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I am building a linear power supply, and I would like to know roughly how big my heatsinks need to be. Is there a rule of thumb for the thermal resistance of a heatsink given its dimensions and material? What would be some rough values be for a heatsink like the one on Amazon here? My supply will output a maximum of 80W as heat. I may need a fan in conjunction with a heatsink to keep it below 50°C above ambient temperature, so what would be some rough thermal resistance values for this kind of setup? I am using two TO220-package LM338s to regulate voltage and current.

Any rough values are fine, but explanations are also welcome.

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    \$\begingroup\$ Please don't use amazon for this. Use a proper supplier like Farnell, DigiKey, Mouser, RS etc.. They will all provide full information and data sheets that cover material, dimensions and thermal resistance. Additionally, you get quality traceability using a recognized distributor. You get none of the above buying from amazon, ebay, aliexpress etc.. That one from Amazon says it includes a memory stick in the spec and, for an absolute certainty, that is ridiculous. \$\endgroup\$
    – Andy aka
    Commented Apr 2 at 11:42
  • \$\begingroup\$ You have a fair point, but I would just like some numbers to get into my head so I can understand what scales I’m working on, and what to expect from a heatsink or heatsink/fan combination, and a rough idea of sizes I need to work with here \$\endgroup\$
    – Klumpy7
    Commented Apr 2 at 11:45
  • \$\begingroup\$ How much heat power is going to be produced worst case working? \$\endgroup\$
    – Andy aka
    Commented Apr 2 at 11:50
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    \$\begingroup\$ What do you want to keep below a 50 deg C rise above ambient? The junction temperature (probably not)? The case temperature? The heat sink temperature? \$\endgroup\$
    – SteveSh
    Commented Apr 2 at 12:04
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    \$\begingroup\$ @Klumpy7 Simple solution. Step 1: Go to Farnell, DigiKey, Mouser, RS, etc. Step 2: Find a few heatsinks that looks similar to the no-name one on Amazon, and check their thermal impedances. Problem solved. \$\endgroup\$
    – 比尔盖子
    Commented Apr 2 at 12:18

2 Answers 2

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I don't see any very similar heatsinks on offer to use @Bill Gates' suggestion.

From here are some rules of thumb for volumetric efficiency of competently designed heatsinks (it will vary a bit with the exact design but this is for ballpark estimates):

enter image description here

If we assume natural convection then your 210cm^3 heat sink should have a thermal resistance of 2.4 to 3.8°C/W. Way too high. So it looks like you would need a fan and maybe ducting to get the power dissipation.

At that power level, especially for a one-off, I would be looking at repurposing a PC CPU cooler, either a fan type or an all-in-one water-cooled system.

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  • \$\begingroup\$ PC CPU cooler is my recommendation as well; one for each TO-220. 40W is old hat for a CPU cooler and the heat generating area is very similar, which is a very important consideration at the scales involved. Tap a hole into the bases or glue them on with thermally conductive epoxy. My recommendation is to use a different package or topology: 40W is far above what the LM338 is designed to dissipate in that package so you might be running into internal shutdowns. In any case, it's being operated well outside of its test conditions, so datasheet values may not be accurate. \$\endgroup\$
    – vir
    Commented Apr 2 at 17:15
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My supply will output a maximum of 80W as heat.

and

I am using two TO220-package LM338s to regulate voltage and current.

Take a look at this table from the LM338 data sheet: -

enter image description here

The first thing you have to face is that the current limiter may not be quite going into current limit hence, all the 80 watts is being produced by the LM338 acting as a voltage regulator. Given the thermal resistance figure, that produces a temperature at the internal junction that is 0.7 x 80 above ambient. So, if ambient is 25° C, then the junction will be 56° C warmer at 81° C.

But, it's worse than this because you can't have an infinite heatsink taking all that heat energy away. You might find one that is 0.5° C/watt and, that means that the net thermal resistance to true ambient is 1.2° C/watt. This means that the junction will be warmer than ambient by 96° C. Or, if ambient is 25 ° C, the junction will be at 121° C.

That is very close to the maximum limit in the data sheet: -

enter image description here

So, realistically you should be looking for heatsinks with a thermal resistance that is significantly lower than 0.5° C/watt to be on the safe side.

However, if you can extract heat power by other surfaces on the LM338 there is a little bit of mitigation. See the top table for the other thermal resistances. All these thermal resistances are in parallel. Small gains maybe but they could be useful to have if you can afford the extra heatsink complexity.

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  • \$\begingroup\$ In your first paragraph, shouldn't that be case and not ambient, as in "...internal junction that is 0.7 x 80 above case?" Or are you assuming the case temp is the same as the ambent? \$\endgroup\$
    – SteveSh
    Commented Apr 2 at 12:21
  • \$\begingroup\$ Ok, this is somewhat useful, What do the other terms in the top table mean (Junction, board, case)? And by what degree should I expect a large CPU fan to improve these numbers? \$\endgroup\$
    – Klumpy7
    Commented Apr 2 at 12:34
  • \$\begingroup\$ @SteveSh I'm assuming case is held at ambient by an infinite heatsink. \$\endgroup\$
    – Andy aka
    Commented Apr 2 at 12:53
  • \$\begingroup\$ Klumpy7 - Did you look at the TI application report spra963d that was referenced under that table? I think you'll find a lot of your answers there. \$\endgroup\$
    – SteveSh
    Commented Apr 2 at 12:53

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