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Chapter 10
Capacitors & Capacitance
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10.1 Capacitance (p. 386)
Capacitor
• Consists of 2 conductors separated by an insulator.
• Basic form: parallel-plate capacitor [Fig. 10-1, page 386] which consists of
two metal plates separated by dielectric (e.g. air).
Charging capcitor
• When DC source is connected to two metal plates [Fig. 10-2, page 386],
electrons are removed from +ve plate and an equal number of electrons
will be deposited on -ve plate.
• This leaves top plate positively charges and bottom plate negatively
charged.
• Capacitor can store charge, i.e. remains charged after DC source is
removed.
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Discharging Capacitor
• By shorting a wire/resistor across the two leads.
Definition of Capacitance
• Charge stored depends on applied voltage
Q = CV or C = Q/V
unit: Farads, F
• C is defined as the capacitance of the capacitor
• Example 10-1, [page 387]
a. How much charge is stored on a 10-F capacitor when it is connected to
a 24-volt source?
b. The charge on a 20-nF capacitor is 1.7 C. What is its voltage?
Solution
a. From Equation 10-1, Q = CV. Thus, Q = (10 x 10-6 F)(24 V) = 240 C.
b. Rearranging Equation 10-1, V = Q/C = (1.7 x 10-6 C)/(20 x 10-9 F) = 85 V.
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10.2
Factors Affecting Capacitance (p. 387)
Effect of area
• The more charge put on a capacitor‘s plate for a given voltage, the
greater will be its capacitance.
• Capacitance is directly proportional to plate area. [Fig. 10-4, page 387]
Effect of Spacing
• As space between plated decreases, the force of attraction increases and
pulls more electrons from one plate to other.
• Capacitance is inversely proportional to plate spacing.
Effect of Dielectric
• Capacitance varies for different materials [Table 10-1, page 388].
• This factor is called relative dielectric onstant or relative perrmittivity of a
material [Fig. 10-6, page 388]
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Capacitance of a Parallel Capacitor (p. 389)
C= A/d
(F)
A = plate area
d = spacing
 = absolute dielectric constant (F/m)
For air or vacuum,  =  o = 8.85 x 10-12 F/m
Other materials,  expressed as the product of the relative dielectric constant,
 r times  o , i.e.  =  o  r
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Example 10-3, [page 389]
A parallel-plate capacitor with air dielectric has a value of C = 12 pF. What is the
capacitance of a capacitor that has the following:
a. The same separation and dielectrric but five times the plate area?
b. The same dielectric but four times the area and one fifth the plate spacing?
c. A dry paper dielectric, six times the plate area, and twice the plate spacing?
Solution
a. Since the plate area has increased by a factor of five and everything else remains
the same, C increases by a factor of five. Thus, C = 5(12 pF) = 60 pF.
b. With four times the plate area, C increases by a factor of four. With one fifth the
plate spacing, C increases by a factor of five. Thus, C = (4)(5)(12pF) = 240 pF.
c. Dry paper increases C by a factor of 2.2. The increases in plate area increases
C by a factor of six. Doubling the plate spacing reduces C by one half. Thus,
C = (2.2)(6)(½)(12 pF) = 79.2 pF.
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Quiz
For a given capacitor, it stores Q1 charges when voltage V1 applied across its two
ends. It will store Q2 charges when voltage V2 applied across its two ends.
If V1 = 10V and V2 = 5V, Q1 is greater than Q2 by 2C. Find the a capacitance of
the capacitor.
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10.3
Electric Fields (p. 390)
Electric flux
• Electric fields are force fields that exist in the region surrounding charged
bodies. [Fig. 10-8, page 391]
• The direction of the field is defined as the direction of force on a positive
charge. It is directed outward from the +ve charge and inward toward
the -ve charge.
• Field lines never cross and density of lines indicates the strength of the
field.
Fig. 10-8:
(a) Field about a pair of positive and negative charges
(b) Field of parallel plate capacitor
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Electric field of parallel-plate capacitor is uniform across the gap with some
fringing near edges.
Electric flux lines are represented by
Electric Field Intensity, 
• Strength of an electric field, i.e. force that the field exerts on a small, +ve
test charge, Qt
 = F / Qt unit: newtons / coulomb (N/C)
F = Q Qt / (4   r2) for a point charge, Q
Hence
 = Q / (4   r2)
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Electric Flux Density, D (p. 391)
• represent the density fo flux lnes in space
• independent of the medium
D = 
or
D = Total flux / area
=  /A
• The number of flux lines emanating from a charge is equal to the charge
itself, i.e.  = Q
Figure 10-9 - In the SI system, total flux  equals charge Q. [page 392]
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Figure 10-10 - Work moving test charge Qt is force times distance. [page 392]
Work required to move the charge against the force (F) through distance (d)
W = Fd
Define voltage, V

=
=
=
W / Qt for a test charge Qt
F d / Qt
V / d as  = F / Qt
i.e. electric field strength between capacitor plates
= voltage across the plates divided by the distance between them.
