# Lecture #12 Correction to Lecture 9, Slide 3

```Lecture #12
ANNOUNCEMENTS
• Graduate school workshop tomorrow (9/23)
– 3-4 PM, Wozniak Lounge (Soda Hall)
OUTLINE
–
–
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Capacitors in series and in parallel
Practical capacitors
The inductor
Inductors in series and in parallel
Chapter 6.1-6.3
EECS40, Fall 2003
Lecture 12, Slide 1
Prof. King
Correction to Lecture 9, Slide 3
• If there are no independent sources in a circuit, VTh = 0.
– If there are dependent sources in the circuit, we need to apply an
external voltage in order to determine RTh.
Example: Circuit used in RTh Calculation Example #2, Lecture 8:
ITEST =
Vx
VTEST − Vx VTEST − 40i∆
+
20
60
VTEST
–
+
RTh =
Applying KCL to node x :
Vx − 40i∆
V −V
− i∆ + x TEST = 0
16
20
− Vx
Definition of i ∆ : i∆ =
80
EECS40, Fall 2003
Lecture 12, Slide 2
⇒ Vx =
VTEST 75
= Ω
I TEST
4
8
VTEST
25
Prof. King
1
Example: Current, Power &amp; Energy for a Capacitor
t
v (V)
1
0
1
2
i (&micro;A)
0
3
i=C
1
2
3
EECS40, Fall 2003
4
5
dv
dt
4
i(t)
v(t)
–
+
1
v(t ) = ∫ i (τ )dτ + v(0)
C0
t (&micro;s)
vc must be a continuous
function of time; however,
ic can be discontinuous.
5
t (&micro;s)
(dc operation), time
derivatives are zero
&AElig; C is an open circuit
Lecture 12, Slide 3
Prof. King
p (W)
i(t)
1
2
3
4
5
–
+
v(t)
0
10 &micro;F
10 &micro;F
t (&micro;s)
p = vi
w (J)
0
t
1
EECS40, Fall 2003
2
3
4
5
Lecture 12, Slide 4
t (&micro;s)
1
w = ∫ pdτ = Cv 2
2
0
Prof. King
2
Capacitors in Parallel
i(t)
i1(t)
i2(t)
+
C1
C2
v(t)
dv
dv
+ C2
dt
dt
dv
dv
i = (C1 + C2 ) = Ceq
dt
dt
i = i1 + i2 = C1
–
+
Ceq = C1 + C2
v(t)
Ceq
i(t)
–
i = Ceq
dv
dt
Equivalent capacitance of capacitors in parallel is the sum
EECS40, Fall 2003
Lecture 12, Slide 5
Prof. King
Capacitors in Series
+ v1(t) – + v2(t) –
i(t)
C1
+
C2
i(t)
Ceq
v(t)=v1(t)+v2(t)
–
dv
d (v1 + v2 )
= Ceq
dt
dt
dv
dv
i = Ceq 1 + Ceq 2
dt
dt
1
1
1
⇒
=
+
Ceq C1 C2
dv1
dv
= C2 2
dt
dt
dv
i
dv2
i
⇒ 1=
and
=
dt C1
dt C2
i = C1
i = Ceq
EECS40, Fall 2003
i
i
+ Ceq
C2
C1
i = Ceq
Lecture 12, Slide 6
Prof. King
3
Capacitive Voltage Divider
Q: Suppose the voltage applied across a series combination
of capacitors is changed by ∆v. How will this affect the
voltage across each individual capacitor?
∆v = ∆v1 + ∆v2
∆Q1=C1∆v1
Q1+∆Q1
C1
v+∆v
+
–
-Q1−∆Q1
Q2+∆Q2
C2
−Q2−∆Q2
∆Q2=C2∆v2
EECS40, Fall 2003
+
v1+∆v1
–
Note that no net charge can
can be introduced to this node.
Therefore, −∆Q1+∆Q2=0
+
v2(t)+∆v2
–
⇒ C1∆v1 = C2 ∆v2
∆v2 =
C1
∆v
C1 + C2
Note: Capacitors in series have the same charge.
