Download Capacitor

Lesson 11:
Capacitors (Chapter 10)
and Inductors (Chapter 11)
1
Learning Objectives
• Define capacitance and state its symbol and unit of
measurement.
• Predict the capacity of a parallel plate capacitor.
• Analyze how a capacitor stores charge and energy.
• Explain and Analyze capacitor DC characteristics.
• Define inductance and state its symbol and unit of
measurement.
• Predict the inductance of a coil of wire.
• Explain Inductor DC characteristics.
• Analyze how an inductor stores charge and energy.
2
Capacitors and Inductors
• For resistive circuits, the voltage-current
relationships are linear and algebraic.
− Resistors can only dissipate energy; they cannot store
energy and return it to a circuit at a later time.
• This is not the case for capacitors and inductors.
Capacitors and Inductors are dynamic elements.
− The voltage-current relationships are non-linear and
differential.
− They are dynamic because they store energy.
3
Capacitor
• A capacitor is passive element designed to store
energy in its electric field. This energy can then be
provided to a circuit at a later time.
− Capacitors consist of two conductors (parallel plates)
separated by an insulator (or dielectric).
− Capacitors accumulate electric charge.
− Conductive plates can become charged with opposite
charges
4
Capacitor
• Electrons are pulled from top plate (creating positive charge).
• Electrons are deposited on bottom plate (creating negative
charge).
5
Capacitor
• While charging a capacitor:
− The voltage developed across the capacitor will increase as charge is
deposited.
− Current goes to zero once the voltage developed across the capacitor
is equal to the source voltage.
6
Definition of Capacitance
• Capacitance (C):of the capacitor: is a measure of the
capacitor’s ability to store charge.
• Unit is the farad (F).
• Capacitance of a capacitor
− Is one farad if it stores one coulomb of charge when the
voltage across its terminals is one volt.
Q
C  ( Farads )
V
7
Effect of Geometry
• Increased surface area means increased Capacitance.
• Larger plate will be able to hold more charge.
8
Effect of Geometry
• Reduced separation distance means increased
Capacitance.
• As plates are moved closer together:
− Force of attraction between opposite charges is greater.
• Capacitance:
− Inversely proportional to distance between plates.
9
Effect of Dielectric
• Substituting a dielectric material for the air gap will
increase the Capacitance.
• Dielectric Constant  (epsilon) is calculated by using the
relative dielectric constant and the absolute dielectric
constant for air:
 = r 0 (F/m)
where 0 = 8.854 x 10-12 F/m and r is the
relative permittivity or dielectric constant (see table
10.1 on pg 401 of text)
10
Capacitance of Parallel-Plates
• Putting the factors together, capacitance of parallel-plate
capacitor is given by
C 
where
A
A
or
d
d
(farads, F)
r is the relative dielectric constant
0 = 8.854 x 10-12 F/m
A is area
d is the distance between plates
11
Example Problem 1
A capacitor consists of 2” x 2” plates separated by 1/32”. No dielectric is used.
a) Determine the capacitance.
b) Determine the charge on the plate if 48 V is applied across the plate.
A
A
C  or
d
d
(farads, F)
0  8.854*1012 (F/m)
and because no dielectric is used: r = 1
Convert plate distance (d) to meters:
1
1m
1
d  in 
* in
32
39.37in 32
d  7.94*104 m
Now calculate Capacitance (C) in Farads (F):
C or
A
d
F
0.0026m 2
C = 8.854*10
*1*
m
7.94*104 m
C = 2.89*1011 (F) = 28.9 (pF)
12
Calculate area (A) in meters:
1m
A  (2in) 2  2in 
* 2in  0.051m
39.37in
Finally calculate the charge (Q) in Coulombs (C):
A  (0.051m) 2  0.0026m 2
Q
C  ( Farads )
V
Q  C *V  28.9 pF * 48V  1.38nC
12
Example Problem 2
A parallel-plate capacitor has dimensions of 1 cm x 1.5 cm and separation of
0.1 mm. What is its capacitance when the dielectric is mica?
A
A
C  or
d
d
Because mica is used as the dielectric;
(farads, F)
0  8.854*1012 (F/m) and r = 40
Convert plate distance (d) to meters:
d  0.1mm  0.0001m
Now calculate Capacitance (C)
in Farads (F):
A
C or
d
0.015m 2
12 F
C = 8.854*10
*40*
m
0.0001m
C = 5.31*107 (F) = 531 (nF)
Calculate area (A) in meters:
A  1cm *1.5cm  1.5cm 2  0.015m
13
Dielectric Voltage Breakdown
• High voltage will cause an electrical discharge
between the parallel plates.
• Above a critical voltage, the force on the electrons
is so great that they become torn from their orbit
within the dielectric.
• This damages the dielectric
material, leaving carbonized
pinholes which short the plates
together.
14
Dielectric Voltage Breakdown
• The working voltage is the maximum operating voltage of
a capacitor beyond which damage may occur.
• This voltage can be calculated using the material’s
dielectric strength (K) measured in kV/mm.
Dielectric Strength (K)
V  Kd
15
Material
kV/mm
Air
3
Ceramic (high Єr)
3
Mica
40
Myler
16
Oil
15
Polystyrene
24
Rubber
18
Teflon
60
Capacitance and Steady State DC
• In steady state DC, the rate of change of voltage is
zero, therefore the current through a capacitor is zero.
• A capacitor looks like an open circuit with voltage
vC in steady state DC.
16
Capacitors in Series and in Parallel
• Capacitors, like resistors, can be placed in series and
in parallel.
• Increasing levels of capacitance can be obtained by
placing capacitors in parallel.
• Decreasing levels of capacitance can be obtained by
placing capacitors in series.
17
Capacitors in Series and in Parallel
Capacitors in parallel are combined in
the same manner as resistors in series.
Capacitors in series are combined in the
same manner as resistors in parallel.
1
C

