# Ideal Transformer

Because we are dealing with balanced 3-phase motors we can achieve considerable simplification by developing single-phase models, it being understood that any calculations using the equivalent circuit (e.g. torque or power) will yield ‘per phase’ values which will be multiplied by three to give the total torque or power.

A quasi-circuit model of an iron-cored transformer is shown diagrammatically in Figure 1. This represents the most common application of the transformer, with the primary drawing power from an a.c. constant voltage source (V_{1}) and supplying it to a load impedance (Z_{2}) at a different voltage (V_{2}).

In the real transformer there would not be a big hole in the middle (as in Figure 1) because the primary and secondary windings would fill the space, each winding having the same total volume of copper.

The primary winding has N_{1} turns with total resistance R_{1}, and the secondary winding has N_{2} turns with total resistance R_{2}, and they share a common ‘iron’ magnetic circuit with no air-gap and therefore very low reluctance.

## Ideal Transformer – No Load Condition

We will begin by asking how the ideal transformer behaves when its primary winding is connected to the voltage source as shown in Figure 1, but the secondary is open circuited. This is known as the no-load condition. Since the secondary has no current it cannot influence matters, so we can temporarily remove it to yield the simpler model shown in Figure 2.

What we want to know is what determines the magnitude of the flux in the core, and how much current (known for obvious reasons as the magnetising current, I_{m}) will be drawn from the voltage source to set up the flux.

We apply Kirchoff’s voltage law to relate the applied voltage (V_{1}) to the voltage induced in the primary winding by the pulsating flux (E_{1}) and to the volt-drop across the primary resistance, yielding

V_{1} = E_{1} + I_{m}R_{1 }…….(equation 1)

We also apply Faraday’s law to obtain the induced e.m.f. in terms of the flux, i.e.

We have of course seen similar equations several times before.

We can simplify equation (1) further in the case of the ‘ideal’ transformer, which not unexpectedly is assumed to have windings made of wire with zero resistance, i.e. R_{1} = 0.

So we deduce that for an ideal transformer, the induced e.m.f. is equal to the applied voltage, i.e. E_{1} = V_{1}.

We really want is to find out what the flux is doing, so we must now use the result that E_{1} = V_{1 }to recast equation (2) in the form

Equation (3) shows that the rate of change of flux at any instant is determined by the applied voltage, so if we want to know how the flux behaves in time we must specify the nature of the applied voltage.

We will look at two cases: the first voltage waveform is good for illustrative purposes because it is easy to derive the flux waveform, while the second represents the commonplace situation, i.e. use on an a.c. supply.

Firstly, we suppose that the applied voltage is a square wave. The rate of change of flux is positive and constant while V_{1} is positive, so the flux increases linearly with time. Conversely, when the applied voltage is negative the flux ramps down, so the overall voltage and flux waves are as shown in Figure 3(a).

It should be evident from Figure 3 that the maximum flux in the core (ɸ_{m}) is determined not only by the magnitude of the applied voltage (which determines the slope of the flux/time plot), but also by the frequency (which determines for how long the positive slope continues).

Normally we want to utilise the full capacity of the magnetic circuit, so we must adjust the voltage and frequency together to keep ɸ_{m }at its rated value.

For example, in Figure 3(b) the voltage has been doubled, but so has the frequency, to keep ɸ_{m} the same as in Figure 3(a): if we had left the frequency the same, the flux would have tried to reach twice its rated value, and in the process the core would have become saturated.

Turning now to the everyday situation in which the transformer is connected to a sinusoidal voltage given by V_{1} = V sin ωt, we can integrate equation (3) to obtain an expression for the flux. The flux is then given by

Where ɸ_{m} = V/ωN_{1} = V/2πfN_{1} ……..(equation 5)

Typical primary voltage and flux waves are shown in Figure 4.

A special feature of a sine function is that its differential (gradient) is basically the same shape as the function itself, i.e. differentiating a sine yields a cosine, which is simply a sine shifted by 90^{o}; and similarly, as here, differentiating a (-)cosine wave of flux yields a sinewave of voltage.

