Tutorial 10 B  Transformer Action 

Learning Objectives 
To learn how transformers behave when off load and on load. To learn how transformers are used to match a source with a load. 

Key Questions 
How do transformers behave off load? How do transformers behave on load? How do we match a source and a load using a transformer? 

Transformers off Load The transformer in the labpack in the picture in the previous tutorial has a resistance of 11.0 ohms (as measured with an ohmmeter). If this were connected across a 230 V dc supply, it would take a current of nearly 21 A and certainly blow the fuse in the plug. When connected to the mains with no load, the current is actually very small. Why is this?
Consider a transformer of N_{1} turns on the primary connected to a voltage source of V_{1} volts. There is a secondary of N_{2} turns which supplies a noload voltage of V_{2} volts. The two coils are linked by a flux of F Weber which is common to both. The flux is set up by a small current I_{0} amps in the primary, which is in phase with the flux. Lenz's Law applies so that V_{1} is opposed by and induced EMF of E_{1} volts and V2 is opposed by an induced EMF of E_{2} volts.
We know that the voltage leads the current by 90 degrees. We will assume that the inductor is perfect. The phasor diagram looks like this:
The induced EMF E1 is in antiphase to V_{1} (as is the EMF E2 with V_{2}). In a perfect transformer there would zero resultant voltage. There would be no copper losses either. The current I_{0} is used to magnetise the core i.e., to provide the flux.
In a real transformer there are iron losses. We will continue to ignore the copper losses. There is a magnetising component, I_{M}, which is in phase with the flux. Then there is a core loss component, I_{C}, which is due to eddy currents and hysteresis. So our phasor diagram looks like this:
The current lags the voltage by an angle of f_{0}.
The relationships between the currents I_{0}^{ }, I_{C}, and I_{M} are as follows. For the vector sum:
The magnetising current:
The core loss component:
We can work out the power factor at no load:
The iron losses are given by:
None of these are easy to measure. But we can use easily measured quantities to work these out. Let's do a worked example see how:
Loading a Transformer Consider a transformer that has negligible resistance in both the primary and secondary. The number of primary turns is the same as the number of secondary turns. It is a 1:1 transformer. You may be thinking what is the point of that, but they are used as isolation transformers, for example for a razor socket in the bathroom. We can say that there is negligible voltage drop:
V_{1} = E_{1} and: V_{2} = E_{2}
The secondary has a current, I_{2} amps flowing in it. The resulting secondary EMF will try to reduce the core flux, as it opposes the change (Lenz's Law). The result of this is that the primary EMF is reduced, which allows a bigger current to flow through the primary. The increased current leads to an increased magnetomotive force to balance out the change in the flux. This restores the flux to what was before.
The key point is that the flux remains constant.
Let's use a phasor diagram to show this rather complex idea:
The current I_{2} lags the voltage V_{2} by f_{2}. There is a reflected increase in current in the primary which we will call I_{1}' ("Ioneprime"). This also lags the voltage V_{1} by phase angle f_{2}. When there was no load, we saw that there was the current vector I_{0}, which was the resultant of the magnetising current I_{M} and the core loss current I_{C}. It lags the voltage V_{1} by f_{0}.
Therefore the primary current I_{1} is the vector sum of the currents I_{2} and I_{0}. It lags the voltage V_{1} by the phase angle f_{1}.
In this argument, the number of turns was equal for the primary and secondary. If they are not equal, we need to work out the values of V_{1}, V_{2}, I_{1}', and I_{2} using the transformer equations. We will use phasor diagrams to help us in the calculations. Let us look at a worked example:


Transformer MatchingTransformers are normally associated with stepping up and stepping down commercial mains electricity. However they also have an important use in impedance matching. In my notes on Electrical Principles, we saw that for maximum power transfer from a Thévenin source that the load matched the internal resistance. Click HERE if you want to revise it.
For an alternating current, reactance is important as a result of the inductance of coils of wire. The load may have a certain impedance.
For maximum power transfer in an AC circuit, the impedance of the load must match the internal impedance of the supply. We can achieve this with a transformer, and the turns ratio (N_{1} ÷ N_{2}) is given by:
With stereo radio receivers, it is essential to use a decent aerial. If there isn’t one, the radio won’t pick up anything, or at best it will be very hissy, which is not pleasant to listen to. There are two kinds of aerial available: · A balanced ribbon aerial; · An unbalanced antenna connected by a coaxial cable.
The input impedance for a balanced aerial is 300 ohms, while that for an unbalanced aerial is 75 ohms. If you connect a ribbon aerial to a 75 ohm socket, the signal will be weak. You need a matching transformer called a balun (from balancedunbalanced).
We can work out the turns ratio, knowing that the source resistance is 300 ohms and the load (input impedance) is 75 ohms:
The schematic is like this:
Proof We know that for an ideal transformer, the voltage is related by:
We also know that current is related by:
Use Ohm’s Law to get the effective resistance, R', in the primary coil:
If we have a load resistance, R¸ we know that it is given by:
So we can write:
In words, we can rewrite this as:
So we can make R' equal to the internal resistance of the source. Impedance matching in an audio systemThere are many incidences of where a source might not be properly matched with its load. Consider an audio amplifier that can drive a 2 ohm load to 50 W. We can work out the current easily enough.
Most loudspeakers are 8 ohms. So if we connect the supply directly to the loudspeaker, we get only 32 W. You can see that the possible power transfer is a lot less.
So we need to increase the voltage to the loudspeaker so that it transfers 50 W of power. We know that:
So rearranging: V^{2} = PR = 50 W × 8 W = 400 V^{2}
V = 20 V
So we need to increase the voltage from 10 V to 20 V using a 1:2 turns ratio. So if the primary has 1000 turns, the secondary has 2000 turns. We can show the consistency of the result using the equation:
In other words we can say that the primary has half the number of turns as the secondary.


