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I am having a slight confusion in the no load situation of the flyback transformer. Consider the high pulse at the gate of the mosfet turning on the switch. At this case the current will increase slowly, which induces an emf on the primary and the secondary side. But no current flows in the secondary due to the diode.

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Now considering the off time of the mosfet with no load. The inductor on the secondary(or the secondary coil) will oppose this change in the current and will induce a very high voltage V=Ldi/dt. Since there is no load therefore no current will be flowing and hence the inductor will continue to give a very high voltage so that some current pass. As Vout/n=Vin, a large voltage will be induced on the primary as well therefore a high voltage of Vin+Vo/n will be developed on the switch. This will for sure burst out the mosfet.

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Secondly the capacitors at the output also won't be able to handle such a voltage. and should be damaged altogether. I know that in SMPS we have the feedback that prevents this behavior, but will the mosfets and capacitors burst without a feedback or load??

Finally, one thing that surprised me was this tutorial where such high voltage didn't caused any issue in the mosfet. Kindly help me in this confusion as I am new into flyback transformers and kindly correct me in case of incorrect arguments.

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I know that in SMPS we have the feedback that prevents this behavior, but will the mosfets and capacitors burst without a feedback or load??

The capacitors will sap up the energy and prevent the output rising near-instantaneously but, if this keeps progressing then yes, they will "burst".

Finally, one thing that surprised me was this tutorial where such high voltage didn't caused any issue in the mosfet. Kindly help me in this confusion as I am new into flyback transformers and kindly correct me in case of incorrect arguments.

Your question should be self-contained such that if the link to the video gets deleted, your question still contains the necessary information to pose the question.

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  • \$\begingroup\$ Thanks for the comment. Yes the capacitor should burst out in case the voltage limits out of its capacity. This means that capacitor will absorb the excess current and will help the voltage on the capacitor to increase until the voltage gets constant. \$\endgroup\$
    – kam1212
    Commented Oct 25, 2023 at 13:26
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    \$\begingroup\$ But remember, without a load, that output voltage will keep rising @kam1212 \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2023 at 15:09
  • \$\begingroup\$ Right. Means that the current flow in the secondary will remain almost same and hence in order to make that flow through the capacitor, a much higher voltage will be developed. \$\endgroup\$
    – kam1212
    Commented Oct 25, 2023 at 19:03
  • \$\begingroup\$ Energy keeps being dumped into the output capacitor and this only ends two ways; you stop this happening (with a control loop) or you expect things to smoke pretty rapidly usually. \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2023 at 19:10

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