capacitors - iowa state university
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EE 201 capacitors – �1
CapacitorsConsider an open circuit:
+
–voc
i = 0
–
++
–
++
––
E
+Q
–Q –
++
–
++
––
+ ++
– ––
voc 2voc
+2Q
–2Q
2E
double the voltage
Q / voc
E / vocThere is some energy “stored”.
There was a bit of current flow when the voltage changed.
Effect is weak for dangling wires. (But not zero!)
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EE 201 capacitors – �2
Change the geometry – have parallel plates with area A
Capacitance
–
++ ++ + ++
– –– – ––
E+Q
–Q
voc
same voltage• much more charge • much more electric field • much more energy stored
Q = CV E =V
d
increase charge with better dielectric material and more area.
increase field (and hence Q) by moving plates closer together
C → capacitance
farads (F) = C / V
air: εo = 8.85x10–12 F/m
other materials: ε = εr εo
relative dielectric: εr = constant
4 = �E$
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EE 201 capacitors – �3
2 plates, each with area A.
Parallel-plate capacitor
d+Q
–Q & =�$G
Example: A = 1 cm2, d = 0.001 cm, air dielectric
Wow. Very small value, and it is already a fairly large area plate.
Higher values?
• higher dielectric material between the electrodes • thinner dielectric • winding or stacking to get larger surface area into a smaller volume.
Other configurations are possible, but parallel-plate is most common.
Values range from 10 pF to 100 µF, (and higher). A 1-F capacitor is huge and quite rare.
& =
��.� � �����)�FP
� ��FP��
�.���FP = �.�� � �����) = ��.�S)
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EE 201 capacitors – �4
Capacitor current
+
–vC
iC
Note that passive sign convention.
At DC, ic = 0. (It’s just a fancy open circuit.)
However, some current must flow when voltage is changing. Otherwise, the charge would not change.
4 = &Y&Current only flows when voltage is changing.
G4GW = L& = &GYF
GW
–
++ ++ + ++
– –– – ––
E+Q
–QvC
C
GY&GW � GE
GW
In Maxwell’s equations, the current due to changing field is called displacement current.
GEGW � G4
GW
As current flows, the capacitor charge increases or decreases.
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EE 201 capacitors – �5
L& = &GYFGW
Capacitor voltage cannot change instantaneously.
t
vC
infinite current!GYFGW � �
Note, though, that current can change instantaneously.
also
can’t happen.
Y& (W) =�&
� W
�L& (W�) GW� + Y& (�)
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EE 201 capacitors – �6
Capacitor energyAn energy storage device
• Charge the cap to some voltage. Charge (and energy) stays. Remove it later.
• Ideal capacitor dissipates no energy – no heat generated. • Real capacitors do show some leakage. (Large resistor in parallel.)
Usually negligible.
When charging a capacitor, the power being delivered is given by:
The energy delivered by the source, and hence the energy stored in the capacitor is (assuming vC = 0 at t = 0 and vC(tf) = VC.)
3& (W) GW = &Y&GY&
( =��&9
�&
( =
� WI
�3& (W) GW = &
� 9&
�Y&GY&
3& (W) = Y& (W) L& (W) = &Y&GY&GW
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EE 201 capacitors – �7
Combinations of capacitors
C1 C2 C3Ceq
–vC+i1 i2 i3
ieq LHT = L� + L� + L�
&HT = &� + &� + &�
C1 C2
C3Ceq
iC
–
+ vC1
+vC2
– +vC3
–
+
–veq
YHT = YF� + YF� + YF�
GYHTGW =
GYF�GW +
GYF�GW +
GYF�GW
L&&HT
=L&&�
+L&&�
+L&&�
�&HT
=�&�
+�&�
+�&�
&HTGY&GW = &�
GY&GW + &�
GY&GW + &�
GY&GW
Capacitors combinations are exactly opposite those of resistors.
Parallel
Series
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EE 201 capacitors – �8
ExampleA voltage source connected across a capacitor has a ramp (or triangle) shape as a function of time. It ramps from 0 V to 5 V in 10 ms and then ramps back down to 0 V in another 10 ms. What is the current in the capacitor?
10 ms < t < 20 ms: vC(t) = (–500 V/s)·(t – 10 ms) + 5 V
= –(500 V/s)·t + 10 V
0 < t < 10 ms: vC(t) = (5V/10ms)·t = (500 V/s)·t
vC
t
5 V
10 ms 20 ms
+–
iC
–vC+
Vramp C
We can write expressions for the voltage as a function of time.
1 µF
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EE 201 capacitors – �9
0 ms < t < 10 ms: iC (t) = CdvCdt = (1 μF) · (500 V/s) = +0.5 mA
10 ms < t < 20 ms: iC (t) = CdvCdt = (1 μF) · (�500 V/s) = �0.5 mA
iC
t
0.5 mA
–0.5 mA
The current is constant for each portion – positive current during the upward ramp and negative during the downward ramp.
We find the current by taking the derivative of the voltage and multiplying by the capacitance.
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EE 201 capacitors – �10
Example
What is vo for the circuit at right?
Start as always: write a node equation at the inverting input.
iR = iC + i–
vs � v�R = Cdvc
dt + i�
For an op-amp with a feedback loop: v– = v+, so v– = 0 in this case (virtual ground). And, for an ideal op amp: i– = 0. Also, note vC = 0 – vo.
vsR = �Cdvo
dt
vo (t) =1RC
� t
0vs (t�) dt� + vo (0) A circuit that integrates!
–++
–
–+
RC
iR
iC
vC
vovS
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EE 201 capacitors – �11
Example
The voltage across the capacitor at right is a sinusoid: vs(t) = Vmsin(ωt). What is the capacitor current?
iC (t) = CdvC (t)dt = [ωCVm] cos (ωt)
Interesting. The current has also a sinusoidal form, but it is shifted by 90°. Also, the magnitude of the current waveform depends on the frequency of the oscillation – faster oscillation leads to bigger currents. The frequency-dependent amplitude and the phase shift will have far-reaching implications when we study sinusoidal circuits in more detail.
= Im cos (ωt) = Im sin (ωt � 90�)
+–
iC
–vC+
Vmsin(ωt) C1 µF
Vm = 5 V; ω = 377 rad/s
Im = 1.88 mA