Deciphering Coss of power MOSFETs

[@gregandark8571]

Thank you very much,awesome video as always !

[@sambenyaakov]

Thanks

[@topphemlig1191]

Absolutely the best explanation of Coss that I have ever come across - thank you!

[@sambenyaakov]

Thanks

[@chanh-tintruong3083]

Thank you so much for your explanation. I have seen for soft switching condition, some paper use energy-related 0.5*Li^2 > 0.5CU^2, in another use charge related Ql>Qc, in this presentation you have mentioned 2 types of capacitor energy-related and time-related. What condition we should use when we consider the ZVS condition for MOSFET. Thank you so much!

[@sambenyaakov]

The requirement is that the communication will be carried out by circuit itself before the switch is activated. The conditions can be formulated in term of energy or charge , the results should be the same (if calculated correctly😊)

[@chanh-tintruong3083]

@@sambenyaakov Thank you so much for your answer!

[@EPandDSP]

Congratulation!! how to calculate Co(er) when it is not specificated?

[@sambenyaakov]

Well, no way to calculate. But any decent datasheet will give Coo

[@catalin3407]

Sam, I am glad you brought up the Coss topic with the energy it dissipates in a hard switched configuration. However, I have found an IEEE article that states that the energy of the Coss should NOT be accounted into the switching losses of the MOSFET because we are already accounting it in the well known approximation Psw=1/2 * Vdd*Id*(ton+toff)*fsw. The logical reason is pretty simple: At turn OFF, we are overestimating the current through the channel of the transistor (hence overestimating the power loss at turn OFF), in other words, the current through the transistor is not "Id" but less than "Id" because part of the current flows into the Coss, charging it. Only part of "Id" flows through the channel of the transistor causing power loss. At turn ON, we are underestimating the current through the transistor (hence underestimating the power loss at turn ON) because the total current through it is "Id" + current from the discharge of the Coss. All in all, the above compensate each other resulting that the usual "Psw" formula does not need to be added to it the loss of the Coss (1/2 *Coss*Vdd^2*fsw). In summary, the article states this: Psw=1/2 * Vdd*Id*(ton+toff)*fsw + 1/2 *Coss*Vdd^2*fsw

[@k7iq]

I definitely liked learning more about the nonlinear behavior of the Coss/Cds and the spice table simulation. It will be interesting to hear Sam's take on that article, Catalin. Maybe it matters because the current going into the resistance of the MOSFET cannot be measured because it is strictly internal to the MOSFET package since a pure capacitance does not dissipate power or energy ?

[@sambenyaakov]

Hi Catalin, thanks for the discussion. Please note that the paper you cited is from 2009, at the time that Coss losses were naively estimated as Coss*V^2/2. To my understanding, this is not the only mistake in the paper. I would like to answer your question in detail but would prefer to do that in the LinkedIn IEEE Power Plectronics group so others can benefit. If Ok with you please log to www.linkedin.com/groups/2091456/2091456-6362073231909744640 (you may need to “be accepted” to the group if you are not already a member), and copy your comment to “The elusive Coss of power MOSFETs: How many output capacitances are there?” post, and I will answer there. Thanks.

[@k7iq]

Another power group. Thanks Sam ! I am surprised that I am not in that group already. I ran across this paper today relative but interestingly shows even something else about Coss... Coss hysteresis loss -the discovery... Phenomenon first reported APEC 2014 by Enphase training.ti.com/sites/default/files/docs/HVI_Coss_ZVS_15-Sept_complete.pdf I may post this link to that group when (if) I get accepted as well.

[@sambenyaakov]

You WILL be accepted. See you there

[@catalin3407]

I will join that group and post again the question there as suggested. In the mean time, here is an app note from OnSemi which shows clearly that the current through the channel is overestimated and underestimated at turn OFF and turn ON respectively, shown clearly in figure 3. www.onsemi.com/pub/Collateral/AN-8230.pdf.pdf

[@chiefrunningfist]

Great Vid professor! In a cascode configuration would the Coss be halved (series capacitors?) compared to a standalone transistor of the same make and model? Thanks

[@sambenyaakov]

Not really. In linear cascode there is a BW increase but in switching, the capacitors still need to be charged and discharged.

[@kokotmkokot4926]

Hello, professor! Thank you for your work! In the power mosfet driver lecture, we considered the gate current by the formula: Ceq=Cgs+Cgd(1+Vo/Vgs), I=Ceq*Vgs/Tr. But here we do not see the dependence of the input capacitance Ciss on voltage Vds. I'm confused.

[@sambenyaakov]

You are absolutely correct. This is a very crude approximation primarily to show what is going on and the Miller effect. One should use the Q as a function of Vgs curve (which is also VDS deoendent). Thanks for pointing this out. Might help others.

[@kokotmkokot4926]

​@@sambenyaakov Thank you so much for the answer! With your permission I will ask more. Correct expression for datasheet calculations: Cgd=Crss, Cgs=Ciss-Crss, Cds=Coss-Crss? Sorry, I am writing through a translator Google.

[@kaceandre7832]

I dont mean to be so off topic but does someone know a method to log back into an instagram account..? I was dumb lost my login password. I would appreciate any help you can give me.

[@davisnoah7472]

@Kace Andre Instablaster ;)

[@kaceandre7832]

@Davis Noah thanks so much for your reply. I got to the site on google and I'm in the hacking process atm. Seems to take quite some time so I will reply here later when my account password hopefully is recovered.

[@akaok7337]

Класс!)

[@catalin3407]

Hi again Sam. At 30:21 for the charge and at 30:27 for the energy, you are plotting them against voltage, which means you are doing a sweep. I do not see how the sweep can be able to compute energy or charge. I saw in your other video (around 21:36 of this kzbin.info/www/bejne/mHK8k3ujqryfptU) that you are plotting against time and I understand how... but in this video I can not see how are you able to plot it directly against voltage ! Could you help please ?

[@sambenyaakov]

Hi, This is by changing the variable of the X axis .

[@catalin3407]

Can you tell me what is the simulation setup ?

[@sambenyaakov]

The simulation is time domain (Transient) but then you click on the X axis of the probes plot and change to axis variable.

[@catalin3407]

Ohh, so it was that easy... THANK YOU VERY MUCH !!