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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

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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).

enter image description here

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:

enter image description here

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

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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/

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  • \$\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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