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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