[SOLVED] Parallel diode effects on the rated dynamic (Irrm, trr, Qrr) datasheet parameters.

[lawcuenca]

Hi,
I would like to know what happens to the diode Irrm, trr and Qrr when they are paralleled in a single case? Thanks!

Regards,
Law

 

[Orson Cart]

If you have the specs for the individual diodes in the package you can expect the paralleled pair to give you twice the Qrr, much the same trr, and much the same Irrm (half each) depending on reverse di/dt just before diode turn off.
Regards, Orson Cart.

 

[lawcuenca]

Hello Mr. Cart,
Let us use these figures for a single device as an example:
trr=9ns
Irrm=24A
Qrr=100nC

Does it mean that when I configure two diodes with the same ratings, the results will be:
trr approximately = 9ns
Irrm = 24A (with each diode carrying 12A)?
Qrr=200nC (same as capacitors in parallel)

Is this correct?

Thanks,
Lawrence

 

[Orson Cart]

Irrm depends on the forward current prior to turn off, so if we assume each device has the same forward current prior to turn off then each device will have nearly the same reverse peak as specified for that forward current, with a given di(reverse)/dt.
Generally, faster reverse di/dt gives better sharing at turn off, but there may be the odd instance of grossly mismatched die in the package. The general assumpton is the diodes will be quite closely matched. So yes, you are generally correct.
Regards, Orson Cart.

 

[mtwieg]

I don't see how Qrr can double while trr and Irrm stay the same, given that Q=I*t. You shouldn't be able to change one parameter without another changing too.

 

[Orson Cart]

Irrm per diode, the total Irrm will be nearly double. The peak reverse current is dependant on the slope and shape of Irev before turn off starts to occur, Qtotal = I (ave) * t, we are talking about peak Irrm (m for max) - which is related to Iave by a factor of 2 for a triangular wave shape and ta = tb = 0.5Trr, Regards, Orson Cart.

 

[mtwieg]

Qrr=Irravg*trr, and Qrr=0.5*Irrm*trr, assuming a triangular shape, right? That is for a given diode. If you put two in parallel, then assuming they share everything evenly, including equal di/dt, so their individual Irrm should be halved from what it would be from one diode under similar conditions. Assuming trr is the same too, Qrr per diode should also be halved. So the total parallel combination should have the same trr, Irrm, and Qrr as one diode.

However, trr usually gets shorter with increasing di/dt, so the above analysis won't be true. The total trr will be longer, so Qrr will be greater as well. It should depend on the curve of trr vs di/di for the diodes. Also I'm quite sure that Qrr is roughly proportional to the square root of di/dt, so in a parallel situation, I'd expect Qrr to increase by a factor of 1.41.

 

[lawcuenca]

Now, this is getting confusing. Each one of you three has a different take on this. Do you have any substantial data or writing regarding this so that we can all verify? Thanks.

Cheers,
Law

 

[mtwieg]

Well, I'm not really sure we're contradicting each other. Let's just try an example. Since trr, Irrm, and Qrr are not constants, and vary with reverse di/dt, we should refer to the device datasheet to see exactly what happens. I'll use this diode as an example:
https://www.microsemi.com/datasheets/40DQ120B_S(G)_B.PDF

And lets say that before switching one of these diodes is conducting 40A, and the fall time is 100ns, so the di/dt is -400A/us. According to its curves, trr will be ~340ns, Irrm will be ~14A, and Qrr will be ~2750nC.

Now say we put two of those diodes in parallel and put 40A through both of them, so each is carrying 20A. Assume the fall time is still 100ns, so the di/dt for each diode is now -200A/us. Now according to the curves, each diode will have trr = 370ns, Irrm = 8A, and Qrr=1600nC. Both diodes combined should give trr=370ns, Irrm=16A, and Qrr=3200nC.

So your overall Irrm and Qrr have increased by about 15% with the same currents and switching speeds. I don't think there are really easy ways to estimate this, you should look at the actual datasheet for answers.

 

[lawcuenca]

I see. What you guys said are really very helpful. I appreciate it. Thanks!

Cheers,
Law