1 K. A. Saaifan, Jacobs University, Bremen 9. The RLC Circuit The RLC circuits have a wide range of applications, including oscillators and frequency filters This chapter considers the responses of RLC circuits The result is a second-order differential equation for any voltage or current of interest We consider the following analysis The Natural Response of a Parallel RLC Circuit The Natural Response of a Series RLC Circuit The Complete (Natural and Step) Response of RLC Circuits
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1K. A. Saaifan, Jacobs University, Bremen
9. The RLC Circuit
The RLC circuits have a wide range of applications, including oscillators and frequency filters
This chapter considers the responses of RLC circuits
The result is a second-order differential equation for any voltage or current of interest
We consider the following analysis
The Natural Response of a Parallel RLC Circuit
The Natural Response of a Series RLC Circuit
The Complete (Natural and Step) Response of RLC Circuits
2K. A. Saaifan, Jacobs University, Bremen
9.1 The Source-Free Parallel Circuit
iL0−iL0iL0
=I0
vR
1L∫t0
t
vd tiL t0Cdvdt
=0
Cd2v
dt2 1Rdvdt
1Lv=0
vR
iLtiC t=0iL(t) iC(t)
Obtaining the differential equation for a parallel RLC circuit
Apply KCL
Differentiate both sides with respect to time
Two initial conditions
The capacitor voltage cannot change abruptly
The inductor current cannot change abruptly iC 0=−iL0−iR0
=−I0−V 0
R
iC 0=CdvC tdt ∣t=0
vC 0−=vC 0=vC 0
=V 0 1
dvC tdt ∣t=0=
−I0−V 0/RC
2
3K. A. Saaifan, Jacobs University, Bremen
Substitute into the ordinary differential equation, we got the characteristic equation of s determined by the circuit parameters
The characteristic equation has two roots
Thus, the natural response has the following form
where the constants A1 and A2 are determined using the initial conditions
Cs21Rs
1L=0 C
d2v
dt2 1Rdvdt
1Lv=0
Cs21Rs
1Lv=0
s1,2=−1
2RC± 1
2RC 2
−1LC
Definition of frequency terms
The resonant frequency
The damping coefficient
0=1
LC
=1
2RC
Solution of the differential equation
We assume the exponential form of the natural response isv t=Aest
v t=A1es1tA2e
s2t
K. A. Saaifan, Jacobs University, Bremen
The roots of the characteristic equation can be expressed as
s1=−2−02
s2=−−2−02
Three types of natural response
4
Response Criteria Solutions
Overdamped >α ω0real, distinct roots
s1, s2
Underdamped <α ω0complex, conjugate
roots s1, s2*
Critically damped =α ω0real, equal roots
s1, s2
K. A. Saaifan, Jacobs University, Bremen 5
9.2 The Overdamped Parallel RLC Circuit
The condition of overdamped response ( ) implies that
The roots of the characteristic equation s1 and s2 are distinct negative real numbers
The response, v(t) , can be seen as a sum of two decreasing exponential terms as
0
0=1
LC=6 =
12RC
=3.5
s1=−1 s2=−6
v t=A1e−tA2e
−6 t
Finding values for A1 and A2
For the shown circuit, we determine
The general form of the natural response
From the initial conditions v(0)=0 and iL(0)=-10 A
v 0=A1A2=0 (1)
−A1−6A2=420 2
v t=A1es1tA2e
s2t 0 t∞
dvC tdt ∣t=0=
−I0−V 0/RC
=A1s1A2s2
K. A. Saaifan, Jacobs University, Bremen
The final numerical solution is
Graphing the response
The maximum point can be determined as
We determine the time
Then
v t=84 e−t−1e−6t V
dv tdt
=0=84−1e−tmax−−61e−6 tmax=0
tmax=0.358 s
v tmax=48.9 V
K. A. Saaifan, Jacobs University, Bremen
Find an expression for vC(t) valid for t > 0 in the circuit
Compute the initial conditions (t < 0)
The capacitor acts as open circuits
The inductor acts as short circuits
vC 0−=150 200
300200=60 V
iL0−=
−150300200
=−300 mA
0=1
LC=100000 =
12RC
=125000
s1=−50000 s2=−200000
After the switch is thrown (t > 0)
The capacitor is left in parallel with a 200 Ω resistor and a 5 mH inductor
K. A. Saaifan, Jacobs University, Bremen
Solve the capacitor voltage Since α > ω0, the circuit is overdamped and so we expect a capacitor voltage of the form
vC t=A1e−50000 tA2e
−200000 t
Finding values for A1 and A2
From the initial conditions vC(0)=60 V and iL(0)=-0.3 A
Solving, A1 = 80 V and A2 = −20 V, so that
vC 0=A1A2=60 1
−50000A1−200000A2=0 2
vC t=80e−50000 t−20e−200000t t0
dvC tdt ∣t=0=
−I0−V 0/RC
=A1s1A2s2
9K. A. Saaifan, Jacobs University, Bremen
(a) Sketch the voltage vR(t) = 2e−t −4e−3t V in the range 0<t<5 s(b) Estimate the settling time(c) Calculate the maximum positive value and the time at which it occurs