Download RLC Circuits

Module 14 - Circuits with Resistors,
Inductors, and Capacitors (RLC)
1
One Reactive Element…
Capacitor and Resistor
Exponential Forms
vC
vC = Vo et /RC
t
vC
vC = Vo (1  et /RC)
t
Inductor and Resistor
Exponential Forms
iL
iL = Io et R/L
t
iL
iL = Io (1  et /RC)
t
2
Capacitor, Inductor, and Resistor
• Behavior Very Different
• Typical Response
RLC NETWORK:
sinusoid
vC or iL

decaying exponential
1
0.8
0.6
0.4
0.2
0
t
-0.2
-0.4
-0.6
-0.8
0
10
20
30
40
50
60
70
80
90
100
Let’s look at a circuit example…
3
Series RLC Circuit
+ vR –
t=0
R
Vo
i(t)
Initial Conditions
+
C
vC
vC = 0
L
–
+
vL
iL = 0
–
Objective: Find i(t)
4
+ vR –
t=0
Objective: Find i(t)
R
Governing Equation:
KVL: Vo = vR + vC + vL
Vo
i(t)
+
C
vC
L
vL
–
+
–
Inductor Eqn:
Ohm’s Law:
R i + vC + L
di
= Vo
dt
Q: How do we take care of vC term?
d
A: Take
of the entire equation!
dt
d i dvC
d2i
dV
R
+
+ L 2 = o
dt
dt
dt
dt
5
Transform some more….
d i dvC
d2 i
dV
R
+
+L 2 = o
dt
dt
dt
dt
R
+ vR –
Vo
Capacitor equation
i
di
R
+
dt
C
+
C
vC
L
vL
–
+
i(t)
–
d2i
dVo
+L 2 =
dt
dt
Rearrange + Divide by L
1 dVo
R di
d2 i
i
+
+
== 0
2
L dt
L dt
LC
dt
Note condition at t = 
d2
( ddt and dt2 = 0 )
zero
i=0
Makes sense:
Capacitor is a dc open
6
1
R di
d2 i
i
+
+
=0
2
L
L
LC
dt
dt
Here’s our differential equation
Let’s solve it!
(Same approach as before:)
Try* solution of the form est
i = A est
Note:
d est
st
=
s
e
dt
• A can be anything…
s2
Aest
+s
d2 est 2 st
dt2 = s e
1
R
st
Ae +
Aest = 0
LC
L
Will be determined by initial conditions
• s must satisfy:
s2
1
R
+s
+
=0
L LC
* (Suggested by mathematicians who came before us…)
7
R
+ vR –
1
R
2
s +s
+
=0
LC
L
Vo
+
C
vC
L
vL
–
+
i(t)
Characteristic Equation (of the circuit)
–
Quadratic Equation:
ax2 + bx + c = 0
Root #1
Root #2
x = –b – b2 – 4ac
2a
x = –b + b2 – 4ac
2a
a
R
Root #1 –
+
2L
1

b
R 2 1
–
LC
2L
R
L
1
c
R
Root #2 –
–
2L
LC

R 2 1
–
LC
2L
8
Root #1
R
–
s1 = 2L +

Root #2
R 2 1
–
LC
2L
R
–
–
s2 =
2L
+ vR –
t=0
R
Vo

R 2 1
–
LC
2L
i(t)
+
C
vC
L
vL
–
+
–
i = A est
Two solutions: Which one is correct?
9
Root #1
R
–
s1 = 2L +

R 2 1
–
LC
2L
Root #2
R
–
–
s2 =
2L

R 2 1
–
LC
2L
i(t) = A est
Two cases to consider:
1
R
>
LC
2L

Positive Number
1
R
<
LC
2L

Negative Number = IMAGINARY
= REAL
Let’s look at these cases one at a time…
10
Root #1
R
–
s1 = 2L +

Root #2
R 2 1
–
LC
2L
1
R
>
LC
2L

R
–
–
s2 =
2L
Positive Number

R 2 1
–
LC
2L
= REAL
10 k
R
Vo
5V
R
= 417
2L
1
LC = 83,300
i(t)
s1 = –116
C
1 F
L
12 H
s2 = – 717
11
10 k
R
i(t)
Vo
5V
s1 = –116
+
C
1 F vC
L
12 H
–
s2 = – 717
Most General Solution: Linear Superposition
i(t) = A1e s1t + A2e s2t
Must satisfy the Initial Conditions:
i(0) = 0
vC(0) = 0
12
R
i(t)
Vo
+
C
L
vC
vC(0) = 0
vL
iL(0) = 0
–
+
–
Ohm’s Law:
i(0) = 0
By KVL: vL(0) = Vo
vR(0) = 0
(The voltage that’s left over after vC = 0 and vR = 0)
Initial Conditions:
i(0) = 0
This is good!
Vo
di
vdtC(0)== 0
L
t=0
(The initial condition on vC
translates into an IC for di/dt )
2nd order differential equation  Need I.C. for i and
di
to solve.
dt
13
Find A1
Vo = 5 V
s1 = –116 sec–1
s2 = – 717 sec–1
s1 = –116 sec–1
s2 = – 717 sec–1
L = 12 H
i(t) = A1e s1t + A2e s2t
i(0) = 0
Vo
di
=
dt
L
t=0
A1 + A2 = 0
s1A1 + s2A2 = Vo
L
Solve for A1
( s2A1 + s2A2 = 0 )
Vo
(s1–s2)A1 =
L
Vo
A1 =
(s1–s2) L
= 0.7 mA
14
Find A2
s1 = –116 sec–1 s2 = – 717 sec–1
A1 + A2 = 0
A1 = 0.7 mA
A2 = –0.7 mA
i(t) = A1e s1t + A2e s2t
This term decays much more quickly
i(t) = 0.7 (e
–166t
–e
–717t
)
mA
15
i(t) = 0.7 (e
–717t
–e
) mA
0.40
0.30
0.20
0.10
60 t (ms)
59.4
54.0
50
48.6
43.2
40
37.8
30
32.4
27.0
20
21.6
16.2
10
10.8
0
5.4
0.00
0.0
Current (mA)
0.50
–166t
Time (ms)
Overdamped Response
16
0.80
0.80
i1(t) = 0.7 e
–166t
0.60
0.60
–166 t
i(t) = 0.7 (e
0.40
0.40
0.20
0.20
00
0.00
0
-0.20
–0.20
–717 t
–e
) mA
i1(t) + i2(t)
10
20
30
40
50
60
–-0.40
0.40
–-0.60
0.60
i2(t) = – 0.7 e
–717 t
– 0.8
-0.80
17
Root #1
R
–
s1 = 2L +

