






Syllabus Sections: Capacitance 3e.1 Understand the factors influencing the capacitance of a capacitor; area and separation of the plates, permittivity of dielectrics and formula C = K A /d. Note the wording of the syllabus: Understand the factors influencing the capacitance... It does not say Understand and apply the formula so no calculations using the formula shown below are required. In the Intermediate Licence course you learned that a capacitor consists of two metal plates separated by an insulating material. In its simplest form is two metal plates set parallel to each other with a separation gap of insulating material. This material is called a dielectric.
The capacitance is the word used to describe the ability of the two plates to store an electrical charge. The capacitance is proportional to the area of the plates (marked A in the diagram) and inversely proportional to the distance between the plates (marked d in the diagram) and also on the properties of the dielectric between the plates. The dielectric can be as simple as air but other materials used are paper, rubber, polythene, mica and ceramics. Thus the bigger the area the bigger the charge the capacitor can hold but the bigger the gap the smaller the charge that can be held. So the ideal high value capacitor will have large plates and a very small gap between them.

Dielectric material 
Use 
Breakdown voltage 
Stability 
Frequency 
Air 
Mainly used for variable tuning capacitors capacitance values up to say 1000pf 
High 
Good 
up to say 300MHz 
Electrolytic semi liquid compound 
Used for very high value capacitors in Power supplies for smoothing 
High 
Good 
up to 1MHz 
Paper 
Used in high voltage capacitors 
High 
Good 
up to 20MHz 
Mica  This is the best dielectric but is expensive 
Good 
Very Good 
up to and higher than 200MHz 
Polyester 
The usual type of capacitor when critical values are not required such as audio coupling and decoupling. Tolerance typically 20% 
Medium 
Fair 
up to about 100 kHz 
Ceramics 
High dielectric constant and has good temperature stability 
Medium 
Good 
up to 150  200MHz 
3e.4 Understand the charging and discharging of a capacitor in a CR circuit and the meaning of the time constant T=CR.
Charge and discharge of Capacitors in CR circuits
CR circuits mean a Capacitor and a Resistor linked in series.
In this section we are going to use the maths notation that if two items need to be multiplied together then they are written together without the times (X) symbol. Thus V = I x R would be written as V = IR
Charge Q (measured in Coulombs) on a capacitor
Consider the circuit arrangement shown below
and the following equations:
The charge Q on a capacitor is equal to the Potential Difference (V) in VOLTS across the capacitor TIMES its capacitance (C) in FARADS
Q
= CV
To put some figure on this a capacitor of 1F will store 1C of charge for an applied p.d. of 1V
thus
1C = 1F x 1V
The charge in Coulombs, is equal to the current in amps X the time that the current has flowed in SECONDS
Q = Amps x Seconds
Now looking at the diagram above before S1 is closed, V_{R} = zero, V_{C} = zero
Then at the instant S1 is closed, no current has flowed into the capacitor, therefore there is no charge Q and V_{C} = zero.
V_{R} = 12V and by Ohm's law current (I) starts to flow and I = V_{S} / R Amps at that instant.
A short time after S1 is closed a current is flowing, therefore C is charging, V_{R} must be falling because V_{C} + V_{R} = 12V
VR is now V_{S } V_{C}.
If V_{R} is less then I_{R} is less and the rate of charge of the capacitor falls, until eventually the capacitor is charged to 12V and no current flows then V_{R} = zero.
The graph of V_{C} with respect to time is shown and is known as an EXPONENTIAL CURVE
You will need to have a basic mathematical appreciation of the foregoing paragraphs in order to be able to understand existing circuits involving C and R, and to be able to select values to give correct function in your own circuits.
With Q = Amps x Seconds and Q = CV then
IT = CV thus T = CV/I, from ohm's law R = V/I so T = CR
So the equation is simply :
T (in seconds) = C (in farads) x R ( in ohms)
where T is the time taken for a capacitor to charge to 66% of the applied voltage V_{S}


