Re: Yet another Chernobyl IRF510 "event" #ubitx


IW4AJR Loris <lorisbollina@...>
 

Hi Don,
 
you could try to calculate the right size of the heatsink using a calculator you can find on the internet, eg
 
https://www.heatsinkcalculator.com/heat-sink-size-calculator.html
or
https://www.allaboutcircuits.com/tools/heat-sink-calculator/
 
I fully agree with Jerry, the problem is almost certainly an excess temperature of the junction, in the calculations never exceed 150 ° C of the junction, possibly keep a margin of 20/30 ° C in order to dissipate any temperature peaks, remember that the power peaks do not dissipate immediately, but take some time to dissipate and this can cause the junction to overheat.
 
Just to give my opinion without pretending to be right, but I think you will have to use a much larger heatsink than the one used in order to get the power you want from the two IFR510s, if you look around, you will see that for a linear from 50W usually a single heatsink is used for the two IFR530s, normally large 4x10 inch with fins of at least 2 inches, the best linear ones also add a fan or two.
 
Again as my simple judgment, I believe that the µBITX was not born to give similar output powers, much better to provide external linear perhaps with MOS more suitable for the purpose.
 
Ciao from IW4AJR Loris

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