Equation (5) shows that the amplitude of the flux wave is proportional to the applied voltage, and inversely proportional to the frequency. As mentioned above, we normally aim to keep the peak flux constant in order to fully utilise the magnetic circuit, and this means that changes to voltage or frequency must be done so that the ratio of voltage to frequency is maintained.

This is shown in Figure 4 where, to keep the peak flux (ɸ_{m}) in Figure 4(a) the same when the frequency is doubled to that in Figure 4(b), the voltage must also be doubled.

Although equation (5) was developed for an ideal transformer, it is also applicable with very little error to the real transformer, and is in fact a basic design equation.

For example, suppose we have a transformer core with a cross-sectional area 5 cm x 5 cm, and we decide we want to use it as a 240 V, 50 Hz mains transformer. How many turns will be required on the primary winding?

We can assume that, the flux density in the core will have to be limited to say 1.4 T to avoid saturation. Hence the peak flux in the core is given by

The peak voltage (V) is the r.m.s (240) multiplied by √2; the frequency (f) is 50, so we can substitute these together with ɸ_{m} in equation (5) to obtain the number of turns of the primary winding as

We cannot have a fraction of a turn, so we choose 309 turns for the primary winding. If we used fewer turns the flux would be too high, and if we used more, the core would be under-utilised.

The important message to take from this analysis is that under sinusoidal conditions at a fixed frequency, the flux in a given transformer is determined by the applied voltage.

Interestingly, however, the only assumption we have to make to arrive at this result is that the resistance of the windings is zero: the argument so far is independent of the magnetic circuit, so we must now see how the reluctance of the transformer core makes its presence felt.

We know that although the amplitude of the flux waveform is determined by the applied voltage, frequency and turns, there will need to be an MMF (i.e. a current in the primary winding) to drive the flux around the magnetic circuit: the magnetic Ohm’s law tells us that the MMF required to drive flux ɸ around a magnetic circuit that has reluctance R is given by MMF = Rɸ.

But in this section we are studying an ideal transformer, so we can assume that the magnetic circuit is made of infinitely permeable material, and therefore has zero reluctance. This means that no MMF is required, so the current drawn from the supply (the ‘magnetising current’, I_{m}) in the ideal transformer is zero.

To sum up, the flux in the ideal transformer is determined by the applied voltage, and the no-load current is zero.

This hypothetical situation is never achieved in practice, but real transformers (especially large ones) come close to it. Viewed from the supply, the ideal transformer at no-load looks like an open circuit, as it draws no current.

We will see later that a real transformer at no-load draws a small current, lagging the applied voltage by almost 90^{o}, and that from the supply viewpoint it therefore has a high inductive reactance, known for obvious reasons as the ‘magnetising reactance’. An ideal transformer is thus seen to have an infinite magnetising reactance.

## Ideal Transformer – No Load Condition

We now consider the secondary winding to be restored, but leave it disconnected from the load so that its current is zero, in which case it can clearly have no influence on the flux. Because the magnetic circuit is perfect, none of the flux set up by the primary winding leaks out, and all of it therefore links the secondary winding.

We can therefore apply Faraday’s law and make use of equation (3) to obtain the secondary induced e.m.f. as

There is no secondary current, so there is no volt-drop across R_{2} and therefore the secondary terminal voltage V_{2} is equal to the induced e.m.f. E_{2}. Hence the voltage ratio is given by

V_{1}/V_{2} = N_{1}/N_{2} ………(equation 7)

This equation shows that any desired secondary voltage can be obtained simply by choosing the number of turns on the secondary winding.

For example, if we wish to obtain a secondary voltage of 28 V in the mains transformer discussed in the previous section, the number of turns of the secondary winding is given by

N2 = V_{2}N_{1}/V_{1} = 28 x 309/240 = 36 turns:

It is worth mentioning that equation (7) applies regardless of the nature of the waveform, so if we apply a square wave voltage to the primary, the secondary voltage would also be square wave with amplitude scaled according to equation (7).

