Electronic – Transformer Design Using AP Method and Construction

If the transformer is ideal, then the problem will be the same as if you attached any \$50\Omega\$ source to a \$25\Omega\$ load. The transformer is not relevant. That is, a portion of any power arriving at the transformer from either direction will be reflected. This could be power generated by your source, or power reflected by your load.

Note that the power reflection, in itself, doesn't reduce your output power. The reflected power will bounce around until it finds a place to go, either radiated as EM energy, or converted to heat by resistive components. However, your voltage source might not like it very much, depending on what it is. The reflected power might result in voltages outside of its specifications, or if it operates on feedback, it may become unstable. The ideal voltage source in your schematic, however, won't care.

If the transformer isn't ideal, then you are introducing additional impedances into the circuit. The windings of the transformer have some capacitance. There is some magnetic flux not shared by the primary and secondary which appears as a series inductor. And of course, the windings have some resistance. The extent to which these are significant depend on many parameters, like the construction of the transformer and the operating frequency.

If you can't alter your source or load to match, then you could use a transformer with a turns ratio of:

$$ T = \sqrt{\frac{R_S}{R_L}} = \sqrt{\frac{50\Omega}{25\Omega}} = \sqrt{2} \approx 1.4 $$

Another possibility is an L-section of a capacitor and inductor to achieve a narrow-band match, just as you would if the transformer were not there.

From ABB's Transformer Handbook, 3e:

3.9 Short Circuit Impedance

Users have sometimes particular requirements regarding the short-circuit impedance. Such requirements may be determined by:

• parallel operation with existing units
• limitation of voltage drop
• limitation of short circuit currents

The transformer designer can meet the requirements in different ways:

• The size of the core cross-section. A large cross-section gives a low impedance, and vice versa.
• A tall transformer gives low impedance and vice versa.

For each transformer there is, however, a smaller range which gives the optimum transformer from an economic point of view, that is the lowest sum of the manufacturing costs and the capital value of the losses.

The 'short circuit impedance' mentioned above is the transformer's percent impedance. The above quote says that the transformer's impedance can be varied by changing the construction of the core.

Note that a transformers' impedance is mostly inductive "leakage reactance", i.e. magnetic impedance. Therefore, the difference between a 1% transformer and a 5% transformer is mostly to do with the design of the transformer's magnetic core. The 1% transformer would require much more iron core than the 5% transformer, and would be physically larger to match, which explains the higher cost.

From J&P Transformer Book, 12e:

In Chapter 1 it was explained that the leakage reactance of a transformer arises from the fact that all the flux produced by one winding does not link the other winding. As would be expected, then, the magnitude of this leakage flux is a function of the geometry and construction of the transformer....

Since reactance is a result of leakage flux, low reactance must be obtained by minimising leakage flux and doing this requires as large a core as possible. Conversely, if high reactance can be tolerated, a smaller core can be provided.

The conductor resistance (i.e. copper winding resistance) is typically small, 1/10th of the total impedance or less. The guideline given in AS3851 is that power transformers of less than 10 MVA may be considered to have X/R = 10.