I've added a small 12V LED light strip to my 3D printer which has 24VDC output. The LEDs have current limiting resistors and with the spec sheet I estimated the current draw to be 200mA. I had a L7812CV laying around, and I looked at its datasheet (https://www.st.com/resource/en/datasheet/l78.pdf). It is a 12VDC regulator supporting 40VDC max input and 1.5A max current. Since my estimated current was 200mA, I assumed there would be so little heat generated that it wasn't worth computing. I plugged it in with an ammeter inline. The LEDs were the expected brightness, and it was indeed drawing 200mA at 12V, but within a minute or so the L7812CV was too hot to touch, so I turned it off and added a resistor in series to reduce the current to 60mA until I figure out what was wrong.

My question is, how could I have known by looking at the datasheet that using the L7812CV to drop the voltage from 24V to 12V at 200mA would have produced temps above 130F/55C?

General Description

Electrical Characteristics


2 Answers 2


The L7812 is a linear regulator. The power dissipated by any linear regulator is a minimum of Pd= (Vin-Vout)Iout. In this case, that would be 2.4W. That alone is enough power to heat the junction up by 120°C above ambient (assuming no heatsink).

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In addition there is a term Vin * Iq where Iq is the quiescent current. Iq is a maximum of 8mA so that power is an additional 200mW or so (typically it's more like 5mA so about 120mW).

That gives a total of more than 2.5W typically, resulting in a rise of more than 125°C above ambient. \$\Delta T = \text P_{\text d}\cdot \text R_{ \text {thJA}}\$.

That is a safe power dissipation given a moderately large heatsink, however it's pretty wasteful (less than 50% efficient).

A switching regulator would typically be much more efficient, but is a more complex and sensitive to proper layout and choice of inductor:

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This particular one is optimized for higher output currents than your 200mA.


The power being shed in the regulator is simply the current times the voltage drop in the regulator. So 0.2A x (24V-12V) = 2.4W. That’s quite a lot, and would need a heat sink to keep it cool.

You can consider using a DC-DC step-down which will be more efficient. There are versions that fit in the same footprint as the LM7812.

You can also consider an LED constant-current driver module. Here’s a bunch of them from Mean-Well: https://www.meanwell-web.com/en-gb/dc-dc-led-drivers/

  • \$\begingroup\$ Might as well tell the OP how to use Watts dissipated with the thermal resistance in the datasheet to estimate temperature. \$\endgroup\$
    – DKNguyen
    Commented Sep 23, 2021 at 5:13
  • \$\begingroup\$ Thanks @hacktastical, I accepted the other answer because it showed where to find the heat per watt for the package style, but I appreciate the concise (and sort of embarrassingly obvious in hindsight) point that I'm getting rid of 12V*0.2A=2.4W somewhere, and since the current is constant, the only place to go is heat. Also, thanks for the Mean Well DC-DC driver recommendations! \$\endgroup\$
    – SteveK
    Commented Sep 23, 2021 at 14:03

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