On 24 Feb 2003 at 22:15, Herbert Graf wrote:

> > The higher frequency case is interesting though. I have to admit
> > that I can't quite visualize what is happening in terms of voltage
> > drop and current flow during diode turn OFF that causes a high
> > instantaneous peak power. (Assumption is that because turn-on time
> > of a "slow" diode is the same as a fast diode, the extra power
> > dissipation must be happening when the the diode turns off ie, when
> > the inductive load is turned on gain). Can you explain?
>
>         IIRC it deals with minority carrier storage effects. It takes
>         a while for
> the junction to "get rid of" the stored charge, during which time the
> diode is conducting current, usually at many times the normal forward
> drop. For more info pretty much any power electronics book covers it,
> I can't find the text I often refer to right now. TTYL

Thanks, I kind of see now but will look up some books. I realise the
voltage accross the diode must change from low (on) to high (off) and
dissipation will be high in the in between state. The hard part is
knowing what the current is and it's direction when the transistor
turns on and the diode is trying to turn off.

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Brent Brown, Electronic Design Solutions
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