I've salvaged a MOSPEC F12C20C Dual Power Rectifier from an old PSU. I've been looking for example circuits that would use this component, and an explanation of how it works but I've only found its datasheet and not much more.

If I understand correctly, its function is to rectify an AC current to make it DC, is that correct? How is this different from the typical diode bridge? (power dissipation capabilities maybe?)

According to the datasheet, this component is the equivalent of two diodes package together, is that so? If that's the case, how is it that it can achieve rectification with only two diodes and not the usual four?

Do you know of any example circuits I could build on a breadboard to test this component and play with it so I can better understand how it works?

Bonus: mine has the letters MNB next to the diodes schematic it has on it. Is it part of the name or is it just a serial number or something?


  • 1
    \$\begingroup\$ Fancy name aside, this product is nothing more than two diodes together in one package with a common cathode. It doesn't have any special functionality beyond that. Based on the datasheet, the diodes can definitely handle a bit more current than your generic general purpose diode for sure. You would need two more diodes (perhaps the common-anode version of the same part, F12C20A) to make a true full-wave AC-DC rectifier. \$\endgroup\$
    – Dan Laks
    Jan 15, 2015 at 0:20

1 Answer 1


You can use this as a full-wave rectifier if you have a centre-tapped tranformer:


simulate this circuit – Schematic created using CircuitLab

The maximum current depends on how good a heat sink you have, but several A are not a problem with a small heatsink, and the maximum AC voltage would be more than 50V (the PIV rating is 200V, so if everything was perfect 140VAC would be okay. The output will be `1.414 times the input voltage AC minus a bit for the diodes, so a 12:12 centre-tapped tranformer (24V from end to end) would give you about 16VDC, give or take depending on capacitor and load.


Your Answer

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

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