Download ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance

ECE2262 Electric Circuits
Chapter 6: Capacitance and Inductance
Capacitors
Inductors
Capacitor and Inductor Combinations
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CAPACITANCE AND INDUCTANCE
Introduces two passive, energy storing devices: Capacitors and Inductors
LEARNING GOALS
CAPACITORS
Store energy in their electric field (electrostatic energy)
Model as circuit element
INDUCTORS
Store energy in their magnetic field
Model as circuit element
CAPACITOR AND INDUCTOR COMBINATIONS
Series/parallel combinations of elements
RC OP-AMP CIRCUITS
Integration and differentiation circuits
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6.1 Capacitors
Electronic Symbol:
3
• A capacitor consists of two conductors separated by a non-conductive region.
The non-conductive region (orange) is called the dielectric
Charge separation in a parallel-plate capacitor causes an internal electric field.
• A dielectric reduces the electric field and increases the capacitance.
Because the conductors (or plates) are close together, the opposite charges on the
conductors attract one another due to their electric fields, allowing the capacitor to
store more charge for a given voltage than if the conductors were separated, giving
the capacitor a large capacitance.
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• In the hydraulic analogy, a capacitor is analogous to a rubber membrane
sealed inside a pipe. This animation illustrates a membrane being repeatedly
stretched and un-stretched by the flow of water, which is analogous to a capacitor
being repeatedly charged and discharged by the flow of charge.
https://en.wikipedia.org/wiki/File:CapacitorHydraulicAnalogyAnimation.gif
• In the hydraulic analogy, charge carriers flowing through a wire are
analogous to water flowing through a pipe. A capacitor is like a
rubber membrane sealed inside a pipe. Water molecules cannot pass through
the membrane, but some water can move by stretching the membrane. The
analogy clarifies a few aspects of capacitors.
• The more a capacitor is charged, the larger its voltage drop; i.e., the more it "pushes
back" against the charging current. This is analogous to the fact that the more a
membrane is stretched, the more it pushes back on the water.
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• The capacitance describes how much charge can be stored on one plate of a
capacitor for a given "push" (voltage drop).
A very stretchy, flexible membrane corresponds to a higher capacitance than
a stiff membrane.
• A charged-up capacitor is storing potential energy, analogously to a
stretched membrane.
C : Capacitance of 1 Farad = 1 Coulomb of charge on each conductor causes
a voltage of 1 Volt across the device.
1C = 1F ! 1V
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Current ! Voltage Relations
A
C!
d
The charge on the capacitor is proportional to the voltage across
q = Cv
(
C : Capacitance in Farads (F) - common values µ F , pF pico = 10 !12
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)
Example
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Voltage ! Current Relation
d ( C v)
dv
dq
=C
Since q = C v and i =
then we have i =
.
dt
dt
dt
i (t ) = C
d v (t )
dt
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i (t ) = C
dv ( t )
dt
In circuits with with time-independent voltage ( DC circuits) the capacitor behaves
as open circuit (capacitor blocks dc).
In analyzing a circuit containing dc voltage sources and capacitors,
we can replace the capacitors with an open circuit and calculate
voltages and currents in the circuit using our many analysis tools.
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Example
C = 5µ F
i=C
dv
dt
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Current ! Voltage Relation
i (t )
v( t )
d v(t )
Since i ( t ) = C
then
dt
d v (! )
"t0 i (! ) d! = C "t0 d! d!
t
t
= C " dv (! ) = = C {v ( t ) ! v ( t 0 )}
t
t0
1 t
v (t ) = v (t 0 ) + " i (! ) d !
C t0
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t ! t0
1 t
v (t ) = v (t 0 ) + " i (! ) d !
C t0
for any time t ! t 0
v ( t 0 ) - initial value (initial state) representing the voltage due to the charge
that accumulates on the capacitor from the past to t = t 0 .
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!t /2
Example Suppose that C = 1F and v ( 0 ) = 1V . The current source i ( t ) = e , t > 0 .
Determine v ( t ) .
i (t )
v( t )
{
t
1 t
v ( t ) = v ( 0 ) + " i (! ) d! = 1+ # e!" /2 d" = 1! 2 e!" /2
0
C 0
{
!t /2
= 1+ 2 1! e
}
{
}
t
0
}
#% 1+ 2 1! e!t /2 t > 0
for t ! 0 ! v ( t ) = $
t "0
%& 1
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{
}
#% 1+ 2 1! e!t /2 t > 0
v (t ) = $
t "0
%& 1
Note: i ( t ) has a jump, but v ( t ) is continuous !
