INC 112 Basic Circuit Analysis

Week 10

RLC Circuits

RLC Circuits

Similar to RL and RC circuits, RLC circuits has two partsof responses, namely: natural response and forced response.

Force Response:Similar to RL and RC, a step input causes a step output.

Natural Response:Different and more difficult than RL, RC.

Source-free RLC Circuits

We study the natural response by studying source-free RLC circuits.

LR+

v(t)-

iR(t)C

iC(t)iL(t)

Parallel RLC Circuit

0)(1)(1)(

0)(

)(1)(

0

2

2

tvLdt

tdv

Rdt

tvdC

dt

tdvCdttv

LR

tv

iii CLR

Second-orderDifferential equation

This second-order differential equation can be solved by assuming solutions

The solution should be in form of

0)(1)(1)(

2

2

tvLdt

tdv

Rdt

tvdC

stAetv )(

If the solution is good, then substitute it into the equation will be true.

011

0)11

(

011

2

2

2

Ls

RCs

Ls

RCsAe

AeL

AseR

eCAs

st

ststst

which meanss=??

0112 L

sR

Cs

LCRCRCs

LCRCRCs

1

2

1

2

1

1

2

1

2

1

2

2

2

1

Use quadratic formula, we got

tseAtv 11)( Both and tseAtv 2

2)( are solution to the equation

Therefore, the complete solution iststs eAeAtv 21

21)(

LCRCRCs

1

2

1

2

12

2,1

From

Define resonant frequencyLC

10

Damping factorRC2

1

Therefore,20

22,1 s

in which we divide into 3 cases according to the term inside the bracket

Solution to Second-order Differential Equations

1. α > ω0 (inside square root is a positive value) Overdamped case

2. α = ω0 (inside square root is zero) Critical damped case

3. α < ω0 (inside square root is a negative value) Underdamped case

1. Overdamped case , α > ω0

7H6Ω+

v(t)-

iR(t)1/42f

iC(t)iL(t)

Initial conditionvc(0) = 0, iL(0) = -10A

Find v(t)

5.32

1

RC 6

10

LC

α > ω0 ,therefore, this is an overdamped case

20

22,1 s s1 = -1, s2 = -6

Therefore, the solution is in form oftt eAeAtv 6

21)(

Then, we will use initial conditions to find A1, A2

From vc(0) = 0 we substitute t=0

210

20

10)0( AAeAeAv

From KCLAt t=0

4206

0642

1)10(

0

0)(

)10()0(

0

21

0

621

0

AA

eAeAR

dt

tdvC

R

v

iii

t

tt

t

CLR

Solve the equation and we got A1 = 84 and A2 = -84

and the solution is

)(848484)( 66 tttt eeeetv

t

v(t)

)(84)( 6tt eetv

2. Critical damped case , α = ω0

7H8.573Ω+

v(t)-

iR(t)

1/42f

iC(t)iL(t)

Initial conditionvc(0) = 0, iL(0) = -10A

Find v(t)

45.21

2

10

LCRC

α = ω0 , this is an critical damped case s1 = s2 = -2.45

The complete solution of this case is in form of

stst eAteAtv 21)(

Then, we will use initial conditions to find A1, A2

From vc(0) = 0 we substitute t=0

20

20

1 )0(0)0( AeAeAv

Find A1from KCLat t=0

0)(42

110

0)45.2(42

1)10(

0

0)(

)10()0(

0

1

0

45.21

45.21

0

A

eAetAR

dt

tdvC

R

v

iii

t

tt

t

CLR

Therefore A2 =0 and the solution is reduced totteAtv 45.2

1)(

Solve the equation and we got A1 = 420 and the solution is

t

v(t)

ttetv 45.2420)(

ttetv 45.2420)(

3. underdamped case , α < ω0

from20

22,1 s

The term inside the bracket will be negative and s will be a complex number

define 220 d

Then djs 2,1

and

)()(

)(

21

)(2

)(1

tjtjt

tjtj

dd

dd

eAeAetv

eAeAtv

)()( 21tjtjt dd eAeAetv

Use Euler’s Identity sincos je j

)sincos()(

sin)(cos)(()(

)sincossincos()(

21

2121

2211

tBtBetv

tAAjtAAetv

tjAtAtjAtAetv

ddt

ddt

ddddt

)sincos()( 21 tBtBetv ddt

3. Underdamped case , α < ω0

7H10.5Ω+

v(t)-

iR(t)

1/42f

iC(t)iL(t)

Initial conditionvc(0) = 0, iL(0) = -10A

Find v(t)

22

1

RC 6

10

LC

α < ω0 ,therefore, this is an underdamped case

2220 d

and v(t) is in form )2sin2cos()( 212 tBtBetv t

Then, we will use initial conditions to find B1, B2

From vc(0) = 0 we substitute t=0

Find B2from KCLat t=0

0)2(42

110

02sin22cos242

1)10(

0

0)(

)10()0(

0

2

0

22

22

0

B

teBteBR

dt

tdvC

R

v

iii

t

tt

t

CLR

Therefore B1 =0 and the solution is reduced to

1210 )0sin0cos()0( BBBev

)2sin()( 22 tBetv t

we got and then the solution is

t

v(t)

29722102 B

)2sin297)( 2 tetv t

)2sin297)( 2 tetv t

Natural responseat different time

Mechanical systems are similar to electrical systems

A pendulum is an example of underdamped second-order systems in mechanic.

t

displacement(t)

Series RLC

0)(1)()(

0)(1)(

)(

0

2

2

tiCdt

tdiR

dt

tidL

dttiCdt

tdiLRti

vvv CLR

LR +

vL(t)-

+ vR(t) -

C

i(t)

-vC(t)

+

0)(1)()(

2

2

tiCdt

tdiR

dt

tidL

Series RLCParallel RLC

0)(1)(1)(

2

2

tvLdt

tdv

Rdt

tvdC

tsts eAeAtv 2121)(

LCRCRCs

1

2

1

2

12

2,1

20

22,1 s

RC2

1

LC

10

tsts eAeAti 2121)(

LCL

R

L

Rs

1

22

2

2,1

20

22,1 s

L

R

2

LC

10

Procedure for Solving RLC Circuits

1. Decide that it is a series or parallel RLC circuit. Find α and ω0. Then, decide which case it is (overdamped, critical damped, underdamped).

2. Assume the solution in form of (natural response+ forced response)

fstst VeAteA 21

ftsts VeAeA 21

21

fddt VtBtBe )sincos( 21

Overdamped

Critical damped

Underdamped

3. Find A, B, Vf using initial conditions and stable conditions

Example

S1 30

t = 0 sec

+

Vc

-

1/27f3H4A 5AiL(t)

Find vc(t)

52

L

R 31

0 LC

9,120

22,1 s

This is overdamped case, so the solution is in form

ftt VeAeA 9

21

ftt

C VeAeAtv 921)(

Consider the circuit we found that the initial conditions will bevC(0) = 150 V and iL(0) = 5 A and the condition at stable point will bevC(∞) = 150 V and iL(∞) = 9 A

fVAA 21150Using vC(0) = 150 V , we got

Using vC(∞) = 150 V , we got fV 00150

Therefore, Vf = 150, A1+A2 = 0

Consider the circuit, we find that iC(0) = 4A

21

02

01

921

9108

)9(27

14)0(

)9(27

1)()(

AA

eAeAi

eAeAdt

tdvCti

C

ttC

A1 = 13.5, A2 = -13.5

Therefore, 1505.135.13)( 9 ttC eetv