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Heatsink Surface Area

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How can I determine the approximate thermal resistance of a heatsink if I know its
surface area?  ::)
but it doesn't answer my question.

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You need to know a lot more than just its surface area. Which metal? How smooth? How thick? Convection or radiation?
That's why real heatsinks are selected from a list of their thermal resistance.

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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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The metal is NOT important!! A heatsink of a particular size, whether it's made of copper or aluminum, IS THE SAME!!  Now if you wanted to talk about thermal capacity, that's a different story.

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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.

Indilus,
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?

-Mike

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Thermal conduction and capacity are material related. Thermal "convection" is not. Yes, the copper heat sink will take longer to heat up, but it will reach the same final temperature as the aluminum heatsink

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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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