Beautifully explained, professor. Just a point for the viewers: The simulation is more identical to an output stage of a buck converter, wherein the reverse recovery effect of a diode is shown on turn on of a low side mosfet with a diode considered across inductor.
Can you explain how to calculate the switching losses of MOSFET in a buck boost converter ?? What are the external parameters we have to consider to calculate accurate switching losses??
Hand calculations of switching losses is difficult due to the non linear behavior of capacitances and reverse recovery of the diode, not to mention parasitic inductances in the transistor and interconnect.
@@sambenyaakov There is quite a lot of material on the Internet on calculating the thermal losses of field-effect transistors, and practically none on bjt. Could you tell me where to find such a calculation method? I would like to learn how to calculate on paper,(having all the necessary transistor parameters) because it is not always possible to use simulators. Thanks anyway!
Great lecture as usual! Can you please explain how to find the efficiency of a converter (e.g. inverter) from these Es or Psw time-domain graphs? A general discussion on efficiency calculation will be really great (using simulation waveform, not from the formulas).
Thank you professor for this great video! Just a small question about the Id overshoot during the turn-on process. Is part of this current overshoot caused by the current charing the Coss of M2 (High-side MOSFET)? And as I know most of these commerial SPICE models did not have the reverse recovery modeled.
Yess about the Coss. Newer models have-include the reverse recovery. See also 140. Krihely, N. and Ben-Yaakov, S., Modeling and evaluation of diode reverse recovery in discrete-transition simulators, IEEE Energy Conversion Congress and Expo, ECCE-2010, 4514-4020, Atlanta, GA, 2010 www.ee.bgu.ac.il/~pel/pdf-files/conf141.pdf
Thank you sir for this video. I have a doubt related to the turn off losses. The energy 13uJ during turn off will not completely contribute to turn off losses right? As some part of it is stored in Cds and will not be dissipative.
@@sambenyaakov From 15:37 to 15:44, the highlighted turn off loss represents the actual turn off loss (dissipative)+ power drawn by Cds to charge from 0 to Vds (non dissipative)
Thank you so much professor Yaakov! As always outstanding quality.
Many thanks!
Beautifully explained, professor.
Just a point for the viewers:
The simulation is more identical to an output stage of a buck converter, wherein the reverse recovery effect of a diode is shown on turn on of a low side mosfet with a diode considered across inductor.
This is the case of a half bridge of an inverter when the phase current is coming in. There is life beyond the Buck converter😊
@@sambenyaakov got it, however you said it's a synchronous buck in the video.. and that took me a while to engulf..😅
Indeed, I should not have mentioned Buck. Thanks for bringing this up.
Sir, This is really wonderful. I have benefited a lot from your good illustration and perfect explanation. With kind regards
Thanks.
Can you explain how to calculate the switching losses of MOSFET in a buck boost converter ?? What are the external parameters we have to consider to calculate accurate switching losses??
Hand calculations of switching losses is difficult due to the non linear behavior of capacitances and reverse recovery of the diode, not to mention parasitic inductances in the transistor and interconnect.
Great visual explanation thanks
Thanks.
Prof. can you share this simulation model??
Sorry, no time look for it and it may have been changed.
Thank you so much. As always, a wonderful lecture. Please tell me, how can I calculate the power loss in a bjt transistor?
Thanks
is this in a linear mode or switching?
@@sambenyaakov switching, professor. Half bridge power supply.
@@SMV1972 he best is by simulation. Otherwise you need to know the CE voltage drop and rise and fall times.
@@sambenyaakov There is quite a lot of material on the Internet on calculating the thermal losses of field-effect transistors, and practically none on bjt. Could you tell me where to find such a calculation method? I would like to learn how to calculate on paper,(having all the necessary transistor parameters) because it is not always possible to use simulators.
Thanks anyway!
@@SMV1972 It is a bit a lost art as BJTs are not uses that much for switching anymore (IGBTs are still in use), will see.
Great lecture as usual! Can you please explain how to find the efficiency of a converter (e.g. inverter) from these Es or Psw time-domain graphs? A general discussion on efficiency calculation will be really great (using simulation waveform, not from the formulas).
I am not sure that I foolow you. Efficiency is Pout/Pin and Pin = Pout +Ploss
Thank you professor for this great video! Just a small question about the Id overshoot during the turn-on process. Is part of this current overshoot caused by the current charing the Coss of M2 (High-side MOSFET)? And as I know most of these commerial SPICE models did not have the reverse recovery modeled.
Yess about the Coss. Newer models have-include the reverse recovery. See also 140. Krihely, N. and Ben-Yaakov, S., Modeling and evaluation of diode reverse recovery in discrete-transition simulators, IEEE Energy Conversion Congress and Expo, ECCE-2010, 4514-4020, Atlanta, GA, 2010 www.ee.bgu.ac.il/~pel/pdf-files/conf141.pdf
@@sambenyaakov Thank you Sir! I already have this paper in my reference list for the topic of "diode reverse recovery".
@@nickname8708 👍
Excellent 👍
Thanks
Thank you sir for this video. I have a doubt related to the turn off losses. The energy 13uJ during turn off will not completely contribute to turn off losses right? As some part of it is stored in Cds and will not be dissipative.
Which minute in video are you referring to?
@@sambenyaakov From 15:37 to 15:44, the highlighted turn off loss represents the actual turn off loss (dissipative)+ power drawn by Cds to charge from 0 to Vds (non dissipative)
@@akshaylaturkar I believe Eoff stored in Coss should be deducted.
Great explanation, can you do one on the IGBTs?
Thanks. Perhaps.
Always good ! :)
Thanks