I have a LM317T-based simple circuit and I'm using a 12V battery to power it. I connected the Vin to + 12V and made a constant current circuit as shown here. Now, I'm using a variable resistor of 100k instead of a fixed-value resistor and have connected a LED in place of RL of the diagram.

Now before connecting the + 12V to Vin, I set the resistor value to the maximum, 100k ohms. This is a really large value so the current flow to the LED should be so low that it shouldn't light at all. But the LED lights up very dimly. And if I reduce resistance on the variable resistor, the LED becomes brighter as current flow increases, as expected.

Even if I remove the 100k resistor and keep the path between Vout and Adjust open, the LED still lights up very dim. Is this normal? Why is this happening? Can anyone explain this behavior of LM317T?


1 Answer 1


Yes, this is normal. The internal circuits in any three-terminal regulator require a certain amount of "quiescent" current to operate correctly (some more than others), and this current flows through the "adjust" or reference pin.

When used in a current regulator configuration, this quiescent current becomes the minimum output current. Even "low power" regulators (Iq << 1 mA) will pass enough current to light an LED visibly.

  • \$\begingroup\$ What could be the voltage between 'adjust' and negative terminal of the 12 V battery in this case? I don't have a multi-meter yet, so I can't measure it for the moment. \$\endgroup\$
    – e_cobra
    Oct 28, 2017 at 18:29
  • \$\begingroup\$ It would be the battery voltage, reduced by the drop across the regulator. In a current source configuration, that voltage drop would be approximately the nominal output voltage of the regulator plus its rated dropout voltage (minimum Vin-Vout). \$\endgroup\$
    – Dave Tweed
    Oct 28, 2017 at 18:49

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