ece 201, section 3 lecture 22 - purdue university
TRANSCRIPT
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
10/17/2012 ECE 201-3, Prof. Bermel
Problem 6
10/17/2012 ECE 201-3, Prof. Bermel
Problem 9
10/17/2012 ECE 201-3, Prof. Bermel
Problem 12
10/17/2012 ECE 201-3, Prof. Bermel
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