T is called the Time Constant, and equally applies to the charge of a capacitor and the discharge of a charged capacitor via a resistor where similarly discharge by 66% of V_{S} occurs in Time T = CR
Example:
How long will it take to charge a capacitor of one microfarad to 66% of the applied voltage, via a resistor of 2 megohms ?
T = CR
Using the exponential notation
T = 1 x 10^{6} x 2 x 10^{6}
the 10^{6} and 10^{6} cancel each other out and the answer is
1 x 2 = 2 Seconds
T = CR applies to timer circuits like the well known NE555 timer, PSU smoothing circuits, decoupling circuits, delay and debounce circuits.
Let's work through another example as it may help you to understand it better.
Whilst under steady state conditions the perfect capacitor presents an infinite resistance to DC voltages however when the capacitor is in a timing circuit the capacitor offers no resistance, it allows current to flow as follows:
At the instant S1 is closed the current will be 1milliamp (by Ohm's Law I= V/R = 100 /100,000 = 1/1000  1 mA
C starts to charge and builds up a voltage.
When the charge on C reaches 50VC, there is 50V left across R and thus the current I is now 50/100,000 = 1/2000 = 0.5mA and thus C is charging slower.
The slow changing voltage curve is called an EXPONENTIAL curve.
At a time T = C (in FARADS ) X R (in OHMS) C = 10 micro F R = 100k
so
T = 10 X 10^{6} X 100 X 10^{3}
T = 10^{1} x 10^{6 }X 10^{2} X 10^{3}
T = 10^{6} X 10^{6}
T = 1 SECOND, the voltage on C will be ABOUT 2/3 of the initial voltage of 100 V = 66 Volts.
Recall the dangers of stored charges on large or high voltage capacitors.
Large value and high voltage capacitors have some dangers to be considered when carrying out repairs or constructing equipment.
Low voltage and large value capacitor  high current flow if shorted
The ability to store a great deal of electrical energy is just what you required when building say a low voltage power supply but that same energy even at say 18 volts could destroy a small screw driver if dropped across the terminals or even if it were not destroyed it could heat up to such an extent that it caused a fire.
High voltage and small or large value capacitor  danger of electric shock
When we come to high voltage those are considered to be anything above 50V. In a linear amplifier power supply the voltage is likely to be of the order of 2000V. If energy of this voltage level was stored in a capacitor when the equipment was turned off, a lethal voltage capable of giving an electric shock would exist without even taking into account the amount of current that could be available!
Recall that large value resistors can be used to provide leakage paths for these stored charges.
So what should we do to make these large value or high voltage capacitors safe ?
Leakage paths resistor
They are fitted with a large value resistor say 1k ohms and this will provide a "leakage" path to take away the charge in the form of heat when the current flows through the resistor when the equipment if turned off. It may take several minutes to discharge down to a safe level so do not be in a rush to delve inside equipment.
In the Foundation Licence course you learned that a resistor restricts the flow of current. You also learned in the Intermediate Licence course that a capacitor can store charges and that these charges can be released into the circuit when needed.
So what happens when the circuit that was using this charge is turned off. The capacitor remains charged to the voltage it reached when it was being used in circuit. If the stored voltage was high say 240V then that would be the same level of voltage as the mains supply and thus equally dangerous.
Here is where the humble resistors comes into play. If a large value resistor say 100K then by ohm's law
240 / 100,000 = 0.0024 or
a current of only 2.4mA would flow and thus over time the voltage would be leaked away to a safe level in the form of heat in the resistor. Such a small level of current flowing whilst the capacitor is "in circuit" would have no effect on the operation of the circuit as the capacitor might be supplying several amps for short intervals.
3e.5 Recall and apply the formulae for calculating the combined values of capacitors in series and in parallel
When capacitors are linked in series the total value can be calculated from the formula:
The special case is when there are two capacitors linked in series then the simplified formula can be used :
When capacitors are linked in parallel the total value can be calculated from the formula: