So, some time ago I bought some LM317T voltage regulators. Today, I made a 50mA constant current circuit like this:

enter image description here Now, before you tell me that my circuit is wrong and that it should be wired in another way, let me tell you: it worked. I mean, I tested it with a 150 ohm load and even connecting the output directly to my multimeter, without load and it read 53mA in both cases (probably due to the resistor tolerance, it was supposed to be 50mA).

Some time later, I found out that the pinout was wrong and connected it like this:

enter image description here It started overheating a lot suddenly. And it blew my 400mA fuse in the multimeter. Later on, I measured about 1.6A of current.

I'm sure it's a LM317T, look for yourself:

enter image description here

Can anyone explain me what is happening? My regulator behaves like this:

  • Pin 1 - Vin
  • Pin 2 - Adj
  • Pin 3 - Vout

I tested it with 4 of my regulators and they all behaved like this! Is there something wrong with the resistor network?

  • 2
    \$\begingroup\$ Where did you put the multimeter when you measured the current? Please show a schematic instead of those meaningless cartoon drawings. \$\endgroup\$
    – The Photon
    Jan 23, 2018 at 18:41
  • \$\begingroup\$ Pinout is 1: ADJ 2: Vout 3:Vin \$\endgroup\$
    – Whit3rd
    Jan 23, 2018 at 23:01
  • 1
    \$\begingroup\$ Fritzing should give you the option of creating a schematic instead of a wiring diagram. The wiring diagram does not show the scheme of the circuit and so we can't understand what it's supposed to do. \$\endgroup\$
    – Transistor
    Jan 23, 2018 at 23:54

1 Answer 1


Neither of your circuits is correct. The correct circuit looks like this:


simulate this circuit – Schematic created using CircuitLab

enter image description here

If you connect it incorrectly (as in the first circuit) it may act as a zener, resistor, zener + resistor, etc. and give you some voltage drop or whatever. I doubt you tested the I-V characteristics thoroughly over a range of loads in order to conclude that it was "working".

The second of your circuits will current limit at an amp or two, because of the internal limiting in the chip. The resistors do nothing of value.

  • 1
    \$\begingroup\$ Yeah, it was just the good old "hooking up things backwards" and having no experience with pinouts, schematics or datasheets. I'm very late, but thanks. \$\endgroup\$
    – Mark42XLII
    Nov 10, 2020 at 10:16

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