I have a computer being powered off a Lipo battery, and I would like the ability to transfer the power supply from the Lipo to a fixed DC source (eg. power supply) without having to turn it off and back on again. The computer draws around 100w at 12-24v.

  • I considered having a capicticter hold up the voltage for me but to get even .1 sec of runtime I would need a .5F cap which is much too expensive and big.

  • Can I just connect the 12v power supply and the battery in parallel with diodes on the positive lines?

  • What are some other options?

  • 2
    \$\begingroup\$ Yes, you can connect both sources together with a diode on each leg, as long as you verify each voltage source works by itself with the computer okay. If you use Schottky diodes, the voltage drop should be about 0.3v. \$\endgroup\$ – tcrosley Apr 18 '15 at 3:01

TL;DR - If you are doing this all manually, go ahead and just use one diode for each input. What follows is a design discussion for DC input stages.

It needs to be noted that with the right physical connections, if the voltages are otherwise guaranteed to be fairly close to each other, there is no reason not to simply place the sources in parallel for a while until you disconnect one.

Using a diode in line with each power source will give you almost exactly what you want, as tcrosley says (+1, tc). Any significant voltage difference will cause one diode to reverse bias, and leads to a seamless transition when one or the other is removed. In the event that the two power sources were completely identical and redundant, we would stop here with basically no additional consideration.

But we are talking about two non-identical power channels, and picking which one is running is important to the health of your battery. Normally, battery operated devices draw from the battery more or less directly, and having the charger attached means that it will supply the voltage as a matter of physics doing it's job. But with a two channel diode circuit, having your normal source attached carries a risk - if, for any reason, it is lower than the battery voltage, the battery will run instead.

It doesn't sound like that is a problem for you, as you seem to be manually and intentionally going from one source to the other, but I'll assume there's the possibility of using it as redundancy.

Ensuring that your main source has a higher voltage automatically solves your problem, but you may not have much control of that. You could use a zener diode to artificially hold the battery voltage lower without much effort, but this means a fairly constant leakage current, which is undesirable for a battery. With this setup, you must also remove the battery from the system in order to charge it, or else your normal DC source will reverse bias.

Another alternative is to use a boost converter to raise the voltage of the main source just a little, if you otherwise can't tweak it. But this is perhaps overcomplicated for the desired end result.

The most traditional method of quickly solving this problem, then, is to skip the diodes altogether, and instead operate the system on the battery voltage by default. At that point you can simply attach an appropriate voltage regulated source in parallel across the terminals of the battery and let physics do the rest. The battery will float on the source when it is plugged in, and provide power when it is not, with essentially no controlling circuitry at all. At these voltages, the battery is hot swappable with no effort on your part, as long as the other voltage source is already on line.

Since we really want to get more life and power storage out of the battery, the most optimal thing to do is to attach an appropriate battery management controller to the circuit to regulate the current into the battery. Chips for this are readily available, but personally I wouldn't build one for fun. A hobby hack for this would be to purchase an appropriate charger for the battery in question, and maybe install it in another enclosure or gut it for parts.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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