From: firstname.lastname@example.org (Bob Wilson)
Subject: Re: Thermal resistance--does this sound right?
Date: Fri, 06 Sep 2002 02:49:39 -0000
Organization: Your Organization
X-Newsreader: WinVN 0.99.9 (Released Version) (x86 32bit)
In article <3D781177.email@example.com>, firstname.lastname@example.org says...
>I found a figure of k = 0.004 W C^-1 m^-1 for heatsink grease.
>Using the formula for thermal resistance = L/(k A) and for a
>10cm x 1cm strip of contact area between two anodized aluminum blocks,
>with a rough guess of about 1 mil or 25 microns average spacing, I get
>=(2.5E-5)/((0.004 W C^-1 m^-1)(0.001 m^2)) = 6.25 C/W
>That seems awful high. I have usually figured about 1-2 C/W for typical
>TO-220 or TO-3 packages clamped to a heatsink with some grease, even
>with only about 1-2 cm^2 of area. The gap is really a wild guess, and
>hard to control.
That is out by an order of magnitude or more. Typical actual values are way
under 1 degree per watt for a thermally-greased joint. Even a TO3 with no
grease at all will only have a thermal resistance of less than the figure
>What do people usually come up with?
>I have some scientists who want to cool a gadget, which rises about 14C
>above ambient with 40W of dissipation, and there is about a 10 cm^2 area
>to attach a heatsink. That's already a 0.35 C/W thermal resistance to
>ambient, so with a 6.25 C/W minimum thermal resistance for whatever we
>might attach to it in parallel with the original 0.35 C/W, that doesn't
>offer much hope of getting it any cooler, unless we use a water cooled
>heat sink block.
Nearly ANYTHING will give you less than 1 degree per Watt, for the large
area you are referring to (assuming the heat is evenly distributed). Even no
grease at all (just air) will do it. So if you want to get a value of (say)
less than 1 degree/Watt, use anything you want; grease, sil-pads