# Flyback diode choice

I am curious about how to determine the turn ratio to get the minimum power loss. Assuming n=Np/Ns Vin=380, Vo=40, different turn ratios can get the same output because Vo =VinD/(n(1-D)).

1. When the n is small ids increase, Vd decrease, ids decrease, Vds increase.
2. When the n is large ids decrease, Vd increase, ids increase, Vds decrease.

Above these two situation, how do I choose the MOSFET and diode to get the minimum total loss?

• Component resistance depends on load min~max current but Vd= Vf ( at ~ 10% rated current)+ I*Rd not inverse or negative as you proposed unless you are stating inverse peak voltages from transformer load impedance ratio. Define specs for power range, ripple V , tolerances and efficiency then L, f are variables too Apr 19, 2021 at 14:30
• THis has to be < 100W with no PFC Apr 19, 2021 at 16:58
• @TonyStewartEE75 I got it, thanks .
– EEC
Apr 19, 2021 at 17:01
• ok here's one of mine webench.ti.com/appinfo/webench/scripts/… Apr 19, 2021 at 17:17
• @EEC - Hi, Did you make a mistake in your last edit? You removed so much information from your question, that the answer then made no sense (e.g. you removed the schematic etc.). Therefore I have reversed (rolled-back) your edit, to "undo" the damage. (If you did intend to make that edit, it is now too late to make such major changes to the question, which could be considered vandalism, especially as it already has an answer.) Thanks. Apr 21, 2021 at 1:50

Multiple paths exist to determine the turns ratio of a transformer in a flyback converter. One of them is to start from the $$\BV_{DSS}\$$ of the selected MOSFET and apply a derating factor for a safe operation. The derating factor is important to make sure that in any worst-case situation (open loop, short circuit, over-voltage situation etc.) the drain-source voltage always remains in a safe zone. For instance, serious power supply manufacturers apply a 10-15% derating factor (I've seen up to 20% with Japanese designs) to the selected MOSFET. For a universal mains application (85 - 265 V rms input), people usually select 600- or 650-V MOSFETs. If you choose a derating factor of 15%, for the first one, then you will determine the transformer turns ratio and adjust the various protections (like the $$\RCD\$$ clamp for instance) so that the voltage across the MOSFET never exceeds 600 x 0.85 = 510 V:

Once you have that number on hand, you can determine the transformer turns ratio as described in my APEC seminar, The Dark Side of the Flyback Converter. You need to select the clamp voltage that you accept and the reflected voltage. When you have these numbers (this is explained in the seminar), then you can have a first value for the turns ratio:

As you can see, the turns ratio affects the MOSFET maximum voltage at turn off but it also impacts the secondary-side diode by imposing a blocking voltage during $$\t_{on}\$$. Therefore, as any converter design, you need to go through several iterations in which you keep the MOSFET voltage under control and makes sure the rectifier is not oversized. Once choices are made, always exercise the converter to check that voltage limits on the semis are guaranteed to be safe across all the possible scenarios (low and high line, short circuit at the board ends, broken optocoupler etc.)

• Hi Verbal Kint If I can guarantee the Voltage spike would not exceed the BVDD for example maybe n =1~5 these values can't break the MOSFET, how to choose the best turn ratio to get the minimum loss.
– EEC
Apr 19, 2021 at 15:49
• You can try to change the turns ratio and see how it affects rms currents and this is why I said it is an iterative process. However, the method I described leads to excellent results and is also used by the big names in the industry. Apr 19, 2021 at 16:53
• The first one is to make sure the voltage across the MOSFET would not exceed the BVDD, and then I need to calculate the component loss in different turn ratios right?
– EEC
Apr 19, 2021 at 17:03