EE101: RLC Circuits (with DC sources)
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Series RLC circuit
i
V0
R
L
VR
VL
C
VC
M. B. Patil, IIT Bombay
Series RLC circuit
R
L
VR
VL
i
V0
KVL: VR + VL + VC = V0 ⇒ i R + L
C
di
1
+
dt
C
Z
VC
i dt = V0
M. B. Patil, IIT Bombay
Series RLC circuit
R
L
VR
VL
i
V0
KVL: VR + VL + VC = V0 ⇒ i R + L
C
di
1
+
dt
C
Z
VC
i dt = V0
Differentiating w. r. t. t, we get,
R
di
d 2i
1
+L 2 +
i = 0.
dt
dt
C
M. B. Patil, IIT Bombay
Series RLC circuit
R
L
VR
VL
i
V0
KVL: VR + VL + VC = V0 ⇒ i R + L
C
di
1
+
dt
C
Z
VC
i dt = V0
Differentiating w. r. t. t, we get,
di
d 2i
1
+L 2 +
i = 0.
dt
dt
C
d 2i
R di
1
i.e.,
+
+
i = 0,
dt 2
L dt
LC
a second-order ODE with constant coefficients.
R
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
V
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
KCL: iR + iL + iC = I0 ⇒
1
1
V+
R
L
R
Z
iC
iL
L
V dt + C
C
V
dV
= I0
dt
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
1
1
V+
R
L
Differentiating w. r. t. t, we get,
KCL: iR + iL + iC = I0 ⇒
R
Z
iC
iL
L
V dt + C
C
V
dV
= I0
dt
1 dV
1
d 2V
+ V +C
= 0.
R dt
L
dt 2
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
1
1
V+
R
L
Differentiating w. r. t. t, we get,
KCL: iR + iL + iC = I0 ⇒
R
Z
iC
iL
L
V dt + C
C
V
dV
= I0
dt
1 dV
1
d 2V
+ V +C
= 0.
R dt
L
dt 2
d 2V
1 dV
1
i.e.,
+
+
V = 0,
dt 2
RC dt
LC
a second-order ODE with constant coefficients.
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
* A series RLC circuit driven by a constant current source is trivial to analyze.
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
* A series RLC circuit driven by a constant current source is trivial to analyze.
Since the current through each element is known, the voltage can be found in a
straightforward manner.
Z
1
di
i dt .
VR = i R, VL = L , VC =
dt
C
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
* A series RLC circuit driven by a constant current source is trivial to analyze.
Since the current through each element is known, the voltage can be found in a
straightforward manner.
Z
1
di
i dt .
VR = i R, VL = L , VC =
dt
C
* A parallel RLC circuit driven by a constant voltage source is trivial to analyze.
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
* A series RLC circuit driven by a constant current source is trivial to analyze.
Since the current through each element is known, the voltage can be found in a
straightforward manner.
Z
1
di
i dt .
VR = i R, VL = L , VC =
dt
C
* A parallel RLC circuit driven by a constant voltage source is trivial to analyze.
Since the voltage across each element is known, the current can be found in a
straightforward manner.
Z
1
dV
V dt .
iR = V /R, iC = C
, iL =
dt
L
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
i
I0
R
L
VR
VL
iR
C
VC
V0
R
iL
L
iC
C
V
* A series RLC circuit driven by a constant current source is trivial to analyze.
Since the current through each element is known, the voltage can be found in a
straightforward manner.
Z
1
di
i dt .
VR = i R, VL = L , VC =
dt
C
* A parallel RLC circuit driven by a constant voltage source is trivial to analyze.
Since the voltage across each element is known, the current can be found in a
straightforward manner.
Z
1
dV
V dt .
iR = V /R, iC = C
, iL =
dt
L
* The above equations hold even if the applied voltage or current is not constant,
and the variables of interest can still be easily obtained without solving a
differential equation.
