Converting thermal conductivity to Thermal resistance.

[cupoftea]

Hi,
This sil pad has thermal conductivity of 6.6 w.m/m^2.k

https://4donline.ihs.com/images/VipMasterIC/IC/BQST/BQST-S-A0001288110/BQST-S-A0001288110-1.pdf?hkey=6D3A4C79FDBF58556ACFDE234799DDF0

So if we have a bar of it which is 1 metre long and 1 m^2 in area, and we have a 6.6w source at the source end, then we see a 1 degc drop at the other end.

Regarding the sil pad..
Its size is 15mm by 20mm.
As such its area is 300e-6 m^2
Its thickness is 0.178mm

Since its 6.6 w.m/m^2.k, then this means its also 1 w.m/m^2. (0.15K) [dividing top and bottom by 6.6]

So for a metre squared block, 1 metre long, the temperature would be 0.15degC lower at the end of the block, if there was 1 Watt at the "heat source" end.

So anyway, we only have 0.178mm, not one metre, so in fact, the temperature will be much nearer to the source temp.
In fact, it will be 0.000178 * 0.15degC = 26.7e-6 degC. (ie that s the temp difference)

But we also have less area, so we must scale the temperature figure upward for that.
So we must multiply degC by 1/300e-6....so we get 26.7e-6 * 1/(300e-6) = 0.089degc/WATT.

Bu this sounds too low.
Would you agree?

 

[FvM]

The calculation is right, but the assumptions are wrong.
1. SIL pad thermal conductivity is not 6.6W/m according to datasheet.
2. The effective area isn't 300 sqmm. It's maximal the metal tab area of your device package, e.g. TO-247
3. The effective thermal conductivity of a thin pad is lower than calculated according to bulk conductivity, need to consider surface roughness etc.

The datasheet has effective thermal resistance values for TO-220, you can scale it to TO-247 by considering the metal areas.

 

[dick_freebird]

Because heat source is much smaller than heat sink
you also have to consider spreading, and that gets messy
quick.