The RLC Circuit. Transient Response
Series RLC circuit
The circuit shown on Figure 1 is called the series RLC circuit. We will analyze this circuit in order to determine its transient characteristics once the switch S is closed.
S + vR - + vL - R L + Vs C vc -
Figure 1
The equation that describes the response of the system is obtained by applying KVL around the mesh
vR + vL +=vc Vs (1.1)
The current flowing in the circuit is
dvc iC= (1.2) dt
And thus the voltages vR and vL are given by
dvc vR ==iR RC (1.3) dt
di d 2vc vL ==L LC (1.4) dt dt 2
Substituting Equations (1.3) and (1.4) into Equation (1.1) we obtain
d 2vc R dvc 11 ++vc =Vs (1.5) dt 2 L dt LC LC
The solution to equation (1.5) is the linear combination of the homogeneous and the particular solution vc =vcph+vc
The particular solution is
6.071/22.071 Spring 2006, Chaniotakis and Cory 1 vcp = Vs (1.6) And the homogeneous solution satisfies the equation
dv2 c R dvc 1 hh+ +vc =0 (1.7) dt 2 L dt LC h
Assuming a homogeneous solution is of the form Aest and by substituting into Equation (1.7) we obtain the characteristic equation
R 1 ss2 + +=0 (1.8) L LC
By defining
R α = : Damping rate (1.9) 2L And
1 ω = : Natural frequency (1.10) ο LC
The characteristic equation becomes
22 ss+ 2αω+ο =0 (1.11)
The roots of the characteristic equation are
22 s1=−α + α−ωο (1.12)
22 s2 =−α − α−ωο (1.13)
And the homogeneous solution becomes
s1ts2t vch =+A1e A2e (1.14)
The total solution now becomes
s1ts2t vc =+Vs A1e +A2e (1.15)
6.071/22.071 Spring 2006, Chaniotakis and Cory 2 The parameters A1 and A2 are constants and can be determined by the application of the dvc(0t = ) initial conditions of the system vc(t = 0) and . dt
22 The value of the term α −ωο determines the behavior of the response. Three types of responses are possible:
1. α = ωο then s1 and s2 are equal and real numbers: no oscillatory behavior Critically Damped System
2. α > ωο . Here s1 and s2 are real numbers but are unequal: no oscillatory behavior Over Damped System s12tst vc =+Vs A12e +A e
22 22 3. α < ωο . α −=ωωοj ο−α In this case the roots s1 and s2 are complex
22 22 numbers: sj1,=−α + ωαο− s2=−α−jωο−α. System exhibits oscillatory behavior Under Damped System
Important observations for the series RLC circuit.
• As the resistance increases the value of α increases and the system is driven towards an over damped response.
1 • The frequency ω = (rad/sec) is called the natural frequency of the system ο LC or the resonant frequency.
R • The parameter α = is called the damping rate and its value in relation to ω 2L ο determines the behavior of the response
o α = ωο : Critically Damped
o α > ωο : Over Damped
o α < ωο : Under Damped
L • The quantity has units of resistance C
6.071/22.071 Spring 2006, Chaniotakis and Cory 3 Figure 2 shows the response of the series RLC circuit with L=47mH, C=47nF and for three different values of R corresponding to the under damped, critically damped and over damped case. We will construct this circuit in the laboratory and examine its behavior in more detail.
(a) Under Damped. R=500Ω
(b) Critically Damped. R=2000 Ω
(c) Over Damped. R=4000 Ω
Figure 2
6.071/22.071 Spring 2006, Chaniotakis and Cory 4 The LC circuit.
In the limit R → 0 the RLC circuit reduces to the lossless LC circuit shown on Figure 3.
