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I am trying to use a voltage regulator to regulate a 12V, 2A DC Power supply down to ~5.2V with a load current of less than 0.5A.

My schematic is as follows: Schematic

Problem: the regulator is getting quite hot, and voltage is dropping down to 4.3V or so with my fan connected. The output voltage is correct (~5.2V) with no load connected.

This is the datasheet for the regulator. The 10 uF capacitor is a tantalum capacitor (16V), and the 22 uF capacitor is an aluminum capacitor (20V). I do not have a heatsink on the regulator.

Is there something I am misunderstanding about the application? Is there a different regulator that would perform better?

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  • \$\begingroup\$ try adding 100nF ceramic between the input of the regulator and the ground end of the 560 ohms. \$\endgroup\$
    – Jasen Слава Україні
    Mar 2, 2020 at 4:03
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    \$\begingroup\$ at 3.5 W dissipation the LM317T will need a heatsink. \$\endgroup\$
    – Jasen Слава Україні
    Mar 2, 2020 at 4:09
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    \$\begingroup\$ Check datasheet for capacitor recommendations. || Your regulator dissipation is about 14 Watts - you'd been a significant heatsink to handle that. (Power = V x I = (12-5)*2A = 14W. Without a heatsink "you haven't got a show" to use technical language :-). \$\endgroup\$
    – Russell McMahon ♦
    Mar 2, 2020 at 4:09
  • \$\begingroup\$ The datasheet does not show the thermal resistance in C/W for the T model regulator but at say 100C rise you'd need a C/W = delta_T/Power = 100 / 14 ~= 7 C/W heatsink. As a TO3 (large metal ) can has 39 °C/W thermal resistance and a TO220 will be quite a lot higher, you must have a heatsink OR get rid of some heat. || If you add a series resistor in the input you can dissipate most of the heat in the resistor. To give the 317 2v headroom and 1.25V for regulation ... \$\endgroup\$
    – Russell McMahon ♦
    Mar 2, 2020 at 4:15
  • \$\begingroup\$ Vdrop_R = (12-5-3-1.25) = 2.75A. R = V/I = 2.75/2 = 1.4 Ohms. Probably even more at say R=3/2 = 1.5 Ohhms (limit is when Vout is 5V at full load). Power reg is now about P=v*I = (12-5-3) x 2 A = 8W so you STILL need some sort of heatsink \$\endgroup\$
    – Russell McMahon ♦
    Mar 2, 2020 at 4:17

1 Answer 1

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Your regulator dissipation is about 14 Watts - you'd need a significant heatsink to handle that.
Power = V x I = (12-5)*2A = 14W.
Without a heatsink "you haven't got a show" to use technical language :-).

The datasheet table 7.5 page 6 shows the thermal resistance in C/W for the T model regulator as 23 °C/W, - say 25 ° C/W, so say 100C rise you'd need a C/W = delta_T/Power = 100 / 14 ~= 7 C/W heatsink. That's a largish heastink.

If you add a series resistor in the input you can dissipate most of the heat in the resistor. The 317 needs about 3V drop / "headroom" at 2A (IF your version will handle 2A) (see datasheet)

To give the 317 3v headroom for regulation

Vdrop_R = (12-5-3) = 4V
R = V/I = 4/2 = 2 Ohms.
Limit is when Vout is 5V at full load.

Power reg is now about P=V*I = (12-5-4) x 2 A = 6W so you STILL need some sort of heatsink. For 100C rise Rth_sink ~= 100/6 = 16 C/W

The resistor power dissipation is about V x I = 4v x 2A = 8W.
So use 10W minimal and better 20W. 2 x ceramic 10 Watt air cooled cheap would work well.

THIS SE Q&A will tell you more than you want to know about regulator overheating

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