ECE 201, Section 3

Lecture 23

Prof. Peter Bermel

October 19, 2012

Driven Series RLC Circuits

From KVL:

+

+1

=

+ 2Γ

+ =

Γ = /2; = 1/ ; = − Γ = −Γ′

10/19/2012 ECE 201-3, Prof. Bermel

Regime Range Solution Behavior

Under-

damped

Γ < = cos + + Oscillate &

decay

Critically

damped

Γ = = !" + ! + Decay

Over-

damped

Γ > = !$% + !

% + Decay

R

LC

+ -

Vs

Driven Parallel RLC Circuits

From KCL:

&'

+1

&'

+1

&' =

&(

&'

+ 2Γ&'

+ &' =

&

Γ = 1/(2); = 1/ ; = − Γ = −Γ′

10/19/2012 ECE 201-3, Prof. Bermel

Regime Range Solution Behavior

Under-

damped

Γ < &' = cos + + & Oscillate &

decay

Critically

damped

Γ = &' = !" + ! + & Decay

Over-

damped

Γ > &' = !$% + !

% + & Decay

R

LC

Is

IL

Example

10/19/2012 ECE 201-3, Prof. Bermel

• Consider an LC circuit connected to a voltage

source at t=0. How will the current and

voltage vary with time?

L

C

+

-

t=0+

VC

-

Vs

I(t)

Solution

10/19/2012 ECE 201-3, Prof. Bermel

• Employing our solution:

& = !$+ + !

,

-± = −Γ ± Γ −

Γ =/

'= 0; = 1/

-± = ±

C

+

-

t=0+

VC

-

Vs

L

I(t)

Solution

10/19/2012 ECE 201-3, Prof. Bermel

• Employing our solution:

& = !$123 + !

123

&

= !$

123 − !123

& 0 = !$ + ! = 0

= &

= 0 = !$ − !

= 2!$ = −2!& = 123 − 123

2 = sin

C

+

-

t=0

Vs

L

I(t)

Solution

10/19/2012 ECE 201-3, Prof. Bermel

• Given that:

& = sin

We can integrate to obtain:

= 1 6 ′ sin ′

7= 6 ′ sin ′

7 = − cos ′ 8 7 = 1 − cos

C

+

-

t=0

Vs

L

I(t)

Example

• What is the equation of motion for the voltage

vA(t) of the right capacitor? What other

system does this resemble, and where might

you use this design?

10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

• Taking the derivative of Q=CV and V=IR:9 − :

= :

− 9

+: − 9

= 9

10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

9 − : = ;:

+ : − 29 = ;9

Substituting, + : − 2 : + ;<=>

<= ;

<=?

<

Time derivative yields, <=?

<−<=>

<= ;

<@=>

<@

10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

+ : − 2: − 2;:

= ; ;:

+:

− : − 3;:

= ;:

;:

+ 3;:

+ : =

Two first order ODEs became one second order ODE:

“Conservation of misery”

10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

;:

+ 3;:

+ : =

Cf. <@B

<@+/

'

<B

<+

"

' = 0

:

+3

;

:

+1

;: =

;

Thus, Γ =/

'=

C

Dand =

"

' =

"

D10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

Since Γ =C

Dand =

"

D, Γ >

-± = −Γ ± Γ −

= −3

2; ±32;

− 1

;= − 3

2; ±52;

Overdamped circuit10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

• This circuit offers an alternative oscillator

• Application: timing on a microchip!

• Hard to fit inductors on microchips, but relatively

easy to put resistors and capacitors on

• Real clock signals generated in a conceptually

related fashion, but more sophisticated

10/19/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Example

• Consider a circuit in which input voltage

switches from -10 V to +20 V at t=0 (i.e.,

= −10 + 30F()). What is VC(t) at all times?

10/19/2012 ECE 201-3, Prof. Bermel

20 Ω

0.25 H0.1 mF

+

-

+

VC

-

-10+30u(t)

Solution

• Use source transformation to recover original

parallel RLC circuit problem:

& =−10 + 30F

20= −

1

2+3

2F()

10/19/2012 ECE 201-3, Prof. Bermel

20 Ω 0.1 mF+

-

+

VC

--10+30u(t)0.25 H

20 Ω 0.1 mF+

VC

-

0.25 H

Solution

• From earlier definitions:

Γ =1

2 =1

2 ∙ 20 ∙ 10H = 250 = 1

0.25 ∙ 10H = 200• Circuit is overdamped

10/19/2012 ECE 201-3, Prof. Bermel

20 Ω 0.1 mF+

VC

-

0.25 H

[-1+3u(t)]/2

Solution

• For t<0, inductor looks like short and VC(t<0)=0

• For t>0, non-zero voltage temporarily possible; start by solving for IL in overdamped regime:

&' = !$% + !

% + &Here, Γ = 250, and

Γ = Γ − = 250 − 200 = 150

10/19/2012 ECE 201-3, Prof. Bermel

20 Ω 0.1 mF+

VC

-

0.25 H

[-1+3u(t)]/2

Solution

• Given that:

&' ≥ 0 = K7 !$"K7 + !

"K7 + 1= !$

"77 + !H77 + 1

• Since IL(0)=-1/2, and VC(0)=0:

−"

= !$ + ! + 1 !$ = −

C

− !

0.25 −100!$ − 400! = 0 !$ = −4!• Thus, ! = 1/2, and !$ = −210/19/2012 ECE 201-3, Prof. Bermel

20 Ω 0.1 mF+

VC

-

0.25 H

[-1+3u(t)]/2

Solution

&' ≥ 0 = −2"77 +1

2H77 + 1

• Since circuit elements are in parallel:

≥ 0 = &'

= 0.25 −2 −100 "77 +1

2−400 H77

= 50 "77 − H77

10/19/2012 ECE 201-3, Prof. Bermel

20 Ω 0.1 mF+

VC

-

0.25 H

[-1+3u(t)]/2

Solution

10/19/2012 ECE 201-3, Prof. Bermel

0

5

10

15

20

25

-5 0 5 10 15 20 25 30 35 40

Voltage (VC)

Time t (ms)

Homework

• HW #22 due today by 4:30 pm in EE 326B

• HW #23 due Mon.: DeCarlo & Lin, Chapter 9:

– Problem 11(a) [Correction: the second iL (0-) should

be iL(0+)]

– Problem 18

– Problem 27

10/19/2012 ECE 201-3, Prof. Bermel