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Power Delivered to Capacitor
d v(t )
1 t
Since i ( t ) = C
or v ( t ) = v ( t 0 ) + " i (! ) d! then we have
dt
C t0
p (t ) = v (t ) i (t )
= C v(t )
d v(t )
dt
1 t
#
&
= i ( t ) % v ( t 0 ) + " i (! ) d! (
C t0
$
'
Note: If v ( t ) would have a jump then
d v(t )
= ! and p ( t ) = ! but this is
dt
impossible. Hence, the voltage response of a capacitor must be continuous !
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If we encounter circuits containing switches then the idea of
“continuity of voltage” for a capacitor tells us that the voltage across
the capacitor just after a switch moves is the same as the voltage
across the capacitor just before that switch moves.
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Energy Stored in the Capacitor Electric Field
d v(t )
1 d v2 ( t )
p (t ) = C v(t )
= C
2 dt
dt
d !1
$
2
C
v
t
=
(
)
"
%
dt # 2
&
1
w ( t ) = C v2 ( t )
2
1 q2 (t )
Since q ( t ) = Cv ( t ) then also w ( t ) =
.
2 C
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{
}
d 2
d
!
v ( t ) = 2 v ( t ) {v ( t )}
dt
dt
(J)
Example An uncharged capacitor 2 F is driven by a rectangular current pulse.
i (t )
1
t
1
(a)The capacitor voltage
! t /2 0 ! t !1
1 t
1 t
v ( t ) = v ( 0 ) + " i (! ) d! = # 1 ( 0 ! " ! 1) d" = "
.
0
0
2
2
t >1
# 1/ 2
! t /2
(b) The capacitor power: p ( t ) = i ( t ) v ( t ) = "
# 0
0 ! t !1
.
else
!# t 2 / 4
1
2
2
(c) The capacitor energy: w ( t ) = C v ( t ) = v ( t ) = "
2
$# 1 / 4
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0 ! t !1 .
t >1
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Note that power is always positive for the duration of the current pulse, which
means that energy is continuously being stored in the capacitor. When the current
returns to zero, the stored energy is trapped because the ideal capacitor offers no
means for dissipating energy. Thus, a voltage remains on the capacitor after i ( t )
returns to zero.
If at some later time an energy-absorbing device (e.g., a flash bulb) is connected
across the capacitor, a discharge current will flow from the capacitor and,
therefore, the capacitor will supply its stored energy to the device.
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Example
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• Energy is being stored in the capacitor whenever the power is positive
• Energy is being delivered by the capacitor whenever the power is negative
•
! p (t ) dt = 4 µ J - energy stored
# p (t ) dt = !4 µ J - energy delivered
1
0
"
1
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6.2 Inductors
An inductor, also called a coil, choke or reactor, is a passive two terminal
electrical component which resists changes in electric current passing through
it. It consists of a conductor such as a wire, usually wound into a coil. When a
current flows through it, energy is stored temporarily in a magnetic field in the
coil.
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When the current flowing through an inductor changes, the time-varying
magnetic field induces a voltage in the conductor, according to Faraday’s law
of electromagnetic induction.
An inductor is characterized by its inductance, the ratio of the voltage to the rate
of change of current, which has units of henries (H). Inductors have values that
typically range from 1 !H ( 10 !6 H) to 1 H.
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1 t
A capacitor is an integrator of its input current: v ( t ) = v ( t 0 ) + " i (! ) d!
C t0
L Inductance
(in henrys )
An inductor produces a current response that is related to the integral of the
voltage applied to its terminals.
1 t
i ( t ) = i ( t 0 ) + " v (! ) d!
L t0
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Current ! Voltage Relation
L
v( t ) = L
di ( t )
dt
Voltage ! Current Relation
L
1 t
i (t ) = i (t 0 ) + " v (! ) d !
L t0
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Power Delivered to an Inductor
p (t ) = v (t ) i (t )
=
Li ( t )
di ( t )
dt
1 t
#
&
= v ( t ) %i ( t 0 ) + " v (! ) d! (
L t0
$
'
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Energy Stored in the Inductor Magnetic Field
di ( t )
1 di 2 ( t )
! p (t ) = L
Since p ( t ) = i ( t ) L
2
dt
dt
d !1 2 $
! p ( t ) = " Li ( t ) %
dt # 2
&
1 2
w ( t ) = Li ( t )
2
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(J)
DC Circuits
Consider the case of a dc current flowing through an inductor.
