Syllabus Sections:-

Transformers

3j.1 Understand the concept of mutual inductance.

If two coils with their axes aligned on the same line then the current passing though one of the coils creates a magnetic field around itself and also around the second coil which has the mutual effect of INDUCING an EMF in this second coil.

However the current passing in the first coil only induces an EMF in the second coil when the field strength of the magnetic field is changing and this is from and AC source only.

Thus the MUTUAL EFFECT of the magnetic field passing around both coils results in MUTUAL INDUCTANCE and causes the EMF in the second coil.

When all the magnetic field, or magnetic flux as it is sometimes called set up by the fist coil cuts all the turns of wire in the second coil the mutual inductance reaches its maximum possible amount.

The amount of coupling is related to the distance between the coils. If the coils are some distance apart then the coils are said to be loosely coupled. The maximum coupling is achieved when the two coils are wound one on top of the other. The minimum coupling is when they are placed far apart or at right angles to each other.

Understand and apply the formula relating transformer primary and secondary turns to primary and secondary potential differences and currents.

The formulae are :- 

Here we are dealing with AC as we are considering transformers!

Transformers and potential differences

In the formulae :-

Vp = Volts in the primary coil (or the driven coil)

Vs = Volts in the secondary coil

Np = Number of turns in the primary coil

Ns = Number of turns in the secondary coil

Ip = current passing in the primary coil

Is = Current passing in the secondary coil

Re-arranging the formula to it is a little easier to see that the ratio of the input voltage output voltage it equal to

the ratio of the input turns divided by the output turns

This is a simple comparison of input and output.

From the above it is possible to calculate the transformation of voltage from the primary to the secondary such as you would need to do if building a power supply.

the part Ns / Np is called the turns ratio and was introduced to you in the Intermediate Licence course click to check back.

Example: A transformer has 200 turns on the primary and 100 on the secondary and an input voltage of 50 volts. What is the calculated output voltage ?

applying the figures to the formula Vs = 50 x 100/200

answer 25 volts.

Transformers and currents

Re-arranging the formula to it is a little easier to see that the ratio of the input current output current it equal to

the ratio of the output turns divided by the input turns.

This is not quite so straight forward but is based on the Power equation P = V x I.

Thus what is put in as power one side (Vin x In), assuming no losses will come out as power the other (Vout x Iout). This leads to the equation

The equation says that the current in the primary the current in the secondary is dependent upon ratio of secondary turns primary turns.

Putting some figures to this if there is 240 volts on the primary of 100 turns and 24 volts at 10 amps on the secondary - there would be less current at 240 Volts in the primary.

With this in mind let's work it out.

From the equation then 24 / 240 = Ns / 100

From the equation then Ns = 24 x 100 /240 answer = 10 turns Ns

so Ip / 10 = 10 / 100

thus Ip = 10 x 10 / 100 answer 1 amp

If you were given the current that can be supplied from the primary you can find out what current can be taken from the secondary such as you would need to do if building a power supply if you know the turns ratio.

So now using this formula

Example: A transformer can allow a secondary coil current of a maximum of 20 amps and it has 400 turns on the primary and 100 on the secondary what is the required input current capability ?

Applying the figures to the formula Ip = 20 x 100 / 400

answer 5 amps.


3j.2 Understand and apply the formula relating transformer primary and secondary turns to primary and secondary impedances

The formula is :- 

Matching input and output impedances

In a perfect situation -one without reactance losses or leakages the above equation would hold true where:-

  • Zp is the impedance looking into the the primary turns on the coil from the power source

  • Zs is the impedance of the load connected to the secondary

  • Np/Ns is the turns ratio Np primary and Ns secondary.

The primary terminal impedance is determined by the equation being dependent only upon the load resistance and the turns ratio.

Where would this be used?

Say a AF amplifier required a load of 200ohms to operate properly and you are driving an 8ohm speaker then you could, assuming a perfect world, design an impedance transformer to ensure the criteria was met.

by dividing both side by Zs we get

by taking the square root of both sides we get

= = 5

So the primary turns must be 5 times as many as the secondary.

This impedance matching means altering the "load" impedance by the transformer to the required output level.


3j.3 Understand the cause and effects of eddy currents and the need for laminations (or ferrites) in transformers.

Eddy Currents

When an AC supply passes through a coil, a magnetic field is generated. The size or density of the magnetic field can be increased by the use of an iron core such as is used in transformers. Iron being a conductor, the magnetic field causes an electric current to flow in the core. The current flow is called "EDDY CURRENTS".

Effect of the eddy currents

The effect of the eddy currents is a loss of power, as flowing through the resistance of the core they create heat. As you are aware from the Foundation Course if there is not a complete circuit a current could not flow.

Use of Laminations

So what can be done to stop the eddy currents in the iron core of the coil?

The solution is to make the core of thin strips of insulated iron material called "laminations" and for the lamination not to form a continuous loop around the coil. So a break in the lamination is formed and this small air gap is the necessary break in the circuit and the eddy currents cannot now flow.


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