LM5088: About LM5088 power loss in overload

Hello John,

I have studied the article and I am not entirely sure where the author got their figures from?  However I can tell you that at 250 kHz for 100 cycles amounts to 400us in time.  Having a hiccup time of only 500us seems to be very short.  I would make the hiccup time approximately 10 x the cycle by cycle time. 

Therefore say we have a hiccup time of 5ms.

1 period = 1 / 250 kHz = 4us

100 cycles *4us = 400us.

The Ton in current limit is assumed to be 100ns

There is a current limit delay time (in Datasheet) of 280ns (typ).  Total Ton time = 100ns + 280ns = 380ns.

This Ton time occurs for 100 cycles amounting to 100*380ns = 38us of total on time in 400us.

We know that during the hiccup time (5ms) the Diode is Off.  We can therefore add this to the 400us for the total time within the cycle by cycle current limit and hiccup time.  This will amounts to 5400us.

Therefore the diode is on for 362us (400-38)us for a total duration time of 5400us.

Therefore 1-D = 362/5400 = 0.067.  from here you can compute the diode losses as shown in article.

Hope this helps?

David.

Hello John,

As previously mentioned, I believe there is a mistake in the application article.  I think its reasonable to assume a much larger hiccup time than presented in the article, and I have speculated a hiccup time of 5ms.

The method I have detail above is what I would follow when calculating the diode losses when hiccup current limiting.

I hope this helps?

David.