I have a small demo breadboard circtuit based on this design:
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…
https://www.hammfg.com/electronics/transformers/power/266.pdf
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:
Best Answer
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.