Since
di ( t )
v (t ) = L
dt
we see that the voltage across the inductor is directly proportional to the time
rate of change of the current flowing through the inductor.
A dc current does not vary with time, so the voltage across the inductor is zero.
We can say that an inductor is “a short circuit to dc.”
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In analyzing a circuit containing dc sources and inductors, we can
replace any inductors with short circuits and calculate voltages
and currents in the circuit using our many analysis tools.
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Continuity of i ( t )
!
p ( t ) = Li ( t )
di ( t )
dt
di ( t )
we note that an instantaneous change in inductor
dt
current would require infinite power. Since we don’t have any infinite power
sources, the current flowing through an inductor cannot change instantaneously.
Due to p ( t ) = Li ( t )
This will be a particularly helpful idea when we encounter circuits
containing switches. This idea of “continuity of current” for an inductor
tells us that the current flowing through an inductor just after a switch
moves is the same as the current flowing through an inductor just before
that switch moves.
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Example The independent current source generates zero current for t < 0 and a
pulse i ( t ) = 10te!5t A for t > 0 in the following circuit
(a) i ( t )
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(b) v ( t ) = L
di ( t )
= 0.1! 10e"5t (1" 5t ) = e!5t (1! 5t ) , t > 0
dt
• At t = 0.2s the voltage changes polarity
• At t = 0 the voltage has a jump, i.e., the voltage can change
instantaneously across the terminals (but not current !)
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(c)
{
} {
}
p ( t ) = i ( t ) v ( t ) = 10te!5t " e!5t (1! 5t ) = 10te!10t ! 50t 2 e!10t W
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(
1 2
1
(d) w ( t ) = Li ( t ) = ! 0.1! 10te"5t
2
2
)
2
= 5t 2 e!10t
• An increasing energy curve indicates that energy is being stored. Thus, energy is being
stored in the time interval 0 to 0.2 sec. This corresponds to the interval when p ( t ) > 0 .
• A decreasing energy curve indicates that energy is being extracted. This takes
place in the time interval 0.2 to ! . This corresponds to the interval when p ( t ) < 0
• Energy is max when i ( t ) is max: wmax = 27.07mJ at t = 0.2 sec.
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37
(e) Integrals
!
0.2
0
p ( t ) dt ,
•!
•
The area
The area
!
0.2
"
!
0
0.2
!
0.2
"
!
0
0.2
"
!
0.2
p ( t ) dt
p ( t ) dt = 27.07mJ
p ( t ) dt = #27.07mJ
p ( t ) dt represents the energy stored in the inductor during [ 0,0.2 ]
p ( t ) dt represents the energy extracted
Energy Stored + Energy Extracted = 0
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Example Find the total energy stored in the circuit (DC circuit)
1
1
wC = Cv 2 , wL = Li 2
2
2
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DC circuit ! capacitors ! open circuit, inductors ! short circuit
We need Vc1 , Vc2 and I L1 , I L 2
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• KCL at A: I L1 + 3 ! I L 2 = 0 ! I L 2 = I L1 + 3
• KVL (big loop): !9 + 6I L1 + 3I L 2 + 6I L 2 = 0 ! 6I L1 + 9I L 2 = 9
! I L1 = !1.2A , I L 2 = 1.8A
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• KVL (first loop): !9 + 6I L1 + VC1 = 0 ! VC1 = 9 ! 6I L1 = 16.2 V
• VC 2 = 6I L 2 = 10.8 V
42
I L1 = !1.2A , I L 2 = 1.8A , VC1 = 16.2A , VC 2 = 10.8A
1
• wC1 = C1VC12 = 2.62 mJ,
2
1
wC 2 = C2VC22 = 2.92 mJ
2
1
1
2
• wL1 = L1 I L1 = 1.44 mJ, wL 2 = L2 I L2 2 = 6.48 mJ
2
2
The total stored energy = 13.46 mJ
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6.3 Capacitor and Inductor Combinations
Series Capacitors
v = v1 + ....+ vN
The charge gained by a plate of any capacitor must come from a plate of an
adjacent capacitor, i.e.,
Q Q
Q
=
+ ....+
CS C1
CN
1
1
1
= + ....+
CS C1
CN
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Example Find C2
C1 = 30 µ F
v1 = 8V
C2
v = 12V
The charge on both capacitors must be the same
! Q = v1C1 = ( 8V ) ! ( 30 µ F ) = 240 µC
! Q = v2C2 ! C2 =
240 µC
Q
Q
= 60 µ F
=
=
4V
v2 v ! v1
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Example
• the capacitors have been charged before they were connected Q1 ! Q2 ! Q3
1
1
1
1
! Ceq = 1µ F
=
+
+
• Equivalent capacitance:
Ceq C1 C2 C3
• Total energy stored: w ==
{
}
1
C1VC12 + C2VC22 + C3VC23 = 31µ J
2
1
!v
+
2
!
