The capacitor is a very common electrical component. It is used to store electrical energy. The term “capacitance” means, the ability to store energy in the form of an electrical charge.
The capacitive effect is of great benefit in electrical / electronic circuitry. For example, controlling AC current, tuning radio receivers, time-delay circuits, separating of AC currents from DC currents, power factor correction, fluorescent lamps and starting of single phase motors.
A capacitor consists of two conducting surfaces or plates, placed very close together but separated by an insulator called a dielectric. See Figure 1. The circuit schematic symbols for capacitors are also shown.
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Figure 1
The unit of Capacitance is the Farad ( F ) and may be defined as:
One Farad is the amount of capacitance that will store a charge of one Coulomb when an EMF of one Volt is applied.
Hence:
Charge = Capacitance x Voltage
Q = Capacitance ( Farads ) x Voltage ( Volts )
Q = C x U
Q = The stored charge in the capacitor and is expressed in Coulombs. Earlier we learned that Coulombs are equal to current ( amperes ) multiplied by
time ( seconds ) or Q = I x t.
C = Capacitance is measured in Farads. It must be remembered that the unit of 1 Farad represents a very large charge. Actual capacitor values will be in Microfarads, Nanofarads or Picofarads where:
1 1
One Microfarad = ———— or — or 10-6 Farads
1,000,000 106
1 1
One Nanofarad = —————— or — or 10-9 Farads
1,000,000,000 109
1 1
One Picofarad = ——————— or — or 10-12 Farads
1,000,000,000,000 1012
If a capacitor were marked with a value of 1000 pF, it could equally have been marked with a value of 1 nF. Similarly, a capacitor of 0.001 µF could have been marked 1 NF. Therefore it can be seen that there are one thousand Picofarads in a Nanofarad, and also that there are 1 thousand Nanofarads in a Microfarad.
It is common practice to represent the prefix “micro” by the Greek letter µ. For example, 10 microfarads may be written 10µF. The value of capacitance is usually clearly marked on the body of the capacitor.
Refer to Figure 2. When the switch is closed, electrons on the upper plate A are attracted to the positive pole of the battery. This leaves a shortage of electrons on plate A, which, is therefore positively charged. At the same time, electrons gather on the lower plate, B, causing it to become negatively charged. Since plates A and B are now charged with opposite polarity, there is a difference of potential between them. When this difference of potential is equal to the battery voltage, no more charge can be placed on the capacitor. Notice that the capacitor voltage has an opposing polarity to that of the battery. When the capacitor cannot be charged any further, we consider it to be, fully charged.
If the switch is now opened the capacitor will remain charged, because there is no path for the excess electrons on plate B to flow to plate A.
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Figure 2
Refer to Figure 3. If a wire link is placed across the plates of a charged capacitor, electrons will flow from B to A. This action discharges the capacitor and returns it to the uncharged state.
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Figure 3
Calculate the charge on a 10µF capacitor when it is connected across a 200V DC supply.
C = 10µF = 10 x 10-6 Farads
U = 200 Volts
Q = C x U
Q = 10 x 10-6 x 200
Q = 0.002 Coulombs
A steady current of 10 Amps flows into a previously discharged capacitor for 20 seconds, when the potential difference between the plates is 600 Volts. What is the capacitance of the capacitor?
I = 10 Amps
t = 20 Seconds
Q = I x t
Q = 10 x 20
Q = 200 Coulombs
Q = C x U
To get C on its own we transform the formula:
Q
C = —
U
200
C = ——
600
C = 0.33 Farads
From the previous exercises it can be seen that the factors affecting the value of the charge on a capacitor depends upon the capacitance and the voltage:
Q = C x U
The greater the capacitance of a capacitor, the greater the charge for the same applied voltage. If 10 Volts is applied to a capacitor, it will charge to 10 Volts, after which no more charging occurs. The charge remains on the capacitor with or without the applied voltage connected.
When the voltage across the capacitor equals the supply voltage, no further current will flow. The capacitor is now fully charged and will remain charged even if disconnected from the supply. See Figure 4.
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Figure 4
The greater the capacitance of a capacitor the greater the charge for the same applied voltage. The factors affecting capacitance are:
If the plate area of a capacitor is increased there is a corresponding increase in capacitance, provided there is no change in the distance between the plates or in the dielectric material. See Figure 5.
