The RLC Circuit. Transient Response Series RLC Circuit

The RLC Circuit. Transient ResponseSeries RLC circuitThe circuit shown on Figure 1 is called the series RLC circuit. We will analyze thiscircuit in order to determine its transient characteristics once the switch S is closed.S vR- vL-LRVsC vc-Figure 1The equation that describes the response of the system is obtained by applying KVLaround the meshvR vL vc Vs(1.1)The current flowing in the circuit isi Cdvcdt(1.2)And thus the voltages vR and vL are given byvR iR RCvL Ldvcdtdid 2 vc LC 2dtdt(1.3)(1.4)Substituting Equations (1.3) and (1.4) into Equation (1.1) we obtaind 2 vc R dvc 11 vc Vs2dtL dt LCLC(1.5)The solution to equation (1.5) is the linear combination of the homogeneous and theparticular solution vc vc p vchThe particular solution is6.071/22.071 Spring 2006, Chaniotakis and Cory1

vc p Vs(1.6)And the homogeneous solution satisfies the equationd 2 vch R dvch1 vch 02dtL dtLC(1.7)Assuming a homogeneous solution is of the form Ae st and by substituting into Equation(1.7) we obtain the characteristic equations2 R1 0s LLC(1.8)By definingR:2Lα Damping rate(1.9)Andωο 1: Natural frequencyLC(1.10)The characteristic equation becomess 2 2α s ωο2 0(1.11)The roots of the characteristic equation ares1 α α 2 ωο2(1.12)s 2 α α 2 ωο2(1.13)And the homogeneous solution becomesvch A1e s1t A2 e s 2t(1.14)vc Vs A1e s1t A2 e s 2 t(1.15)The total solution now becomes6.071/22.071 Spring 2006, Chaniotakis and Cory2

The parameters A1 and A2 are constants and can be determined by the application of thedvc(t 0)initial conditions of the system vc (t 0) and.dtThe value of the term α 2 ωο2 determines the behavior of the response. Three types ofresponses are possible:1. α ωο then s1 and s2 are equal and real numbers: no oscillatory behaviorCritically Damped System2. α ωο . Here s1 and s2 are real numbers but are unequal: no oscillatory behaviorOver Damped Systemvc Vs A1e s1t A2 e s 2 t3. α ωο .α 2 ωο2 j ωο2 α 2 In this case the roots s1 and s2 are complexnumbers: s1 α j ωο2 α 2 , s 2 α j ωο2 α 2 . System exhibitsoscillatory behaviorUnder Damped SystemImportant observations for the series RLC circuit. As the resistance increases the value of α increases and the system is driventowards an over damped response.1(rad/sec) is called the natural frequency of the systemLCor the resonant frequency.The frequency ωο Ris called the damping rate and its value in relation to ωο2Ldetermines the behavior of the responseo α ωο :Critically DampedThe parameter α o α ωο :o α ωο : The quantityOver DampedUnder DampedLhas units of resistanceC6.071/22.071 Spring 2006, Chaniotakis and Cory3

Figure 2 shows the response of the series RLC circuit with L 47mH, C 47nF and forthree different values of R corresponding to the under damped, critically damped andover damped case. We will construct this circuit in the laboratory and examine itsbehavior in more detail.(a) Under Damped. R 500Ω(b) Critically Damped. R 2000 Ω(c) Over Damped. R 4000 ΩFigure 26.071/22.071 Spring 2006, Chaniotakis and Cory4

The LC circuit.In the limit R 0 the RLC circuit reduces to the lossless LC circuit shown on Figure 3.S vL-LC vc-Figure 3The equation that describes the response of this circuit isd 2 vc 1 vc 0dt 2 LC(1.16)Assuming a solution of the form Ae st the characteristic equation isWhere ωο s 2 ωο2 0(1.17)s1 jωο(1.18)s 2 jωο(1.19)1LCThe two roots areAnd the solution is a linear combination of A1e s1t and A2e s 2tvc(t ) A1e jωot A2e jωο t(1.20)By using Euler’s relation Equation (1.20) may also be written asvc(t ) B1cos(ωο t ) B 2sin(ωο t )(1.21)The constants A1, A2 or B1, B2 are determined from the initial conditions of the system.6.071/22.071 Spring 2006, Chaniotakis and Cory5

