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Capacitor
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
The Basic Capacitor
Capacitors are one of the fundamental passive
components. In its most basic form, it is composed of
two conductive plates separated by an insulating
dielectric.
The ability to store charge is the definition of
capacitance.
Conductors
Electronics Fundamentals 8th edition
Floyd/Buchla
Dielectric
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
The Basic Capacitor
VVSS
The charging
process…
Leads

Initially
Source
Fully
Charging
charged
removed
uncharged



++
+++
+++
+++
++
+++

 ++
+
++
+
++
+
+
+
AA +
A

 +
Dielec tric


++
Plates


+
+




+

+  Elec trons



+ B

BB




A capacitor with stored charge can act as a temporary battery.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitance
Capacitance is the ratio of charge to voltage
C
Q
V
Rearranging, the amount of charge on a
capacitor is determined by the size of the
capacitor (C) and the voltage (V).
Q  CV
If a 22 mF capacitor is connected to
a 10 V source, the charge is 220 mC
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitance
An analogy:
Imagine you store rubber bands in a
bottle that is nearly full.
You could store more rubber bands
(like charge or Q) in a bigger bottle
(capacitance or C) or if you push
them in more (voltage or V). Thus,
Q  CV
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitance
A capacitor stores energy in the form of an electric field
that is established by the opposite charges on the two
plates. The energy of a charged capacitor is given by the
equation
W
1
CV 2
2
where
W = the energy in joules
C = the capacitance in farads
V = the voltage in volts
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitance
The capacitance of a capacitor depends on
three physical characteristics.
 A
C  8.85 1012 F/m  r 
 d 
C is directly proportional to
the relative dielectric constant
and the plate area.
C is inversely proportional to
the distance between the plates
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitance
Find the capacitance of a 4.0 cm diameter
sensor immersed in oil if the plates are
separated by 0.25 mm.  r  4.0 for oil 
 r A 
C  8.85 10 F/m 

d


2
2
3
2
The plate area is A  πr   0.02 m  1.26 10 m
12


The distance between the plates is 0.25 103 m
  4.0  1.26 103 m2 
C  8.85 1012 F/m 

0.25 103 m

Electronics Fundamentals 8th edition
Floyd/Buchla

  178 pF


© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor types
Mica
Mica capacitors are small with high working voltage.
The working voltage is the voltage limit that cannot
be exceeded.
Foil
Mic a
Foil
Mic a
Foil
Mic a
Foil
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor types
Ceramic disk
Ceramic disks are small nonpolarized capacitors They
have relatively high capacitance due to high r.
Lead wire soldered
to silver elec trode
Solder
Ceramic
dielec tric
Electronics Fundamentals 8th edition
Floyd/Buchla
Dipped phenolic c oating
Silv er elec trodes deposited on
top and bottom of c eram ic disk
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor types
Plastic Film
Plastic film capacitors are small and nonpolarized. They
have relatively high capacitance due to larger plate area.
High-purity
foil elec trodes
Plastic film
dielec tric
Outer wrap of
polyester film
Capac itor sec tion
(alternate strips of
film dielec tric and
Lead wire
foil elec trodes)
Solder c oated end
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor types
Electrolytic (two types)
Electrolytic capacitors have very high capacitance but
they are not as precise as other types and tend to have
more leakage current. Electrolytic types are polarized.
+
_
Al electrolytic
Ta electrolytic
Symbol for any electrolytic capacitor
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor types
Variable
Variable capacitors typically have small capacitance
values and are usually adjusted manually.
A solid-state device that is used as a variable
capacitor is the varactor diode; it is adjusted with an
electrical signal.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor labeling
+ ++ +
47VTTMFVTT
Capacitors use several labeling methods. Small
capacitors values are frequently stamped on them such
as .001 or .01, which have implied units of microfarads.
Electrolytic capacitors have larger
values, so are read as mF. The unit is usually
stamped as mF, but some older ones may be
shown as MF or MMF).
. 022
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitor labeling
A label such as 103 or 104 is read as 10x103
(10,000 pF) or 10x104 (100,000 pF)
respectively. (Third digit is the multiplier.)
When values are marked as 330 or 6800, the
units are picofarads.
222
2200
What is the value of
each capacitor? Both are 2200 pF.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Series capacitors
When capacitors are connected in series, the total
capacitance is smaller than the smallest one. The
general equation for capacitors in series is
CT 
1
1
1
1
1


