I'm building a circuit that will use a LM7805 and LM7812 to produce 5VDC and 12VDC respectively, with the input power being ~16VDC from an external wall wart.

I already fried one of each regulator because I'm apparently not capable of keeping track of polarity.

I figure I might as well prevent that. I see in the datasheets that I can use an 1N4007 diode across the LM7812 for reverse-bias protection, which I understand is intended for transient states basically during shutdown.

I also read that I can use a Schottkey diode for reverse polarity (e.g., if I connect the p/s incorrectly again) protection.

Can I do both? Is there anything I need to know about selecting either the regular diode or the Schottkey? Overall, I'm sizing the circuit to handle ~1A of 12VDC and <100mA of 5VDC (just enough to power an ATTiny and one DO). I've literally never bought a diode before, so do I just find a Schottkey diode rated >20V and 1A then call it a day? I don't believe I care about forward voltage drop in my application...anything else?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Are you sure you fried it? They are fairly idiot proof. \$\endgroup\$ Commented Mar 28, 2017 at 13:23
  • \$\begingroup\$ Your circuit is wrong. The input capacitor to regulate voltage should not be connected that way and you will also need another capacitor on the 7805 \$\endgroup\$
    – 12Lappie
    Commented Mar 28, 2017 at 13:31
  • \$\begingroup\$ @StainlessSteelRat: Well, smoke came out of the 7805 and the 7812 just isn't working. Maybe it was DOA. Now that I think about it, I might have put +V from pin 2-1 (not 3-1) \$\endgroup\$
    – RobotAndy
    Commented Mar 28, 2017 at 13:37
  • \$\begingroup\$ @12Lappie: The input cap is wired exactly as per the spec sheet. .33uF on the Vin to common. If I place another input cap on the second 7805 then I basically have two .33uF caps in parallel between by Vin and Common. That can't be right? \$\endgroup\$
    – RobotAndy
    Commented Mar 28, 2017 at 13:39
  • \$\begingroup\$ D3 is mainly if you intend on virtually shorting C2 and having a big cap on the 12V output. Otherwise it's optional, but it won't hurt anything unless you put it in with the wrong polarity or it fails short, in which case whatever is connected to +12 could be damaged. \$\endgroup\$ Commented Mar 28, 2017 at 14:18

2 Answers 2


Yes, you can do what you propose.

I would recommend larger output capacitors in parallel with the 100 nF they have now. The regulator cannot react instantaneously to step increases in load current so the capacitors are there to provide the current until it does. Capacitors with a low ESR are accordingly good for this, try ceramic parts. Details of how to calculate the values can be readily found on the internet. From experience (gulp), I would use the following. On the 5 V rail, put a 10 uF and a 100 nF close to the regulator, put a 10 uF and at least one 100 nF very close to the load. On the 12 V rail, put a 10 uF and a 100 nF close to the regulator and a 100 nF close to the load.

Incidentally, the reverse diode across the regulator is used to discharge the regulator's output decoupling capacitor when the supply is removed and the regulator has a light/no load. If the regulator's load will discharge the output capacitor anyway then you don't need that diode.

  • \$\begingroup\$ Capacitor sizing seems like magic to me. Can you tell me why you suggest a different size? Note that .1uF is the spec-sheet recommended value. The load on the 5V will be an ATTiny, the load on the 12V will be a .6A cooling fan. \$\endgroup\$
    – RobotAndy
    Commented Mar 28, 2017 at 16:12
  • \$\begingroup\$ @RobotAndy, please see revised answer. \$\endgroup\$
    – TonyM
    Commented Mar 28, 2017 at 18:47

You should connect the LM7805 Input to LM7812 output. In fact LM7805 can work with maximum 12V on input and you'll burn it with 16V. So you can connect 16V to the INPUT of LM7812, next you connect the output of LM7812 to the INPUT of LM7805. Of course all the current provided by LM7805 comes from LM7812, that will supply all the currents in your circuit. So you should check LM7812 power dissipation is within safe limits.

  • \$\begingroup\$ Where did you get maximum 12V input for LM7805? All datasheets I have read state an AMR of 30V and sometimes 35V \$\endgroup\$
    – Bart
    Commented Jul 30, 2020 at 7:27

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