4
!
1
=
0
v
=
!3V
!
! wCeq == Ceq v 2 = 4.5 µ J
•
2
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Parallel Capacitors
The total stored charge: QT = Q1 + ....+ QN
! vCT = vC1 + ....+ vC N
C p = C1 + ....+ CN
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Example Find each capacitor voltage
20µF
30 µ F
+V1 !
40 µ F
400V
V2
+
!
+
9 µF
V3
!
70 µ F
+
V4
!
• 20 || 40 ! C1 = 20 + 40 = 60 µ F
• C2 = ( 30 ! 70 ) || 9 =
30 ! 70
+ 9 = 21 + 9 = 30 µ F
30 + 70
48
C1 = 60 µF
20µF
+V1 !
40 µ F
400V
C2 = 30 µ F
30 µ F
V2
+
!
+
9 µF
V3
!
70 µ F
+
V4
!
60 ! 30
= 20 µ F ! • Qeq = Ceq ! V = 20 ! 10 "6 ( 400 ) = 8mC
• Ceq = C1 ! C2 =
60 + 30
8mC
8mC
= 133V • V2 =
= 267V
• V1 =
60 µ F
30 µ F
•charge on 30 µ F & 70 µ F = Qeq ! charge on 9 µ F = 8 ! 267V " 9 µ F = 5.6mC
(
• V3 =
5.6mC
= 187 V,
30 µ F
V4 =
5.6mC
= 80 V
70 µ F
49
)
Series Inductors
v ( t ) = v1 ( t ) + ...+ vN ( t ) = L1
di ( t )
di ( t )
di ( t )
+ ...+ LN
L
+
....+
L
=( 1
N)
dt
dt
dt
LS = L1 + ....+ L N
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Parallel Inductors
#
&
#
1 t
1
i ( t ) = i1 ( t ) + ...+ iN ( t ) = $i1 ( 0 ) + " v (! ) d! ' + ...+ $iN ( 0 ) +
L1 0
LN
%
(
%
#1 t
1
i ( t ) = {i1 ( 0 ) + ...+ iN ( 0 )} + $ " v (! ) d! + ...+
LN
% L1 0
!1
1 $ t
i ( t ) = i ( 0 ) + " + ...+
% (0 v (' ) d'
LN &
# L1
51
&
"0 v (! ) d! '(
t
&
"0 v (! ) d! '(
t
1
1
1
= + ...+
L p L1
LN
52
Example
• Initial values: i1 ( 0 ) = 3A , i2 ( 0 ) = !5A
• The voltage at the terminal v ( t ) = !30e!5t mV for t ! 0
L1 ! L2 60 ! 240
=
= 48mH
(a) Equivalent inductance: Leq =
L1 + L2 60 + 240
(b) Initial current: i ( 0 ) = i1 ( 0 ) + i2 ( 0 ) = 3 + ( !5 ) = !2A (current goes up)
53
1
(c) Find i ( t ) : i ( t ) = i ( 0 ) +
Leq
" v (! )d!
t
0
1
= !2 +
48 " 10 !3
${
t
0
}
!30e!5# " 10 !3 d# = !2 +
= !2.125 + 0.125e!5t A, t ! 0
54
(
)
1 !5t
e !1
8
(d) Find i1 ( t ) , i2 ( t )
1
1 t
i1 ( t ) = i1 ( 0 ) + " v (! )d! = 3 +
60 ! 10 "3
L1 0
${
t
0
}
"30e"5# ! 10 "3 d#
= 2.9 + 0.1e!5t A, t ! 0
1
i2 ( t ) = i2 ( 0 ) +
L2
1
!5
+
v
!
d
!
"0 ( ) =
240 " 10 !3
t
$ {!30e
t
0
!5 #
= !5.025 + 0.025e!5" A, t ! 0
55
}
" 10 !3 d#
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SUMMARY
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58
■ In dc steady state, a capacitor looks like an open circuit and
an inductor looks like a short circuit.
■
• The voltage across a capacitor cannot change instantaneously
• The current flowing through an inductor cannot change
instantaneously.
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