Using the formula PowerDissipated = (Vin - Vout) * Iout where:
Vin = 12 V
Vout = 2 V
Iout = 450 mA
Power dissipated = (12 - 2) x 0.45 = 4.5 W

And heat sink value = (MaxTemp - AmbientTemp) / Power dissipated
Power Dissipated = 4.5 W
MaxTemp = 60
AmbientTemp = 20
Heat sink value = (60 - 20) / 4.5 = 8.88

So I need a heat sink value less than 8.88 °C/W - is that correct?

I'm using a linear regulator LM317 by the way

  • \$\begingroup\$ What's the MaxTemp=60C referenced to? Is that the case? The die? \$\endgroup\$
    – SteveSh
    Commented Mar 29, 2021 at 23:50
  • 1
    \$\begingroup\$ Note that 60 C is rather too hot to touch but we'll below what is acceptable in many cases even allowing for junction-case and cade-sink thermal resistances. \$\endgroup\$
    – Russell McMahon
    Commented Mar 30, 2021 at 0:18
  • \$\begingroup\$ I know you probably know that (otherwise you wouldn't explicitly mention this being a linear regulator), but this is a classical use case of switch-mode regulators. The money you save in not needing a heat sink can quickly outgrow the cost of a Buck regulator. \$\endgroup\$
    – mmmm
    Commented Mar 30, 2021 at 6:17
  • \$\begingroup\$ This answer of mine is liable to be of interest. \$\endgroup\$
    – Russell McMahon
    Commented Mar 30, 2021 at 7:26

2 Answers 2


You're on the right track but there are two numbers you're missing: maximum junction temperature and the thermal resistance between the junction and case. Going off of ST's datasheet, the maximum junction temperature is 125C and thermal resistance between the junction and case is 5 C/W. I'd go with an ambient temperature of at least 30 (wouldn't want your circuit to burn up on a hot day) and put a reasonable buffer in for additional resistance between the case and heatsink (which you can reduce with thermal grease). With a maximum junction temp of 115, ambient of 30, that means you need a heatsink with a natural convection resistance to ambient of 13.9 C/W or less. There are plenty of heatsinks out there with that level of performance but some of the little dinky stamped aluminum ones are up around 30 C/W so make sure you check first. Also: those numbers are assuming you're using a TO-220 package.

  • 1
    \$\begingroup\$ I agree with your 13.9 deg C number. But that's for the entire case to ambient path. You can't allocate all of that to the heatsink itself, because you're going to have some thermal resistance between the case and the heatsink, because that interface is not going to be perfect. \$\endgroup\$
    – SteveSh
    Commented Mar 29, 2021 at 23:55

Without seeing the datasheet it is hard to be certain, but usually you would calculate the maximum thermal resistance as

\$\Theta_{TOTAL} = \frac{T_{DIEMAX}-T_{AMBIENT}}{P_D}\$

Where \$T_{DIEMAX}\$ is the maximum allowed temperature of the die in the packaged regulator. Then you can calculate the maximum thermal resistance for the heatsink as

\$\Theta_{HS} = \Theta_{TOTAL} - \Theta_{JC} - \Theta_{CS}\$

where \$\Theta_{JC}\$ is the junction-to-case resistance of the device package and \$\Theta_{CS}\$ is the resistance from device case to the heat sink (perhaps a silicon insulator and thermal paste).


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