[Amps] Anodizing aluminum, painting etc.

[Will Matney]

Peter,

I agree on the math. The problem I see on using a black heatsink is that even 
though all heat is infrared, and there's a concentrated heat transfer from the device 
to the heatsink, using black supposedly helps this transfer. The problem arises in 
the day time, or around any source of infrared radiation in that the heatsink would 
absorb this radiation becoming hotter than would a silver surface which would reflect 
it off of it. The heatsink though would still radiate it's heat if silver. To my opinion, 
using a good thermal heatsink compound with a silver heatsink works probably better 
or just as good. If black works like they say for the device, why not make a black 
silicone grease for the thermal compound? Most I've seen is gray or white! That then 
is in between the black heatsink, or an insulator is used which Is generally not black 
either. On top of that, to my opinion the anodizing creates a surface insulation the 
same way it does on the foil of an electrolytic capacitor. I just don't think anodizing a
heatsink is as good as some make it out to be when you start picking it apart and
investagating it with reason. The best place to run a black heatsink would be in the
dark away from any infrared sources that it could absorb heat from. I place more faith 
in the mass of the heatsink and the way the fins are designed rather than its color.
That's just my opinion on it.

Best,

Will

*********** REPLY SEPARATOR  ***********

On 8/15/06 at 8:21 AM Peter Chadwick wrote:

>If you had a heat sink of zero mass, infinite conductivity and zero
>thermal resistance btween sink and air, it would work perfectly, no matter
>what size it was. So mass itself doesn't matter: the implication is that
>greater mass equates to greater area and lower thermal resistance. After
>all, which is going to give best results - 500 grams (OK, 1 pound in the
>US!) of depleted uranium or 500 grams (1 pound) of aluminium? The
>aluminium obviously has a greater volume, and thus a greater surface area.
>In this imperfect world, the mass times the specific heat tells you how
>many calories are needed to raise the sink temperature above ambient by
>some amount. The power being dissipated at 4.2 Joules/calorie tells you
>how long it takes to do it. The Theta sink-to-air tells you how much heat
>the sink is losing. The complication then  occurs because the sink is not
>generally at equal temperature all over. In any case, all you're really
>interested in at the end of the day is Theta-junction-to-ambient. From
>memory, you can end up with set of simultaneous second order differential
>equations trying to work it all out from first principles, and those are
>things that I avoid!
>
>73
>Peter G3RZP
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