9 THE RLC CIRCUIT
9
49
The RLC Circuit
9.1
Summary and Review Exercises
• Two storage devices which can not be combined will result in second-order differential equation. examples are
R
L
vs +
−
is
C
L C
prlc-nosrc.m4
srlc.m4
R1
vs +
−
R
R2
L1
R
is
L2
C1
C2
iscrc.m4
srcrlrl.m4
• The natural response of a 2nd order system can represented by the 2nd order ODE of the form:
dy
d2 y
+ 2α
+ ωo2 y = 0
2
dt
dt
or
d2 y
dy
+ 2ζωo
+ ωo2 y = 0
dt2
dt
Here ζ = damping factor, ωo = undamped natural frequency in rad/sec, α = neper frequency in rad/sec and it
is referred to exponential damping factor .
ζ = ωαo
• Let y(t) = Kest (eigenfunction of the system), substitute in the ODE we get:
Kest (s2 + 2ζωo s + ωo2 ) = 0, and the characteristic equation of the system will be
s2 + 2ζωo s + ωo2 = 0
two solutions (roots)
s1 , s2 = ωo (−ζ ±
• Response for RLC (t ≥ 0) (series or parallel) can be :
p
ζ 2 − 1)
– Overdamped, α > ωo ≡ ζ > 1 ⇒ s1 , s2 real, distinct and the solution will be
y(t) = K1 es1 t + K2 es2 t , K1 , K2 are real
– Critically damped, α = ωo ≡ ζ = 1 ⇒ s1 , s2 real, repeated, s1 = s2 = −ζωo = −α and the solution will be
y(t) = K1 es1 t + K2 tes1 t = (K1 + K2 t)e−αt , K1 , K2 are real
– Underdamped, α < ωo ≡ ζ < 1 ⇒ s1 , s2 complex conjugates, s1 = s∗2 , s1 , s2 = −ζωo ± jωo
solution will be
√ 2
√ 2
y(t) = K1 es1 t + K2 es2 t = K1 e−ζωo t ejωo 1−ζ t + K2 e−ζωo t e−jωo 1−ζ t
p
1 − ζ 2 and the
9 THE RLC CIRCUIT
50
Recall that α = ζωo and let β = ωo
p
1 − ζ 2 we can re-write the solution for this case as
y(t) = K1 e−αt ejβt + K2 e−αt e−jβt
The above solution can be put in terms of sin and cos using Euler’s formula as follows:
y(t) = A1 e−αt cos(βt) + A2 e−αt sin(βt) = Ce−αt sin(βt + φ)
and since our circuit is real, hence its natural response must be real and therefore, A1 , A2 , C, and φ must
be real.
– Oscillatory, α = 0 ≡ ζ = 0 ⇒ s1 , s2 = ±jωo imaginary and the solution will be
y(t) = K1 cos(ωo t) + K2 sin(ωo t) = K cos(ωo t + φ), K1 , K2 , K, φ are real
• Response for RLC circuit needs two initial conditions to determine the two unknowns.
• For series RLC circuits,
i
R
+ vR −
+
L vL
−
C
−vC+
RLC-sourcefree.m4
di
KVL: +vR + vL + vC = 0, substitute for vR , vL to get Ri + L dt
+ vC = 0. Differentiate the last equation we get
2
dvC
dvC
d i
di
d2 i
di
R dt + L dt2 + dt = 0, substitute for i = C dt to get R dt + L dt2 + C1 i = 0, re-arrange the terms and divide by
L to put equation into the canonical form
d2 i R di
1
+
+
i=0
2
dt
L dt LC
its characteristic equation s2 + 2αs + ωo2 = 0, with α =
R
2L
and ωo =
√1
LC
• For parallel RLC circuits,
iR
R
iL
L C
iC
+
vC
−
prlc-sourcefree.m4
KCL: −iR − iL − iC = 0, vC = vR = RiR = vL = L didtL , we get vRC + iL + C dvdtC = 0, then take the derivative we
2
2
get R1 dvdtC + didtL + C ddtv2C = 0 = R1 dvdtC + L1 vC + C ddtv2C , re-arrange terms and divide by C to get:
d 2 vC
1 dvC
1
+
+
vC
dt2
RC dt
LC
its characteristic equation s2 + 2αs + ωo2 = 0, with α =
1
2RC
and ωo =
√1
LC
• Overdamped response is of the form of the sum of two exponentials, e.g. A1 es1 t + A2 es2 t
• Critically damped response has the form e−αt (A1 t + A2 ).
• Underdamped response is an exponentially damped sinusoid, e−αt (B1 cos ωd t+B2 sin ωd t), ωd = β as defined above
• During transient of an RLC circuit, energy transfers between energy storage elements to the extent allowed by
the resistive component(s) of the circuit, which acts to dissipate the energy initially stored.
