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I have not been able to find a 100% clear answer to this question, but here is what I have found:

From my understanding, the answer is it comes from the primary side inductor trying to resist the change in current when the MOSFET switch opens. The voltage at the drain pin of the FET (pin two in diagram) increases to the point where it is much higher than our input voltage, and current flows from higher potential to lower potential.

Is that correct?

enter image description here

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  • \$\begingroup\$ what do you mean by "reflected voltage"? \$\endgroup\$
    – FrancoVS
    Dec 18, 2021 at 8:12
  • \$\begingroup\$ Rev A tinyurl.com/y58vc54p Can you press the switch to goggle it and pump up the output voltage? Then stop the simulation and view the plots, restart and watch the current switch from primary To secondary \$\endgroup\$ Dec 18, 2021 at 9:08

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The flyback converter is an indirect energy-transfer dc-dc converter: first, during the on-time, it stores energy in the primary-side inductance and, because of the diode, the secondary is decoupled from the primary. In this configuration, the voltage across the primary is the input voltage \$V_{in}\$ while, across the secondary you have \$-NV_{in}\$.

enter image description here

Then, when the main switch opens, the energy stored in the magnetizing inductance transfers to the secondary and the diode now conducts. Across the secondary you now have the output voltage (neglecting the diode forward drop) and this voltage flies back or reflects to the primary via the transformer turns ratio \$1:N\$. The voltage across the switch is now \$V_{in}+\frac{V_{out}}{N}\$. In reality, this level is affected by parasitics such as the leakage inductance and other switching mechanisms. You will find more details in my APEC 2011 seminar, The Dark Side of the Flyback Converter.

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  • \$\begingroup\$ To be slightly clearer: "during the on-time, it stores energy in the primary-side inductance" --> during the on-time, energy is added via the primary winding and is stored in the transformer's magnetic field. The diode prevents the secondary from extracting energy during this interval. "when the main switch opens, the energy stored in the magnetizing inductance transfers to the secondary and the diode now conducts" --> "... the energy stored in the magnetic field is extracted by the secondary winding because the diode becomes forward biased" \$\endgroup\$
    – jp314
    Dec 18, 2021 at 18:21
  • \$\begingroup\$ @jp314, merci, your English is better than mine : ) More seriously, you're right but I did not want to enter too much into the details for this short explanation. \$\endgroup\$ Dec 18, 2021 at 18:29
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The reflected voltage (ideally) is just the output voltage plus a diode drop times the turns ratio Np/Ns, like you would expect in any transformer.

In reality there is some leakage inductance which will cause an additional spike, and there will be some ringing because of capacitance, and some kind of snubber is usually used to deal with that (relatively small) amount of energy per switch operation.

So the total maximum voltage across the switch is the input voltage + reflected voltage + spike voltage.

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Not really. When the switch gets turned off, the secondary side of the transformer has a voltage around Vd(on) + Vo. The transformer induces the secondary side voltage into primary side with this equation: Vpri = Vsecondary(N1/N2). That's how you get the reflected voltage. I hope I could help.

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The way it works in the circuit shown is as follows: As soon as Q 1 is closed, a current flows on the primary side into the turn drawn with the point, thereby building up a magnetic field in the core, which induces a voltage on the secondary side, also with the polarity Plus at the point. However, since no current can flow away because the diode 1 blocks, energy remains stored as a magnetic field in the coupled inductance. As soon as the switch Q1 opens, the magnetic field will collapse and the change in the magnetic field will induce a voltage. This time the positive pole will be on the one not marked with the point since it.

I hope that answers the question. It is always the case that a change in the magnetic field always induces a voltage on both sides of the coupled inductance.

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  • \$\begingroup\$ I think it kind of makes sense now. I guess it is confusing because the primary side is generating the magnetic field with Vin, but it also gets affected by the changing magnetic field. \$\endgroup\$ Dec 19, 2021 at 1:11
  • \$\begingroup\$ @RGBEngineer surely it makes sense. You should adjust your mindset as voltage is never related to a magnetic field. The relationship is always the flow of current to the magnetic field and voltage to the electric field. These are the terms that correspond to each other and that also makes understanding a lot easier. \$\endgroup\$
    – arnisz
    Dec 19, 2021 at 14:30
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There is no such thing as the magnetic field collapsing. The primary and secondary winding share the same core and therefore have the same flux. When the primary switch opens, the MMF (synonymous with flux) is maintained at value just before switch off by a current flow in the secondary such that Ippk.Np=Isec.Nsec.

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