Bredhurst Receiving and Transmitting Society

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.

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 a 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.

The Dielectric

The property of the dielectric is called permittivity or the dielectric constant.

It is the dielectric material which determines :-

• the losses (as heat) which increase as the frequency gets higher(see moe below)

• The variation in the value of the capacitor with a rise in temperature or due to aging

• the maximum voltage the capacitor may be used at as known as the working voltage

• the voltage at which the capacitor would be destroyed also know as the breakdown voltage (see more in the next section).

The formula C = K x A / d  relates the Capacitance (C) to the Dielectric constant (K) to the overlapped area of the plates (A in sq inches) and the distance between the places (d in inches) and gives the answer in farads.

Capacitors are rated for the maximum continuous voltage that should be applied. Note that the voltage applied will differ as to whether it is DC of AC.

The effect of applying too much PD is to destroy the capacitor and that might occur in spectacular fashion with a small explosion !!!

What actually happens is that the charge on the plates flashes over or makes a hole in the plates. Depending upon the capacity of the capacitor will determine how devastating the destruction will be !!!

3e.3 Recall that different dielectrics are used for different purposes, e.g. air, ceramic, mica and polyester; and that with some dielectrics, losses increase with increasing frequency.

In addition to ceramic, mica, polyester, and air spaced as the dielectric in capacitors the electrolytic capacitors have a dielectric of a semi liquid conducting compound between their plates which can be of aluminium or tantalum foil. This dielectric is often impregnated paper as a very thin insulating layer.

Losses increase with increasing frequency

As has been mentioned before the maximum frequency that a capacitor can operate at is determined by the dielectric material. As a general rule losses in energy increase with increasing frequency.

3e.4 Understand the charging and discharging of a capacitor in a CR circuit and the meaning of the time constant T=CR.

Capacitance

Charge and discharge of Capacitors in CR circuits

CR circuits mean a capacitor and a resistor linked in series.

In the section we are going to used 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 on a capacitor

Consider the circuit arrangement shown below

and the following equations:-

Equation A

The charge Q on a capacitor is equal to the voltage (V) in VOLTS across the capacitor X its capacitance (C) in FARADS

Q = CV

Equation B

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

Time Constant

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 63% of the applied voltage V_S

T is called the Time Constant, and equally applies to discharge of a charged capacitor via a resistor where discharge by 63% of V_S occurs in Time T = CR

Example:

How long will it take to charge a capacitor of one microfarad to 63% of the applied voltage, via a resistor of 2 megohms ?

T = CR  

Using the exponential notation

T = 1 x 10-⁶ x 2 x 10⁶

the 10-⁶ and 10⁶ 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 de-bounce 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 10yF r = 100k

so

T = 10 X 10^-6 X 100 X 10³

T = 10⁶ 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 volts 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 lethal voltage in 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 so 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:-

When capacitors are linked in parallel the total value can be calculated from the formula:-