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Capacitance,
C
=
=
=
=
= Q/V
/V
A D / d
(D / ) (A / d)
A/ d
• EXAMPLE 10-4, [page 392] Suppose that the electric field intensity
between the plates of a capacitor is 50 000 V/m when 80 V is applied
a. What is the plate spacing if the dielectric is air? If the dielectric is
ceramic?
b. What is  if the plate spacing is halved?
Solution
a.  = V/ d, independent of dielectric. Thus, d = V/ξ = 80V/50x103V/m
= 1.6x10-3 m
b. Since  = V/d, will double to 100 000 V/m.
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10.4 Dielectrics (p. 393)
Voltage breakdown
• If voltage applied across capacitor is increased beyond a critical value,
force on electronics is so great that they are torn from orbit.
• This is called dielectric breakdown.
• Electric field intensity at breakdown is called dielectric strength of a
material. E.g. air = 3kV / mm
• Table 10-2, [page 393]
MATERIAL
kV/mm
Air
3
Ceramic (high r)
3
Mica
40
Mylar
16
Oil
15
Polystyrene
24
Rubber
18
Teflon
60
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Voltage rating
• because of dielectric breakdown, capacitors are rated for max operating
voltage (called working voltage).
EXAMPLE 10-5, [page 394] A capacitor with plate dimensions of 2.5 cm by
2.5cm and a ceramic dielectric with  r = 7500 experiences breakdown at
2400 V. What is C?
Solution From Table 10-2 dielectric strength = 3 kV/mm. Thus, d = 2400
V/3000 V/mm = 0.8 mm = 8 x 10-4 m. So
C= r o A/d
= (7500) (8.85 x 10-12) (0.025m)2/(8x10-4m)
= 51.9 nF
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10.5
Nonideal Effect (p. 394)
Leakage current
• charged capacitor will discharge after disconnected from source
• small leakage current will pass through dielectric when capacitor
connected to a source.
• Charge leaks through the dielectric [Fig. 10-12, page 394]
R = hundreds of Mohm
• Figure 10-12 - Leakage current. [page 394]
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Equivalent series resistance
• resistance (RS) may develop in capacitor’s leads as its internal
connections begins to fail
• cause problems in high-frequency circuits
• dissipation factor
D = R S / XC
where XC = 1/2fC
Dielectric absorption
• after shorting two leads of a capacitor, a residual voltage remains
• Disadv.: upset circuit voltage levels
Temperature coefficient
• +ve / zero / -ve temperature coefficient imply capacitance increases / no
change / decreases with increasing temp.
• in parts per million (ppm)
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10.6
Types of Capacitors (p. 395)
Fixed capacitor [Fig. 10-13, 10-14, page 396]
Variable capacitor [Fig. 10-19, page 399]
Figure 10-13: Stacked capacitor construction. The stack is
compressed, leads attached, and the unit coated
with epoxy resin or other insulting material.
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Fixed capacitor
Ceramic:
relative permittivity = 30 to 7500
• extremely high permittivity permit small packaging but characteristics vary
wifely with temp. and operating voltage. Use in limited temp applications
where small size and cost are important
• many surface mount capacitors use ceramic dielectrics
Plastic film:
• film/foil [Fig. 10-14, page 396] use metal foil use metal foil separated by
plastic film
• metallized-film have foil material vaccum deposited directly onto plastic film
Mica:
• low in cost with low leakage and good stability
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Electrolytic
• provide large apacitance up to hundred thousand microfarads
• relatively low cost
• leakage is relatively high and breakdown voltage is relatively low
Surface mount: extremely small and provide high packaging density
Variable capacitor
• used in radio tuning circuits
• stationary plates and set of movable plates which are ganged together and
mounted on a shaft
• as shaft is rotated, the effective area change
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10.7
Capacitors in parallel and series
• capacitors in parallel [fig. 10-20, page 400]
. Voltage is the same across each.
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Example 10-6, [page 401]
Example 10-6 A 10-F, a 15-F, and a 100-F capacitor are connected in
parallel across a 50-V source. Determine the following:
a. Total capacitance.
b. Total charge stored.
c. Charge on each capacitor.
Solution
a. Cr = C1 + C2 + c3 = 10F + 15 F + 100 F = 125 F
b. QT = CTV = (125 F)(50 V) = 6.25 mC
c. Q1 = C1V = (10 . Q1 = C1V = (10 F)(50 V) = 0.5 mC)(50 V) = 0.5 mC
Q2 = C2V = (15 F)(50 V) = 0.75 mC
Q3 = C3V = (100 F)(50 V) = 5.0 mC
Note Q1 + Q2 + Q3 = (0.5 + 0.75 + 5.0) mC = 6.25 mC, which checks with (b).