Capacitance is directly proportional to plate area;
C µ a |
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Figure 5
When two capacitors are placed in parallel, the plate area is increased and so the capacitance is increased. See Figure 6.
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Figure 6
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(2) Plate Spacing
The capacitance of a capacitor changes when the distance between the plates changes. It increases when the plates are brought closer together and decreases when they are moved further apart.
Refer to Figure 7. Plates ( a ) have more capacitance than plates ( b ).
Capacitance is inversely proportional to the distance between the plates;
1 |
Where d = distance between plates
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Figure 7
Refer to Figure 8. When two capacitors are connected in series, the distance between the plates has increased so the capacitance has decreased.
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Figure 8
Using the same plates fixed a certain distance apart, the capacitance will change if different insulating materials are used for the dielectric. The effect of different materials is compared to that of air - that is, if the capacitor has a given capacitance when air is used as the dielectric, other materials used instead of air will multiply the capacitance by a certain amount called the “dielectric constant”.
Changing the Dielectric Material changes the capacitance. See Figure 9.
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Figure 9
For example, some types of oiled paper have a dielectric constant of 3; and if such oiled or waxed paper is placed between the plates, the capacitance will be 3 times greater than it would be if the dielectric was air.
Different materials have different dielectric constants and so will alter the capacitance when they are placed between the plates to act as the dielectric. Listed below are the Dielectric Constants for typical materials
Air 1.0
Quartz 3.4 to 4.2
Glass 5.1 to 8.0
Mica 7.0 to 8.0
When a capacitor is fully charged and immediately disconnected from the supply, the capacitor will remain charged.
If the capacitor is now shorted out by a piece of conductor the energy stored in the capacitor will be dissipated in the form of a spark / crack of the discharging current.
The energy stored in the capacitor is measured in joules ( symbol W). The larger the capacitance value, the greater the energy stored by the capacitor, for a given voltage.
Capacitors can be divided into two types, polarised, and non-polarised.
Polarised types include the standard aluminium electrolytic and tantalum electrolytic capacitors. They are widely used in power supplies. Both types have positive and negative terminals and must be correctly connected in order to maintain the dielectric action.
See Figure 10.
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Figure 10
Non-polarised capacitors such as the polypropylene, polycarbonate, polyester, polystyrene, mica and ceramic types can be connected either way round. They all have extremely good dielectric properties. See Figure 11.
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Figure 11
General-purpose capacitors use wax or oil impregnated paper as the dielectric. Two long rectangular aluminium foils, separated by two slightly larger strips of the impregnated paper, are rolled up. They are then inserted into an insulated cylinder and sealed at the ends. A lead is brought out from each plate to enable the device to be connected to a circuit. Refer to Figure 12.
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Figure 12
Capacitors using mica dielectric have a capacitance range from a few pF to 0.02µF. These are usually precision capacitors, with high working voltages and excellent long-term stability. Ceramic and mylar type capacitors each exhibit certain advantages in particular circuit applications. Different capacitor types usually derive their names from the types of dielectric used. See Figure 13.
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Using normal construction, capacitors above 2µF become very bulky and cumbersome. The electrolytic capacitor has a large capacitance within a package, which is much smaller than if normal construction were used.
The dielectric of electrolytic capacitors consists of a thin film of oxide formed by electrochemical action directly on a metal foil plate. The other plate consists of a paste electrolyte.
See Figure 14.
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Figure 14
The large capacitance is a result of the oxide dielectric layer being extremely thin and the effective plate area being much increased by etching. An electrolytic capacitor is a polarised component, which means it must be connected into a circuit according to the plus and minus markings on its case. If it is connected wrong the capacitor is usually destroyed and may explode. They range in values from 1µF to 10,000µF.
Ceramic disk capacitors for general applications usually have a tolerance of ± 20%.
Paper capacitors usually have a tolerance of ± 10%.
For closer tolerances, Mica and Ceramic tubular capacitors are used. These have tolerance values of ± 2 to 20%.
Silver plated Mica capacitors are available with a tolerance of ± 1%.
The tolerances may be less on the minus side to make sure there is enough capacitance, particularly with electrolytic capacitors, which have a wide tolerance. For instance, a 20µF electrolytic with a tolerance of -10%, + 50% may have a capacitance of 18 to 30 µF. However, the exact capacitance value is not critical in most applications of capacitors.
