I try to regulate a 12V into 3.7V using an LM317 voltage regultor.

I used R1=220Ohm and R2=470Ohm, using this circuit:

this diagram

It regulates a 3.9V (close enough for me,) but it heats up. I can't even touch it while it is powering an ESP8266 ( which draws a max of 200mA.)

Why is it getting hot?

  • 2
    \$\begingroup\$ You have to put a heatsink on it, or use a lower input voltage. \$\endgroup\$ Jul 4, 2020 at 19:22
  • \$\begingroup\$ Use a switching buck/step down regulator. \$\endgroup\$
    – Passerby
    Jul 4, 2020 at 19:27
  • \$\begingroup\$ @SpehroPefhany I need to use a 12v power supply to light up a led strip. Using 2 power sources is not an option \$\endgroup\$
    – guyd
    Jul 4, 2020 at 19:27
  • \$\begingroup\$ Then use a switching regulator module or add a heat sink to dissipate all your wasted energy from using a linear regulator. \$\endgroup\$ Jul 4, 2020 at 19:29
  • \$\begingroup\$ @SpehroPefhany can you please ref ? \$\endgroup\$
    – guyd
    Jul 4, 2020 at 19:36

1 Answer 1


Yes, under the conditions you are using the regulator, it will heat up a lot, as it is a linear regulator. It will dissipate all excess power as heat.

Since the voltage over the regulator input and output is about 8V, and the maximum current draw is about 0.2A, the regulator will have to dissipate maximum of 1.6W as heat.

The junction to ambient thermal resistance is about 25 degrees C/W, so as an extremely crude approximation, chip case can be approximated to be about 40 degrees higher than ambient room temperature.


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