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I m back again on this document ! :D The document teaches me a lot of things and I think it will help me to better understand the losses across the switching components. I really appreciate if someone could tell me if my reasoning is wrong. I spent a lot of time on this document !

Here it is : https://pdfs.semanticscholar.org/c061/313303ee8c231b069be4f0d67d9e4cddcbf0.pdf (the document is really insteresting)

Here are my questions :

enter image description here enter image description here enter image description here

Here are my questions :

"Ipeak is the current value when the switch opens" It means that the peak current through Lp and Lleak is maximal at the switch opening. According to Lenz's Law and Faraday's Law, if the current is going to be lower than Ipeak after the switch opening, then the voltage across Lp and Ll will reverse and the inductances will begin to discharge their relative energy previously accumulated during Ton. Do you agree ?

Nevertheless, it seems if I correctly understand the document that Lp and Ll are not reversed just after opening the switch. So the current through Lp and Ll must grow up and this current is provided by Vin as the energy accumulated into the inductance is still growing up.

We now have a capacitor which is connected to Vin, So Vin = VLp + VLleak + VClump. (Please tell me If i do a mistake, My reasoning is probably false). The slope of the current through Lp and Lleak will reduce as the voltage Vlp and Vleak will reduce as the voltage across VClump will increase. It will increase rapidly as Ip (=Ileak) is already high depending on the value of Clump.

So far, the voltage across the secondary winding was oriented so that the diode was not be able to conduct as Lp did not reverse his voltage. (energy was being accumulated and the current Ip was still growing up)

enter image description here

"The drain voltage increases until the voltage across Lp reverse" . Then "As both Lp and Lleak are energized, Lleak forces a current into the lump capacitance which continues its charge." What I have to understand ? The drain voltage continues to grow up so the the voltage across Lp is not reversed. Nevertheless it seems to be reversed there is a mention as "Lp and Lleak are energized". Besides, Lp is reversed as it is written " The secondary diode becomes biased". So at this moment, the current through Ip and Ileak goes down and the energy accumulated into the inductance is decreasing. When does Lp reverse ?

It seems that the voltage across Lp reverse when the voltage across VClump is equal to Vin. Vin source cannot more provide charge to the capacitor as its voltage is equal to Vin. So no energy is provided by the voltage source Vin and the inductance Lp and Lleak have to reverse as there is no more energy from the voltage source. The formula is obtained thanks to the Kirchoff's voltage law.

Then as it is mentionned, Lleak provides charges to Clump thanks to the energy accumulated previously. So the drain voltage is still increasing whereas the voltage across Lp is reversed. The drain voltage stop increasing when its voltage is equal to Vin + Vclamp.

Between the moment where VClump is equal to Vin and the moment where Vclump is equal to Vin + Vclamp, the energy into the circuit is only provided by the Lp and Lleak inductance. So the current through the secondary is proportionnal to Ip-Ileak. (No ?) As the two inductors was energetically charged with a current equal to "Ipeak" (E = 1/2*(Lleak + Lp)*Ipeak²), the current through the secondary at t = 0 (when the voltage across the two inductors reverse) is null as Ileak(t=0) is equal to Ip(t=0). At t=0+, the current Ip is different of Ileak and the current begins to rise at the secondary (Is this wrong ?). After a certain time the energy of Lleak and Lp has charged the Clump capacitor to the voltage Vin+Vclamp. And then the slope of current is constant as the voltage across Lleak is constant as it is mentionned.

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Thank you very much :)

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1 Answer 1

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Nevertheless, it seems if I correctly understand the document that Lp and Ll are not reversed just after opening the switch. So the current through Lp and Ll must grow up and this current is provided by Vin as the energy accumulated into the inductance is still growing up.

Your 2nd sentence is incorrect. The current can't grow. The current through the inductors now flows into the capacitor and its voltage rises at an initial rate governed by: -

$$I = C\dfrac{di}{dt}$$

Where \$I\$ is that initial current flowing though the inductors and now flowing in the capacitor called \$C_{lump}\$.

Please tell me If i do a mistake

That's your first mistake as I see it so resolve that in your head and see if the rest of your confusion evaporates because I don't want to go through the rest in case your other comments/question are founded on this false premise.

I would also urge you to use a simulation tool in order to get a better picture of what is actually happening.

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  • \$\begingroup\$ Thank you for your comment. Can you explain me why the current can't grow ? If it can't, so the voltage reverse ? I will try do a simulation ! \$\endgroup\$
    – Jess
    Feb 19, 2020 at 16:31
  • \$\begingroup\$ The current can only grow if the circuit is receiving energy from a power source and it isn’t because the MOSFET (or switch) has disengaged. \$\endgroup\$
    – Andy aka
    Feb 19, 2020 at 16:49
  • \$\begingroup\$ This is not the case according to me there is a path when the capacitor is dischard then once the capacitor is charged there is no path. ? \$\endgroup\$
    – Jess
    Feb 19, 2020 at 20:31
  • \$\begingroup\$ If there is still energy in the inductor then the capacitor charges to a voltage higher than the supply voltage. That keeps happening until all the residual energy in the inductor is gone (either through normal flyback action transferring energy to the secondary or through leakage inductance energy becoming depleted). If there was no secondary load, the capacitor would eventually take all the original energy of the inductor then, the inductor starts taking this energy back and you get a series of damped sinewave oscillations. Get a simulator!! \$\endgroup\$
    – Andy aka
    Feb 19, 2020 at 20:53
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    \$\begingroup\$ Remember, this site is not a training school but a question and answer site. \$\endgroup\$
    – Andy aka
    Feb 19, 2020 at 20:56

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