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I'm designing a voltage regulator using the LM317, and I've seen people using 2 different configurations when it comes to the potentiometer. I tested both on a simulator and on a breadboard, and both worked. So, is there a best way to use such regulator? enter image description here

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    \$\begingroup\$ Worth mentioning the difference in behavior should the potentiometer wiper lose contact: The top one will output it's maximum voltage, the bottom one close to 1.25V... \$\endgroup\$ Commented Apr 24, 2021 at 19:42
  • \$\begingroup\$ The datasheet shows your 240 ohms resistor only for the more expensive LM117. Use 120 ohms and re-calculate the value for the pot so that the output voltage does not rise when there is no load. \$\endgroup\$
    – Audioguru
    Commented Apr 24, 2021 at 19:48

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I've never seen the second circuit recommended in any datasheet. In the TI one, you will find this note:

the LM317-N was designed to minimize \$I_{ADJ}\$ and make it very constant with line and load changes.

Which means that the resistor in series with the ADJ pin will not make the load and line regulation noticeable worse. On the other hand, from the same datasheet, you see that the ADJ pin current varies considerably with temperature:

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and ST datasheet also documents this:

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Both datasheets list 100 \$uA\$ as a worst case. TI shows a typical 50 \$uA\$ and ST a typical 65 \$uA\$.

The ADJ pin current is an error term in the output voltage calculation (see below, from the TI datasheet). Putting a large resistor in series with this current changes the formula and also makes the output voltage adjustment drift more with temperature. Even with small temperature differences (10 - 20 deg.), a couple of % change in the output voltage should be expected.

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Summary: if you don't mind the small non-linearity while adjusting the voltage and also having to fine tune the voltage adjustment every time you change the load and the component heats up or cools down, the second circuit is ok.

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The second one does not draw the minimum required current for regulation (3.5mA typical, 10mA maximum) from the LM317, so the output voltage without sufficient load will tend to rise to near the input voltage.

Also the relatively small (< 100uA) ADJ current will cause the voltage to drift significantly as the regulator heats (and with ambient changes). Vangelo has posted an answer showing the typical temperature variation. It will also degrade the line regulation since the ADJ pin current varies with Vin.

The first one, if R1 is sufficiently low for the LM317 in use, will draw a constant current that exceeds that minimum, which also minimizes the wasted power at higher voltage settings compared to a simple resistive load.

As @Audioguru mentions in a comment, the commonly-shown 240 ohm value results in 5.2mA nominal (5mA minimum) which is okay typically but does not meet datasheet worst-case conditions. It's probably okay with 9V in according to the original NS datasheet:

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    \$\begingroup\$ I mentioned a 120 ohm s resistor because one of the pots will create a voltage range as high as 37VDC where some LM317 ICs must have a minimum operating current of 10mA when the output voltage is set low. \$\endgroup\$
    – Audioguru
    Commented Apr 25, 2021 at 0:52
  • \$\begingroup\$ @Audioguru Agreed, there is no guarantee with > 120 ohms. Worst-case is high Vin, minimum load and cold temperature. \$\endgroup\$ Commented Apr 25, 2021 at 1:35
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The first is correct.

Infact, under any potentiometer condition/position, R1 is always greater than zero and closes the mandatory, for the LM317, negative feedback circuit.

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Both are ok, but the one above has the advantage of not giving unnecessary resistance to the bias current of pin ADJ. In thee second one, in extreme case, the resistance seen by ADJ may go to 0. Then I would look at two points:

  1. You should consider changes with temperature and tolerances: potentiometers have usually a significant temperature coefficient, so that you may want to minimize the amount that is trimmed with a potentiometer in favor of more stable chip resistors. That means putting a resistor in series towards the ground, and trimming only a smaller resistance value.
  2. Bias current into pin ADJ and ways to minimize the consequential voltage drop or to keep it constant, so it can be compensated in the overall transfer ratio.
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    \$\begingroup\$ Both are ok even when it comes to a full load operation e.g. a load draining 1A? They both changed the voltage when I spin the potentiometer, but I was testing on a no-load scenario. \$\endgroup\$ Commented Apr 24, 2021 at 19:44
  • \$\begingroup\$ I honestly do not see a problem at full load. Only, the second scheme can go to 0 ohm seen by ADJ and that may be a problem; then accuracy of output voltage is influenced by the amount of resistance seen by DJ ad how this resistance changes with temperature. If you do not need full range of adjustment, use smaller potentiometer and put a slack resistor beneath it. If you need full rage changes, but accurate, you can think of a rotary switch changing a fixed value resistor ad the potetiometer take care of fie adjustment. \$\endgroup\$
    – andrea
    Commented Apr 24, 2021 at 19:54

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