So my senior project is due Thursday and because of my extreme wits and knowledge I accidentally connected A 7912 Voltage regulator the incorrect way because I was unwary of the different pin-out. I went to radio shack and they did not have any 7912 regulators. Is there any common household objects that have a 7912 or equivalent that someone may know about? I understand this is a long shot but I really need this part my project will not work without it. Thanks

EDIT Here is a Picture of the schematic, The Negative 12 volts are used for a reference for a LT6105. enter image description here

  • \$\begingroup\$ I am Basically building the power supply scene here kerrywong.com/2013/11/24/… \$\endgroup\$ – liljoey112 May 11 '16 at 12:24
  • \$\begingroup\$ If it's only for a reference voltage with a very small current, maybe the shop has a 79L24 (absolute max. input voltage 40 V) and a 79L12. The 79L series are 100 mA regulators in TO-92 packages. Possibly also available in a "variety bag". \$\endgroup\$ – Andrew Morton May 11 '16 at 14:51

The thing about both the 7812 and 7912 is that the pin out is different as you are aware but there is a strong possibility that the 7912 you connected incorrectly is not actually damaged.

Pin 1 for the 7812 is the most positive pin i.e. Vin & Pin 1 for the 7912 is also the most positive pin i.e. 0V

Pin 2 is the most negative for the 7812 i.e. 0V and pin 2 for the 7912 is also the most negative pin i.e. -Vin: -

enter image description here

Pin 3 is the output for both and sits naturally at a voltage somewhere between pin 1 and pin 2.

So, given the above, there is a strong case for assuming the regulator is undamaged and therefore reusable.

  • \$\begingroup\$ When The Regulator is in the circuit it just passes the original voltage right though so on the output pin i get around 36 volts. \$\endgroup\$ – liljoey112 May 11 '16 at 12:17
  • \$\begingroup\$ Well there's your basic problem - the absolute maximum input voltage you can put on a 7912 is |35 volts| - if you have |36 volts| on the output then you'll likely have even more on the input side. Read the data sheet. The diode droppers in your circuit, at low currents will not drop the supply voltage below 35 volts. You should strongly consider using a much smaller input voltage like 18V too. \$\endgroup\$ – Andy aka May 11 '16 at 12:26
  • \$\begingroup\$ I wish i could but the power supply I am making is a 30+ 30- The -12 is only to reference to a LT6105. I also posted the link to the website in my first comment. So you are telling me that the regulator is just passing the voltage though because it is over volted? \$\endgroup\$ – liljoey112 May 11 '16 at 12:43
  • \$\begingroup\$ I cannot be sure but over-driving at maybe 37 or 38 volts is a good reason you destroyed it. \$\endgroup\$ – Andy aka May 11 '16 at 13:28
  • \$\begingroup\$ Well I remeasured and it is actually outputting -28volts the input is ~36 \$\endgroup\$ – liljoey112 May 11 '16 at 14:57

If your 7912 negative regulator is blown and you need a quick fix, replace the whole thing with a off the shelf 12 V power supply.

The power supply output will be isolated, so can float arbitrarily. It may be labeled +12 V and GND, but all it's really doing is putting a 12 V difference between two leads. Connect the positive of the two leads to your ground, and the negative will be your -12 V supply.

If this is just for a short demo, then you may be able to use a few batteries.

  • \$\begingroup\$ I Just feel guilty doing that for all they judges know who I am presenting to I could just Put a 7805 in there and say look it works and stuck a bunch of wires in there and they would be beyond impressed :p \$\endgroup\$ – liljoey112 May 11 '16 at 13:26

Something like this should suffice for a quick setup. D2.. are the diodes already shown on your schematic. D1 is a 13V 1W Zener and could be made up from some combination of zeners in series. Q1 is a PNP power transistor (TO-220). R1 is a 2W power resistor and could be made from combining other values.


simulate this circuit – Schematic created using CircuitLab

It's not short-circuit proof so take care not to short it.

Alternatively (and generally a better solution), if you can find a 7905 negative regulator you can put a ~7V zener in the 'GND' line to change the output voltage to -12V from -5.

  • \$\begingroup\$ I have a few 2n6491 transistors laying around would that work? \$\endgroup\$ – liljoey112 May 11 '16 at 15:02
  • \$\begingroup\$ Should work fine. Datasheet \$\endgroup\$ – Spehro Pefhany May 11 '16 at 15:03
  • \$\begingroup\$ So my local radioshack has a in472a they sell it in a pack of two could I use these 2 in series and therically have double the current capacity? \$\endgroup\$ – liljoey112 May 11 '16 at 15:23
  • \$\begingroup\$ You need about 3 in series \$\endgroup\$ – Spehro Pefhany May 11 '16 at 15:47

That's a negative regulator. You could build something similar if you don't need high performance.

If you can live with ~ 0.5 V accuracy on the output voltage, and your load current is not too large (say < 100 mA), then a 12 V zener (or stack of 3*4.7 V zeners etc.) and a PNP emitter follower may work for you.

Basically, assuming your negative input voltage is ~ -20 V You can use nearly any PNP transistor -- say 2N3906; if current is > 20 mA, you might need a heatsink, or find a power PNP transistor. Connect collector to negative input supply. Connect emitter to the -12V output. Connect a 1k resistor from negative input to the PNP's base, and the ~12 V zener from the base to GND (be careful with the polarity; if you get it wrong, you'll get no output).

  • \$\begingroup\$ Hmm This might work, but the input voltage is ~-36 volts would that be too much for the Zener to Drop? and I only need the -12 volts to be referenced to a LT6105. And for a quick fix just so it works tomorrow this might work correct? \$\endgroup\$ – liljoey112 May 11 '16 at 12:21
  • \$\begingroup\$ The LT6105 consumes < 200 uA, so you'd still be OK. For -36 V, keep the zener current at ~ 1 mA, so use a (36-12)/1m = 24 k resistor -- use a 27k. At 200 uA load, you won't need a heatsink. \$\endgroup\$ – jp314 May 12 '16 at 2:56

You didn't post your circuit. Assuming the 7912 in your circuit is to generate the -12V reference to your common ground and you are also having a 7812 regulating the supply to +12V, what probably you can do is, firstly, split the power supply at the transformer secondary.

Your transformer must be center tapped 24V type, commonly denoted as 12,0,12.

What I mean by split the power supply at the transformer secondary is to actually split the center tap carefully to make 12, 0 and another 12, 0.

Now bridge rectify these two secondaries individually and then use two 7812 to create regulated and isolated two 12V supplies. Now connect them in series and take the center tap as your common ground. This leaves one end as +12 and the other end as -12.

Your problem solved without need to have the 7912.


Not the answer you're looking for? Browse other questions tagged or ask your own question.