# How much power does LM7805 dissipate?

I'm using an LM7805 in voltage regulator mode.

• Vin is 15V
• Vout is 5V (regulated)
• The attached load draws 200mA

How to calculate how much power the LM7805 would dissipate in this setup (and whether or not I need a heat sink)? Couldn't understand it from looking at the datasheet.

P.S. I'm asking because it gets quite hot, which I didn't expect. Checked the attached load, it is around 0.2A, so well within the limit of 1.5A.

• For a linear regulator, Iout is roughly equal to Iin. All voltage dropped across the pass device is dissipated as heat. – M D Mar 30 at 10:55
• Wait, this has been asked multiple times, see here. – Unknown123 Mar 30 at 12:38
• Also there is more detailed datasheet here – Unknown123 Mar 30 at 23:37

Take a look at package thermal data section in your datasheet page two.

Depending on the package of the regulator you are using, lets say the thermal junction-ambient coefficient $$\\theta_{JA}\$$ is roughly $$\20°C/W\$$.

You've got $$\P = V . I = (15-5) .(0.2) = 2W\$$ dissipated as heat.

If your ambient temperature is $$\25°C\$$, then the regulator would heat up more or less into $$\65°C\$$.

It is quite hot for sure.

• 2W is about the limit for use without a heatsink. – Jasen Mar 30 at 11:48
• Yes, I'm very linearizing here for simplification. The ambient will also heat up which eventually also increase the regulator temperature. Also, in the NOTE 1 below the package thermal table in the datasheet OP linked there are maximum power dissipation formula and careful warning statement which would be useful for OP while learning this topic. – Unknown123 Mar 30 at 11:56
• Also note that, different manufacture or series may have slightly different specification such as $R_{\\ThetaJA} = 23.9°C/W$. Thus, uA7805 may behave slightly different with LM340-5 or LM7805, see here at Thermal Information section page four. – Unknown123 Mar 31 at 0:25