For vc(t 0) Vo and fordvc (t 0) 0 (no current flowing in the circuit initially) wedthave from Equation (1.20)A1 A2 Vo(1.22)jωo A1 jωo A2 0(1.23)AndWhich giveA1 A2 Vo2(1.24)And the solution becomesVo jωot jωο te e2 Vo cos(ωot )vc(t ) ()(1.25)The current flowing in the circuit isdvcdt CVoωο sin(ωο t )i C(1.26)And the voltage across the inductor is easily determined from KVL or from the elementdirelation of the inductor vL LdtvL vc Vo cos(ωot )(1.27)Figure 4 shows the plots of vc (t ), vL(t ), and i (t ) . Note the 180 degree phase differencebetween vc(t) and vL(t) and the 90 degree phase difference between vL(t) and i(t).Figure 5 shows a plot of the energy in the capacitor and the inductor as a function oftime. Note that the energy is exchanged between the capacitor and the inductor in thislossless system6.071/22.071 Spring 2006, Chaniotakis and Cory6

(a) Voltage across the capacitor(b) Voltage across the inductor(c) Current flowing in the circuitFigure 46.071/22.071 Spring 2006, Chaniotakis and Cory7

(a) Energy stored in the capacitor(b) Energy stored in the inductorFigure 56.071/22.071 Spring 2006, Chaniotakis and Cory8

Parallel RLC CircuitThe RLC circuit shown on Figure 6 is called the parallel RLC circuit. It is driven by theDC current source Is whose time evolution is shown on Figure 7.iR(t)RIsiC(t)iL(t)LC v-Figure 6Is0tFigure 7Our goal is to determine the current iL(t) and the voltage v(t) for t 0.We proceed as follows:1.2.3.4.Establish the initial conditions for the systemDetermine the equation that describes the system characteristicsSolve the equationDistinguish the operating characteristics as a function of the circuit elementparameters.Since the current Is was zero prior to t 0 the initial conditions are: iL(t 0) 0Initial Conditions: v(t 0) 0(1.28)By applying KCl at the indicated node we obtain6.071/22.071 Spring 2006, Chaniotakis and Cory9

Is iR iL iC(1.29)The voltage across the elements is given byd iLdt(1.30)v L d iL R R dt(1.31)dvd 2iL LC 2dtdt(1.32)v LAnd the currents iR and iC areiR iC CCombining Equations (1.29), (1.31), and (1.32) we obtaind 2iL 1 d iL 11 iL Is2dtRC dt LCLC(1.33)The solution to equation (1.33) is a superposition of the particular and the homogeneoussolutions.iL (t ) iL p (t ) iLh (t )(1.34)iL p (t ) Is(1.35)The particular solution isThe homogeneous solution satisfies the equationd 2iLh1 d iLh1 iLh 02dtRC dtLC(1.36)By assuming a solution of the form Ae st we obtain the characteristic equations2 11 0s RCLC(1.37)Be defining the following parameters6.071/22.071 Spring 2006, Chaniotakis and Cory10

1: Resonant frequencyLCωο (1.38)Andα 1: Damping rate2RC(1.39)The characteristic equation becomess 2 2α s ωο2 0(1.40)s1 α α 2 ωο2(1.41)s 2 α α 2 ωο2(1.42)The two roots of this equation areThe homogeneous solution is a linear combination of e s1t and e s 2 tiLh (t ) A1e s1t A2 e s 2 t(1.43)iL(t ) Is A1e s1t A2 e s 2 t(1.44)And the general solution becomesThe constants A1 and A2 may be determined by using the initial conditions.Let’s now proceed by looking at the physical significance of the parameters α and ωο .6.071/22.071 Spring 2006, Chaniotakis and Cory11

The form of the roots s1 and s2 depend on the values of α and ωο . The following threecases are possible.1. α ωο : Critically Damped System.s1 and s2 are equal and real numbers: no oscillatory behavior2. α ωο : Over Damped SystemHere s1 and s2 are real numbers but are unequal: no oscillatory behavior3. α ωο : Under Damped Systemα 2 ωο2 j ωο2 α 2 In this case the roots s1 and s2 are complex numbers:s1 α j ωο2 α 2 , s 2 α j ωο2 α 2 . System exhibits oscillatory behaviorLet’s investigate the under damped case, α ωο , in more detail.For α ωο , α 2 ωο2 j ωο2 α 2 jωd the solution is α tiL(t ) Is eN( A1e jωd t A2e jωd t )Decaying(1.45)OscillatoryBy using Euler’s identity e jωd t cos ωd t j sin ωd t , the solution becomes α tiL(t ) Is eN( K1 cos ωd t K 2 sin ωd t )Decaying(1.46)OscillatoryNow we can determine the constants K1 and K 2 by applying the initial conditionsiL(t 0) 0 Is K1 0 K1 IsdiL 0 α K1 (0 K 2ωd ) 0dt t 0 K2 αωd(1.47)(1.48)IsAnd the solution is6.071/22.071 Spring 2006, Chaniotakis and Cory12