T
1

1
C C
1
 ..... 
2
C  C C
1
C
T
1
2
 ....  C n
n
As an example - Solve for total charge and the voltage across each capacitor for the series circuit:
Q Q Q
T
1
2
 ....  Q
n
Q
CT  T ( Farads )
V
 1
1
1 
QT  CT *V  


 *60V
200

F
50

F
10

F


QT  480  C
18
QT
480  C

 2.4V
C1 2000  F
Q
480  C
V2  T 
 9.6V
C2
50  F
Q
480  C
V3  T 
 48V
C3 10  F
V1 
Example Problem 3
What is the total capacitance (CT) of the circuit shown?
Capacitors in parallel are
combined in the same
manner as resistors in
series.
C  C C
T
C
eq
 C 2  C 3  4 F  2  6 F
1
2
 ....  C n
Capacitors in series are
combined in the same manner
as resistors in parallel.
Now, as you would with resistors in
parallel, add Ceq to C1 (remember
these are capacitors):
1
1
1
 1
 1
1 
1 
 
CT  



C C 
3

F
6

F


eq 
 1
CT  2  F
C
19

T
1

1
C C
1
 ..... 
2
1
C
n
Example Problem 4
Assume that the circuit shown has been ‘on’ for a long time. Determine the
voltage across and the charge on the capacitors?
Remember: A capacitor looks like an open
circuit with voltage vC in steady state DC
To calculate the voltage, without
knowing the current, use VDR:
VC1  E
R1
2
 72V *
R1  R2
2  7
VC1  16V
Use KVL to solve for VC2:
VC2 = E – VC1 = 72V – 16 V = 56V
Now solve for the charges:
Q
( Farads )
V
Q1  C1 *VC1  2  F *16V  32 C
C
NOTE: R3 is effectively negated by the C2 (open
therefore no current flow thus no voltage across R3
20
Q2  C2 *VC 2  3 F *56V  168C
Power and Work
• The energy (or work) stored in a capacitor under
steady-state conditions is given by:
1
2
W  CV
2
21
Example Problem 5
Using the same circuit from Example 4, determine the energy stored at each
capacitor.
VC1  E
R1
2
 72V *
R1  R2
2  7
VC1  16V
VC2 = E – VC1 = 72V – 16 V = 56V
1
W  CV 2
2
1
1
W1  C1V12  (2  F ) *(16V ) 2  256  J
2
2
1
1
W2  C2V2 2  (3 F ) *(56V ) 2  4.7mJ
2
2
22
Inductor
• An inductor is a passive element designed to store
energy in its magnetic field.
− Inductors consist of a coil of wire, often wound around a
core of high magnetic permeability.
23
Faraday’s Experiment #4:
Self-induced voltage

