I am building a 12v power supply based on this.

Main changes are use of two 2,200uF filter caps instead of one, and 7812 (12v) regulator instead of 7809. enter image description here Plugged it in the first time and the voltage regulator went up in smoke before I noticed and unplugged it.

Having now replaced it, it still gets very hot very quickly after being plugged in, but I can't see what I have done wrong with the wiring. Voltage straight off the rectifier is about 17v. With the rest of the circuit connected, voltage on pin 1 (input) of the regulator reads about 11v, and the output around 1v. Within a couple of seconds the regulator is too hot to touch.

First thought was that regulator pinout was wrong, but it is 7812 - pin 1 v(in), pin 2 ground, pin 3 v(out).

Any ideas what would cause behaviour like this?

This is a picture of my implementation:

Rats nest

The terminals 1 & 2 are the input rail, terminals 3, 4, 5 & 6 are output - the copper bar (and outer two terminals) is the negative rail. The regulator connects the three - Pin 1 in, pin 2 negative, pin 3 out.

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    \$\begingroup\$ Unplug the circuit and check for any short circuit across the output. Simply, measure the resistance across the output and if it is less than, say, 100Ohms then this means that you have a short. One of the common mistakes is to connect the L78xx regulator IC wrong. You say that you have checked the pinout but you should also check the load. Maybe your load causes the short circuit (or low resistance). \$\endgroup\$ Commented May 4, 2017 at 5:57
  • \$\begingroup\$ If you built it as suggested (using strip board) then there's an excellent chance that you have connected something incorrectly. Also note that electrolytic capacitors are polarized - if you put one in backwards, bad things will happen \$\endgroup\$
    – JRE
    Commented May 4, 2017 at 8:08
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    \$\begingroup\$ On a related issue, you are going to drop 5V at up to 2A. The regulator is linear, which means that those 10W are going to heat the IC. You will need a heatsink. \$\endgroup\$ Commented May 4, 2017 at 8:19
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    \$\begingroup\$ Upload a picture of your implementation so we can see if there's something wrong about it. \$\endgroup\$ Commented May 4, 2017 at 8:46
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    \$\begingroup\$ Disconnect output of 7812 from your rat's nest. Measure output voltage. If it is 12V, then your problem is on the rest of your circuit. If not 12V, then the problem is on the input side. \$\endgroup\$ Commented May 7, 2017 at 19:52

2 Answers 2


Your implementation is quite messy as seen from the picture. There could easily be a short circuit anywhere in your circuit. I'd recommend looking for short circuits at the output (or failed components that behave as short circuits).

Specifically verify that the output protection diode isn't reversed, because it looks like it is and that would provide a short circuit at the output.

Output protection diode

One very important thing you should note about your implementation is that your 7812 isn't heatsinked at all. For the sake of comparison, take a look at how he author of the original design deals with heatsinking, as seen in the pictures of the page you've linked:

enter image description here

As you can see, the author has bolted the regulator to the case so it can serve as a heatsink - a clever move.

Your voltage drop will be 5 V. Even at your nominal expected 500 mA current draw that's a 2.5 W power dissipation. The TO-220 junction-to-ambient thermal resistance is 65 C/W, so your temperature rise would be 162.5 C. At 25 C ambient temperature that'll mean a junction temperature of 187.5 C, far above the 125 C maximum rating. That's enough to either trigger its internal thermal protection or fry it, whatever happens first. You need a heatsink no matter what.

If you happen to have a short circuit at the output, the situation can be even worse. The maximum current could peak at the datasheet maximum's 2.2 A, meaning 10 W dissipation in the 7812, or a whopping temperature rise of 650 C. Probably that's why you fried the first one.

  • \$\begingroup\$ Thanks Eric - I on closer inspection I think you are right about the protection diode. Cheers for the pointer about the heat sink, I will be doing a similar thing to what the author did - this implementation was purely a very quick and unfinished prototype. Heatsink, enclosure and a bit more heat shrink is still to come. \$\endgroup\$
    – cf-m
    Commented May 8, 2017 at 0:02

Double check the orientations of all your components. looking at that picture, I suspect that either the capacitors are all the wrong way round, or else the protection diode is the wrong way round.

If it's the diode, then it's effectively shorting the output.


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