9 THE RLC CIRCUIT
51
Forced Circuit: Consider the series RLC circuit
i
R
+
vS
L
−
C
−vC+
rlc.m4
get
di
Writing KVL for the mesh, −vS + R ∗ i + L dt
+ vC = 0 and substituting for i using the relation iL = i = C dvdtC we
−vS + R(C
d
dvC
dvC
) + L (C
) + vC = 0
dt
dt
dt
which can be written as
R dvC
1
vS
d 2 vC
+
+
vC =
dt2
L dt
LC
LC
the characteristic equation is the same as before since it characterizes the natural (force-free) response of the circuit.
The same equation can be obtained via the s−operator method as follows:
R
d
Replace dt
by s while (.)dt with 1s .
−vS + Ri + sLi + vC = 0 and i = sCvC . Substitute for i in the first ODE we get:
−vS + R(sCvC ) + sL(sCvC ) + vC = 0 which gives (LCs2 + RCs + 1)vC = vS
(s2 +
R
1
vS
s+
)vC =
L
LC
LC
and the differential equation can be written as
d 2 vC
R dvC
vC
vS
+
+
=
dt2
L dt
LC
LC
Using PSpice for the series RLC circuit with DC source VS=1 volts, R=400 Ω , L=1mH, C=5nF:
tr11-21.cir: Transient in a series RLC with battery (TR11-21 page 628)
Date/Time run: 12/27/106 06:32:59
Temperature: 27.0
(A) tr11-21
1.4V
1.2V
1.0V
0.8V
Transient in a series RLC with source
*(TR11-21 page 628)
VS
1
0
dc
1
L
1
2
1e-3 ic=0
R
2
3
400
C
3
0
5e-9 ic=0
.tran 0.3u
30u
uic
.probe
.end
R
1
Solving the above series RLC circuit, α = 2L
and ωo = √LC
In Matlab, the following is the solution for the transient in a series RLC circuit forced by a step function
0.6V
0.4V
0.2V
0V
0s
5us
10us
15us
20us
25us
30us
V(3)
Date: December 27, 2006
Time
Page 1
%series_rlc.m
%Transient in RLC series circuit forced by unit step
%solution vC(t)=vn(t)+vf(t)=natural response+forced response
Time: 07:09:08
9 THE RLC CIRCUIT
52
%here forced response (capacitor voltage) due to DC voltage will be vf(t)=Vs
Vs=1; R=400; L=1e-3; C=5e-9;
omega=1/sqrt(L*C);
alpha=R/(2*L);
zeta=alpha/omega;
s1=omega*(-zeta+sqrt(zeta^2-1));
s2=omega*(-zeta-sqrt(zeta^2-1));
%solve for the capacitor voltage vC(t):
%given the initial condition vC(0)=i1, iL(0)=i2;
%[K1; K2]=[1, 1; s1, s2]\[vC(0)-Vs; iL(0)/C]%t=0, K(1)+K(2)+Vs=vC(0)
i1=0; %vC(0)=0 (continuity)
i2=0; %iL(0)=0 (continuity)
K=[1,1; s1,s2]\[i1-Vs; i2/C];
t=0:0.5e-6:30e-6;
vC=K(1)*exp(s1*t)+K(2)*exp(s2*t)+Vs;
plot(t,vC); xlabel(’time in sec’); ylabel(’vC(t)’);
title(’Series RLC circuit forced by unit step’);
grid;
print -deps series_rlc.eps
and vC (t) will be
Series RLC circuit forced by unit step
1.4
1.2
1
225500
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
time in sec
2.5
3
3.5
−5
x 10
9 THE RLC CIRCUIT
53
Example on two-storage elements circuit
DS7eex9.3.1 page 372: Two inductors in two meshes
8Ω
2H
vs +
−
4Ω
1H
ds7ep372.m4
Writing the two mesh equations, we have
−vs + 8i1 + 2
di1
+ 4(i1 − i2 ) = 0
dt
4(i2 − i1 ) + 1 ∗
di2
=0
dt
using the s operator,
(12 + 2s)i1 − 4i2 = vs
−4i1 + (4 + s)i2 = 0
solving the last two equations we get i2 =
2vs
s2 +10s+16
or equivalently
(s2 + 10s + 16)i2 = 2vs
roots are s1 = −1, s2 = −8 and the natural response is i2 = A1 e−2t + A2 e−8t
Time constants for the circuit are: 12 , 81 seconds.
Example: HKD6e Practice Problem 9.7 on page 290.
In the circuit, vs (t) = 10 + 20u(t), find iL (t) at t = 0.1sec.
iLL = 15.625H
vs +
−
50Ω
+
R vC
−
C = 1mF
hkd6epp9.7.m4
Figure 8: HKD6e PP9.7
KV L : −vs + L didtL + vC = 0
KCL : iL − vRC − C dvdtC = 0
−vC
−vs + sLiL + vC = 0, ⇒ iL = vssL
vC
vs −vC
2
C
iL − R − CsvC = 0 ⇒ sL − vRC − CsvC = 0 ⇒ vs − vC − sLv
R − LCs vC = 0
vs
s
1
2
(LCs2 + sL
R + 1)vC = vs ⇒ (s + RC + LC )vC = LC
s
1
s
1
2
2
Characteristic equation: s + RC + LC = 0, ⇒ s + 50(1e−3)
+ 15.625(1e−3)
= 0 ⇒ s2 +20s+64 = 0, ⇒ s1 = −16, s2 = −4.
