I'm planning to make a power supply with the LM2596 (adjustable voltage variant). In the datasheet I read that efficiency for VIN=12V, VOUT=3V, ILOAD=3A is 73%. That means, if I'm not mistaken, that 3 × 3 / 73 × 27 ≈ 3.33W will be wasted as heat.

When a heatsink is rated as 20K/W, does that mean that for every watt that is dissipated as heat, the temperature of the heatsink rises with 20K? When the heatsink is connected properly, is this the value that I should calculate with, or is the temperature of the component always higher than that of the heatsink, s.t. I should actually calculate with for example 22K/W?

In this example then, the temperature rise with a 20K/W would be 3.33 * 20 = 66.7K, which is still OK because the chip can handle up to 125°C according to the datasheet.

Is this correct? I'm a little bit worried about being OK with my component getting around 90°C, even though the datasheet says it's fine.

  • 1
    \$\begingroup\$ Can you live with max ambient temperature + 67 degrees? Sounds toasty to me. That 125 degree max you can take with a big pinch of salt. Check max power dissipation at 125 degrees and you will notice somehting drastic. Also factor in the thermal resistance of the interface between the heatsink and your device. \$\endgroup\$
    – winny
    Mar 29 '17 at 12:24
  • \$\begingroup\$ @winny thanks, the thermal resistance of the interface is part of my question. What would be a reasonable estimate, when the device is mounted directly on the heatsink for example? From figure 1 of the datasheet I understand that the voltage does not change much when the temperature rises. But if it is still a bad idea considering surrounding components or otherwise, that would constitute an answer to my question. \$\endgroup\$
    – user17592
    Mar 29 '17 at 12:29
  • \$\begingroup\$ About 0.5 K/W? I would also aim for 90 degree C maximum for the component. \$\endgroup\$
    – winny
    Mar 29 '17 at 12:33
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    \$\begingroup\$ EEVblog on youtube done a pretty decent tutorial on calculating heatsinks if I remember right. Went through everything from calculating heat dissipated right down to how to select a heatsink.... I can try and find the link if you want? Or you could just search EEVblog heatsink in youtube! \$\endgroup\$
    – MCG
    Mar 29 '17 at 12:36
  • \$\begingroup\$ @MCG thanks, that was very helpful. The link, for reference, is: youtube.com/watch?v=8ruFVmxf0zs \$\endgroup\$
    – user17592
    Mar 29 '17 at 15:11

Just for reference, I'm going to use °C/W because it's what I'm used to.

In the datasheet for your regulator under the thermal characteristics there are typically two values.
\$R_{\theta JA}\$ - Junction to ambient
\$R_{\theta JC}\$ - Junction to case

If not using a heatsink then the \$R_{\theta JA}\$ is the value you use to calculate how warm the regulator is going to be.
When using a heatsink you take your \$R_{\theta JC}\$ and add this to your heatsink rating (20°C/W in this case) also remembering to add in any other thermal resistances. The thermal pad between the device and the heatsink is one such example.

Now this 22°C/W is in a perfect world where it has a super perfect thermal connection so take your result as an estimate rather than a "It will always be 60°C"

When you calculate the temperature, you are correct in thinking this will be the temperature of both the component and heatsink. However, remember that there's a lot of difference between a TO-220 package and a block of aluminium and it will take time for both items to reach the same temperature.

Edit: Because I remembered I had it to hand, here's a thermal image of a PCB I'm working on. This is about 5 minutes after switch on, notice the component itself is 43°C and yellow on the thermal image but the surrounding large copper heatsink pad for the device is a purple and closer to 20°C.

P.S - Please ignore the white hot supernova approaching from the right, this is a work in progress board

enter image description here


The chip datasheet will tell you the thermal resistance from 'junction to case', which for a device on a heat-sink will be the thermal resistance to the heatsink (or at least to the device side of the thermal pad if you have one.).

You are trying to get heat from the junction (i.e. the semiconductor) to air, which passes through several resistances - junction to case, case to heatsink (i.e. thermal pad), heatsink to air. Each of these will be specified by the manufacturer of the respective part, and you need to add them up and then multiply them by the power flowing through that resistance.

Section 11.3 of your datasheet covers 'thermal considerations' and has some real graphs based on real heatsinks - you could use those to check your assumptions.

(I think your heatsink is probably too small, but that's just a feeling)


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