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I am currently reading a document about leakage inductance in a flyback converter from ON SEMICONDUCTOR . Here it is : https://pdfs.semanticscholar.org/c061/313303ee8c231b069be4f0d67d9e4cddcbf0.pdf (the document is really insteresting)

Here are the schematics : enter image description here

And here is what I do not understand (in yellow): enter image description here enter image description here enter image description here

I do not understand why the current through the diode is null for a certain time. When the switch opens, due to lenz's law the voltage accross Lp reverse and as the leakage inductance was energized, the current across Lp and Ll is the same. So if the current through Lp is not null, as the current to the secondary is proportionnal to the current through Lp, it is not null... So there is something like the voltage across Lp and Ll are opposed in order to iLl(t) = -iLp(t), but why the voltage across Lp and Ll would be opposed ? It would mean that the leakage inductance by "discharging its energy" is still charging the energy of the primary inductance ... Why it would be ? and why it would not be the inverse ? I M LOST ...

Thank you very much,

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    \$\begingroup\$ ill just wait for Verbal Kint :) \$\endgroup\$ – JonRB Feb 15 at 9:36
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    \$\begingroup\$ Your misunderstanding is with; "So if the current through Lp is not null, as the current to the secondary is proportionnal to the current through Lp, it is not null", the current to the secondary is not proportional to the current through \$\ L_p\$, the current to the secondary is \$\ \frac{I_p-I_{leak}}{N}\$, where \$\ I_p\$ is the current through \$\ L_p\$, \$\ I_{leak}\$ is the current through \$\ l_{leak}\$, and \$\ N\$ is the number of windings. if you read the circuit diagram it is obvious that if \$\ I_p=I_{leak}\$ then no current flows in the secondary. \$\endgroup\$ – Vinzent Feb 15 at 10:38
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    \$\begingroup\$ Ask the author directly @Verbal Kint \$\endgroup\$ – G36 Feb 15 at 12:10
  • \$\begingroup\$ Thank you for your comment Vinzent. I understood :) \$\endgroup\$ – Jess Feb 15 at 15:33
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So if the current through Lp is not null, as the current to the secondary is proportionnal to the current through Lp, it is not null

The current in the secondary is NOT proportional to the current in the primary; the voltage on the secondary is proportional to the voltage on the primary. Of course any secondary current modifies this but, until the primary voltage has risen above the bus voltage (Vin), even an ideal diode in the secondary won’t conduct.

I do not understand why the current through the diode is null for a certain time.

Due to the parasitic capacitance lumped at the drain, there has to be some time before the primary voltage rises to be bigger than Vin.

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    \$\begingroup\$ Are we done with this question and answer now or do you need clarification? \$\endgroup\$ – Andy aka Feb 16 at 20:13
  • \$\begingroup\$ Done with this question :D But I have a new one about this document :D electronics.stackexchange.com/questions/481511/… \$\endgroup\$ – Jess Feb 17 at 7:53

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