• Voltage is induced across coil when i is changing.
• When i is steady state, the voltage across coil returns to zero.
24
Faraday’s Law
• From these observations, Faraday concluded:
A voltage is induced in a circuit whenever the flux
linking the circuit is changing and the magnitude of
the voltage (electro-magnetic force (emf) is
proportional to the rate of change of the flux
linkages over change in time.
d
emf 
dt
25
EMF in an Inductor
26
Counter EMF
• Induced voltage tries to counter changes in current so it is
called counter emf or back voltage.
• Back voltage ensures that any current changes in an inductor
are gradual.
• An inductor RESISTS the change of current in a circuit.
• Acts like a short circuit when current is constant.
27
Inductance:
Inductor Construction
• The level of inductance (L) is dependent on the area
within the coil, the length of the unit, to the number
of turns of wire in the coil and the permeability of the
core material.
28
Example Problem 6
For the inductor shown (the core is simply air), determine the
inductance:
  0 *  r
Convert diameter (d) to meters:
1
1m
1in
d  in 
*
 6.35mm
4
39.37in 4
Calculate area (A) in meters:
d
 (6.35mm)

 31.7  m
4
4
Calculate length (l) in meters:
A
2
l  1in  1in *(
2
1m
)  25.4mm
39.37in
 Wb 
where, μ for air  4 *107 

 A* m 
L
N 2A
(H)
l
(4 *107 ) * (1002 ) * (31.7 *10 6 )
L
=15.6H
3
25.4*10
29
Inductor Calculations
• Induced emf (e) in an inductor can be calculated:
di
eL
dt

(volts, V)
• L is the self-inductance of the coil. Units are Henry
(H).
• The inductance of a coil is one Henry if changing its
current at one Ampere per second induces
a potential difference of one Volt across the coil.
30
Example Problem 7
The current in a 0.4 H inductor is changing at the rate of 200
A/sec. What is the voltage across it?
di
eL
dt
(volts, V)
200 A
e  0.4( H )
(volts, V)
sec
e  80
(volts, V)
31
Inductance and Steady State DC
• In steady state DC, the rate of change of current is
zero, therefore the induced voltage across an inductor
is also zero.
• An inductor looks like a short circuit in steady state
DC.
32
Inductors in Series and in Parallel
• Inductors, like resistors and capacitors, can be placed
in series or in parallel.
• Increasing levels of inductance can be obtained by
placing inductors in series, while
• Decreasing levels of inductance can be obtained by
placing inductors in parallel.
33
Inductors in Series and in Parallel
Inductors in Series
L L L
T
1
2
Inductors in Parallel
1
 ...  L n
L

T
1

1
L L
1
 .... 
2
1
L
n
• Inductors in series are combined in the same way as
resistors in series.
• Inductors in parallel are combined in the same way as
resistors in parallel.
34
Example Problem 8
Determine I1 for steady state DC.
In a steady state, the inductor
effectively looks like a short.
Therefore, all current will flow
through the inductor and NOT
through R2.
V
I1 
R1
(A)
10V
I1 
=5A
2
35
Inductor Energy Storage
i
di
1 2
W   pdt  L  i dt L  idi  Li
0
0 dt
0
2
t
t
1
W 
L * i2
2
36
(J )
(J)
QUESTIONS?
37