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
-
L
R
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
i=C
dvc
dt
(1.2)
And thus the voltages vR and vL are given by
vR = iR = RC
vL = L
dvc
dt
di
d 2 vc
= LC 2
dt
dt
(1.3)
(1.4)
Substituting Equations (1.3) and (1.4) into Equation (1.1) we obtain
d 2 vc R dvc 1
1
+
+
vc =
Vs
2
dt
L dt LC
LC
(1.5)
The solution to equation (1.5) is the linear combination of the homogeneous and the
particular solution vc = vc p + vch
The particular solution is
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vc p = Vs
(1.6)
And the homogeneous solution satisfies the equation
d 2 vch R dvch
1
+
+
vch = 0
2
dt
L dt
LC
(1.7)
Assuming a homogeneous solution is of the form Ae st and by substituting into Equation
(1.7) we obtain the characteristic equation
s2 +
R
1
=0
s+
L
LC
(1.8)
By defining
R
:
2L
α=
Damping rate
(1.9)
And
ωο =
1
: Natural frequency
LC
(1.10)
The characteristic equation becomes
s 2 + 2α s + ωο2 = 0
(1.11)
The roots of the characteristic equation are
s1 = −α + α 2 − ωο2
(1.12)
s 2 = −α − α 2 − ωο2
(1.13)
And the homogeneous solution becomes
vch = A1e s1t + A2 e s 2t
(1.14)
vc = Vs + A1e s1t + A2 e s 2 t
(1.15)
The total solution now becomes
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The parameters A1 and A2 are constants and can be determined by the application of the
dvc(t = 0)
initial conditions of the system vc (t = 0) and
.
dt
The value of the term α 2 − ωο2 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
vc = Vs + A1e s1t + A2 e s 2 t
3. α < ωο .
α 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 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
(rad/sec) is called the natural frequency of the system
LC
or the resonant frequency.
The frequency ωο =
R
is called the damping rate and its value in relation to ωο
2L
determines the behavior of the response
o α = ωο :
Critically Damped
The parameter α =
o α > ωο :
o α < ωο :
•
The quantity
Over Damped
Under Damped
L
has units of resistance
C
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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
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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
d 2 vc 1
+
vc = 0
dt 2 LC
(1.16)
Assuming a solution of the form Ae st the characteristic equation is
Where ωο =
s 2 + ωο2 = 0
(1.17)
s1 = + jωο
(1.18)
s 2 = − jωο
(1.19)
1
LC
The two roots are
And the solution is a linear combination of A1e s1t and A2e s 2t
vc(t ) = A1e jωot + A2e − jωο t
(1.20)
By using Euler’s relation Equation (1.20) may also be written as
vc(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.
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For vc(t = 0) = Vo and for
dvc (t = 0)
= 0 (no current flowing in the circuit initially) we
dt
have from Equation (1.20)
A1 + A2 = Vo
(1.22)
jωo A1 − jωo A2 = 0
(1.23)
And
Which give
A1 = A2 =
Vo
2
(1.24)
And the solution becomes
Vo jωot − jωο t
e +e
2
= Vo cos(ωot )
vc(t ) =
(
)
(1.25)
The current flowing in the circuit is
dvc
dt
= −CVoωο sin(ωο t )
i=C
(1.26)
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
= −Vo cos(ωot )
(1.27)
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.
iR(t)
R
Is
iC(t)
iL(t)
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.
2.
3.
4.
Establish the initial conditions for the system
Determine the equation that describes the system characteristics
Solve the equation
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(t = 0) = 0
Initial Conditions: ⎨
⎩ v(t = 0) = 0
(1.28)
By applying KCl at the indicated node we obtain
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Is = iR + iL + iC
(1.29)
The voltage across the elements is given by
d iL
dt
(1.30)
v L d iL
=
R R dt
(1.31)
dv
d 2iL
= LC 2
dt
dt
(1.32)
v=L
And the currents iR and iC are
iR =
iC = C
Combining Equations (1.29), (1.31), and (1.32) we obtain
d 2iL 1 d iL 1
1
+
+
iL =
Is
2
dt
RC dt LC
LC
(1.33)
The solution to equation (1.33) is a superposition of the particular and the homogeneous
solutions.