−16t
−4t
−16t
Natural Response: vCn (t) = A1 e
+ A2 e
and iLn (t) = B1 e
+ B2 e−4t .
Here forcing function (excitation) vs (t) = 10 + 20u(t), ⇒ forced response is DC (constant voltage).
Initial conditions: iL (0− ) = 10/R = 10/50 = 0.2 = iL (0+ ), and vC (0− ) = 10 = vC (0+ ).
At t = ∞, iL (∞) = (10 + 20)/50 = 0.6A
9 THE RLC CIRCUIT
54
To find iL (0.1) we need B1 , B2 .
iL (t) = iLn (t) + iLf (t) = B1 e−16t + B2 e−4t + 0.6, iL (0− ) = iL (0+ ) = 0.2 = B1 + B2 + 0.6
20
from KVL: t = 0+ , −30 + L didtL + 10 = 0, ⇒ didtL |t=0+ = 20
L = 15.625 = 1.28
−16t
−4t
Taking the derivative of iL (t) = −16B1 e
− 4B2 e , at t = 0+ , didtL |t=0+ = 1.28 = −16B1 − 4B2 and using the
other equation for B1 , B2 previously derived B1 + B2 = −0.4 we get:
B1 = 0.027, B2 = −0.427 and at t = 0.1 sec, iL (0.1) = 0.027e−1.6 −0.427e−0.4 +0.6 = 5.45e−3−0.286+0.6 = 0.3195A.
9 THE RLC CIRCUIT
55
Another example:
i
R1
v
R
+
vS
iL
−
L
+
vC
C −
Dorfp381.m4
In the figure, vs = 10u(−t) + 6e−3t u(t). R = 6, R1 = 4, L = 1, C = 41
Initial conditions: at t = 0− , i = iL = 10/(4 + 6) = 1A and vC = iL ∗ R = 1 ∗ 6 = 6V .
The node equation
dvC
vC − v s
− iL − C
=0
−
R1
dt
and KVL on the right mesh
−L
diL
− RiL + vC = 0
dt
d
v
v
s
Let s = dt
the two equations become Ril + LsiL = v ⇒ iL = R+sL
and v−v
R1 + R+sL + Csv = 0. Substituting for iL
we get, (v − vs )(R + sL) + R1 v + CsvR1 (R + sL) = 0, collect the terms for v in one side and for the forcing function
vs in the other side, we get
R1 + R
R + sL
L + CRR1
s+
)v =
vs
(s2 +
CR1 L
CR1 L
CR1 L
Substitute for R = 6, R1 = 4, L = 1, C =
1
4
we get (s2 + 7s + 10)v = (6 + s)vs or equivalently
d2 v
dv
dvs
+7
+ 10v =
+ 6vs
dt2
dt
dt
First we find the forced response which is due to an exponential input vs (t) = 6e−3t , t > 0 hence it will have the same
form, i.e. vf = Be−3t (Recall that ”complex” exponential is an eigenfunction of linear systems).
The forced solution (response) must satisfy the differential equation.
Substituting we get (−3)(−3)Be−3t +7(−3)Be−3t +10Be−3t = −18e−3t +36e−3t from which B = −9, and vf = −9e−3t .
Solving for the roots of the characteristic equation s2 + 7s + 10 = 0 we get s1 = −2, s2 = −5 and the natural
response becomes vn = A1 es1 t + A2 es2 t = A1 e−2t + A2 e−5t
To find A1 , A2 we need to find two equations in the two unknowns A1 , A2 . This must be obtained via the initial
conditions of the system. v = vf + vn
One equation can be formed due to the continuity of the voltage across the capacitor, i.e. v(0− ) = v(0+ ) = 6 =
A1 + A2 + B ⇒ A1 + A2 = 15.
dv
+
The other equation can be formed if we know dv
dt |t=0 . Note that dt for the capacitor voltage is related to the
current flowing into the capacitor, i.e. iC = C dvdtC . At t = 0+ , the following equation can be obtained (Node equation)
dvC
−
+
s
iL + v−v
R1 + C dt = 0 and using the continuity of the current in the inductor iL (0 ) = iL (0 ) = 1, v(0) = 6, vs (0) = 6,
dv
then dt = −4.
−2t
Taking the derivative of equation v = vf + vn = A1 e−2t + A2 e−5t − 9e−3t gives dv
− 5A2 e−5t + 27e−3t
dt = −2A1 e
and substituting for t = 0, then −4 = −2A1 − 5A2 + 27 and using A1 + A2 = 15 one can find A1 = 44/3, A2 = 1/3.
−2t
+ 31 e−5t − 9e−3t
The complete solution v = vn + vf = 44
3 e