ECED 2000 – Electric Circuits
Chapter 8 – Response of RLC Circuits (2nd order)
• Natural Response of a Parallel RLC circuit
̶ Overdamped, Underdamped & Critically Damped
• Step Response of a Parallel RLC Circuit
• Natural & Step Response of a Series RLC Circuit
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ECED 2000 – Electric Circuits
Chapter 8 – Response of RLC Circuits (2nd order)
RLC Circuits (Second Order Circuits)
- involve two energy storage systems
(Create second order Differential Equations)
• Transfer energy from one storage to another and back again
• L stores energy in Magnetic field from the current
• C stores energy in Electric field from stored charge
• Resistor is always dissipating energy
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ECED 2000 – Electric Circuits
8.1 – Natural Response of a Parallel RLC Circuit
Exponential solution:
v = Ae st ,
A & s constants
Replacing it in equation:
Applying KCL:
Ae st  s 2 +

t
v 1
dv
+ ∫ v ⋅ dτ + I 0 + C
=0
R L0
dt
1
1 
s+
=0
RC
LC 
Characteristic equation:
1
1
s +
s+
=0
RC
LC
2
Differentiating w.r.t. time t :
1 dv 1
d 2v
+ v+C 2 = 0
R dt L
dt
Dividing by C and re-arranging:
d 2v
1 dv 1
+
+
v=0
2
dt
RC dt LC
Roots:
s1, 2 = −α ± α 2 − ω02
where
1
α=
,
2 RC
1
ω =
LC
2
0
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ECED 2000 – Electric Circuits
8.1 – Natural Response of a Parallel RLC Circuit
If α ≠ ω0 :
If α = ω0 :
v(t ) = A1e s1t + A2 e s2t ,
v(t ) = D1te −αt + D2 e −αt .
Next… Find constants, e.g. A1 & A2,
from initial conditions v(0
+
)
&
( )
( )
dv 0 +
iC 0 +
=
dt
C
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ECED 2000 – Electric Circuits
8.1 – Natural Response of a Parallel RLC Circuit
Three cases set by the discriminant D
3
(1) D > 0 : roots real and unequal
• Overdamped (OD) case, α > ω0
v(t)
(2) D = 0 : roots real and equal
• Critically damped (CD) case, α = ω0
2
1
(3) D < 0 : roots complex and unequal
• Underdamped (UD) case, α < ω0
ζ=α/ω0
Second order equations are all about the energy flow
• In Overdamped the energy is lost very fast
Spring analogy: The mass moves slowly to the rest point
• Critically Damped the loss rate is smaller
Spring analogy: Just enough for one movement up and down
• For Underdamped spring moves up and down
• Energy is transferred from the mass to the spring and back again
• Loss rate is smaller than the time for transfer
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ECED 2000 – Electric Circuits
8.1 – Natural Response of a Parallel RLC Circuit
Example 8.1
R = 200Ω, L = 50mH, C = 0.2µF
a) Find roots of characteristic equation
b) Classify response in OD, UD or CD
c) Repeat a) & b) for R = 312.5Ω
d) What value of R makes it CD?
