I have a mini water-pump/motor hooked up to an LM338 (similar to LM317, but capable of higher current output).

I set up the LM338 to create a 6.1-V output, from an 18-V input; this was done by setting R1 = 120-Ohm, and R2 = 470-Ohm in this circuit:

CASE 1: Using a DC bench Power-supply unit == SUCCESSFUL

When I power the regulator input with a power-supply unit (rated up to 3-A at 18-V), the motor runs off the regulator output very well.

CASE 2: Using a Laptop power adapter == UNSUCCESSFUL

But when I instead power the regulator input with a Universal laptop power adapter (rated for up to 6.5-A at 18-V), the motor exhibits this behavior: it ramps up and then within 1-2 seconds, slows down, and then after 3-4 seconds, runs much slower or not at all.

I also noticed the following in CASE 2:

  • when I initially touch the motor's V+ cable to the regulator output, it's creating a little spark at the contact.

  • the regulator gets really, really hot to the touch.

  • \$\begingroup\$ wild and probably wrong guesses: the laptop power adapter isn't really capable of its stated output, because it has failed in some way... or 2) it is a switching supply and some how that freaks out your regulator... try adding another capacitor... \$\endgroup\$ Commented Jan 16, 2015 at 1:21
  • 1
    \$\begingroup\$ @GradyPlayer: No luck with the capacitors; also, just noticed the regulator is getting extremely hot in Case 2 (the adapter). I'm thinking of testing a different adapter. \$\endgroup\$
    – boardbite
    Commented Jan 16, 2015 at 1:51
  • \$\begingroup\$ Are you just exceeding the power dissipation of the regulator and having some internal protection circuit shutting off turning on etc? Datasheet mentions something about some internal limiting. Would be interesting to measure voltage at input and output with a scope. \$\endgroup\$ Commented Jan 16, 2015 at 2:28
  • \$\begingroup\$ Almost certainly Eric Gunnerson got it right, but please tell us how much current is your motor/load actually drawing. The datasheet will should explain how to calculate dissipated power based on (Vout-Vin)*current. Also beware that motors can create back EMF and fry the regulator by raising its output above the input. So I would add protection diodes too. There's surely a schematic showing that in the datasheet of LM317/LM338 etc. \$\endgroup\$ Commented Jan 16, 2015 at 9:19

1 Answer 1


Here's what's going on.

The LM317 is what is known as a linear regulator. You are going from 18 volts to 6 volts. To do that with a linear regulator, the 12 volt difference is converted into heat. The regulator is trying to dissipate that, but it can't, so it's getting hot and going into thermal shutdown.

My guess is that the bench power supply is running a lower voltage than the laptop power supply

You have a few options.

You could reduce the input voltage to the regulator, to something like 9 volts. That would reduce the dissipation drastically, though it might not be enough. Note that linear regulators have what is called the "dropout voltage", which is a delta between the input and output voltage. The datasheet says it's about 3 volts, so 9 volts would be about the minimum.

You could add a heat-sink to the regulator; that would allow it to dissipate more heat. It would still get hot, but it would let the heat get away more easily.

You could use a different kind of power supply - known as a switch power supply (in this case, running in "buck" mode) that can convert from 18V to 6V at a much higher efficiency. They are relatively cheap.

Or, you could find a 6V power supply / adapter that will give you the current that you need. That would be the simplest to do.


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