Electronic – Thermal resistance of anodized aluminium interface

heatsink

I'm considering extruded aluminium enclosures for a power amplifier, but need to do something with as much as 255W of heat. Currently I have a sufficient heat sink, attached directly to the transistors. If I attached the transistors to the anodized aluminium enclosure, and the heat sink to that (all with appropriate thermal grease, of course), what thermal performance can I expect? Can I estimate the additional thermal resistance due to the interface between the parts?

Best Answer

The coating (on both sides) has a thickness of about 50um or less (according e.g. to this paper) and a thermal conductivity of 0.5 ... 1.5 W/Km (let's use 1.0 W/Km in the simple calculation below).

The extruded aluminium housing has a thickness of 1.5mm and a thermal conductivity of up to 200 W/Km.

The thermal grease layer has a more or less undefined thickness (also changing over time) in the range of 50 - 100um. As a rule of thumb designers often set the layer thickness to 100um and its thermal conductivity to 1 W/Km.

This gives a thermal resistance power module "case to sink" of

$$ \begin{align} R_{th} &= {2 \cdot \num{50e-6} \over A} + {\num{1.5e-3} \over 200 \cdot A} + {2 \cdot \num{100e-6} \over A} \\ &= {\num{100e-6} + \num{7.5e-6} + \num{200e-6} \over A} \\ &= {\num{0.3e-3} \over A} \end{align}$$

with \$A\$ being the semiconductor module case size (footprint) in square-meters.

Without the housing (power module directly on heatsink) your actual thermal resistance is about 3 times smaller:

$$ R_{th} = {\num{100e-6} \over A} = {\num{0.1e-3} \over A} $$

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