Post on 27-Dec-2015
EMLAB
1
Chapter 6. Capacitance and in-ductance
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2Contents
1. Capacitors
2. Inductors
3. Capacitor and inductor combinations
4. RC operational amplifier circuits
5. Application examples
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Power supply board
PC motherboard
inductor capacitor
inductor
capacitor
Cell phone
Usage of inductors and capacitors
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41. Capacitors• Capacitance is defined to be the ratio of charge to voltage difference.• Used to store charges• Used to store electrostatic energy
0V
0Qq
SV SV
SV SVV EIf the voltage difference between the terminals of the capacitor is equal to the supply voltage, net flow of charges becomes zero.
V
QC
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Electrons(-) are absorbed.(+) charges are generated
Electrons(-) are generated. (+) charges are absorbed.
Generation of charges : battery
e2ZnZn 2
234 HNH222NH e
Electrons are generated via electro-chemical reaction.
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• Used to store charges• Used to store electrostatic energy• Slow down voltage variation
Usage of capacitors
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7Type of capacitors
t
0
)(1
)( dttiC
t
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8Frequently used formulas on capacitors
dt
dC
dt
dqi
t
t
t
t
tt
diC
t
diC
diC
diC
0
0
0
)(1
)(
)(1
)(1
)(1
0
2
2
1)()()()()( C
dt
d
dt
tdtCtittp
22 )]([2
1
2
1)()()( tCdC
d
dditW
tt
q
C
Energy :
Power :
Voltage :
Capacitance :
Current :
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9
i
t
t0
0
tdi
C )(
1
iυ relation of capacitors
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10
)(ti
Example 6.2
The voltage across a 5-μF capacitor has the waveform shown in Fig. 6.4a. Deter-mine the current waveform.
dt
dCi
mst
mstt
mstt
t
80
8696102
24
60106
24
)(3
3
mst
mstmA
mstmA
ti
80
8660
6020
)(
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• Capacitor voltage cannot change instantaneously due to finite current supply.
Properties of capacitors
SV
i
t
dt
dCi
SR
CC
• In steady state, capacitor behaves as if open circuited.
SV
iSR
0dt
dCi DC
)()( 00 tt
0t
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12Example 6.3
Determine the energy stored in the electric field of the capacitor in Example 6.2 at t=6 ms.
][1440)]([2
1)( 2 JtCtW
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The current in an initially uncharged 4μF capacitor is shown in Fig. 6.5a. Let us de-rive the waveforms for the voltage, power, and energy and compute the energy stored in the electric field of the capacitor at t=2 ms.
Example 6.4
mst
mst
mstt
ti
40
428
20102
16
)(3
mst
msttdx
msttxdx
tt
t
40
42824)8(4
1
2010001084
1
)(0
0
23
mst
mstt
mstt
ttitp
40
426416
208
)()()(
3
mst
msttt
mstt
dxxptWt
40
421012810648
202
)()( 1292
4
0
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142. Inductors
dt
diLtL )(
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15Two important laws on magnetic field
Current generates magnetic field (Biot-Savart Law)
inducedV
Time-varying magnetic field generates induced electric field that opposes the variation. (Faraday’s law)
Current
Current
B-field
B-field
V
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Biot-Savart Law Faraday’s Law
rrRRr
B
,4
ˆ2C R
Id
LidS
aB
dt
dV
ind
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17Self induced voltage
dt
diL
dt
dV
ind
• The induced voltage is generated such that it opposes the applied magnetic flux.
• The inductor cannot distinguish where the applied magnetic flux comes from.• If the magnetic flux is due to the coil itself, it is called that the induced voltage
is generated by self-inductance.
