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Recently I am working on power loss calculation, there are two components I need to calculate, the MOSFET and Diode, and I have some questions I would like to ask you.

  1. What's the most accurate calculation in MOSFET switching loss and driving loss? For switching loss, I think \$\frac12 VI\$ is too simple.
  2. How to calculate the diode reverse recovery loss?

can someone offer me some paper or material?

Marcus Müller
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EEC
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    "most accurate": You *really* don't want to describe your MOSFET on a quantum level, but that's the most accurate model that science has these days. So, you'll have to be slightly more specific for which purpose your accuracy has to suffice. **Why** do you think your current model is too easy? Also, where does that factor of 1/2 come from? – Marcus Müller Jun 02 '21 at 15:27
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    1) You don't need (nor want) "most accurate", what you need is "accurate enough". 2) explain **why** you think 1/2 VI is too simple. 3) thick books have been written about switching losses in switching converters, did you read any of those books? – Bimpelrekkie Jun 02 '21 at 15:31
  • @MarcusMüller Because the switching loss is related to the Coss, and the Coss value is not a constant value, why can use simple equation to calculate the switching loss? – EEC Jun 02 '21 at 15:56
  • @Bimpelrekkie Yes, I read some books, and most of the book use 1/2*V*I, but I think this calculation is not very accurate, what do you think? – EEC Jun 02 '21 at 15:56
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    *but I think this calculation is not very accurate* I asked already **why** you think the method is inaccurate. I think the 1/2 VI method is **accurate enough** for general purposes. Feel free to prove that I am wrong and more accuracy is needed. – Bimpelrekkie Jun 02 '21 at 16:25
  • @Bimpelrekkie the 1/2 VI is based on the area of triangular, but this just approximation so I think this method is not accurate. – EEC Jun 02 '21 at 17:48
  • @Bimpelrekkie why do you think the 1/2VI is accurate enough? Do you compare with real circuit power loss? – EEC Jun 02 '21 at 17:51
  • *why do you think the 1/2VI is accurate enough?* Like you wrote: "most of the book use 1/2*V*I". Also **you** made the claim that that's not accurate enough so it is also up to **you** to backup that claim with facts/reasons why that is so. For calculating the losses, I would just follow the books (use 1/2 VI method) to get an estimate and then **simultate the circuit** to get a more accurate number. Then the **real implementation on a PCB** will show me the real numbers. Then again, efficiency varies with load, voltages and temperature so why would I need an **accurate** value? – Bimpelrekkie Jun 02 '21 at 19:07
  • @Bimpelrekkie Thanks for you answer, I think I can do some simulation to prove it – EEC Jun 03 '21 at 16:13

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I usually use 1/2VI as an upper bound for power and then find the average switching time. This is calculating the worst case power value which what most designers care about (and is easy to calculate); if the power is too much for the temperature rating of the package, then a new design is needed (either more thermal heatsinking or a different package).

If I need a better number for the power dissipation I use LT Spice, which when used properly gives reasonable numbers for power dissipation. (in addtion power can be plotted by ctrl clicking on a part)

Voltage Spike
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  • Like you said, using LT spice can get the more accurate switching loss, but this is not convenient, when you design the converter, do you use Mathcad or a calculation tool to predict the power loss? – EEC Jun 02 '21 at 15:59
  • It depends, if I'm doing PWM from a GPIO with a single mosfet, then I'm mostly worried about max power dissipation, so I use excel or matlab to do a quick calculation. If I'm designing a swticher I usually use off the shelf controllers from TI or Liner and use their tools since there are many calculations that I've got do do anyway to size components and calculate power not only on the fets but on inductors also. – Voltage Spike Jun 02 '21 at 16:09
  • Turn-on loss linked to the capacitance lumped at the drain implies the formula \$P=\frac{1}{2}C_{oss}V_{DS}^2F_{sw}\$ assuming that \$C_{oss}\$ dominates the cap. In reality, the capacitance variation is nonlinear and the above formula leads to an *overkill* as Lemmy would yell. Look at my [APEC 2011 seminar](https://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202011.pdf) where I show on slide 17 how to determine switching losses. Regarding the diode, look [here](https://electronics.stackexchange.com/questions/192831/calculation-of-diode-switching-losses?rq=1). – Verbal Kint Jun 02 '21 at 17:37
  • @VerbalKint the turn-on loss and turn-off loss are the same? – EEC Jun 02 '21 at 17:53
  • No, here this loss is incurred when the MOSFET turns on. When the MOSFET turns off, the external capacitance delays the \$v_{DS}(t)\$ rise and snubs the MOSFET, reducing the overlap between voltage and current if the capacitance is large enough. You can try to predict these switching losses but so many parasitics are at work that the best is to measure losses on a real board. – Verbal Kint Jun 02 '21 at 18:36
  • @VerbalKint Thanks for your answer – EEC Jun 03 '21 at 16:14