Electronic – Is RF Transformer Impedance Significantly Affected by Wire Size

You should possibly consider looking at the transformer in two ways; one without a load and one with a load on the secondary.

Without a load on the secondary, the transformer is just an inductor and if you have components (such as L1 and R1) in series with the primary, the voltage developed on the primary will not be the full AC amount from your generator. It's a simple case of calculating the impedances and volt-drops. This is with the secondary unconnected remember.

The primary has inductance like any other coil but, for a transformer to more effective, it is desirable for the primary's self inductance to be high in power applications. If you looked at how much current flowed into the primary (secondary open circuit) you would find that the current was small compared to when driving a load on the secondary and it may have an inductance of several henries.

With 10 henries inductance, at 50Hz the impedance is 3142 ohms and from 230VAC would take a current of 73mA - that current through R1 (10 ohm) hardly drops any voltage.

It's a different matter when there is a load on the secondary. If the turns ratio is 1:1 and you have 100 ohms on the secondary, it is reasonable to argue that the impedance presented to the primary circuit is also 100 ohms. This assumes power out is close to power in. In fact the impedance relationship between primary and secondary is related to turns ratio squared. For instance if it is a 10:1 step-down transformer with a load of 100 ohms, the equivalent impedance at the primary is 10k ohms i.e. 10 x 10 x 100.

In summary, for a power transformer, you'd like the primary inductance to be infinite but that is impractical so you live with something that doesn't take too much current when the secondary is open circuit. The off-load current that flows is real current taken from the AC power and if everyone had low-impedance transformers the electricity companies would be supplying a load of current that doesn't get them revenue. This is a slight exaggeration but not far off the truth. On industrial sites power factor correction is used to minimize this effect but that's a whole new story!

And if your transformer primary was 100 ohm impedance you'd be seeing something less than half your AC voltage applied. If R1 was zero then you'd see exactly half.

As regards saturation I've shown the equivalent circuit of a transformer below. Note that saturation is caused by the current flowing through the magnetizing inductor which is nothing to do with load current: -

Here is a good document from Elliott Sound Products and please note what it says about maximum flux density therefore saturation:

Why doesn't the core saturate more under load conditions? Imagine two coils sharing the same magnetic core. Ignore magnetization currents and losses. The primary is 100 turns and the secondary is 10 turns. If the secondary load current is 10A, the primary current must be 1A and the ampere-turns is therefore the same on both coils. Are these ampere-turns additive or subtractive? They are subtractive and this can easily be seen with dot notation....

If current is flowing into the dot on the primary, current is flowing out of the dot on the secondary and this produces opposing fluxes in the magnetic material. When you think about this you have to be consistent and use the right-hand rule to see that the two fluxes oppose and cancel.

Because the dots are at the top on both coils, they are wound in the same direction and the currents are flowing in (primary) and out (secondary) therefore due to the RH rule the fluxes (due to ampere-turns) are cancelled.

To maximize power into the resistors the inductive reactance needs to be cancelled by the capacitive reactance. This is because the circuit is series-tuned (despite the presence of the transformer).

You know the frequency (1kHz) and you know the inductance (1uH) and the natural resonant frequency for an LC circuit (series or parallel) is: -

\$F_N = \dfrac{1}{2\pi\sqrt{LC}}\$

So then you transpose this to get the required capacitor value that would tune the circuit.

But your cap value is 1uF and this needs to be transformed to the value you have just worked out using the resonance formula. Does it need to increase from 1uF or decrease?

I suspect it needs to increase and so you have to decide by how much. Let's say it needs to increase ten thousand times to (say) 10,000\$\mu F\$, what should the turns ratio be to obtain this impedance transformation?

The hint in your question tells you that it is the square of the turns ratio that governs how this works so, if the required impedance transformation is 10,000 then the actual turns ratio is the square root of this.

I'll leave you to decide whether the turns ratio is 10,000:1 from left to right or 1:10,000. By the way 10,000 isn't the right answer - you need to calculate this.