Confused about the core flux waveform (and by extension the magnetising current waveform) of a current-fed push-pull DC-DC converter

Question

I was going through the above converter and cannot understand how the flux in the transformer core changes (and by extension the magnetizing current).

According to the text I'm reading, there are three states for the current-fed push-pull topology:

• The first state is when SW1 is on and SW2 is off.
• The second state is when SW1 is off and SW2 is on.
• The third state is when both the switches are on.

So, just to have a place to start, let's assume I turn on SW1 first. Some current must be drawn by the transformer to establish the flux in the core and a constant positive voltage has to appear across P1 due to this flux increasing.

Here comes the part I do not know how to approach. Now, as the flux in the core keeps increasing, I turn on SW2. The book (which did not even consider the flux aspect of the transformer) says that in this state, no voltage is induced across the winding, which must imply that the flux in the core stays the same as it was just before I turned on SW2.

How can I explain why the flux in the core stays the same even after turning on SW2?

Please excuse me if this question sounds silly. I'm new to these topologies.

Answer 1

How can I explain why the flux in the core stays the same even after turning on SW2?

• SW1 turns on and current through P1 ramps up at say "X" amps per microsecond
• Some time later SW2 turns on and the current through P2 also ramps up at "X" amps per microsecond.
• But, both P1 current and P2 current are in opposite directions due to the dot notation
• This means that from the point when P2 turns on, the core flux remains constant

Answer 2

How can I explain why the flux in the core stays the same even after turning on SW2?

Turning on SW2 bring P1- and P2+ to the same voltage. There is thus zero voltage across the windings on the core. Therefore the magnetising current and hence flux does not change.

This also brings VLx- to the Vs- voltage, so the current through VLx now rises rapidly.