Thanks for comment. Nice to read.

Thanks. This is for the top notch designers 😊

Thanks😊

Thank you! 😃

🙏👍

Hi Michael, welcome to power r electronics and thanks for comment and encouragement. These keeps me going.

Thanks.

Thanks for taking the time to comment.

Thanks

Thanks Fariham

👍🙏

Thanks

​@@sambenyaakov was researching this concept in the context of a synchronous buck converter which is hard switching ~20 amps into ~1 mH loads at 54 V. Until your video (and a few others from nexperia), I didn't realize the impacts of the Qrr of the lowside mosfet during a half-bridge turnoff. The key takeaways for our application I hope aren't misinterpreted are: 1. Selecting a MOSFET with a lower Qrr will reduce ringing and EMI -all other things equal 2. Qrr implies some current 'shoot through' when switching on the highside MOSFET. This current is in addition (and proportional to?) the switched load current 3. An RC snubber on the switch node will reduce negative voltage and Vds stresses (ie reduce avalanche concerns??) caused by Qrr of the lowside fet

@@kylehagen1476 Use a Schottky diode if you

Thank you.

@@sambenyaakov Sir do you have Fast Recovery Diode structural analysis in your channel even with the comparison of GPP (Glass Passivated) structure. I couldnt find it. Do you have a plan to upload a FRED Diode Analysis? I would appreciate to watch it.

@@SefaOralz Good subject will try

Thanks

Thanks for comment

Thanks for comment.

At turn ON the transistor current is limited by the leakage inductance, which is the slope in the plot, and the transistor is already Rds(on)

See kzfaq.info/get/bejne/rZqTq6-B0Zi7o5s.html

@@sambenyaakov Thanks, I checked that video. And I still stick to my point that Q1 and Q4 gets ZVS during turn on however turn off is having hard switching for primary

@@vbidawat93 This is not hard switching because there is a minimum overlap between voltage and current . I call this pseudo ZVS. Hard switching is when the transistor turns on against a conducting diode , e.g. a buck converter with diode when the high side is turned on. But then, we are talking about notation, you don't have to stick to the convention, call it as you wish😊

Thanks

Just had a second look at your question and realized that my answer needs amendment. The MOSFET is used to turn on and off the correct. Indeed, in this case there are-additional losses due to the series diode but it could be a Schottky diode of low voltage. This solution is being used in high voltage applications in which case the efficiency penalty is not that high.

Deadtime is generaly not related to the reverse recovery. An example for deadtime need is in a syncronous Buck convertr after the high side transistor is turned off. This is followed by the mid point voltagd swing until the diode is catching. This is the required deadtime until the lower transistor is turned on.

Thanks

Thanks Hamza

This is the resonant current when the switching frequency is above resonance.

See references 42 and 45 in www.ee.bgu.ac.il/~pel/seminars/seminar9.pdf

So many thanks.

Good question. I am planning to prepare a detailed answer in aviseo to be posted here. In short, the losses are really not in the diode but the diode is causing it - so it is considered as "diode loss"

@@sambenyaakov ok! the question arises because that should mean a lot for the thermal design... if transistor and diode chips are separated, such as in IGBTs and in the fast diode example you show, it means these "diode losses" do not in fact contribute to heating of the diode

@@Dahmac Indeed. BTW there is some extra heating of the diode due to the high peak current but most of the energy is absorbed by others.

😊

A function of forward current.

@@sambenyaakov and reverse bias voltage right ?

Thanks. I got it translated by friend not Google

Sorry, I do not follow.

Power loss due to diode is the forward current in conduction mode and the reverse recovery loss as shown in slide.

@@sambenyaakov but in the boost converter the diode reverse recovery will affect the switching power loss of mosfet right?

Thanks for comment. Welcome to join: www.linkedin.com/groups/13606756/

Seem to be a real problem. I din't have of hand any good solution to offer.

PSIM is not the right tool. You can calculate lossed by Qrr*f*V

@@sambenyaakov thanku sir

And I think that you are mistaken, unless they have copied this video😊

Yes, but how would you recycle the energy stored in stray inductance?

Not so. In section 2 they calculate the DIODE extra loss while I am calculation the TOTAL system loss. The diode extra loss is relatively small even when when compared to its loss in the forward conduction.