When two or more capacitors are connected in parallel the plate area is increased and so the capacitance is increased. See Figure 15.
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Figure 15
Therefore the total capacitance ( CT ) is the sum of the individual capacitances in a parallel.
CT = C1 + C2 + . . . . . CN |
When the group is connected to a supply U, the capacitors will each store a charge, and we will refer to these as Q1 and Q2 respectively. The total stored charge QT will be the sum of the individual charges:
QT = Q1 + Q2
As U is the same in a parallel circuit:
CT = C1 + C2
Two capacitors of capacitance 2µF and 5µF are connected in parallel to a 20V DC supply.
Calculate:
(a) The equivalent capacitance of the group
(b) The total charge
(c) The charge on each capacitor.
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(a) CT = C1 + C2
CT = 2 + 5
CT = 7µF
(b) The total charge:
QT = CT x U
QT = 7 x 10-6 x 20
QT = 140 x 10-6 Coulombs
The charge on each capacitor is found by the formula: Q = C x U
As the capacitors are in parallel the voltage across them is the same.
Charge on 2µF:
Q1 = 2 x 10-6 x 20 = 40 x 10-6 Coulombs
Charge on 5µF:
Q2 = 5 x 10-6 x 20 = 100 x 10-6 Coulombs
Check: QT = Q1 + Q2
140 x 10-6 = 40 x 10-6 + 100 x 10-6
Five capacitors of capacitance 20µF, 100µF, 50µF, 300µF and 40µF respectively are connected in parallel to a 1000V supply.
Calculate:
(a) The equivalent capacitance of the group
(b) The total charge
(c) The charge on each capacitor.
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(a) CT = C1 + C2 + C3 + C4 + C5
CT = 20 + 100 + 50 + 300 + 40 = 510µF
QT = CT x U
QT = 510 x 10-6 x 1000 = 0.51 Coulombs
(c) Q1 = C1 x U
Q1 = 20 x 10-6 x 1000 = 0.02 Coulombs
Q2 = C2 x U
Q2 = 100 x 10-6 x 1000 = 0.1 Coulombs
Q3 = C3 x U
Q3 = 50 x 10-6 x 1000 = 0.05 Coulombs
Q4 = C4 x U
Q4 = 300 x 10-6 x 1000 = 0.3 Coulombs
Q5 = C5 x U
Q5 = 40 x 10-6 x 1000 = 0.04 Coulombs
QT = Q1 + Q2 + Q3 + Q4 + Q5
QT = 0.02 + 0.1 + 0.05 + 0.3 + 0.04
QT = 0.51 Coulombs
Consider the effect of connecting two similar capacitors in series. The plate area remains the same, but the thickness of the dielectric increases. See Figure 16.
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Figure 16
1
Capacitance µ —————
distance ( d )
1
Distance µ —————
Capacitance
If all the distances between the plates of the capacitors are combined, there would be in effect, one capacitor of distance dT ( d1 + d2 = dT ).
1 1 1 1 |
Note:
The total capacitance in a series circuit is calculated in the same manner as the total resistance in a parallel circuit.
The total capacitance in a parallel circuit is calculated in the same manner as the total resistance in a series circuit.
Just as current is the same in all resistors in a series circuit; charge is the same in all capacitors in a series circuit. This is because the same charging current must flow in all parts of a series circuit for the same time.
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QT = Q1 = Q2 = Q3 ( Coulombs )
Calculate the total capacitance of three capacitors of values 10µF, 30µF and 60µF connected in series.
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1 1 1 1
— = — + — + —
CT C1 C2 C3
1 1 1 1
— = — + — + —
CT 10 30 60
1 6 + 2 + 1 9
— = ——————— = —
CT 60 60
60
CT = — = 6.66µF ( to 2 decimal places )
9
This example makes it clear that the total capacitance of a string of series connected capacitors is less than that of the smallest individual capacitor. This also applies to resistors in parallel.
The total charge can now be calculated for the previous example when connected to a 200 Volt supply.
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QT = CT x U ( Coulombs )
QT = 6.66 x 10-6 x 200
QT = 1333 x 10-6 Coulombs
Since the capacitors are connected in series, the charge on each is the same as the total charge, i.e. 1333 x 10-6 Coulombs.