 ... 
C1 C2 C3
CT
The total capacitance of two capacitors is
CT 
1
1
1

C1 C2
…or you can use the product-over-sum rule
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Series capacitors
If a 0.001 mF capacitor is connected
in series with an 800 pF capacitor,
the total capacitance is 444 pF
C
1
0
.0
0
1µ
F
Electronics Fundamentals 8th edition
Floyd/Buchla
C
2
8
0
0p
F
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Parallel capacitors
When capacitors are connected in parallel, the total
capacitance is the sum of the individual capacitors.
The general equation for capacitors in parallel is
CT  C1  C2  C3  ...Cn
If a 0.001 mF capacitor is
connected in parallel with
an 800 pF capacitor, the
total capacitance is 1800 pF
Electronics Fundamentals 8th edition
Floyd/Buchla
C
1
C
2
0
.0
0
1
µ
F
8
0
0
p
F
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitive reactance
Capacitive reactance is the opposition to ac by a
capacitor. The equation for capacitive reactance is
1
XC 
2πfC
The reactance of a 0.047 mF capacitor when a
frequency of 15 kHz is applied is 226 W
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitive reactance
When capacitors are in series, the total reactance is the sum of the
individual reactances. That is,
X C(tot )  X C1  X C2  X C3    X Cn
Assume three 0.033 mF capacitors are in series with a 2.5 kHz
ac source. What is the total reactance?
The reactance of each capacitor is
1
1

 1.93 kW
2πfC 2π  2.5 kHz  0.033 μF 
X C(tot )  X C1  X C2  X C3
XC 
 1.93 kW  1.93 kW  1.93 kW  5.79 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitive reactance
When capacitors are in parallel, the total reactance is the reciprocal
of the sum of the reciprocals of the individual reactances. That is,
X C(tot ) 
1
1
1
1
1


  
X C1 X C2 X C3
X Cn
If the three 0.033 mF capacitors from the last example are
placed in parallel with the 2.5 kHz ac source, what is the total
reactance?
The reactance of each capacitor is 1.93 kW
X C(tot ) 
Electronics Fundamentals 8th edition
Floyd/Buchla
1
1

 643 W
1
1
1
1
1
1


+
+
X C1 X C2 X C3 1.93 kW 1.93 kW 1.93 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Capacitive Voltage Divider
Two capacitors in series are commonly used as a capacitive voltage
divider. The capacitors split the output voltage in proportion to their
reactance (and inversely proportional to their capacitance).
What is the output voltage for the capacitive voltage divider?
1
1

 4.82 kW
2πfC1 2π  33 kHz 1000 pF 
1
1


 482 W
2πfC2 2π  33 kHz  0.01 μF
X C1 
XC2
X C (tot )  X C1  X C 2
 4.82 kW  482 W  5.30 kW
Vout
 XC2 
 482 W 

Vs  

1.0 V = 91 mV
X

 5.30 kW 
 C (tot ) 
Electronics Fundamentals 8th edition
Floyd/Buchla
1.0 V
f = 33 kHz
C1
1000 pF
C2
Vout
0.01 µF
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Summary
Capacitive phase shift
When a sine wave
is applied to a
capacitor, there is a
phase shift between
voltage and current
such that current
always leads the
voltage by 90o.
Electronics Fundamentals 8th edition
Floyd/Buchla
VC
0
90o
I
0
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Sinusoidal response of RC circuits
When both resistance and capacitance are in a series
circuit, the phase angle between the applied voltage and
total current is between 0 and 90, depending on the
values of resistance and reactance.
VR
VC
V R leads VS
V C lags V S
R
C
VS
I
I leads V S
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Impedance of series RC circuits
In a series RC circuit, the total impedance is the phasor
sum of R and XC.
R is plotted along the positive x-axis.
XC is plotted along the negative y-axis.
 XC 

 R 
  tan 1 
R
R


Z is the diagonal
XC
XC
Z
Z
It is convenient to reposition the
phasors into the impedance triangle.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Impedance of series RC circuits
Sketch the impedance triangle and show the
values for R = 1.2 kW and XC = 960 W.
Z
1.2 kW 
2
+  0.96 kW 
 1.33 kW
  tan 1
0.96 kW
1.2 kW
 39
Electronics Fundamentals 8th edition
Floyd/Buchla
2
R = 1.2 kW