Lecture 12, Slide 7
Prof. King
Application Example: MEMS Accelerometer
• Capacitive position sensor
used to measure acceleration
(by measuring force on a
proof mass)
g1
g2
FIXED OUTER PLATES
EECS40, Fall 2003
Lecture 12, Slide 8
Prof. King
4
Sensing the Differential Capacitance
– Fixed electrodes are biased at +Vs and –Vs
– Movable electrode (proof mass) is biased at Vo
Circuit model
Vs
Vo = −Vs +
C1
Vo
C2
–Vs
EECS40, Fall 2003
C1
C − C2
Vs
( 2Vs ) = 1
C1 + C2
C1 + C2
εA ε A
−
Vo g1 g 2 g 2 − g1 g 2 − g1
=
=
=
Vs εA + εA g 2 + g1
const
g1 g 2
Lecture 12, Slide 9
Prof. King
Practical Capacitors
• A capacitor can be constructed by interleaving the plates
with two dielectric layers and rolling them up, to achieve
a compact size.
• To achieve a small volume, a very thin dielectric with a
high dielectric constant is desirable. However, dielectric
materials break down and become conductors when the
electric field (units: V/cm) is too high.
– Real capacitors have maximum voltage ratings
– An engineering trade-off exists between compact size and
high voltage rating
EECS40, Fall 2003
Lecture 12, Slide 10
Prof. King
5
The Inductor
• An inductor is constructed by coiling a wire around some
type of form.
+
vL(t)
iL
_
• Current flowing through the coil creates a magnetic field
or flux that links the coil: LiL
• When the current changes, the magnetic flux changes
&AElig; a voltage across the coil is induced:
Note: In “steady state” (dc operation), time
derivatives are zero &AElig; L is a short circuit
EECS40, Fall 2003
vL (t ) = L
Lecture 12, Slide 11
diL
dt
Prof. King
Symbol:
L
Units: Henrys (Volts • second / Ampere)
(typical range of values: &micro;H to 10 H)
Current in terms of voltage:
1
vL (t )dt
L
t
1
iL (t ) = ∫ vL (τ )dτ + i (t0 )
L t0
diL =
iL
+
vL
–
Note: iL must be a continuous function of time
EECS40, Fall 2003
Lecture 12, Slide 12
Prof. King
6
Stored Energy
Consider an inductor having an initial current i(t0) = i0
di
p (t ) = v(t )i (t ) = Li (t )
dt
t
i
i
dι
w(t ) = ∫ p (τ )dτ = ∫ Lι dτ = ∫ Lιdι
dτ
t0
i0
i0
w(t ) =
1 2 1 2
Li − Li0
2
2
EECS40, Fall 2003
Lecture 12, Slide 13
Prof. King
Inductors in Series
v = Leq
+ v1(t) – + v2(t) –
v(t) +
–
L1
i(t)
L2
v(t) +
–
di
dt
i(t)
Leq
+
v(t)=v1(t)+v2(t)
–
v = L1
di
di
di
di
+ L2 = (L1 + L2 ) = Leq
dt
dt
dt
dt
Leq = L1 + L2
Equivalent inductance of inductors in series is the sum
EECS40, Fall 2003
Lecture 12, Slide 14
Prof. King
7
Inductors in Parallel
+
i1
i(t)
+
i2
i(t)
v(t) L2
L1
Leq
v(t)
–
–
t
t
i = i1 + i2 =
t
1
1
vdτ + i1 (t0 ) + ∫ vdτ + i2 (t0 )
∫
L1 t0
L2 t0
i=
1
vdτ + i (t0 )
Leq t∫0
 1 1 t
i =  +  ∫ vdτ + [i1 (t0 ) + i2 (t0 )]
 L1 L2  t0
1
1 1
⇒
= +
with i (t0 ) = i1 (t0 ) + i2 (t0 )
Leq L1 L2
EECS40, Fall 2003
Lecture 12, Slide 15
Prof. King
Summary
Capacitor
Inductor
dv
dt
1
w = Cv 2
2
i=C
v=L
w=
v cannot change instantaneously
i can change instantaneously
Do not short-circuit a charged
capacitor (-&gt; infinite current!)
n cap.’s in series:
EECS40, Fall 2003
1 2
Li
2
i cannot change instantaneously
v can change instantaneously
Do not open-circuit an inductor with
current (-&gt; infinite voltage!)
n
1
1
n ind.’s in series:
=∑
Ceq i =1 Ci
n cap.’s in parallel: Ceq =
n
∑C
i =1
i
di
dt
n ind.’s in parallel:
Lecture 12, Slide 16
n
Leq = ∑ Li
i =1
n
1
1
=∑
Leq i =1 Li
Prof. King
8
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