# How can a constant secondary side current cause a primary current to flow in a transformer?

This is a question which was originally asked by my power electronics professor but left unanswered.

Suppose you are given the following system (kindly borrowed from this question)

Suppose that:

1. The inductance L is so big that the load current is practically constant (remove Vx and think of the load as an ideal current source).
2. The diodes are ideal.
3. The voltage source is a secondary side of an ideal transformer (no leakage reactance no resistance and no magnetizing current).

As you can see, the secondary current is a square wave. How can a square wave (i.e. a constant current on a half period basis) cause a primary current in the transformer to flow?

We can see the secondary current as a Fourier series, that is as an infinite sum of sinewaves. This means that, at least on the transformer part of the circuit which is linear, we can apply the superposition principle and therefore obtain a lot of AC circuits to be solved and then combined. It's the fundamental frequency and the harmonics which are causing primary current to flow. If it were pure DC it wouldn't work.

In reality of course, the transformer has a leakage reactance which will smooth out the edges of the square wave, providing a (perhaps more evident) time varying current on both its sides.

• Would you maybe expect to rectify secondary Vi into coil then Idc will continue to rise with primary current as a staircase result or if open a staircase voltage with zero crossing BEMF from current rectifiers ? ( with ideal no Bsat or BDV on insulation) Mar 30 '18 at 14:39
• I'm not sure I understand your comment. Could you clarify? Mar 30 '18 at 20:28 Here a rectifed +-100Vp sine 50Hz drives a full bridge to 1H 1:10 step-up
See the coil current rise with DC steps ( shaped by sine ). This is what I meant by staircase rise in current. Meanwhile AC current increases as well. Then diode drop increases slightly. 