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
A general RLC circuit (with one inductor and one capacitor) also leads to a
second-order ODE. As an example, consider the following circuit:
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
A general RLC circuit (with one inductor and one capacitor) also leads to a
second-order ODE. As an example, consider the following circuit:
i
V0
R1
L
C
V
R2
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
A general RLC circuit (with one inductor and one capacitor) also leads to a
second-order ODE. As an example, consider the following circuit:
i
V0
R1
L
C
V
di
+V
dt
dV
1
i =C
+
V
dt
R2
V0 = R1 i + L
R2
(1)
(2)
M. B. Patil, IIT Bombay
Series/Parallel RLC circuits
A general RLC circuit (with one inductor and one capacitor) also leads to a
second-order ODE. As an example, consider the following circuit:
i
R1
L
C
V0
V
R2
di
+V
dt
dV
1
i =C
+
V
dt
R2
V0 = R1 i + L
Substituting (2) in (1), we get
ˆ
˜
ˆ
˜
V0 = R1 CV 0 + V /R2 + L CV 00 + V 0 /R2 + V ,
00
0
V [LC ] + V [R1 C + L/R2 ] + V [1 + R1 /R2 ] = V0 .
(1)
(2)
(3)
(4)
M. B. Patil, IIT Bombay
General solution
Consider the second-order ODE with constant coefficients,
dy
d 2y
+a
+ b y = K (constant) .
dt 2
dt
M. B. Patil, IIT Bombay
General solution
Consider the second-order ODE with constant coefficients,
dy
d 2y
+a
+ b y = K (constant) .
dt 2
dt
The general solution y (t) can be written as,
y (t) = y (h) (t) + y (p) (t) ,
where y (h) (t) is the solution of the homogeneous equation,
d 2y
dy
+a
+ by = 0,
dt 2
dt
and y (p) (t) is a particular solution.
M. B. Patil, IIT Bombay
General solution
Consider the second-order ODE with constant coefficients,
dy
d 2y
+a
+ b y = K (constant) .
dt 2
dt
The general solution y (t) can be written as,
y (t) = y (h) (t) + y (p) (t) ,
where y (h) (t) is the solution of the homogeneous equation,
d 2y
dy
+a
+ by = 0,
dt 2
dt
and y (p) (t) is a particular solution.
Since K = constant, a particular solution is simply y (p) (t) = K /b.
M. B. Patil, IIT Bombay
General solution
Consider the second-order ODE with constant coefficients,
dy
d 2y
+a
+ b y = K (constant) .
dt 2
dt
The general solution y (t) can be written as,
y (t) = y (h) (t) + y (p) (t) ,
where y (h) (t) is the solution of the homogeneous equation,
d 2y
dy
+a
+ by = 0,
dt 2
dt
and y (p) (t) is a particular solution.
Since K = constant, a particular solution is simply y (p) (t) = K /b.
In the context of RLC circuits, y (p) (t) is the steady-state value of the variable of
interest, i.e.,
y (p) = lim y (t),
t→∞
which can be often found by inspection.
M. B. Patil, IIT Bombay
General solution
For the homogeneous equation,
dy
d 2y
+a
+ by = 0,
dt 2
dt
we first find the roots of the associated characteristic equation,
r2 + a r + b = 0 .
Let the roots be r1 and r2 . We have the following possibilities:
M. B. Patil, IIT Bombay
General solution
For the homogeneous equation,
dy
d 2y
+a
+ by = 0,
dt 2
dt
we first find the roots of the associated characteristic equation,
r2 + a r + b = 0 .
Let the roots be r1 and r2 . We have the following possibilities:
* r1 , r2 are real, r1 6= r2 (“overdamped”)
y (h) (t) = C1 exp(r1 t) + C2 exp(r2 t) .
M. B. Patil, IIT Bombay
General solution
For the homogeneous equation,
dy
d 2y
+a
+ by = 0,
dt 2
dt
we first find the roots of the associated characteristic equation,
r2 + a r + b = 0 .
Let the roots be r1 and r2 . We have the following possibilities:
* r1 , r2 are real, r1 6= r2 (“overdamped”)
y (h) (t) = C1 exp(r1 t) + C2 exp(r2 t) .
