n t =B e B e

24 4 3. The complete response v t =vf t B e t B e 9 t il t =ilf t A e t A e 9 t = 9 A e t A e 9 t =50 B e t B e 9 t 4. Solve for the values of the unknown constants v t t=0 =v 0 50 B B = dv t i 0 = dt t=0 B 9 B =08 For t=0- (the left-hand current source is off) t v t = B e 3.5B e 9 t il t t=0 =il 0 9 A A= 5 dil t vl 0 = dt t=0 L A 9 A = 40 il t = e t 4.5 e 9 t vl 0 =0 V vr 0 = 50 V v 0 =50 V il 0 = 5 A ir 0 = 5 A i 0 =0 A For t=0+ (the left-hand current source is on) vl 0 = 0 V vr 0 = 30 V v 0 =50 V il 0 = 5 A ir 0 = A i 0 =4 A

25 9.7 THE LOSSLESS L IRUIT The resistor in the RL circuit serves to dissipate initial stored energy When this resistor becomes 0 in the series RL or infinite in the parallel RL, the circuit will oscillate Example: Assume the shown circuit with the following initial conditions v 0 =0 V il 0 = A 6. We find R 0 = =3 s = =0 s L L. So d=3 s, the voltage is simply t v t =e [B cos 3 t B sin 3 t] 3. We use the initial condition v 0 =B B =0 dv t il 0 B = = =3 B dt t=0 4. Thus, we have obtained a sinusoidal response v t = sin 3 t V 5

26 Homework Assignment 8 P9., P9.6, P9., P9.3, P9.6, P9.0, P9.6, P9.7, P9.35, P9.37, P9.46 P9.50, P9.5, and