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Pabo 12th May 2008 07:37 PM

Tekko and Dave

Just consider that the current has to leave the channel and enter the parallell diode in a matter of nano seconds if it is going to give an improvement on reverse recovery. If the parallell diode has 0,2V lower drop and the stray inductance and internal inductance inside the MOSFET/diode is 10nH you can easily calculate that it takes 20A of current 1 microsecond to "move" to the other diode. Has anyone here seen an improvement on EMI, efficiency or THD by doing this?

Pafi 13th May 2008 08:15 AM


IMHO your post a little difficult to understand, I think you could tell it in a simpler way: the diode with smaller forward voltage drop will conduct most of the current, hence it will determine recovery speed. And this will be the body diode in this case, if you just parallel them.

Eva 13th May 2008 10:38 AM

Forward voltage drop is not the only parameter that matters, actually parasitic inductances also determine current sharing between the body diode and any external diode.

Pafi 13th May 2008 01:51 PM

You are saying something!

Pabo 14th May 2008 08:31 AM


The body diode only conducts after the channel has been turned off if the current is flowing from source to drain (i.e. freewheeling). Immediately when the channel is turned off the current will flow through the body diode. This is of course a simplification as the body diode is actually the channel itself when not turned on. In order for the current to move to an external diode which is connected in series a certain time is needed. In a class d amplifier it has to happen within the dead time which is usually very short. So if the voltage drop is even half of the body diode it will take much longer than the dead time itself hence giving no improvement in reverse behaviour.

I am missing something?

peranders 14th May 2008 08:59 AM

I have tested this with 475 A and a separate diode didn't help much so you are pretty much right.

Pafi 14th May 2008 09:05 AM


I am missing something?
Yes, the resistance of the channel. The whole reverse current cannot flow on the channel typically. The output current can be more then 10 A, the Rdson can be more then 0,12 ohm (-> 1,2 V drop), so body diode can conduct during freewheeling easily.

Pafi 14th May 2008 09:27 AM


What is "475 A"?

And how could you test this:

if the voltage drop is even half of the body diode
Where did you get from a >200 V hyperfast diode with <0,4 V forward voltage?

Pabo 14th May 2008 10:22 AM


OK, in that case I suppose it helps. I usually use FETs with much lower rds and if the rds is that high the body diode is usually OK.

phase_accurate 14th May 2008 11:37 AM

Has anyone ever tried to use a very small series resistors in the drain path ? This should also help to reduce reverse-recovery stress. It does however not speed-up anything but it would help to reduce the current-peak. If dimensioned properly it would not give more conduction losses than a series diode.



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