I have a small demo breadboard circtuit based on this design: enter image description here

I use this transformer, it's connected for 115V input on primary, and 2x24V output, and I'm using only one of these outputs at the moment...


So, what happens is that for some dummy load that I have, which works on 12v and consume just 130mA, my voltage from this PSU which is adjusted to 12v drop to 2.8V when I attach that load?!

What's wrong with this and how's that I have so much voltage drop?

I triple checked all the parts and connections.

Voltage which comes on rectifier is 27VAC.

Voltage at C1 is 37.4VDC.

Voltage on ADJ pin on LM317 without load is 3.1V, and when I attach load of 130mA it drop to 0.3V.

Voltage on C1 stay still at 37.4V with and without load.

Note, only different part on my breadboard is that I didn't use D1-D4, I used actual rectifier, but that shouldn't make any difference.

And for R2 I used 10k (I didn't have 12k or any other resistor co combine them to actual 12k).

I added these silly breadboard images, maybe someone will see what I couldn't...



I tried to draw how I connected it: demo.png

  • 1
    \$\begingroup\$ We need more information: for example Does the voltage drop immediately or take a few seconds. What was the voltage at C1? Do you have a lower load for example a 1K resistor? Do you ave cheat sink on IC1? Is it getting hot? \$\endgroup\$ Commented Nov 28, 2016 at 0:20
  • \$\begingroup\$ It drop right away. I don't have heat sink at the moment. I did have earlier but it was the same situation. And voltage on C1 is 37VDC \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 0:23
  • 1
    \$\begingroup\$ Can you adjust the voltage to 12V OK? What is the voltage at the Adj pin of IC1? \$\endgroup\$ Commented Nov 28, 2016 at 0:29
  • \$\begingroup\$ Yes, I can adjust it to 12V. I didn't measure, give me a few min to check it again. \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 0:30
  • 1
    \$\begingroup\$ Does the voltage on C1 remain near 37 volts when your 130 mA load is connected to the supply? \$\endgroup\$ Commented Nov 28, 2016 at 0:35

3 Answers 3


You are probably dissipating too much power in the regulator and sending it into thermal shutdown.

Input voltage of 37.4V - 12V output gives 25.4V across the regulator. 27.4V * 130ma load gives you 3.3 Watts power dissipated in the regulator. The LM317 in a bare TO-220KCT package has a thermal junction-to-ambient resistance of about 38 degrees C per Watt -- in this case that gives you a junction temperature of 125 degrees above ambient. Assuming a 20 deg C room temp, the junction will be around 145 degrees C. This is getting really close to the absolute maximum junction temperature (150C), and well over the recommended max of 125C. Depending on airflow, thermal connection to the leads, etc, the junction temp could easily be high enough to cause thermal overload shutdown.

Use an adequate heatsink.

  • \$\begingroup\$ Hi Paul, I put a big heatsink, but I still have the same results. It looks like something is wrong with wiring. \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 3:21

It looks like from the photo you have IC1 incorrectly wired. The pinout is Adjust, output, input when looking from top i.e. plastic side. It looks like you have the middle pin as adjust.

enter image description here


Edit - improve description of pinout.

The voltage across R1 should be 1.25V of the regulator is working properly. A faint possibility is that the regulator is oscillating due the long wire lengths you are using. I would put C2 as close to IC1 as possible and try to reduce wire lengths, especially the ground wires. If you had an oscilloscope it would be good to look at the various voltages to ensure they are clean. I have never had a problem with the LM317 although I did once have one with a 7805 (fixed output regulator).


This circuit should regulate properly but the breadboard is not the ideal implementation - your wires are rather long.

The voltage across IC1 is rather large for a 12V output - it will dissipate a lot of heat, a lower voltage transformer would be better. C1 is also rather small for a high current output - the ripple will be too large at high currents. At 1Amp output the ripple will be ~10V.

You need to be careful where you measure the voltages as the drop along wires may be significant. IC1 will regulate the voltage between the bottom end of R1 and the output pin of IC1 - if there are voltage drops along the wires it will not know about them and will not be able to compensate. make sure you are not sharing the load current along with your measurement wires. Connect the voltmeter directly to IC1 output and the bottom of R1.

  • \$\begingroup\$ No, red wire (which continue from same line where's yellow from pot) from that image that I shared goes on pin 1 for ADJ. Purple wire goes to input and green from output where I attach my load... \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 1:24
  • \$\begingroup\$ I tried here to draw what I have on my breadboard: ![demo.png](postimg.org/image/yyagtczrj) \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 1:49
  • \$\begingroup\$ Across R1 I have 3.2v \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 3:25
  • \$\begingroup\$ OK, it look like I had one issue...on board...wrong (empty) hole, other end of R1 I didn't hookup on OUT :) Now I have drop from 12v to 11.7v for load of 141mA \$\endgroup\$
    – ShP
    Commented Nov 28, 2016 at 3:32
  • \$\begingroup\$ That's wrong. The regulator should regulate it to ~1.25v. I would check the voltages at each end of the various wire to ensure you don't have a bad connection or miswiring - you could also do that with an ohmmeter. For example do the pins of R1 really connect to the pins of IC1 etc. \$\endgroup\$ Commented Nov 28, 2016 at 3:32

The built-in overtemperature protection, located ondie near the series power transistor, can react faster than a DVM can get a reading.

The thermal timeconstant, for 0.3mm separation,[lots of wafers are 0.3mm as run through the fab, so they don't break during handing] is 1.14 milliseconds.

To heat the tab, underneath, if 1mm thick, is 11.4 millisecond. If 2mm thick, thermal tau is 4X longer at 45 milliseconds. That has the entire structure, die surface + die bulk + vertically on down to backside of the tab, heated.

To heat the tab, laterally, say 2cm long, needs about 4 seconds.

Since silicon and copper have nearly the same Thermal Diffusivity, these are good numbers.


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