iL (t ) = iL p (t ) + iLh (t )
(1.34)
iL p (t ) = Is
(1.35)
The particular solution is
The homogeneous solution satisfies the equation
d 2iLh
1 d iLh
1
+
+
iLh = 0
2
dt
RC dt
LC
(1.36)
By assuming a solution of the form Ae st we obtain the characteristic equation
s2 +
1
1
=0
s+
RC
LC
(1.37)
Be defining the following parameters
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1
: Resonant frequency
LC
ωο ≡
(1.38)
And
α=
1
: Damping rate
2RC
(1.39)
The characteristic equation becomes
s 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 are
The homogeneous solution is a linear combination of e s1t and e s 2 t
iLh (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 becomes
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 .
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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
α 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 behavior
Let’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 t
iL(t ) = Is + eN
A1e jωd t + A2 e− jωd t )
(
Decaying
(1.45)
Oscillatory
By using Euler’s identity e ± jωd t = cos ωd t ± j sin ωd t , the solution becomes
−α t
iL(t ) = Is + eN
K1 cos ωd t + K 2 sin ωd t )
(
Decaying
(1.46)
Oscillatory
Now we can determine the constants K1 and K 2 by applying the initial conditions
iL(t = 0) = 0 ⇒ Is + K1 = 0
⇒ K1 = − Is
diL
= 0 ⇒ −α K1 + (0 + K 2ωd ) = 0
dt t =0
⇒ K2 =
−α
ωd
(1.47)
(1.48)
Is
And the solution is
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⎡
⎤
⎢
⎞⎥
α
−α t ⎛
iL(t ) = Is ⎢1 − eN
sin 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 ⎟ the
B1 ⎠
⎝
solution becomes
iL(t ) = Is − Is
⎛
ωο −α t
α ⎞
e cos ⎜ ωd t − tan −1
⎟
ωd
ωd ⎠
⎝
(1.50)
Recall that ωd ≡ ωο2 − α 2 and thus ωd is always smaller than ωo
Let’s now investigate the important limiting case:
As R → ∞ , α << ω0
α
≈ 0 , e −α t ≈ 1
ωο
And the solution reduces to iL(t ) = Is − Is cos ωot which corresponds to the response of
the circuit
ωd ≡ ωο2 − α 2 ≈ ωο and tan −1
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 ωο t / 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
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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 ) = A1e st + A2e st
= A3e st
(1.51)
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
d 2iLh
d iLh
+ 2α
+ α 2iLh = 0
2
dt
dt
(1.52)
The solution of this equation is1
iL (t ) = A1te −α t + A2 e −α t
(1.53)
Which is a linear combination of the exponential term and an exponential term multiplied
by t.
d ⎛ di
d 2i
di
⎞
⎛ di
⎞
+ 2α
+ α 2i = 0 may be rewritten as ⎜ + α i ⎟ + α ⎜ + α i ⎟ = 0 , by
2
dt ⎝ dt
dt
dt
⎠
⎝ dt
⎠
dξ
di
defining ξ =
+ α i the equation becomes
+ αξ = 0 whose solution is ξ = K1e−α t . Therefore
dt
dt
di
d
eα t + eα tα i = K1 which may be written as (eα t i ) = K1 . By integration we obtain the solution
dt
dt
−α t
−α t
i = K1te + K 2 e
1
The equation
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Summary of RLC transient response
ωο
α
Critically
Damped
Series
1
ωο =
LC
R
α=
2L
Parallel
1
ωο =
LC
1
α=
2RC
α = ωο
Response: A1te −α t + A2 e −α t
α < ωο
Under
Damped
Response: eN ( K1 cos ωd t + K 2 sin ωd t )
Decaying
−α t
Oscillatory
Where ωd ≡ ωο2 − α 2
Over
Damped
α > ωο
Response: A1e s1t + A2 e s 2t
Where s1, 2 = −α ± α 2 − ωο2
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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
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L
20