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Capacitors in series [Fig. 10-21, page 401]
• Charge is the same on each.
1  1  1  ... 1
Cr C C
C
N
1
2
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EXAMPLE 10-7
(p. 402)
Refer to Figure 10-22(a):
a. Determine Cr .
b. If 50 V is applied across the capacitors, determine Q.
d. Determine the voltage on each capacitor.
Solution
a. 1/Cr = 1/C1 + 1/C2 + 1/C3 = 1/30μF + 1/60μF + 1/20μF
= 0.0333 x 106 + 0.0167 x 106 + 106 = 0.1 x 106
Therefore,
Cr = 1/(0.1 x 10-6) = 10 F
b. Q = CTV = (10 x10-6F)(50 V) = 0.5 mC
c. V1 = Q/C1= (0.5 x 10-3 C) / (30 x 10-6F) = 16.7V
V2 = Q/C2 = (0.5x10-3C)/(60x10-6F) = 8.3V
V3 = Q/C3 = (0.5x10-3C)/(20x10-6F) = 25.0V
Check: V1 + V2 + V3 = 16.7 + 8.3 + 25 = 50 V.
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EXAMPLE 10-8
[page 403]
For the circuit of Figure 10-23(a), determine CT
Refer to Figure 10-23: Systematic reduction.
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Quiz:
Consider capacitors of 1F, 1.5F, and 10F. If equivalent capacitance is equal to
10.6F. How are the capacitors connected?
Solution:
To yield 10.6 F, the 10 F capacitor must be connected in parallel with some combination
of the 1 F and 1.5F capacitors. Only the connection shown below works.
10.8
Capacitor current and voltage [page 404]
(Refer to Fig. 10-25, page 405)
During charging
• movement of electrons constitutes current
• current lasts for capacitor to be charged
• no current pass through dielectric
• capacitor voltage builds as charge deposited on plates
• as capacitor voltage increases, charging current decreases
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Capacitor V-1 relationship, [page 405]
q = C VC
ic = dq / dt = d (C VC) / dt
ic = C d Vc / dt
Current through a capacitor is equal to C times the rate of change of voltage
across it.
Example 10-9 [page 406] A signal generator applies voltage to a 5-µF
capacitor with a wavefrom as in Figure 10-27(a). The voltage rises linearly from
0 to 10V in 1ms, falls linearly to -10V at t=3ms, remains constant until t=4ms,
rises to 10V at t=5ms, and remains constant thereafter.
a. Determine the slope of vc.
b. Determine the current and sketch its graph.
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Solution
a. We need the slope of vc during each time interval where slope = rise/run =  v/t.
0 ms to 1 ms: v = 10 V; t = 1 ms; Therefore, slope = 10 V/1 ms 10 000 V/s.
1 ms to 3 ms: Slope = -20 V/2 ms = -10 000 V/s.
3 ms to 4 ms: Slope = 0 V/s.
4 ms to 5 ms: Slope = 20 V/1 ms = 20 000 V/s.
b.
ic = Cdvc/dt = C times slope. Thus,
0 ms to 1 ms: i = (5 x 10-6F) (10 000 V/s) = 50 mA.
1 ms to 3 ms: i = -(5 x 10-6F) (10 000 V/s) = -50 mA.
3 ms to 4 ms: i = (5 x 10-6F) (0 V/s) = 0 mA.
4 ms to 5 ms: i = (5 x 10-6F) (20 000 V/s) = 100 mA.
Refer to Figure 10-27, page 406
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10.9
Energy stored by a capacitor [page 407]
• an ideal capacitor does not dissipate power
• when power is transferred to a capacitor, all of it is stored as energy in the
capacitor’s electric field
Stored energy, W = 1/2 C V2
10.10
Capacitor failures and troubleshooting [page 408]
Failures:
• short internally
• leads open
• dielectric leaky
(Noted: if electrolytic capacitor is connected with its polarity reversed, it may
explode.)
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Capacitors fail because of
• misapplication
• excessive voltage, current, or temperature
• aging
Basic testing with an ohmmeter
• out-of-circuit tests with analog ohmmeter
- detect opens and shorts
- leaky dielectrics
(Noted: discharge capacitor first before measurement)
• for normal capacitor, the ohmmeter reading should be low initially
and gradually increase to infinity.
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Capacitor testers [page 408]
• some digital multimeter can measure capacitance
• LCR (inductance, capacitance, resistance) analyzer can determine
capacitance as well as detect opens and short
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