9. The RLC Circuit - Electrical Engineeringtrsys.faculty.jacobs-university.de/wp-content/uploads/2015/02/... · 9. The RLC Circuit The RLC circuits ... frequency filters ... The Natural

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


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


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


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


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


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


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


(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



We compute the settling time as follows

dvR tdt

=0

2−1e−tmax−4−3e−3 tmax=0

tmax=ln 62

=0.895 s

vR tmax=544 mV

vR tsettling=vRtmax

100 tsettling=5.9 s2e−tsettling−4e−3 tsettling=5.44 mV

2e−tsettling=5.44 mV


9.3 Critical Damping

The condition of a critical damping ( ) implies that

The roots of the characteristic equation s1 and s2 are equal and negative real numbers

For repeated roots, the response, v(t) , can be seen as

=0

s1=s2=−

0==6 s−1

s1=s2=−6 s−1



The general form of the natural response

From the initial conditions v(0)=0 and iL(0)=-10 A v t=A1te

−6tA2e−6 t

v 0=A2=0

A1−A2=10C

=420

v t=A1te−tA2e

− t

dvC tdt ∣t=0=

−I0−V 0/RC

=A1−A2


The solution is



We determine the time tmax

Then

The settling time

dv tdt

=0

v tmax=63.1 V

v t=420te−6 t V

=420e− tmax420 tmax −e−t=0

420e− tmax 1−tmax=0

tmax=1=0.408 s

vtmax 100

=420tsettlinge−6 tsettling


Find R1 such that the circuit is critically damped for t>0 and R2 so that v(0)=2 V

For t < 0



v 0−=5R1R2

R1R2=2 V

0=1

LC=15810 =

12R1C

=1

2R1×10−9


The current source has turned itself off and R2 is shorted

The capacitor is left in parallel with R1 and a 4 H inductor

Since the critically damping implies that , we have

0=

R1=31625 31625R2

31625R2=0.4 R2=0.4


9.4 The Underdamped Parallel RLC Circuit

The condition of a critical damping ( ) implies that

The roots of the characteristic equation s1 and s2 are complex conjugate numbers

where is the natural resonant frequency

For complex conjugate roots, the response, v(t) , can be seen as

The derivative of v(t) is

0

v t=e− tA1ejd tA2e

−jdt

s1=−2−02

d=02−2

s1=−jd

s2=−−jd s2=−−2−02

=e−t[A1cosdtjsind tA2cosdt−jsind t]

=e−t[A1A2cosd tj A1−A2sind t]

=e−t[B1cosd tB2sindt]

dvtdt

=−e−t[B1 cosd tB2sindt]

e−t[B1−dsind tB2dcosdt]

=e−t[−B1dB2cosd t−B2dB1sind t]


Finding values for B1 and B2


The natural response is

From the initial conditions v(0)=0 and iL(0)=-10 A


d=02−2=2 s−1

=1

2RC=2 s−1 0=

1LC

=6 s−1

v t=e−2t[B1 cos2tB2 sin 2t]

v 0=B1=0

2B2=420

v t=2102e−2tsin 2t

dvC tdt ∣t=0=

−I0−V 0/RC

=−B1dB2



The voltage oscillates (~ωd) and approaches to the final value (~α)

The voltage response has two extreme points (minimum and maximum points)

v t=2102e−2tsin 2t


Find an expression for vC(t) valid for t > 0 in the circuit

Compute the initial conditions (t < 0)



vC 0−=3 48×100

10048=97.3 V

iL0−=3 100

48100=2.027 A

0=1

LC=4.99 s−1

=1

2RC=1.2 s−1


The current source is off

The capacitor is left in parallel with a 48 Ω resistor and a 10 H inductor

Since , the circuit is underdamped

where

0

vC t=e− t[B1 cosd tB2 sind t]

d=02−2=4.75 s−1


Finding values for B1 and B2

The natural response is

From the initial conditions vC(0)=97.3 and iL(0)=2.027 A


v 0=B1=97.3

4.75B2=240−2.027−97.3 /100

vC t=e−1.2 t[B1 cos 4.75tB2 sin 4.75t]

vC t=e−1.2 t[97.3 cos4.75t−151.57 sin 4.75t] V

dvC tdt ∣t=0=

−I0−V 0/RC

=−B1dB2


9.5 The Source-Free Series Circuit

RiLdid t

1C∫t0

t

id tvC t0=0

vC 0=V 0

Ld2i

d t2Rdidt

1Ci=0

RivLtvC t=0

Obtaining the differential equation of a series RLC circuit

Apply KVL (V0, I0, i(t) must satisfy the passive sign convention)

Differentiate both sides with respect to time

The two initial conditionsThe inductor current cannot change abruptly

The capacitor voltage cannot change abruptlyi 0=I0 1

vL0=−vC 0−vR 0=−V 0−I 0R

vL0=LdiLtdt ∣t=0

d iLtdt ∣t=0=

−V 0−I0R

L2


Solution of the differential equation

We assume the exponential form of the natural response is

v t=Aest

Substitute into the ordinary differential equation, we got the characteristic equation of s determined by the circuit parameters