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ECED 2000 – Electric Circuits
8.2 – The Overdamped Voltage Response
α > ω0 → s1, s2 real & different, s1, 2 = −α ± α 2 − ω02
v(t ) = A1e s1t + A2 e s2t , t ≥ 0
1. Find roots of characteristic equation, s1 & s2, from
R, L & C
2. Find v(0+) & dv(0+)/dt = iC(0+)/C using circuit
analysis
3. Find A1 & A2 by solving:
I. v(0+) = A1 + A2
II. dv(0+)/dt = s1 A1 + s2 A2
4. Replace values of s1, s2, A1 & A2 in equation for v(t)
above
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ECED 2000 – Electric Circuits
Example 8.2
v(0+)= 12V, iL(0+)= 30mA
8.2 – The Overdamped Voltage Response
v(t ) = −14e −5, 000t + 26e −20,000t V
a) Find initial currents
b) Find dv(0+)/dt
c) Find v(t) expression
d) Sketch v(t), 0 ≤ t ≤ 250ms
Example 8.3
Calculate branch currents
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ECED 2000 – Electric Circuits
8.2 – The Underdamped Voltage Response
α < ω0 → s1, s2 complex & different, s1, 2 = −α ± jωd
where, ωd = ω02 − α 2
v(t ) = A1e
e
± jω d t
v(t ) = e
−αt
e
jω d t
+ A2 e
−αt
e
− jω d t
= cos(ωd t ) ± j sin (ωd t ),
−αt
, t≥0
Leonhard Euler
[( A1 + A2 )cos(ωd t ) + j ( A1 − A2 )sin (ωd t )]
Using new constants B1 = A1 + A2 & B2 = j ( A1 ̶ A2 ) :
v(t ) = B1e
−αt
∗
A1 = A2
cos(ωd t ) + B2 e
−αt
sin (ωd t ), t ≥ 0
thus B1 & B2 are real → v(t) is also real
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ECED 2000 – Electric Circuits
8.2 – The Underdamped Voltage Response
It can also be shown that:
v(t ) = B ⋅ e
−αt
Damped Frequency
⋅ cos(ω d t − φ ), t ≥ 0
where B = B12 + B22 and φ = tan −1 (B2 B1 )
Damping Factor
e −αt
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ECED 2000 – Electric Circuits
8.2 – The Underdamped Voltage Response
α < ω0 → s1, s2 complex & different, s1, 2 = −α ± jωd
where, ωd = ω02 − α 2
v(t ) = B1e
−αt
cos(ωd t ) + B2 e
−αt
sin (ωd t ), t ≥ 0
1. Find damping factor α & damped oscillation frequency ωd
from R, L & C
2. Find v(0+) & dv(0+)/dt = iC(0+)/C using circuit analysis
3. Find B1 & B2 from:
I. v(0+) = B1
II. dv(0+)/dt = ̶ α B1 + ωd B2
4. Replace values of α, ωd, B1 & B2 in equation for v(t) above
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ECED 2000 – Electric Circuits
Example 8.4
V0 = 0, I0 = ̶ 12.25mA
8.2 – The Underdamped Voltage Response
a) Find roots of characteristic
equation
b) Find v(0+) & dv(0+)/dt
c) Find v(t) expression for t ≥ 0
d) Sketch v(t), 0 ≤ t ≤ 11ms
v(t ) = 100e −200t sin (979.8t ) V
As R → ∞, α → 0,
ωd = ω0 = 1,000:
v(t ) = 98 sin (1,000t ) V
Choice of Underdamped
versus Overdamped
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ECED 2000 – Electric Circuits
α
8.2 – The Critically Damped Voltage Response
1
= ω0 → s1, s2 real & equal, s1 = s2 = −α = −
2 RC
v(t ) = D1te −αt + D2e −αt , t ≥ 0
1. Find damping factor α from R & C
2. Find v(0+) & dv(0+)/dt = iC(0+)/C using circuit analysis
3. Find D1 & D2 from:
I. v(0+) = D2
II. dv(0+)/dt = D1 ̶ αD2
4. Replace values of α, D1 & D2 in equation for v(t) above
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ECED 2000 – Electric Circuits
Example 8.5
V0 = 0, I0 = ̶ 12.25mA
8.2 – The Critically Damped Voltage Response
a) Find R that makes it CD
b) Find v(t) expression for t ≥ 0
Since V0 = D2 = 0:
v(t ) = D te −αt
1
d) Sketch v(t), 0 ≤ t ≤ 7ms
v(t ) = 98,000te −1000t V
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ECED 2000 – Electric Circuits
8.3 – Step Response of a Parallel RLC Circuit
This is like the voltage equation
of the natural response except
