ECE 201, Section 3

Lecture 22

Prof. Peter Bermel

October 17, 2012

Series RLC Circuits

General solution to RLC circuits:

= + ± = −Γ ± Γ −Γ = /2; = 1/

10/17/2012 ECE 201-3, Prof. Bermel

Regime Value Range Root type Behavior

Undamped Γ = 0 Pure imaginary Oscillates forever

Underdamped 0 < Γ < Complex Oscillate & decay

Critically damped Γ = Pure real Decay

Overdamped Γ > Pure real Decay

R

RLC Circuit

LC

Problem 2

10/17/2012 ECE 201-3, Prof. Bermel

Problem 3

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Problem 6

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Problem 9

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Problem 12

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Problem 13

10/17/2012 ECE 201-3, Prof. Bermel

Example 1

• Consider a circuit which charges a capacitor

for t<0 then switches to an RLC circuit at t=0.

What is VC(t) for R=0, and 180?

10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• For t<0: VC=50 V

• For t>0: I(0)=0, and:

= =

!.∙!. = 10 rad/s

Γ = 2 =

2 ∙ (0.1%) = 5

( = 2Γ =

102 ∙ (5) =

1

10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• From before:

) = ( / = + ) = * + ,-*- + + ,-*-

) = * +,. + +,.Matching boundary conditions:

+ = )−Γ + /′ + −Γ − /′ = 010/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• Substituting:

+ = )−Γ + /′ + −Γ − /′ ) − = 02/′ = Γ + /′ ) = )

2 1 − / Γ′−Γ + /′ ) − + −Γ − /′ = 0−2/′ = Γ − /′ ) = )

2 1 + / Γ′10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• From before:

) = )*2

1 − /Γ

′+,. + 1 + /

Γ

′+,.

) = )* cos. +

Γ

′sin.

10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Example

10/17/2012 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

10/17/2012 ECE 201-3, Prof. Bermel

• Employing our solution:

6 = + ± = −Γ ± Γ −Γ = 7

= 0; = 1/ ± = ±/

C

+

-

t=0+

VC

-

Vs

L

I(t)

Solution

10/17/2012 ECE 201-3, Prof. Bermel

• Employing our solution:

6 = +,8 + +,8969 = / +,8 − +,8

6 0 = + = 0) = 969 = 0 = / −

) = 2/ = −2/6 = ) +,8 − +,8

2/ = ) sin

C

+

-

t=0

Vs

L

I(t)

Solution

10/17/2012 ECE 201-3, Prof. Bermel

• Given that:

6 = ) sin

We can integrate to obtain:

) = 1 : 9′ ) sin ′

!= ): 9′ sin ′

!) = −) cos ′ ; ! ) = ) 1 − cos

C

+

-

t=0

Vs

L

I(t)

Example

• Consider a circuit which charges a capacitor

for t<0 then switches to an RLC circuit at t=0.

What is VC(t) for R=0, and 180?

10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• For t<0: VC=50 V

• For t>0: I(0)=0, and:

= =

!.∙!. = 10 rad/s

Γ = 2 =

2 ∙ (0.1%) = 5

( = 2Γ =

102 ∙ (5) =

1

10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• From before:

) = ( / = + ) = * + ,-*- + + ,-*-

) = * +,. + +,. , where . = − ΓMatching boundary conditions:

+ = )−Γ + /′ + −Γ − /′ = 010/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• Substituting:

+ = )−Γ + /′ + −Γ − /′ ) − = 02/′ = Γ + /′ ) = )

2 1 − / Γ′−Γ + /′ ) − + −Γ − /′ = 0−2/′ = Γ − /′ ) = )

2 1 + / Γ′10/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

Solution

• From before:

) = )*2

1 − /Γ

′+,. + 1 + /

Γ

′+,.

) = )* cos. +

Γ

′sin.

Underdamped for R<2

Critically damped when R=2

Overdamped when R>210/17/2012 ECE 201-3, Prof. Bermel

R

0.1 H0.1 F

+

-

t=0

+

VC

-

20 Ω

50 V

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?

10/17/2012 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)<910/17/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

)< − )= = > 9)=9) + )= − 2)< = >

9)<

9

Substituting, ) + )= − 2 )= + >?@A

?= >

?@B

?

Time derivative yields, ?@B

?−

?@A

?= >

?-@A

?-

10/17/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

) + )= − 2)= − 2>9)=

9= > >

9)=

9+9)=

9

) − )= − 3>9)=

9= >

9)=

9

>9)=

9+ 3>

9)=

9+ )= = )

Two first order ODEs became one second order ODE:

“Conservation of misery”

10/17/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

> 9)=9 + 3> 9)=9 + )= = )

Cf. ?-D?- +

7?D? +

( = 0

9)=9 + 3

>9)=9 + 1

> )= =)>

Thus, Γ = 7 =

EF and =

=F

10/17/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Solution

Since Γ = EF and =

F, Γ > ± = −Γ ± Γ −

= − 32> ±

32>

− 1

>= − 3

2> ±5

2>Overdamped circuit

10/17/2012 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

10/17/2012 ECE 201-3, Prof. Bermel

R

C

+

-

+

VB

-

R

V

C

+

VA

-

Homework

• HW #21 due today by 4:30 pm in EE 326B

• HW #22 due Fri.: DeCarlo & Lin, Chapter 9:

– Problem 8

– For the circuit in the previous problem, find the

instantaneous power absorbed by each of the

elements. What is the sum of these powers?

10/17/2012 ECE 201-3, Prof. Bermel