4QC (Full bridge) transistor duty cycle

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refuser

Joined Dec 11, 2009
1
I'm going through an old exam where we have a 4QC like in this image (don't mind the component values they are not correct).


Now a part of the task is to calculate losses. The duty cycle is 80% but in the solutions they are sharing the flyback diode load equally over the diode and the transistor. So that transistor conducting percentage is 80%+0.5*20% and the flyback diode conducting percentage is 0.5*20%.

What I dont get is how you can divide the flyback diode work over the diode and the transistor (they are IGBTs actually)?
 
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