Can we make heatsinking easy?

[KatieandDad]

Originally Posted by AndrewT

10*30*100mm is 0.03litres

The volume is immaterial.

It is the surface area that dissipates the heat.

Volume only comes into play with MASSIVE heatsinks that are trying to dissipate heat over a large area.

[CharlieLaub]

If you navigate your browser on over to Heat Sink USA and select a profile, it says RIGHT AT THE TOP OF THE PAGE the heat transfer per length of heatsink (or per length specified if not 1").

It doesn't get simpler that that.

-Charlie

[hpp3]

10*30*100mm is 0.03litres

Typo. Sorry.
I was referencing Marra's post about the chunk of aluminum 10x80x300mm. If I recall correctly, I mentioned that I was tired. Thanks for the correction.

The volume is largely immaterial.
It is the surface area that dissipates the heat.

Hmm, really? So I could cut and bend up some old sheet aluminum I have lying around, and it would work? Sweet.
Although I could see problems with that being dimensionally unstable (warps, etc causing surface contact voids) and so probably more prudent to use properly machined aluminum stock.

If you navigate your browser on over to Heat Sink USA and select a profile, it says RIGHT AT THE TOP OF THE PAGE the heat transfer per length of heatsink (or per length specified if not 1").

Sure, except I was having quite the trouble figuring the math for individual chipamps to arrive at that magical number. What would be EVEN COOLER is if they also provided a list of "Recommended to work with..." devices (which was kinda the point of my OP), but I'm sure that list would be prohibitively large and quickly out-of-date, so scratch that idea, I'll just do the bleeping math. *grumble*

Thanks all, i know SO much more about heatsinking than I did when I started.
Really, I mean that.

[454Casull]

Originally Posted by hpp3

Typo. Sorry.
I was referencing Marra's post about the chunk of aluminum 10x80x300mm. If I recall correctly, I mentioned that I was tired. Thanks for the correction.

Hmm, really? So I could cut and bend up some old sheet aluminum I have lying around, and it would work? Sweet.
Although I could see problems with that being dimensionally unstable (warps, etc causing surface contact voids) and so probably more prudent to use properly machined aluminum stock.

Sure, except I was having quite the trouble figuring the math for individual chipamps to arrive at that magical number. What would be EVEN COOLER is if they also provided a list of "Recommended to work with..." devices (which was kinda the point of my OP), but I'm sure that list would be prohibitively large and quickly out-of-date, so scratch that idea, I'll just do the bleeping math. *grumble*

Thanks all, i know SO much more about heatsinking than I did when I started.
Really, I mean that.

If volume is sufficient, surface area is what determines the heat loss to air. If you have a lot of surface area but the fins are too thin, the internal conductive resistance will be too high and much of your surface area will go to waste, lowering your effectiveness.

[hpp3]

If you have a lot of surface area but the fins are too thin, the internal conductive resistance will be too high...

Aha, I just KNEW there had to be some logic to why we all didn't just use some zig-zagged sheet metal for heat sinks, here it is, and I'm left with my head in my hands wondering why I just didn't pick up on it.
Thank you, Sir.

[jaycee]

some commercial products DO use zigzagged sheet metal crimped to a piece of L-angle as a heatsink. It is usually quite inferior to a real aluminium extrusion.

For the chipamps you mention it will probably be adequate, however by the time you have manufactured this yourself (unless you have the tools and materials to hand for little/no cost), you can buy a sensible extrusion already.

For your reference, A 2xLM3886 amp I built uses a 160mm wide, 25mm deep, 70mm high non-anodised extrusion I got from eBay, with 25v supplies. It gets warm when driving a 4 ohm load at decent volume, but not excessively so.

[sofaspud]

If you have a lot of surface area but the fins are too thin, the internal conductive resistance will be too high...

I see that as a simple redundancy, so of course it's true.

[Wavebourn]

I use rough approximate formua, it works for me. Resistance in Celsius degrees per watt equels 50 divided by square root from area in square centimeters.

[AndrewT]

That sqrt & 50 sounds familiar.
What does ESP say on the subject?