39o
Z = 1.33 kW
XC =
960 W
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Analysis of series RC circuits
Ohm’s law is applied to series RC circuits using Z,
V, and I.
V  IZ
V
I
Z
V
Z
I
Because I is the same everywhere in a series circuit,
you can obtain the voltages across different
components by multiplying the impedance of that
component by the current as shown in the following
example.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Analysis of series RC circuits
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasor diagram. The impedance
triangle from the previous example is shown for reference.
The voltage phasor diagram can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
R = 1.2 kW

39o
Z = 1.33 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can only
be drawn for a single frequency because X is a
function of frequency.
R
As frequency changes,
Increasing f



the impedance triangle
Z
X
f
for an RC circuit changes
Z
as illustrated here
because XC decreases
f
X
Z
with increasing f. This
determines the frequency
f
X
response of RC circuits.
3
2
1
3
C3
3
C2
2
C1
1
2
1
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Applications
For a given frequency, a series RC circuit can be used to
produce a phase lag by a specific amount between an
input voltage and an output by taking the output across
the capacitor. This circuit is also a basic low-pass filter, a
circuit that passes low frequencies and rejects all others.
V
R

Vin
C
Vout
VR
f
(phase lag)
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
f
Vin
Vin
(phase lag)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Applications
Reversing the components in the previous circuit produces
a circuit that is a basic lead network. This circuit is also a
basic high-pass filter, a circuit that passes high frequencies
and rejects all others. This filter passes high frequencies
down to a frequency called the cutoff frequency.
C
V
Vout

Vin
(phase lead)
Vin
R
Vout
Vout
VC
Electronics Fundamentals 8th edition
Floyd/Buchla
Vin

(phase lead)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Sinusoidal response of parallel RC circuits
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
Conductance is the reciprocal of resistance.
Capacitive susceptance is the reciprocal
of capacitive reactance.
G
1
R
1
BC 
XC
1
Admittance is the reciprocal of impedance. Y 
Z
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Sinusoidal response of parallel RC circuits
In a parallel RC circuit, the admittance phasor is the sum
of the conductance and capacitive susceptance phasors.
The magnitude can be expressed as Y  G2 + BC 2
 BC 

G 
From the diagram, the phase angle is   tan 1 
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Sinusoidal response of parallel RC circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
 BC 

G 
  tan 1 
Y is the diagonal
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Sinusoidal response of parallel RC circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G
1
1

 1.0 mS
R 1.0 kW
Y  G 2 + BC 2 
BC  2 10 kHz  0.01 mF  0.628 mS
1.0 mS +  0.628 mS  1.18 mS
2
2
BC = 0.628 mS
VS
f = 10 kHz
R
1.0 kW
C
0.01 mF
Y=
1.18 mS
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using
Y, V, and I.
V
I
Y
I  VY Y 
I
V
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Analysis of parallel RC circuits
If the voltage in the previous example is 10 V, sketch the
current phasor diagram. The admittance diagram from the
previous example is shown for reference.
The current phasor diagram can be found from
Ohm’s law. Multiply each admittance phasor by 10 V.
BC = 0.628 mS
Y=
1.18 mS
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 V
=
IC = 6.28 mA
IS =
11.8 mA
IR = 10 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Phase angle of parallel RC circuits
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: BC  2 fC
As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.
Electronics Fundamentals 8th edition
Floyd/Buchla
IC
IS

IR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Equivalent series and parallel RC circuits
For every parallel RC circuit there is an equivalent
series RC circuit at a given frequency.
The equivalent resistance and capacitive
reactance are shown on the impedance triangle:
Req = Z cos 

Z
Electronics Fundamentals 8th edition
Floyd/Buchla
XC(eq) = Z sin 
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 9
Series-Parallel RC circuits
Series-parallel RC circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
series and parallel circuits.
Z1
Z2
For example, the
R1
C1
components in the
R2
C2
green box are in
series: Z1  R12  X C21
The components in
the yellow box are
R2 X C 2
in parallel: Z 2 
R22  X C2 2
Electronics Fundamentals 8th edition
Floyd/Buchla
The total impedance can be found by
converting the parallel components to an
equivalent series combination, then
adding the result to R1 and XC1 to get the
total reactance.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.