We will see later that when the transformer supplies a load the primary and secondary currents are inversely proportional to their respective voltages: so if the secondary voltage is lower than the primary there will be fewer secondary turns but the current will be higher and therefore the cross-sectional area of the wire used will be greater.

The net result is that the total volumes of copper in primary and secondary are virtually the same, as is to be expected since they both handle the same power.

## Ideal Transformer On Load

We now consider what happens when we connect the secondary winding to a load impedance Z_{2}. We have already seen that the flux is determined solely by the applied primary voltage, so when current flows to the load it can have no effect on the flux, and hence because the secondary winding resistance is zero, the secondary voltage remains as it was at no-load, given by equation (7).

The current drawn by the load will be given by I_{2} = V_{2}/Z_{2}, and the secondary winding will therefore produce an MMF of N_{2}I_{2} acting around the magnetic circuit.

If this MMF went unchecked it would tend to reduce the flux in the core, but, as we have seen, the flux is determined by the applied voltage, and cannot change.

We have also seen that because the core is made of ideal magnetic material it has no reluctance, and therefore the resultant MMF (due to both the primary winding and the secondary winding) is zero.

These two conditions are met by the primary winding drawing a current such that the sum of the primary MMF and the secondary MMF is zero, i.e.

N_{1}I_{1} + N_{2}I_{2} = 0, or I_{1}/I_{2} = -N_{2}/N_{1} ……..(equation 8)

In other words, as soon as the secondary current begins to flow, a primary current automatically springs up to neutralise the demagnetising effect of the secondary MMF.

The minus sign in equation (8) serves to remind us that primary and secondary MMFs act in opposition. It has no real meaning until we define what we mean by the positive direction of winding the turns around the core, so because we are not concerned with transformer manufacture we can afford to ignore it from now on.

The current ratio in equation (8) is seen to be the inverse of the voltage ratio in equation (7). We could have obtained the current ratio by a different approach if we had argued from a power basis, by saying that in an ideal transformer, the instantaneous input power and the instantaneous output power must be equal.

This would lead us to conclude that V_{1}I_{1} = V_{2}I_{2}, and hence from equation (7) that

V_{1}/V_{2} = N_{1}/N_{2} = I_{2}/I_{1} ……(equation 9)

To conclude our look at the ideal transformer, we should ask what the primary winding of an ideal transformer ‘looks like’ when the secondary is connected to a load impedance Z_{2}.

As far as the primary supply is concerned, the apparent impedance looking into the primary of the transformer is simply the ratio V_{1}/I_{1}, which can be expressed in secondary terms as

So, when we connect an impedance Z_{2} to the secondary, it appears from the primary side as if we have connected an impedance Z’_{2} across the primary terminals. This equivalence is summed-up diagrammatically in Figure 5.

The ideal transformer effectively ‘scales’ the voltages by the turns ratio and the currents by the inverse turns ratio, and from the point of view of the input terminals, the ideal transformer and its secondary load (inside the shaded area in Figure 5(a)) is indistinguishable from the circuit in Figure 5(b), in which the impedance Z’_{2} (known as the ‘referred’ impedance) is connected across the supply.

We should note that when we use referred impedances, the equivalent circuit of the ideal transformer simply reduces to a link between primary and referred secondary circuits: this point has been stressed by showing the primary and secondary terminals in Figure 5, even though there is nothing between them.

We will make use of the idea of referring impedance from secondary to primary when we model the imperfections of the real transformer, and then we will find that there are circuit elements between the input (V_{1}) and output (V’_{2}) terminals.

Finally, we should note that although both windings of an ideal transformer have infinite inductance, there is not even a vestige of inductance in the equivalent circuit. This remarkable result is due to the perfect magnetic coupling between the windings.

As we will see shortly, the real transformer can come very close to the ideal, but for reasons that will also become apparent, ultimate perfection is not usually what we seek.