Hello I have an LM7812CT power regulator and I want to use it to convert a 28V dc to a 12V dc.

I am using the circuit provided in the data-sheet using 2 capacitors, one 0.33uF connected to the input voltage pin(1)and the ground ,and one 0.1uF connected to the ground and the output pin(3).

My problem is that I get output voltage 3.9-4.0 Volts.

I get this value even with 12V supply to input.

How can i fix it to get my 12V?

Where do you think am I wrong? Is it burned??

Any ideas welcome

edit:: ok guys the LM7812ct was burned after all. I replaced it and everything worked ok!! thnx and keep up the good work

  • \$\begingroup\$ It there any load connected to the output? And please give a diagram, or preferrably a photo, of what YOU are doing. Your intent might be to follow the datasheet, but you probably didn't. \$\endgroup\$ Sep 10, 2012 at 7:23
  • \$\begingroup\$ no load at the output. the rigth link is freedatasheets.com/datasheet-download/…. I am using the circuit at page 22 up \$\endgroup\$
    – ar ar ar
    Sep 10, 2012 at 7:32
  • \$\begingroup\$ Check if you maybe reversed the regulator. I've seen such results when it's reversed for some reason. \$\endgroup\$
    – AndrejaKo
    Sep 10, 2012 at 8:10
  • \$\begingroup\$ If the pins are correctly connected, the input voltage is filtered so as to not have drops below the datasheet specs and you say there's no load, then it's very possible that it is "burned". Or your multimeter has weak batteries or oxidized/dirty/improper connected probes. \$\endgroup\$
    – Vlad
    Sep 10, 2012 at 8:41
  • 1
    \$\begingroup\$ You may post your edit as an answer. You won't be able to accept it right away (couple of days I think), but when you do we'll know that the problem is solved and that we shouldn't bother. \$\endgroup\$
    – stevenvh
    Sep 10, 2012 at 9:16

2 Answers 2


I don't know how you burned the 7812; it has both short-circuit and thermal overload protection.

Anyway, you have 16 V input-output difference, and that's a lot. A 60 mA load will already cause 1 W dissipation. I would suggest to place a power resistor in series with the 28 V to decrease the 7812's voltage drop. You can afford at least 10 V across the resistor, so if you want 1 A output you need a 10 Ω/ 10 W resistor. The 7812 still will dissipate 6 W then, so will need a considerable heatsink.


Any way i have such exprience but i think adding a heat zink may be important


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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