for the non-zero right-hand side
Thus, the characteristic equation
is the same and so are the roots
Applying KCL:
v
dv
iL + + C
=I
R
dt
With:
diL
v=L
dt
Turns into:
L diL
d 2 iL
iL +
+ LC 2 = I
R dt
dt
Dividing by LC and re-arranging:
2
d iL
1 diL
1
1
+
+
iL =
I
2
dt
RC dt LC
LC
A difference exists in how the
forcing term in the right side
impacts the final value iL(∞)
The solution can be written as:
 Natural response
iL (t ) = iL (∞ ) + 

(
)
of
i
t
L


In this case:
iL (∞ ) = I
And the natural response is in one of
the three forms studied: OD, UD or CD
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ECED 2000 – Electric Circuits
8.3 – Overdamped Step Response of a Parallel RLC Circuit
Example 8.6
Initial energy stored is zero, I = 24 mA, R = 400Ω
a) Find iL(0)
b) Find diL(0)/dt
c) Find the roots of the characteristic equation
d) Find the expression of iL(t), t ≥ 0
' s1t
' s2 t
iL (t ) = iL (∞ ) + A1e
iL (t ) = 24 − 32e
−20,000t
+ A2 e
+ 8e
−80,000t
mA
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ECED 2000 – Electric Circuits
8.3 – Underdamped Step Response of a Parallel RLC Circuit
Example 8.7
Initial energy stored is zero, I = 24 mA, R = 625Ω
a) Find the roots of the characteristic equation
s1 = −α + jωd = −3.2 ×10 4 + j 2.4 × 10 4 rad / s
s2 = −α − jωd = −3.2 × 10 4 − j 2.4 × 10 4 rad / s
b) Find the expression of iL(t), t ≥ 0
' −αt
iL (t ) = iL (∞ ) + B1e
cos(ωd t ) + B e
'
2
−αt
sin (ωd t )
iL (t ) = 24 − 24e −32,000t cos(24,000t ) − 32e −32,000t sin(24,000t ) mA
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ECED 2000 – Electric Circuits 8.3 – Comparing Step Response Forms of a Parallel RLC Circuit
Example 8.9
b) Find the time required for iL(t) = 0.9 iL(∞) = 21.6mA
c) Which response would you use in a design that requires
reaching 90% of the final value in the shortest time?
d) Which response would you use in a design that requires
the final value of the current is never exceeded?
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ECED 2000 – Electric Circuits
8.4 – Natural Response of a Series RLC Circuit
Characteristic equation:
s2 +
R
1
s+
=0
L
LC
Roots:
Applying KVL:
t
di 1
Ri + L + ∫ i ⋅ dτ + V0 = 0
dt C 0
Differentiating w.r.t. time t :
di
d 2i 1
R +L 2 + i=0
dt
dt
C
Dividing by L and re-arranging:
d 2i R di
1
+
+
i=0
2
dt
L dt LC
s1, 2 = −α ± α 2 − ω02
where
R
α=
,
2L
1
ω =
LC
2
0
A) Overdamped:
st
i(t ) = A1e 1 + A2 e s2t
B) Underdamped:
−αt
i(t ) = B1e
cos(ωd t ) + B2 e −αt sin (ωd t )
C) Critically damped:
i(t ) = D1te −αt + D2 e −αt
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ECED 2000 – Electric Circuits
8.4 – Step Response of a Series RLC Circuit
Differential equation:
d 2 vC R dvC
1
1
+
+
v
=
V
C
2
dt
L dt LC
LC
A) Overdamped:
vC (t ) = vC (∞ ) + A e
'
1
s1t
+ A2' e s2t
B) Underdamped:
vC (t ) = vC (∞ ) + B1' e −αt cos(ωd t ) + B2' e −αt sin (ωd t )
C) Critically damped:
vC (t ) = vC (∞ ) + D1'te −αt + D2' e −αt
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ECED 2000 – Electric Circuits
8.4 – Underdamped Step Response of a Series RLC Circuit
Example 8.12
Initial energy stored is zero. Find vC(t)
s1 = −α + jωd = −1,400 + j 4,800 rad / s
s2 = −α − jωd = −1,400 − j 4,800 rad / s
vC (t ) = 48 + B1' e −1,400t cos(4,800t ) + B2' e −1, 400t sin (4,800t ), t ≥ 0
( )
+
dv
0
C
vC (0) = 0 = 48 + B ,
= 0 = 4,800 B2' − 1,400 B1'
dt
'
'
Solving, B1 = −48 & B2 = −14 . Substituting in vC(t):
'
1
vC (t ) = 48 − 48e −1, 400t cos(4,800t ) − 14e −1, 400t sin (4,800t ) V , t ≥ 0
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