0
\$\begingroup\$

Trying to calculate thermal resistance before heat sink for cooling MOSFET. MOSFET Rjc is given in datasheet. But heat sink will be glued using thermal glue like HC910 which has thermal conductivity of 1.7 W/m*K

I need to calculate Thermal resistance of glue layer.

So,

Rglue=THICKNESS / ( AREA * K)

K- thermal conductivity

I my case area is 0.01m*0.009m=0,0000054 sq.m K is 1.7 And i have no clue what thickness will be. I took 0.1mm as a wild guess. I know that thermal paste (like artic silver usually go 0.01-0.02mm, but glue is thicker.

So, in my case Rglue = 0.53 K/W

Is my logic correct? Any idea about thickness?

\$\endgroup\$
0
\$\begingroup\$

Sounds like you're gluing a TO-220 package or something similar. 0.53 C/W (or K/W - same units) sounds quite reasonable. If you press real hard on the device while the glue is curing you might get down to 0.4-0.2 C/W, but that's probably as low as you can realistically go unless you have a really runny glue (0.1mm sil pads bottom out ar around 0.25-0.35 C/w with something like 5kg of force on a TO-220 so I'm using that as a guide)

\$\endgroup\$
0
\$\begingroup\$

So, i checked and rechecked. My logic seems to be correct. Also, i found some application notes on several silicon based thermal adhesives and thickness is specified between 0.07 to 0.17 mm, so, my estimate of 1mm more or less correct.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.