My head is ready to explode trying to figure this out.
I have two electrically insulating materials being considered for use
with TO-247AC devices being mounted to a heat sink. They have the
following thermal conductivity and thickness specs:
- Berquist Sil-Pad K10 pad: 1.3 W/m-K conductivity, .006" thick.
- Aavid 4180 Alum. Oxide insulator: 15.06 W/m-K conductivity, 0.080"
thick.
In order to compare their thermal resistances, can I simply divide the
thermal conductivity by the thickness? The units and equations I can
find on the Web make this, what I thought to be very simple, decision
process very difficult.
- K10 thermal resistance = 1.3 / .006 = 216.67
- 4180 thermal resistance = 15.06 / 0.080 = 188.25
Therefore, even though the Aavid aluminum oxide insulator is 13.33
times thicker, it's thermally conductive enough to be the better
choice (thermally, that is)?
Essentially, I'm looking for the best (but still safe) TO-247AC,
electrically insulating, thermal interface material I can find. I'm
currently using Aavid's UltraStick compound (and love it), but it's
not electrically insulating and I need that for this application
(can't prevent touching of the heat sinks).
Thanks!
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