Re: Yet another Chernobyl IRF510 "event" #ubitx

Jerry Gaffke


The IRF510's should be operating in a more linear region when powered from 24v instead of 12v,
so you might get by with less quiescent current than 100ma.
The issue is intermodulation distortion (IMD), so you might create trash
in adjacent channels if you reduce the quiescent current too far.  

There's an awful lot of heat when running at 24v continuously in digital modes. 
Even if you were to reduce the quiescent current to zero (which would make some bad IMD).
The heat does not get conducted easily from IRF510 die to IRF510 tab.
The silicon die might be 100 degrees C hotter than the tab when you push things.
The tab should never get too hot to hold your finger on it, so you need a big heat sink.

Let's assume you are getting 20 watts out from your IRF510's when running at 24 volts,
and that they are 50% efficient.  So there is 20 watts worth of heat to dissipate.
The IRF510 datasheet says that for 1 watt worth of heat to transfer from die to tab,
there will be 3.5 degrees C difference between die and tab. 
When dissipating 20 watts, that's a difference of 3.5*20=70 degrees C (or 126 degrees F).   

Assume the heatsink is getting warm to the touch but not blisteringly hot, say 60 C (140 F).
If you are transmitting at 20 Watts out with 50 percent efficiency, the silicon die
is at 60+70 = 130 C ( 266 F), which is blisteringly hot.  The IRF510 is rated for an
absolute max die temp of 150 C,  but at those temperatures it will fail over time.

If transmitting at high power levels with IRF510's, you want those IRF510 tabs
kept cool with a truly massive heatsink.   From your photo, you don't have one.
A safe bet those die were well over 150 C (302 F) when they decided to fail.

IRF510's are cheap, so have fun and learn lots.

Jerry, KE7ER

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