α α t iL(t ) Is 1 eNsin ωd t cos ωd t ωd Decaying Oscillatory(1.49) B By using the trigonometric identity B1 cos t B2 sin t B12 B22 cos t tan 1 2 theB1 solution becomesiL(t ) Is Is ωο α tα e cos ωd t tan 1 ωdωd (1.50)Recall that ωd ωο2 α 2 and thus ωd is always smaller than ωoLet’s now investigate the important limiting case:As R , α ω0α 0 , e α t 1ωοAnd the solution reduces to iL(t ) Is Is cos ωot which corresponds to the response ofthe circuitωd ωο2 α 2 ωο and tan 1IsLCThe plot of iL(t) is shown on Figure 8 for C 47nF, L 47mH, Is 5A and for R 20kΩ and8kΩ, The dotted lines indicate the decaying characteristics of the response. Forconvenience and easy visualization the plot is presented in the normalized time ωο t / π .Note that the peak current through the inductor is greater than the supply current Is.6.071/22.071 Spring 2006, Chaniotakis and Cory13

(a) For R 8kΩ(b) For R 20kΩFigure 8.6.071/22.071 Spring 2006, Chaniotakis and Cory14

(a) R 20kΩ6.071/22.071 Spring 2006, Chaniotakis and Cory15

(b) R 8kΩFigure 9The energy stored in the inductor and the capacitor is shown on Figure 10.Figure 10. Energy as a function of timeFigure 11 shows the plot of the response corresponding to the case where α ω0 . Thisshows the persistent oscillation for the current iL(t) with frequency ω0 .6.071/22.071 Spring 2006, Chaniotakis and Cory16

Figure 116.071/22.071 Spring 2006, Chaniotakis and Cory17

The Critically Damped Response.When α ωο the two roots of the characteristic equation are equal s1 s2 s. And ourassumed solution becomesiL(t ) A1e st A2e st A3e st(1.51)Now we have only one arbitrary constant. This is a problem for our second order systemsince our two initial conditions can not be satisfied.The problem stems from an incorrect assumption for the solution for this special case.For α ωο the differential equation of the homogeneous problem becomesd 2iLhd iLh 2α α 2iLh 02dtdt(1.52)The solution of this equation is1iL (t ) A1te α t A2 e α t(1.53)Which is a linear combination of the exponential term and an exponential term multipliedby t.d did 2idi di 2α α 2i 0 may be rewritten as α i α α i 0 , by2dt dtdtdt dt dξdidefining ξ α i the equation becomes αξ 0 whose solution is ξ K1e α t . Thereforedtdtdideα t eα tα i K1 which may be written as (eα t i ) K1 . By integration we obtain the solutiondtdt α t α ti K1te K 2 e1The equation6.071/22.071 Spring 2006, Chaniotakis and Cory18

Summary of RLC transient responseωοαCriticallyDampedSeries1ωο LCRα 2LParallel1ωο LC1α 2RCα ωοResponse: A1te α t A2 e α tα ωοUnderDampedResponse: eN ( K1 cos ωd t K 2 sin ωd t ) Decaying α tOscillatoryWhere ωd ωο2 α 2OverDampedα ωοResponse: A1e s1t A2 e s 2tWhere s1, 2 α α 2 ωο26.071/22.071 Spring 2006, Chaniotakis and Cory19

ProblemFor the circuit below, the switch S1 has been closed for a long time while switch S2 isopen. Now switch S1 is opened and then at time t 0 switch S2 is closed.Determine the current i(t) as indicated.S1R1S2R2i(t)Vs6.071/22.071 Spring 2006, Chaniotakis and CoryCL20

Parallel RLC Circuit The RLC circuit shown on Figure 6 is called the parallel RLC circuit. It is driven by the DC current source Is whose time evolution is shown on Figure 7. Is R L C iL(t) v -iR(t) iC(t) Figure 6 t Is 0 Figure 7 Our goal is to determine the current iL(t) and