https://www.eevblog.com/forum/projects/the-performance-of-a-heatsink-with-a-black-surface/Someone here mentions of microholes in heat sinks that are 7 times deeper then wide. What is this kind of patterning called and what diameters are used?
It has to do with blackbody radiation. A blackbody is an 'object' that absorbs all incident radiation, thus it appears black. Whatever is good at absorbing one, more, or all wavelengths (blackbody) is also good at emitting these wavelengths when excited (see sodium lamps etc).
Every body (heatsink in this case) that is at a temperature above absolute 0 (0 K) will emit radiation (photons). This radiation is what say IR thermometers or thermal imagers actually image.
In the case of a blackbody, the radiation energy is given by the Stefan/Boltzmann and is a function of the fourth power of the absolute temperature (Kelvin) of the body. This is the formula:
E = k * T4* A [Joules*sec-1].
k is a constant, in the order of 10-8 and A is area in square meters.
Because the constant is so small, you need a very large area or rather high (>500degC) temperature to have significant energy dissipation in the form of radiation. And by significant I mean comparable with the power levels of what you are trying to cool, in the order of Watts.
So unless your heatsink is huge (football field?) or it is glowing hot, you will not dissipate a lot of heat this way. Better get a different heatsink or better interface (lower thermal resistance) between surfaces or a material with higher thermal conductivity like copper compared to aluminium.
A heatsink painted black is a bad approximation of a blackbody, more like a 'grey' body where the energy radiated by the formula above is multiplied by a constant (emissivity epsilon) which is less than 1. A better approximation would be a heatsink with microholes with depth about 7x their diameter (the whole appears black inside, try it). This is used by Ersa in their IR heating plates for PCB rework.
To conclude, painting heatsinks black does not contribute to their ability to dissipate heat by radiation.
« Last Edit: June 20, 2011, 04:01:46 am by Alex »
IMHO the biggest advantage of the ceramic heatsinks is that they don't conduct. I have used these to increase the cooling on an existing circuit board. They give some measureable improvement.
For power electronics I would love having a ceramic (even just a coating) heatsink/cold plate, unfortunately the matherials that have a high enough thermal conductivity is either horrendously toxic (berilium oxide) stupidly expensive (boron nitride) or a mix of the above
hard to justify over a spacer, unless its R F , but I am really curious about the effect of those holes as described by the other thread.
on the bright side with how hard boron nitride is to produce, at least I don't feel like im being completely screwed over...
are you sure convection is a dominant force with the pore size it has?
In air, with the heatsink exposed to the thermal IR radiation from ambient surfaces above 0 deg C? Conduction and convection will certainly dominate.