S + vL - L + C vc -
Figure 3
The equation that describes the response of this circuit is
dv2 c 1 + vc = 0 (1.16) dt 2 LC
Assuming a solution of the form Aest the characteristic equation is
22 s + ωο = 0 (1.17) 1 Where ω = ο LC
The two roots are
s1= + jωο (1.18)
s2= − jωο (1.19)
And the solution is a linear combination of A1es1t and A2es2t
vc(t)=+A1e jtωo A2e− jωο t (1.20)
By using Euler’s relation Equation (1.20) may also be written as
vc(t)= B1cos()ωοt + B2sin(ωοt) (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 Cory 5 dvc(0t = ) For vc(0t ==)Vo and for = 0 (no current flowing in the circuit initially) we dt have from Equation (1.20)
AA12+ =Vo (1.22) And
jAωo12− jAωo= 0 (1.23)
Which give
Vo AA12== (1.24) 2 And the solution becomes
Vo vc()t =+()e jtωωo e− jοt 2 (1.25)
= Vo cos(ωot) The current flowing in the circuit is
dvc iC= dt (1.26)
=−CVoωοοsin(ω t)
And the voltage across the inductor is easily determined from KVL or from the element di relation of the inductor vL = L dt
vL = −vc (1.27) =−Vocos(ωot)
Figure 4 shows the plots of vc()t ,vL(t),and i()t . Note the 180 degree phase difference between 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 of time. Note that the energy is exchanged between the capacitor and the inductor in this lossless system
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(a) Voltage across the capacitor
(b) Voltage across the inductor
(c) Current flowing in the circuit
Figure 4
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(a) Energy stored in the capacitor
(b) Energy stored in the inductor
Figure 5
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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.
iC(t) iR(t) iL(t) + Is R L C v -
Figure 6
Is
0 t Figure 7
Our goal is to determine the current iL(t) and the voltage v(t) for t>0.
We proceed as follows:
1. Establish the initial conditions for the system 2. Determine the equation that describes the system characteristics 3. Solve the equation 4. Distinguish the operating characteristics as a function of the circuit element parameters.
Since the current Is was zero prior to t=0 the initial conditions are:
⎧iL(0t = )= 0 Initial Conditions:⎨ (1.28) ⎩ vt(0= )= 0
By applying KCl at the indicated node we obtain
6.071/22.071 Spring 2006, Chaniotakis and Cory 9 Is = iR ++iL iC (1.29)
The voltage across the elements is given by
diL vL= (1.30) dt
And the currents iR and iC are
vLdiL iR == (1.31) R Rdt
dv d 2iL iC ==C LC (1.32) dt dt 2
Combining Equations (1.29), (1.31), and (1.32) we obtain
di2 L 11diL 1 ++iL =Is (1.33) dt 2 RC dt LC LC
The solution to equation (1.33) is a superposition of the particular and the homogeneous solutions.
iL()t = iLp()t + iLh(t) (1.34)
The particular solution is
iLp ()t = Is (1.35)
The homogeneous solution satisfies the equation
di2 L 1diL 1 hh+ +iL =0 (1.36) dt 2 RC dt LC h
By assuming a solution of the form Aest we obtain the characteristic equation
11 ss2 + +=0 (1.37) RCLC
Be defining the following parameters
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1 ω ≡ : Resonant frequency (1.38) ο LC
And
1 α = : Damping rate (1.39) 2RC
The characteristic equation becomes
22 ss+ 2αω+ο =0 (1.40)
The two roots of this equation are
22 s1=−α + α−ωο (1.41)
22 s2 =−α − α−ωο (1.42)
The homogeneous solution is a linear combination of es1t and es2t
s1ts2t iLh ()t =+A1e A2e (1.43)
And the general solution becomes
s1ts2t iL()t =+Is A1e +A2e (1.44)
The 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 Cory 11 The form of the roots s1 and s2 depend on the values of α and ωο . The following three cases are possible.
1. α = ωο : Critically Damped System. s1 and s2 are equal and real numbers: no oscillatory behavior
2. α > ωο : Over Damped System Here s1 and s2 are real numbers but are unequal: no oscillatory behavior
3. α < ωο : Under Damped System 22 22 α −=ωωοj ο−α In this case the roots s1 and s2 are complex numbers:
22 22 sj1,=−α + ωαο− s2=−α−jωο−α. System exhibits oscillatory behavior
Let’s investigate the under damped case, α < ωο , in more detail.
