This question isn't so much about my specific case, but rather, I'm looking for a general rule for estimating the max current that can be pulled from heatsink-less LDOs at some voltage difference.

I know this will depend on the device and package size, so lets just assume TO-220 for now since its common, or feel free to constrain other LDO assumptions as necessary to fit your answer. As I understand, increased voltage increases dissipation power, but at what scale? Is it quadratic with the voltage delta?

(As a single example, I was wondering if I can use this 3.3v LDO at 12V. According to the specs, it has a 15V max, but 3.3v to 12v is a pretty big jump and seems likely to get very hot.)

I looked around for some kind of graph like voltage-delta vs watts at some mA value but haven't found anything. The datasheet noted above doesn't have any graphs at all.


2 Answers 2


It’s all in there. First calculate the power loss in watts: Heat=(Vin-Vout)*Iout.

Then check the thermal resistance. For a to-220 you should see two numbers. Tjc is junction to case which only applies if you have a Heatsink (this is usually about 1). Tja is junction to ambient which is probably about 50.

This means it rises ~50 degrees per watt in still air. Assuming 25C ambient and a junction max of 175 you can dissipate 3 watts. But 175C is hot so I consider 1W about the limit for a TO-220 in free air.

If you add a Heatsink redo the math with Tjc plus the thermal resistance of the Heatsink.

Note this math applies to most all power parts like mosfets. They may have a crazy watts rating on the front page but it’s based on an ideal heatsink. Physics is physics so all similarly sized parts are going to have similar thermal impedances to the air.


The power dissipated by this kind of regulator is just \$P_D= Iq\cdot V_{IN} + I_{OUT}(V_{IN}-V_{OUT})\$.

You have to figure out the allowable power dissipation based on the ambient temperature limit and heat sinking, and then the maximum current for a given maximum input voltage can be calculated.

For example (conservative industrial choice) I may think 0.6W is okay for a TO-220 package without heat sink. Iq is 10mA maximum, so there is 120mW maximum without any output, leaving 480mW. That means 55mA would be the maximum continuous output current, given that set of assumptions.

  • \$\begingroup\$ Can I ask about how we get 0.6W? Is that somehow derived from junc-to-ambient 50 °C/W (linked datasheet)? \$\endgroup\$
    – jonathanjo
    Commented Sep 17, 2023 at 22:41
  • 4
    \$\begingroup\$ It's my rule-of-thumb that I calculated years ago. Let's see if I can recreate it. Maximum Tj = 100°C for high reliability. Maximum ambient temperature 70°C. Datasheet thermal resistance J-A 50°C/W, so that would lead to 0.6W. Different assumptions will obviously lead to different results. If you're willing to allow 150°C junction and assume it will never get hotter Ta than (say) 40°C you get to 2.2W, more than 3.5x higher. \$\endgroup\$ Commented Sep 17, 2023 at 22:47
  • \$\begingroup\$ Worth mentioning that you can frequently ignore the Iq·Vin term, as, especially with modern regulators, Iq is likely much smaller than Iout. \$\endgroup\$
    – Hearth
    Commented Sep 18, 2023 at 2:19
  • \$\begingroup\$ @Hearth True. And for some older regulators the Iq goes asymptotic (like hundreds of mA) near the dropout due to the crummy lateral PNP pass elements. Fortunately, that's also where the power dissipation due to voltage drop is minimum. Eg. L4947 \$\endgroup\$ Commented Sep 18, 2023 at 2:41
  • 1
    \$\begingroup\$ @jonathanjo Note that the assumption "it will never get hotter Ta than (say) 40°C" essentially means a breadboard / bare PCB setup. Once you put a 2+W device in a box, Ta will inevitably rise. \$\endgroup\$ Commented Sep 18, 2023 at 12:10

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