The LM317 can produce a Vout from 1.25V to 37V. I have a small piezo pickup I'd like to step down to as close to 1V as possible. The formula for the LM317 is roughly Vout = 1.25(1 + R2/R1). Can I just omit the resistor on the adjust pin so that R2=0, therefore getting me 1.25V? Do I still need to provide a resistor on Vout so that R1 is non-zero?

Am I going about this wrong?

  • \$\begingroup\$ Your formula for the output voltage is wrong. It should be \$V_{out}=1.25\left(1+\dfrac{R_2}{R_1}\right)\$. Now it should be obvious what happens if R2 goes to 0 or R1 goes to infinity (removed and left open circuit). \$\endgroup\$
    – The Photon
    Apr 18 '16 at 0:28
  • \$\begingroup\$ Ah corrected. So R1 -> infinity would effectively render the ratio zero, correct? \$\endgroup\$ Apr 18 '16 at 0:30
  • \$\begingroup\$ Depending on the current needed, adjust the LM317 to a voltage that can be dropped by a simple diode. With a standard silicon diode (1N400x) that has a V<sub>Forward</sub> of 0.7V, go for 1.7V out. With a germanium diode of 0.3V V<sub>Forward</sub>, for for 1.3V. Or simply go with the reference voltage output of 1.25V, the 0.5V difference is minimal. \$\endgroup\$
    – Passerby
    Apr 18 '16 at 3:06
  • \$\begingroup\$ Diode forward voltage drop does change with current level so regulator followed you "diode drop" will not yield a constant voltage across the the 0 to FullScale current loading. \$\endgroup\$ Apr 18 '16 at 4:43

To get down below 1.25V (nominal) with an LM317 you would need to provide a negative supply.

enter image description here

You can substitute an inexpensive LM431 for the LM329.

Frankly, it would be easier to use a regulator that has a lower reference voltage than 1.25V.


The LM317 regulates the voltage on its output such that the ADJ pin arrives at 1.25 volts. It will increase OUT if ADJ is below 1.25 volts, and decrease OUT if ADJ is above 1.25 volts.

If you would like a 1.25 volt output, simply tie OUT to ADJ - no resistor needed.

It can not produce a voltage lower than 1.25 volts, but can work as a current regulator with a single "current sense" resistor, sometimes negating the need for a lower voltage. Or you can make a divider network with two resistors, but the voltage would be load-dependent.

As requested, Here is a datasheet: http://www.ti.com/lit/ds/symlink/lm317.pdf

  • \$\begingroup\$ I would upvote this answer if ut (1) linked to the datasheet, and (b) used the phrase "... 1.25 V output, simply tie ADJ to GND ..." \$\endgroup\$
    – davidcary
    Apr 18 '16 at 3:24
  • \$\begingroup\$ @davidcary, Hi David - Thanks for the suggestions. I'm not sure what you mean by tying ADJ to GND. The LM317 has no ground pin (the device floats), and tying ADJ to the input's ground will cause no regulation at all (the output will equal the input minus a voltage-drop). \$\endgroup\$
    – Mark
    Apr 18 '16 at 7:58
  • \$\begingroup\$ By "tying ADJ to GND", I mean the circuit shown in Figure 16 of the datasheet you linked to, with the variable resistor R2 adjusted to produce the minimum output voltage -- zero ohms producing 1.25 V output, as Photon already pointed out. I agree that the LM317 has no ground pin, but I am mystified at your further comments -- probably one of us is overlooking something very simple. \$\endgroup\$
    – davidcary
    Apr 27 '16 at 4:22

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