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In a push pull converter during turn on a switch magnetizing current and load reflected current build up. So magnetizing inductance store energy. During both switch off state secondary coil does not carry current. But what about the built up magnetizing current? Current in the inductor can not instantly goes to zero. But when both switch off waveform shows primary current is zero. So my question is where does the charged magnetizing current discharges?

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According to this picture, input current/primary current, Iin comprises of secondary current iL and magnetizing current im during sw1 on and sw2 off. What happens to this current when both switch is off? If the current is zero during both switch off, becuase there is no free wheeling path, then how can primary winding of transfromer allow such abrupt current change to zero without creating harm to switch or destroying something?

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    \$\begingroup\$ What waveform are you talking about? Please show a picture of the waveform. Magnetization current will need to be quenched. \$\endgroup\$ – Andy aka Mar 2 at 12:40
  • \$\begingroup\$ Andy aka please check the post, i have added picture \$\endgroup\$ – Zarzisur Mar 2 at 14:18
  • \$\begingroup\$ Run a quick simulation with a loaded push-pull and you'll see that the magnetizing current pauses (it circulates in the secondary but remains flat) when both switches are off. \$\endgroup\$ – Verbal Kint Mar 2 at 14:38
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In a push-pull converter, when both switches are off, there is a net 0-V across the transformer primary side because both diodes are conducting for the output inductor demagnetization phase. The magnetizing current keeps circulating in the secondary side but because there is no voltage across the primary winding, the magnetizing current pauses. This is the flat portion missing in your drawing. The below simulation shows the waveforms:

enter image description here

You can find a comprehensive description of the push-pull converter in the Google preview of this book.

Edit: to avoid confusion, I've added below a SIMPLIS simulation using a simple pulse width modulator. Please note the presence of the output inductor which is necessary in all buck-derived topologies:

enter image description here

As shown in the below sketch, the converter operates in a heavy continuous conduction mode (CCM). As expected, the inductor current equally splits between the two output diodes as long as their \$V_f\$s are similar. Here, these are perfect diodes and the current during their conduction time is half of the inductor current (5 A):

enter image description here

In reality, some imbalance can occur considering variations between diodes.

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  • \$\begingroup\$ You say that when both switches are off, there is a net 0-V across the transformer primary side because both diodes are conducting for the output inductor demagnetization phase - how can both be conducting? That would mean that the secondary has equal positive voltages on each outer wire with respect to the secondary centre-tap and that doesn't appear to make sense. \$\endgroup\$ – Andy aka Mar 4 at 13:02
  • \$\begingroup\$ Please have a look at Fig. 4.49 of the book whose Google preview link is given in my answer. If the magnetizing current is flat, there is no voltage across the mag. inductance. This is what the simulation shows. \$\endgroup\$ – Verbal Kint Mar 4 at 13:06
  • \$\begingroup\$ Your linked document says just below Fig 4.49 This current flows through the diode D1 in the reverse direction and then it says This is possible because one half of the choke current flows through D1 in the forward direction and that doesn't appear to make any sense. \$\endgroup\$ – Andy aka Mar 4 at 13:23
  • \$\begingroup\$ To extinguish the magnetization flux when both primary transistors are off there has to be a current path in the secondary or you get spark-over or diode breakdown but, both diodes can't be forward biased because that implies that both the dotted secondary terminal and the un-dotted secondary terminal are both producing a positive voltage. That's the problem I'm having with your explanation. I can see that if that is possible then the primary voltage would be zero but then that implies there can be no secondary equivalent of the primary magnetization current. \$\endgroup\$ – Andy aka Mar 4 at 13:43
  • \$\begingroup\$ The magnetizing current is not extinguished, it just circulates and keeps constant. \$\endgroup\$ – Verbal Kint Mar 4 at 13:45
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In a push pull converter during turn on a switch magnetizing current and load reflected current build up.

Correct

So magnetizing inductance store energy.

Correct

During both switch off state secondary coil does not carry current.

Incorrect. The residual energy in the primary is in the form of magnetism and that can equally be used in the secondary as it would be in a flyback converter so, one of the secondary half windings will be forward biased and push that energy to the load: -

enter image description here

Either (a) current will flow down the red path or (b) current will flow down the blue path and the flux will become reset.

But when both switch off waveform shows primary current is zero. So my question is where does the charged magnetizing current discharges?

I think the above explains that.


Additional information

I've added a simulation to show what I mean: -

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The circuit is this: -

enter image description here

And this substantiates what I was saying; D1 and D2 do not conduct the secondary flyback current at the same time. If M1 was pulsed on and off then D1 will conduct the flyback current to the load. If M2 is pulsed on and off then D2 conducts the flyback current to the load. And, during that flyback period (just after the pulse ends) the primary voltage doesn't fall to zero volts because it cannot - it stays at a voltage that is determined by the rate of change of the relevant diode (D1 or D2) current.

This is for a push pull converter with a coupling factor of 0.95 and 1:1:1:1 inductors.

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  • \$\begingroup\$ Thanks a lot. I got it.. \$\endgroup\$ – Zarzisur Mar 2 at 14:52

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