The supply voltage is U and the volt drops across the individual capacitors C1, C2 and C3 are U1, U2 and U3 respectively, since the capacitors are all in series:
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UT = U1 + U2 + U3
Q
Q = C x U U1 = —
C1
Q 1333 x 10-6 1333
U1 = — = ———— = —— = 133.30 Volts
C1 10 x 10-6 10
Q 1333 x 10-6 1333
U2 = — = ———— = —— = 44.43 volts
C2 30 x 10-6 30
Q 1333 x 10-6 1333
U3 = — = ———— = —— = 22.21 volts
C3 60 x 10-6 60
Note:
The sum of the three individual volt drops, across the three capacitors equals the supply voltage. The larger volt drop is across the smaller value capacitor and the smaller volt drop is across the larger value capacitor.
Three capacitors of values 6µF, 8µF and 16µF respectively are connected in series to a 100 V DC supply. Calculate:
(1) The total capacitance of the circuit
(2) The total stored charge
(3) The volt drop across each capacitor.
1 1 1 1
(1) — = — + — + —
CT C1 C2 C3
1 1 1 1
— = — + — + —
CT 6 8 16
1 8 + 6 + 3 17
— = —————— = —
CT 48 48
48
CT = — = 2.82µF
17
(2) QT = CT x U
QT = 2.82 x 10-6 x 100
QT = 282 x 10-6 Coulombs
Q 282 x 10-6
(3) U1 = — = ———— = 47.05 Volts
C1 6 x 10-6
Q 282 x 10-6
U2 = — = ———— = 35.35 Volts
C2 8 x 10-6
Q 282 x 10-6
U3 = — = ———— = 17.63 Volts
C3 16 x 10-6
Check:
UT = U1 + U2 + U3
100 = 47.05 + 35.35 + 17.63
100 = 100
Manufacturers specify the safe working voltage on the body of capacitors and this value should not be exceeded. See Figure 17.
If this safe working voltage across a capacitor is exceeded, it is possible that the dielectric may break down causing a short circuit in the capacitor.
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Figure 17
When two equal values of capacitors are connected in series, the working voltage is the sum of the two working voltages of the capacitors. For example, two equal value capacitors intended for 130 Volts maximum supply could be connected in series and placed across a supply of up to 260 Volts.
This is not done in practical circuits. It would be prudent to use two 260 Volt capacitors.
The following is the method used to find the total capacitance of the circuit shown in Figure 18.
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Figure 18
First find the total capacitance of the parallel branch ( CP ):
CP = C2 + C3
CP = 3 + 6 = 9µF
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This 9µF capacitor CP is in series with the 12µF capacitor.
To find the total capacitance of the circuit:
1 1 1
— = — + —
CT C1 CP
1 1 1
— = — + —
CT 12 9
1 3 + 4 7
— = ———— = —
CT 36 36
36
CT = — = 5.14 µF
7
The total charge of the circuit:
QT = CT x U
QT = 5.14 x 10-6 x 200 = 1028 x 10-6 Coulombs
Now we can find the volt drop across the 12µF capacitor:
QT 1028 x 10-6 1028
U1 = — = ————— = —— = 85.66 Volts
C1 12 x 10-6 12
The volt drop across both capacitors in the parallel circuit will be the same ( U2 ):
QT 1028 x 10-6 1028
U2 = — = ———— = —— = 114.2 Volts
CP 9 x 10-6 9
As the 12µF capacitor is in series in the circuit, the total current flows through it and therefore the charge stored on it will equal the total charge QT.
Q1 = U1 x C1
Q1 = 85.66 x 12 x 10-6 = 1028 x 10-6 Coulombs
The stored charge on 3µF capacitor:
Q2 = U2 x C2
Q2 = 114.2 x 3 x 10-6 = 343 x 10-6 Coulombs
The stored charge on 6µF capacitor:
Q3 = U2 x C3
Q3 = 114.2 x 6 x 10-6 = 685 x 10-6 Coulombs
The charge stored by the two capacitors in parallel is equal to the total charge.
QT = Q2 + Q3
QT = 343 x 10-6 + 685 x 10-6
1028 x 10-6 = 1028 x 10-6
Source: http://local.ecollege.ie/Content/APPRENTICE/liu/electrical_notes/LL218.doc
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