Root #2
R 2 1
–
LC
2L
R
–
–
s2 =
2L

R 2 1
–
LC
2L

R R 1 1
< >
2L 2L LC LC
Negative
Number
Positive
Number = IMAGINARY
= REAL
1 k
10
k
R
Vo
5V
R
= 41.7
417
2L
1
LC
= 83,300
(same)
i(t)
C
1 F
L
12 H
Huh?
s1 = – 41.7 +  –81,600
s2 = – 41.7 –  –81,600
18
Huh?
 –81,600
 –81,600
How to deal with this term…
 –81,600  –1
 –81,600
 –1
ij
(Mathematicians)
(Engineers)
(41.7)
s1 = – 42 + j  81,600
s1 = – + j o
s2 = – 42 – j  –81,600
s2 = – – j o
 = 42
o =  81,600 = 286
units: sec–1
19
s1 = – + j o
s2 = – – j o
i(t) = A1e s1t + A2e s2t
e s 1t
=e
(– + jo)t
=e
–t
Huh? (again)
jo
et
n! = n(n–1)(n–2)(n–3)..(1)
e
x
=1+x +
n=0 n=1
e
x
=1+x +
x2
+
2!
x3
3!
+
x4
4!
+
x5
5!
x6
+
6!
+…
n=2
x2
2!
j 2 = –1
x2
x4
+
x3
3!
+
x4
4!
j 4 = +1
x6
+
x5
5!
+
x6
6!
+…
j 6 = –1
x3
x5
jx = 1 –
+ 4! – 6! + … + j x – 3! + 5! – …
e
2!
j2
j4
j6
20
j x = 1 – x2 + x4 – x6 + … + j x – x3 + x5 – …
e
2!
4!
6!
3!
5!
cos x = 1 –
Basic definitions of
sine and cosine
sin x = x –
e
e
jx
j ot
= cos x + j sin x
= cos ot + j sin ot
x2
2!
+
x3
3!
x4
4!
+
–
x5
5!
x6
6!
+…
–…
“Euler’s Equation”
Back to the circuit…
21
s1 = – + j o
 = 42
R
s2 = – – j o
o = 16.9
+ vR –
Vo
Do some linear algebra…
+
C
vC
L
vL
–
+
i(t)
–
s1t = A e (– + jo)t = A e –t cos  t + j sin  t
A1e
cos o ot + j sin oot
1
1
-+-
(–
 –– j
joo))tt
(–
t
––t
s
t
2
2
=
A
e
j sin
cosot ot––j sin
otot
cos
A2e
==AA22ee
21
Soln #1:
Soln #2:
2 (A1+ A2) e –t cos t = K1 e –t cos ot
2 j (A1– A2) e –t sin t
= K2 e –t sin ot
General Solution: i(t) = K1 e –t cos ot + K2 e –t sin ot
22
General Solution:
Equals 1 at t = 0
Equal 0 at t = 0
Equals 1 at t = 0
Equals 1 at t = 0
i(t) = K1 e –t cos ot + K2 e –t sin ot
Initial Condition #1: i(0) = 0
K1 = 0
Vo
d
i
Initial Condition #2:
=
dt
L
t=0
Find derivative of i(t)
Equals 0 at t = 0
K2
–
e
–t
Vo
K2o =
L
Equals 1 at t = 0
Equals 1 at t = 0
–t
sin ot + o e
cos ot
Vo
K2 =
o L
23
Vo –t
sin ot
i(t) =  L e
o
1.5
K2e–t envelope
1.0
0.5
0
0.0
10
20
30
40 50
60
70
80 90
100 t (ms)
–0.5
–1.0
Underdamped Response
Period 2
o
24
Root #1
R
–
s1 = 2L +

R 2 1
–
LC
2L
1
R
=
LC
2L
Root #2
R
–
–
s2 =
2L


R 2 1
–
LC
2L
0
Two Double Roots
R
–
s1 = s2 = 2L
Critically Damped Response
(Boundary between Overdamped and Underdamped Responses)
25
End of This Module
Homework
26