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Heatsink Surface Area


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What type of metal is important. (Al is most common)
Also the shape is important to some degree. If the part
is mounted at one end of a long sink the wattage will be lower
than if the part were mounted in the middle of the same sink.

I don't know of any actual 'formula' to convert the shape, material, and surface area of a
sink into an actual wattage or Efficiency % value.

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Radiation matters more if the heatsink is painted in matte black, but it's a lot less than the heat transferred to the ambient by convection.

I could measure the thermal resistance if I have a contact thermometer... and apply a known voltage to a known value resistor stuck to the heatsink.
I don't need a precise formula.

I took a look at some circuits that mentioned the dissipated power and the minimum surface area required for the heatsink.
After some very simple math I got this result:
A = 10 to 20cm2 per watt,
where A is the surface area of the heatsink.

Here are some examples:
For 30 watts I'll need a heatsink with a surface area of 600cm2.
For 70 watts I'll need a 1400cm2 heatsink.

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Paul_J,  That’s like the test I was talking about. We did it (2 decades ago) in a class on microprocessors of all things!  It was just a side note where the instructor demonstrated how heat sinks have different abilities.  He used the Soldering iron as a fixed and constant heat source.  But he just glossed over the calculations he used. I do remember now that the goal was to get the whole heat sink to the same temp.  A better heat sink
Would maintain a lower temp than a poor one given the same energy input.

    Huh?  I thought copper was about 2 times better than aluminum so for the exact given shape and size?
Diamond              (2300 W/mK)
Pyrolytic Graphite (1950 W/mK)
Silver                    (429 W/mK)
Pure Copper          (401 W/mK)
Pure Aluminum      (237 W/mK).
Of course that is only true of pure metals, Alloys are generally much lower than any pure metal.  So a pure aluminum sink could be the same as a low grade copper sink.
But it is very rare to find affordable copper sinks any more.  So I guess the whole thing is moot.  I have used copper water pipe before.  Flatten out one end, Drill a hole to mount the part and then split the sides of the remaining tube open .  In a pinch it works fine.

Speaking of water pipe…
Has anyone mentioned water blocks yet?


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So, would 30cm2 of heatsink surface be enough for each watt of dissipated power? Even if the heatsink calculated with the above formula is a little bit oversized it doesn't matter. In fact, the cooler a semiconductor runs the longer it lasts.

I could build a unit for testing heatsinks because I have a lot of them  ;D, most coming from broken computer power supplies, but I need a simpler way to determine how much power a heatsink could handle so the junction of the semiconductor attached to it doesn't exceed about 100 degrees Celsius.

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