Quick question (hopefully!). I'm new to electronics, but getting quite interested. Coming from a programming background, it's always interested me...

Anyway. I have a circuit with two DC Power sources, one is 12V the other is 7.2. I have a regulator which turns that down to 5V. Now, the 12V power source isn't always available. The 7.2V however, is always available. But I would like to build a circuit that will use the 12V if it's there, and 'fall back' to the 7.2 if it isn't.

Once I have this sussed, I need to have a look into re-charging the 7.2V battery from the 12V source when it's available, but that is a problem for another day!

It sounds simple, but I'm guessing it isn't? Please help. :)

  • \$\begingroup\$ What chemistry is your 7.2V pack? \$\endgroup\$
    – Joe
    Apr 9, 2014 at 20:41
  • \$\begingroup\$ It's nimh, 1800mah, 6 cell. Just had it lying around from an old RC car \$\endgroup\$ Apr 9, 2014 at 20:51

1 Answer 1


Yes, this is simple. It can be done passively with two diodes. At these low voltages, they can be Schottky diodes, thereby dropping less voltage and being more efficient.

Put a diode from each source to a internal power node. When 12 V is present, the internal power node will be at just under 12 V. The diode to the 7.2 V source is therefore reverse biased, essentially disconnecting the 7.2 V source. When the 12 V source goes away, the internal power node will be at just under the 7.2 V.

A switching power supply can deal with this variation in source voltage and still produce your final intended 5 V reasonably efficiently regardless of which source is providing the power.

  • \$\begingroup\$ Sorry, i'm not sure what you mean by internal power node? Thanks though, i'm just a bit new to this! \$\endgroup\$ Apr 9, 2014 at 20:34
  • 2
    \$\begingroup\$ @DeanThomas by internal power node he means the supply input of your circuit (that is the regulator input according to your description), the diodes should be placed between each supply output and the circuit supply input \$\endgroup\$
    – alexan_e
    Apr 9, 2014 at 22:00
  • \$\begingroup\$ The node would be where the diodes meet, which would be connected to the input of your regulator. Speaking of regulators, I really hope you don't mean a linear regulator. That eats power pretty well. Olin is spot on; after the diodes what you really want to have is a chopper style switch mode buck converter. \$\endgroup\$
    – user39962
    Apr 9, 2014 at 22:12
  • \$\begingroup\$ @SeanBoddy Can you provide some details on the chopper style converter you're referring to? I have an application where one of the power supplies is batteries (needs to be efficient when running off this) and the other supply is rectified mains power so it doesn't matter the efficiency. The currents are typically extremely low (10s of microamps) but can jump kinda high (up to a couple hundred milliamps for about a second) and I'm wondering if those converters you mention would work well... \$\endgroup\$
    – NickHalden
    Apr 9, 2014 at 23:14
  • \$\begingroup\$ They would work extremely well, but now that I know the kind of currents your dealing with, may not be warranted necessarily. Or at least you may not particularly need a high frequency one. They are switched mode power supplies that essentially start and stop charging a capacitor to maintain a specific voltage that is lower than the input. Thanks to modern power BJTs and MOSFETs, they can be wildly more efficient than linear regulators. You might even be able to find what you need off the shelf. \$\endgroup\$
    – user39962
    Apr 9, 2014 at 23:56

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.