I wanted to make a power supply providing roughly 8V (+-2V) for small currents, given a whole range of different DC voltages (let's say about 12-30V).

For that I wanted to use an LM317L (link to pdf).

Setup: I hooked a adjustable benchtop PSU to INPUT of the regulator and GND, put a 100 ohm resistor between ADJ and OUTPUT and a 510 ohm resistor between ADJ and GND.

I choose these resistors to have an output voltage of Vout = 7.6V = 1.25(1 + 510 / 100), while loading the regulator with 7.6 / (510 + 100) = 12.4 mA (the specs say the load needs to be at least 5mA). I omitted the capacitors for this experiment.

Issue: As I slowly increased the input voltage, the output voltage also increased up to about 8V. When increasing the input to about 25V, the output still had a stead 8V. Increasing the input to about 30V made the output increase to about 18V.

When I replaced the regulator with another (LM317L) one, I got a similar erratic behaviour again when the input exceeded 25V. Replacing it with same type regulator in the larger package (LM317 pdf) everything worked fine and I got around 8V for any input of 8-30V. Apart from the amount of output current they can provide, both of them should behave identically as far as I understand.

Can anyone point out what I'm doing wrong or what I possibly misunderstood?

  • \$\begingroup\$ How long leads from power supply to regulator? Did you build this on a breadboard? \$\endgroup\$
    – Justme
    Dec 25, 2020 at 11:52
  • \$\begingroup\$ I tried it on a solderless breadboard as well as directly deadbug soldering it. The leads were some alligator clip cables about 30cm long. \$\endgroup\$
    – flawr
    Dec 25, 2020 at 12:01
  • 1
    \$\begingroup\$ And you have the decoupling capacitors as recommended in the datasheet? \$\endgroup\$
    – Transistor
    Dec 25, 2020 at 12:10
  • 1
    \$\begingroup\$ 20V Vin-Vout, 12.4mA, it's failing at about 250mW internal dissipation, but the same chip in a larger package works .... hmmm ... what's the thermal resistance of that package? come to that, which package? One of them is 290C/watt... \$\endgroup\$
    – user16324
    Dec 25, 2020 at 12:54
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    \$\begingroup\$ To get 100mA out, Vin-Vout would have to be much lower. I'll not go into the precise max power, I prefer to rate these things conservatively and if it fails the finger test, improve the heatsink. One experiment : Have you tried doubling the resistors to see how well it does at 6mA? If I'm right it'll work; if not, look elsewhere. \$\endgroup\$
    – user16324
    Dec 25, 2020 at 14:33

2 Answers 2


There are 3 issues that must be followed given in the datasheet and a couple other comments;

  1. Input cap near the regulator to gnd to suppress input noise
  2. Output cap near regulator to gnd to suppress load-induced noise near or above internal gain bandwidth, GBW
  3. Junction Temperature rise estimate T.rise['C] = Vdrop * I.load * Rja['C/W] of case must be well within thermal cutout temp. Rja may be reduced with a heatsink (Rca, case to ambient), Rja=Rjc+Rca
  • Aim to leave 50% junction temp. rise margin by adding heatsink as required when using large voltage drops

The datasheet shows how to use the LDO as a high gain amplifier. This means ground noise and long wire loops may pick up noise, is amplified greatly and is a source for instability, which you will see when R2 gets larger for higher Vout. This gain is used internally to reduce the error between the ratio of Vout and the internal 1.25V reference between Vout-Vadj.

  1. Avoid ground loops where load currents are shared by input ground wire to R2 and Cinput as this will become amplified voltage and degrades performance.
  2. If R1 =240 Ohms everywhere in App note , why use 100 Ohm?
  • \$\begingroup\$ Thanks for the suggestions! So now I found out that the caps really didn't change anything in my observations: I noticed that with an input voltage of >= 25v, and you current draw was around 10mA, the regulator couldn't do its job anymore. As far as I understand now this is due the junction temperature rise you explained. I wrongly believed that these 100mA mentioned in the datasheet were the minimum it could provide under all permissible factors, but now I understand that this is probably the maximum under ideal conditions. I'm using a bigger package now with a heatsink and it works fine. \$\endgroup\$
    – flawr
    Dec 27, 2020 at 14:15

The problem really is about not using capacitors. And possibly using the breadboard makes things worse.

The datasheet specifically mentions that if the input leads are longer than 6 inches (15cm), the input capacitor is recommended.

The datasheet also mentions the device is stable wihtout any output capacitor, except that there is excessive ringing when the output capacitance is between 0.5 nF and 5 nF, so the breadboard can make things worse.

So it is possible that the output is actually wildly oscillating, but your meter simply reports back a DC voltage value so you don't see the instability and oscillation.

  • \$\begingroup\$ Thanks for elaborating, there is indeed an oscillation starting as soon as we cross 25V at a few thousand hertz, but the amplitude seems to be below 1V. (I checked with an oscilloscope.) But even when using leads that are as short as possible as well as a 0.1uF or 1uF capacitor, the voltage still increases. \$\endgroup\$
    – flawr
    Dec 25, 2020 at 13:09

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