The characteristic equation has two roots

where

Ls2Rs1C=0

=−±2−02

The resonant frequency

The damping coefficient

0=1

LC

=R

2L

s1,2=−R

2L± R2L 2

−1LC


Condition Criteria α ω0 Response

Overdamped >α ω0 where

Underdamped <α ω0 where

Critically damped =α ω0

1LC

R

2L

R

2L

R

2L

1LC

1LC

i t=A1es1tA2e

s2t

i t=A1te−tA2e

−t

i t=e−t[B1 cosd tB2sindt]

Summary of Relevant Equations for Series Source-Free RLC Circuits

d=02−2

s1,2=−±2−02


9.6 The Complete Response of the RLC Circuit

The response of RLC circuits with dc sources and switches consists of the natural response and the forced response:

The general solution is obtained by the same procedure that was followed for RL and RC circuits

v t= vf tForced Response

vntNatural Response

it= if tForced Response

intNatural Response


The Solution Steps of RLC Circuits

Determine the initial conditionsCompute the circuit current, “iL(t), iR(t), and iC(t)”, and voltages,

“vL(t), vR(t), and vC(t)”, at t=0- and t=0+

“Note that the inductor current the capacitor voltage cannot change abruptly, iL(0-)=iL(0)=iL(0+) and vC(0-)=vC(0)=vC(0+) ”

Upon we are confronted with a series or a parallel circuit

= R

2L(series RLC)= 1

2RC(parallel RLC)

0=1

LC

hnt=A1es1tA2e

s2thnt=A1te−tA2e

−t

s1,2=−±2−02

hnt=e−t[B1cosd tB2sin d t]

0

0=0

d=02−2

The forced response

hf(t)

The complete response

hf(t)+hn(t)

Find unknown constants given the initial conditions


+

-

Find an expression for vC(t) and iL(t) valid for t > 0 in the circuit

1. Determine the forced response ( t > 0 )iLf t=−9 A t0vCf t=150 V t0

RivLtvC t=0

Ld

2i

dt2Rdidt

1Ci=0

Ld

2vC

d t2 R

dvCd t

1CvC=0

0=1

LC=3 s−1 =

R

2L=5 s−1

s1=−1 s2=−9

3. Since α>ω0, the response is over-damped, we have

andiLnt=A1e

−tA2e−9 t

vCnt=B1e−tB2e

−9t

Note:Independent current sources → open circuitsIndependent voltage sources → short circuits

+

-2. Determine the natural response

1. Write the differential equation

The differential equation in terms of i reduces to

The differential equation in terms of vC(t) is given as

2. Compute αand ω0


3. The complete responsevC t=vCf tB1e

−tB2e−9 t

iLt=iLf tA1e−tA2e

−9 t

=150B1e−tB2e

−9t =−9A1e−tA2e

−9t

For t=0- (the left-hand current source is off)

For t=0+ (the left-hand current source is on)

4+

-

iR 0−=−5 A vR0

−=−150 V

iR 0=−1 A vR0

=−30 V

iL 0−=−5 A

iC 0−=0 A

vL0−=0 V

vC 0−=150 V

iL 0=−5 A

iC 0=4 A

vL0=−120 V

vC 0=150 V

150B1B2=150 1

dvCtdt ∣t=0=

iC 0

C

4. Solve for the values of the unknown constants

−B1−9B2=108 2

vC t∣t=0=vC 0

iLt∣t=0=iL0

−9A1A2=−5 1

diLtdt ∣t=0=

vL0

L−A1−9A2=−40 2

vC t=15013.5B1e−t−13.5B2e

−9 t

iLt=−9−0.5e−t4.5e−9 t


9.7 THE LOSSLESS LC CIRCUIT

The resistor in the RLC circuit serves to dissipate initial stored energy

When this resistor becomes 0 in the series RLC or infinite in the parallel RLC, the circuit will oscillate

Example: Assume the shown circuit with the following initial conditions

1. We find

2. So , the voltage is simply

3. We use the initial condition

4. Thus, we have obtained a sinusoidal response

iL0=−16

A vC 0=0 V

0=1

LC=3 s−1 = R

2L=0 s−1

d=3 s−1

vt=e−t [B1 cos3tB2 sin 3t ]

v0=B1 B1=0dvC tdt ∣t=0=

−iL0 C

=3B2 B2=2

vt=2sin 3t V


Homework Assignment 8P9.1, P9.6, P9.12, P9.13, P9.16, P9.20, P9.26, P9.27, P9.35, P9.37, P9.46 P9.50, P9.51, and 9.64

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9. The RLC Circuit - Electrical Engineeringtrsys.faculty.jacobs-university.de/wp-content/uploads/2015/02/... · 9. The RLC Circuit The RLC circuits ... frequency filters ... The Natural