22 22 For α < ωο , α −=ωωοοj−α≡jωd the solution is
−α t jtωd− jtωd iL()t =+Is eN ( A1e +A2e ) (1.45) Decaying Oscillatory
± jtωd By using Euler’s identity e=±cosωdtjsinωdt, the solution becomes
−α t iL()t =+Is eN (K1cosωdt +K2sinωdt) (1.46) Decaying Oscillatory
Now we can determine the constants K1 and K2 by applying the initial conditions
iL(0t = )=⇒0Is +K1 =0 (1.47) ⇒=KI1 −s
diL = 0(⇒−αωKK12+ 0+ d )=0 dt t=0 (1.48) −α ⇒=KI2 s ωd
And the solution is
6.071/22.071 Spring 2006, Chaniotakis and Cory 12 ⎡ ⎤ ⎢ ⎥ −α t ⎛⎞α iL()t =−Is ⎢1 eN ⎜cosωdt +sinωdt ⎟⎥ (1.49) ω ⎢ Decaying ⎝⎠ d ⎥ ⎣⎢ Oscillatory ⎦⎥
22⎛⎞−1B2 By using the trigonometric identity Bt12cos +=Bsin tB1+B2cos⎜⎟t−tan the ⎝⎠B1 solution becomes
ωο −α t ⎛⎞−1 α iL(t) =−Is Is e cos⎜ωd t −tan ⎟ (1.50) ωdd⎝⎠ω
22 Recall that ωd ≡ωο −α and thus ωd is always smaller than ωo
Let’s now investigate the important limiting case:
As R →∞ , α << ω0
22 −1 α −αt ωd ≡−ωαο ≈ωο and tan ≈ 0, e ≈ 1 ωο
And the solution reduces to iL(t) = Is − Is cosωot which corresponds to the response of the circuit
Is L C
The plot of iL(t) is shown on Figure 8 for C=47nF, L=47mH, Is=5A and for R=20kΩ and 8kΩ, The dotted lines indicate the decaying characteristics of the response. For convenience and easy visualization the plot is presented in the normalized time ωοt /π . Note that the peak current through the inductor is greater than the supply current Is.
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(a) For R=8kΩ
(b) For R=20kΩ Figure 8.
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(a) R=20kΩ
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(b) R=8kΩ Figure 9
The energy stored in the inductor and the capacitor is shown on Figure 10.
Figure 10. Energy as a function of time
Figure 11 shows the plot of the response corresponding to the case where α << ω0 . This shows the persistent oscillation for the current iL(t) with frequency ω0 .
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Figure 11
6.071/22.071 Spring 2006, Chaniotakis and Cory 17 The Critically Damped Response.
When α = ωο the two roots of the characteristic equation are equal s1=s2=s. And our assumed solution becomes
iL()t =+AestsA e t 12 (1.51) st = Ae3
Now we have only one arbitrary constant. This is a problem for our second order system since 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 becomes
di2 L diL hh+ 2αα+2iL =0 (1.52) dt 2 dt h
The solution of this equation is1
−αt−αt iL()t =A1te +A2e (1.53)
Which is a linear combination of the exponential term and an exponential term multiplied by t.
di2 di dd⎛⎞i ⎛⎞di 1 The equation 2 may be rewritten as , by 2 ++2ααi =0 ⎜⎟+ααii++⎜α⎟=0 dt dt dt ⎝⎠dt ⎝⎠dt di dξ defining ξ =+αi the equation becomes +αξ = 0 whose solution is ξ = Ke−αt . Therefore dt dt 1 di d eeααtt+αi=K which may be written as ()eiαt = K. By integration we obtain the solution dt 1 dt 1 −−αttα iK=12te +Ke
6.071/22.071 Spring 2006, Chaniotakis and Cory 18 Summary of RLC transient response
Series Parallel 1 1 ω ω = ω = ο ο LC ο LC R 1 α α = α = 2L 2RC
Critically α = ωο −αtt−α Damped Response: A12te + A e
α < ωο −α t Under Response: eKN ( 12cosωddt+ Ksinω t) Decaying Damped Oscillatory 22 Where ωd ≡−ωο α
α > ωο Over Response: A eAs12ts+ et Damped 12 22 Where s1, 2 =−α ± αω− ο
6.071/22.071 Spring 2006, Chaniotakis and Cory 19 Problem
For the circuit below, the switch S1 has been closed for a long time while switch S2 is open. Now switch S1 is opened and then at time t=0 switch S2 is closed. Determine the current i(t) as indicated.
S1 R1 S2 R2
i(t) Vs C L
6.071/22.071 Spring 2006, Chaniotakis and Cory 20