# Why does the LM317 generate VOUT of ~1.7V with Adj at ground, instead of 1.25V?

According to the datasheet equation of the LM317, if R2 is 0, then Vout should always be 1.25V. The Wikipedia page of the LM713 says the same thing.

I have five LM317s, and if I ground the Adjustment pin, they all produce ~1.65 to 1.7V on Vout and not 1.25V. I tested with 1k, 10k and 100k load.

My test schematic is trivial: Vin = +5V (from a fancy power supply), ADj = GND, measure GND to Vout, plus .1uF and 1uF caps.

I'm looking for a temperature and Vin independent way to generate 1.25V reference and I thought since I had some of these they would work...

UPDATE: linear regulators need a load (see answer below). However the min current is 10mA, which is quite a bit for a battery op device. I found an LM385 which is a 1.2V diode shunt that only requires 10uA.

• How much current is drawn from output? Why you don't have any caps, the device may now be an oscillator instead of regulator, but you can't see that without an oscilloscope. Mar 29, 2020 at 17:43
• @Justme Caps are for transients. I added them to prevent further digression. Same issue. No oscillation. Assume >10M load so no Current draw. LDOs don't need a load to generate their voltage (e.g., put 12V on a 7805 and you will measure 5V at Vout with no load). Mar 29, 2020 at 17:47
• @PeterT, LM317 does need a load to produce an accurate output voltage. See my answer. Mar 29, 2020 at 17:49

## 1 Answer

Most likely, you didn't provide any load resistor.

The output voltage of the LM317 is specified with a minimum load current of 10 mA.

With lower load current, the output voltage will increase noticeably.

• Thanks! That solved the problem. Mar 30, 2020 at 20:28