Flyback output diode peak curent vs average current

In a DCM flyback with its high peak to average current ratio. When sizing a diode do you rate thediode for the maximum peak current. I have a 2A average current with 12A peaks at the start of the flyback. Repeats every

5uS.

Which current rating should I look for in the diode? 3A? 6A? 15A? All of this makes a difference in the size of the diode package which I'm trying to minimize. The diode datasheets don't have a repetative peak current rating on the nonrepeataive rating.

thanks

Reply to
Mook Johnson
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Put the smallest diode in and see how hot it gets. If you can hold your finger on it it's probably good enough. Some may tell you it should sizzle when you spit on it. This is the backyard engineering way to do it. To do this correctly you would want to calculate losses and determine junction temperature in hottest environent and so forth. To calculate losses you need to know instantaneous current and voltage and then integrate that over a switching cycle. stand by for a bunch of alternate ways of doing. many ways to skin this cat.

regards, Bob N9NEO just say NEO!

Reply to
Yzordderrex

Conventional rectifiers are rated for average forward current, as the forward drop is relatively constant. The power loss vs average current for a range of duty cycles is a fairly common chart provided for design - 0.1 duty assumes a 10:1 peak to average current ratio.

In hard switching, peak current and temperature will affect stored charge in conventional diodes, increasing losses elsewhere.

Schottkys tending to look more resistive and have RMS considerations. Though forward losses reduce as Tj increases, increasing leakage may be signifigant and can result in thermal runaway.

Your concern should be Tj, Rthja, power loss, and maximum ambient, in most cases.

RL

Reply to
legg

As others have said it depends on maximum Tj, which depends on ambient temperature and junction to ambient Rth. A 200A diode will cook itself in your application if its in a vacuum and attached with thin wires.

Reply to
Adam S

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