=
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18Frequently used formulas on inductors
dt
diL
t
t
t
t
tt
dL
ti
dL
dL
dL
i
0
0
0
)(1
)(
)(1
)(1
)(1
0
2
2
1)(
)()()()( Li
dt
dti
dt
tdiLtittp
22 )]([2
1
2
1)()()( tiLdLi
d
dditW
tt
i
NL
Energy :
Power :
Voltage :
Inductance :
Current :
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19Properties of inductors
SV
i
t
i
dt
diL
SR
LL
• In steady state, inductor behaves as if short circuited.
SV
iSR
0dt
diL DC
• Inductor current cannot change instantaneously due to finite current supply.
0t
)()( 00 titi
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Find the total energy stored in the circuit of Fig. 6.8a.
Example 6.5
0815254273
09
36
9
111
11
CCC
CC
VVV
VV
][8.109
62.16],[2.16
21VVVV CC
][8.19
],[2.16
91
2
1
1A
VIA
VI C
LC
L
][44.1)2.1)(102(2
1 23
1mJWL
][48.6)8.1)(104(2
1 23
2mJWL
][62.2)2.16)(1020(2
1 26
1mJWC
][92.2)8.10)(1050(2
1 26
2mJWC
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21Example 6.6
The current in a 10-mH inductor has the waveform shown in Fig. 6.9a. Determine the voltage waveform.
mst
mstt
mstt
ti
40
421040102
1020
20102
1020
)( 33
3
3
3
mst
mstmV
mstmV
t
40
42][100102
1020)1010(
20][100102
1020)1010(
)(3
33
3
33
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22Example 6.7
The current in a 2-mH inductor is
][)377sin(2)( Atti
Determine the voltage across the inductor and the energy stored in the inductor.
][)377cos(508.1)]377sin(2[
)102()( 3 Vtdt
td
dt
diLtL
][)377(sin004.0)]377sin(2)[102(2
1)]([
2
1)( 2232 JtttiLtWL
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23Example 6.8
The voltage across a 200-mH inductor is given by the expression
00
0][)31()(
3
t
tmVett
t
Let us derive the waveforms for the current, energy, and power.
00
0][5)31(200
10 3
0
33
t
tmAtedxexitt x
00
0][)31(5)()()(
6
t
tWetttittp
t
00
0][5.2)]([
2
1)(
622
t
tJettiLtW
t
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24Capacitor and inductor specifications
Standard tolerance values are ; 5%, ; 10%, and ; 20%.
Tolerances aretypically 5% or 10% of the speci-fied value.
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25Example 6.10The capacitor in Fig. 6.11a is a 100-nF capacitor with a tolerance of 20%. If the volt-age waveform is as shown in Fig. 6.11b, let us graph the current waveform for the minimum and maximum capacitor values.
dt
dCi
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266.3 Capacitor and Inductor Combinations
=
N
N
i iS CCCCC
11111
211
N
N
iiP CCCCCC
3211
=
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=
=
N
N
i iP LLLLLL
111111
3211
N
N
iiS LLLLLL
3211
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286.4 RC Operational Amplifier Circuits
Op-amp differentiator
Ci
iRdt
dC o
211 )(
0,0 i
dt
tdCRo
)(112
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29Op-amp integrator
i
dt
dC
R o )(21
1
0,0 idt
tdC
Ro )(
21
1
0)0(
)(1
)0()(1
)(1
0 121
0 121
121
o
t
o
tt
o dxxCR
dxxCR
dxxCR
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30Example 6.17
The waveform in Fig. 6.26a is applied at the input of the differentiator circuit shown in Fig. 6.25a. If R2=1 kΩ and C1=2 μF, determine the waveform at the output of the op-amp.
mstV
mstV
dt
td
dt
tdCRo 105][4
50][4)(10)2(
)( 13112
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31Example 6.18
If the integrator shown in Fig. 6.25b has the parameters R1=5 kΩ and C2=0.2μF, de-termine the waveform at the op-amp output if the input waveform is given as in Fig. 6.27a and the capacitor is initially discharged.
][1.020
][1.002010)20(10)(
1 33
0 121 stt
stttdxx
CR
t
o