I am wondering about catch diodes in the context of switching power supplies. I have read all the books that are the norm in designing a SMPS, but all of them don't or barely touch the subject. My question is about how do you select it's current and voltage rating.

From a buck converter standpoint I have reached the (maybe bad) conclusion, that the current rating of the diode must be equal or larger than the current flowing through the load, since the inductor stores energy then it returns it to the load through the diode. But in an SMPS with a transformer, which ideally only transfers energy, the diode selection would be based on the leakage inductance. If so, how do you select the ratings?


1 Answer 1


The "catch" diode on the primary side of a transformer-coupled SMPS serves the same purpose it does on any other inductive load such as a relay. Its reverse voltage rating needs to be at least as high as the voltage across the primary when the switch is on. Its (peak) forward current rating needs to be at least as high as the peak primary current at the moment the switch turns off. The average forward current and the total power dissipation will depend on the switching frequency and the value of the leakage inductance of the transformer.

Note that this assumes that we're talking about a "forward" converter, in which the transformer itself is not intended to store significant energy. In a "flyback" converter, you do not put a catch diode directly across the primary.


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