* r1 , r2 are complex, r1,2 = α ± jω (“underdamped”)
y (h) (t) = exp(αt) [C1 cos(ωt) + C2 sin(ωt)] .
M. B. Patil, IIT Bombay
General solution
For the homogeneous equation,
dy
d 2y
+a
+ by = 0,
dt 2
dt
we first find the roots of the associated characteristic equation,
r2 + a r + b = 0 .
Let the roots be r1 and r2 . We have the following possibilities:
* r1 , r2 are real, r1 6= r2 (“overdamped”)
y (h) (t) = C1 exp(r1 t) + C2 exp(r2 t) .
* r1 , r2 are complex, r1,2 = α ± jω (“underdamped”)
y (h) (t) = exp(αt) [C1 cos(ωt) + C2 sin(ωt)] .
* r1 = r2 = α (“critically damped”)
y (h) (t) = exp(αt) [C1 t + C2 ] .
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
iL (0− ) = 0 A ⇒ iL (0+ ) = 0 A.
V (0− ) = 0 V ⇒ V (0+ ) = 0 V .
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
iL (0− ) = 0 A ⇒ iL (0+ ) = 0 A.
V (0− ) = 0 V ⇒ V (0+ ) = 0 V .
d 2V
1 dV
1
+
+
V = 0 (as derived earlier)
dt 2
RC dt
LC
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
iL (0− ) = 0 A ⇒ iL (0+ ) = 0 A.
V (0− ) = 0 V ⇒ V (0+ ) = 0 V .
d 2V
1 dV
1
+
+
V = 0 (as derived earlier)
dt 2
RC dt
LC
The roots of the characteristic equation are (show this):
r1 = −0.65 × 105 s −1 , r2 = −0.35 × 105 s −1 .
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
iL (0− ) = 0 A ⇒ iL (0+ ) = 0 A.
V (0− ) = 0 V ⇒ V (0+ ) = 0 V .
d 2V
1 dV
1
+
+
V = 0 (as derived earlier)
dt 2
RC dt
LC
The roots of the characteristic equation are (show this):
r1 = −0.65 × 105 s −1 , r2 = −0.35 × 105 s −1 .
The general expression for V (t) is,
V (t) = A exp(r1 t) + B exp(r2 t) + V (∞),
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
I0
R
iC
iL
L
C
R=10 Ω
V
C=1 µF
L=0.44 mH
I0 = 100 mA
iL (0− ) = 0 A ⇒ iL (0+ ) = 0 A.
V (0− ) = 0 V ⇒ V (0+ ) = 0 V .
d 2V
1 dV
1
+
+
V = 0 (as derived earlier)
dt 2
RC dt
LC
The roots of the characteristic equation are (show this):
r1 = −0.65 × 105 s −1 , r2 = −0.35 × 105 s −1 .
The general expression for V (t) is,
V (t) = A exp(r1 t) + B exp(r2 t) + V (∞),
i.e., V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ) + V (∞),
where τ1 = −1/r1 = 15.4 µs, τ2 = −1/r1 = 28.6 µs.
M. B. Patil, IIT Bombay
Parallel RLC circuit
As t → ∞ , V = L
diL
= 0 V ⇒ V (∞) = 0 V .
dt
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
dV +
Our other initial condition is iL (0 ) = 0 A, which can be used to obtain
(0 ).
dt
+
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
dV +
Our other initial condition is iL (0 ) = 0 A, which can be used to obtain
(0 ).
dt
1
dV +
+
+
iL (0 ) = I0 −
V (0 ) − C
(0 ) = 0 A, which gives
R
dt
+
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
dV +
Our other initial condition is iL (0 ) = 0 A, which can be used to obtain
(0 ).
dt
1
dV +
+
+
iL (0 ) = I0 −
V (0 ) − C
(0 ) = 0 A, which gives
R
dt
+
(A/τ1 ) + (B/τ2 ) = −I0 /C .
(2)
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
dV +
Our other initial condition is iL (0 ) = 0 A, which can be used to obtain
(0 ).
dt
1
dV +
+
+
iL (0 ) = I0 −
V (0 ) − C
(0 ) = 0 A, which gives
R
dt
+
(A/τ1 ) + (B/τ2 ) = −I0 /C .
