thermal conductivity and thermal resistivity of heatsink pads

In thermal conductivity (W/m-K), the "m" in this relationship is really the ratio of the contact area of the interface divided by the thickness of the interface (m^2/m=m).

For example, for a material like your K-10 Sil Pad with its conductivity of

1.3W/m-K, if the Sil Pad is 1m x 1m in area, and it is 1m thick, then the ratio of the area to the thickness is equal to 1m. So, if you take that area/thickness ratio and multiply it with the conductivity, you will have a thermal interface that for every 1.3W of thermal power through that interface, it will develop 1 deg K (or C) of temperature difference across the thickness of that interface (1.3W/C). This is the same as saying that there will be 0.77C/W of thermal resistance.

I think that you have the ability, now, to write down the simple formula to get from thermal conductivity to thermal resistance, as a function of thermal interface material area and thickness. If you have something like Excel, Matlab, or Mathcad, they're very convenient for automating this so you can merely enter a materials conductivity, the interface area and thickness, and let it calculate the thermal resistance. Just make sure you get distances converted to meters so that the number come out accurately.

Also, in case it's not already clear to you, when you have a device that is dissipating (let's say) 4W of thermal power, there is nothing you can do to stop it from dissipating its 4W of thermal power (steady state). The only thing you can do by adding a "better" heatsink is to make it easier to dissipate its power. Making it "easier" to dissipate a device's power results in its temperature being lower -- and temperature is what does the damage.

When you're doing thermal analysis, you need to seek out enough information to calculate the temperature rise from the ambient air, through the heatsink, through the thermal interface material, through the IC case, and finally to the IC junction. Keep the IC junction below its max rating and you're cool.

Bob

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Bob did that: Specific thermal resistance t is the reciprical of the conductance (w/m-k); the thermal resistance of the insulator is

r = t * thickness / area

dimensionally ( (m*k)/w ) * m / m^2 = k/w

So, the thinner the better, and the bigger the footprint the better.

Sil-pads suck. Alumina is pretty good, if you don't give it back because of the thickness. Alumium nitride is superb.

The best insulator is no insulator, just a little silicone heatsink grease. Next best is 0.5 to 1 mil hard anodize plus grease.

Always use grease. It fills small air pockets and squishes down well below 100 microinches.

John

Actually, diamond is the best insulating material for highest heat transfer.

Except that the discussion is about commercially available product mediums, not what one finds in the CRC Handbook.

Show me where Digi-Key (or anyone else) sells diamond transfer plates... That's what I thought.

Especially isotopically pure diamond.

John