(2)
From (1) and (2), we get the values of A and B, and
V (t) = −3.3 [exp(−t/τ1 ) − exp(−t/τ2 )] V .
(3)
(SEQUEL file: ee101 rlc 1.sqproj)
M. B. Patil, IIT Bombay
Parallel RLC circuit
diL
= 0 V ⇒ V (∞) = 0 V .
dt
⇒ V (t) = A exp(−t/τ1 ) + B exp(−t/τ2 ),
As t → ∞ , V = L
Since V (0+ ) = 0 V , we have,
A + B = 0.
(1)
dV +
Our other initial condition is iL (0 ) = 0 A, which can be used to obtain
(0 ).
dt
1
dV +
+
+
iL (0 ) = I0 −
V (0 ) − C
(0 ) = 0 A, which gives
R
dt
+
(A/τ1 ) + (B/τ2 ) = −I0 /C .
(2)
From (1) and (2), we get the values of A and B, and
V (t) = −3.3 [exp(−t/τ1 ) − exp(−t/τ2 )] V .
(3)
(SEQUEL file: ee101 rlc 1.sqproj)
iR
I0
R
iC
iL
L
C
R=10 Ω
V
100
0.8
C=1 µF
0.6
L=0.44 mH
0.4
iL (mA)
iR (mA)
V (Volts)
I0 = 100 mA 0.2
iC (mA)
0
0
0
0.05
0.1
time (ms)
0.15
0.2
0
0.05
0.1
time (ms)
0.15
0.2
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
t=0
Vs
R
L
VR
VL
C
VC
L=1 mH
C=1 µF
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
t=0
Vs
R
L
VR
VL
C
VC
L=1 mH
C=1 µF
(a) Show that the condition for critically damped response is R = 63.2 Ω.
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
t=0
Vs
R
L
VR
VL
C
VC
L=1 mH
C=1 µF
(a) Show that the condition for critically damped response is R = 63.2 Ω.
(b) For R = 20 Ω, derive expressions for i(t) and VL (t) for t > 0 (Assume that
VC (0− ) = 0 V and iL (0− ) = 0 A). Plot them versus time.
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
t=0
Vs
R
L
VR
VL
C
VC
L=1 mH
C=1 µF
(a) Show that the condition for critically damped response is R = 63.2 Ω.
(b) For R = 20 Ω, derive expressions for i(t) and VL (t) for t > 0 (Assume that
VC (0− ) = 0 V and iL (0− ) = 0 A). Plot them versus time.
(c) Repeat (b) for R = 100 Ω.
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
t=0
Vs
R
L
VR
VL
C
VC
L=1 mH
C=1 µF
(a) Show that the condition for critically damped response is R = 63.2 Ω.
(b) For R = 20 Ω, derive expressions for i(t) and VL (t) for t > 0 (Assume that
VC (0− ) = 0 V and iL (0− ) = 0 A). Plot them versus time.
(c) Repeat (b) for R = 100 Ω.
(d) Compare your results with the following plots.
(SEQUEL file: ee101 rlc 2.sqproj)
M. B. Patil, IIT Bombay
Series RLC circuit: home work
i
5V
0V
R
L
VR
VL
Vs
C
VC
L=1 mH
t=0
C=1 µF
(a) Show that the condition for critically damped response is R = 63.2 Ω.
(b) For R = 20 Ω, derive expressions for i(t) and VL (t) for t > 0 (Assume that
VC (0− ) = 0 V and iL (0− ) = 0 A). Plot them versus time.
(c) Repeat (b) for R = 100 Ω.
(d) Compare your results with the following plots.
(SEQUEL file: ee101 rlc 2.sqproj)
R = 20 Ω
8
R = 100 Ω
VC
5
VC
4
VL
0
−4
VR
VL
VR
0
0.2
0.4
time (ms)
0
0.6
0.8
0
0.2
0.4
time (ms)
0.6
0.8
M. B. Patil, IIT Bombay