I have a device I am powering using two batteries. My constraint is that I can never let the device power down. For instance, I have multiple cameras that needs to stay on 24/7 but their isn't a battery that can last that long, therefore my goal is cascade multiple batteries and be able to swap them out without powering down the device.

My current solution is to use two batteries in parallel, one with a higher voltage (Power source 2, 14V) than the other one (Power source 1, 12V). When I need to swap my main power source (Power source 2), would the second battery keep the device running?

In other words, Power source 1 should power the device during the swap. Then, when power source 2 is reconnected, it should power the device, and at the same time recharge power source 1.

enter image description here

Batteries I am thinking of:

Power source 1: https://www.amazon.co.uk/gp/product/B00BWW9WRM/ref=ox_sc_sfl_title_1?ie=UTF8&psc=1&smid=A17M1OH9UAKGE7

Power source 2: https://hobbyking.com/en_us/multistar-high-capacity-4s-20000mah-multi-rotor-lipo-pack.html?___store=en_us

  • \$\begingroup\$ "Please see the Diagram for a visualisation!". To that I say: Please don't force people to leave this site and make a proper circuit diagram under the edit tab. Or just copy and paste the image here. We should not need to leave this site to be able to answer your question. \$\endgroup\$ – Harry Svensson Oct 24 '17 at 11:06
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    \$\begingroup\$ @Harry new users can't use the circuitlab tool or post images. You should not demands things if You dont have a clue about what you are asking. Instead, edit the photos in. \$\endgroup\$ – Passerby Oct 24 '17 at 11:08
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    \$\begingroup\$ And dont downvote a perfectly good question. \$\endgroup\$ – Passerby Oct 24 '17 at 11:12
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    \$\begingroup\$ No, they cannot copy paste the image. NEW USERS CANNOT UPLOAD IMAGES OR USE THE CIRCUIT LAB TOOL WHICH IS BUILT ON THE IMAGE UPLOAD TOOL. They did everything a new user without enough rep can do, they hosted the image elsewhere and linked it here. Then higher rep users review and edit the image in. That's how it works. It's a simple task like i asked you to do. \$\endgroup\$ – Passerby Oct 24 '17 at 15:24
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    \$\begingroup\$ Thanks @Passerby for making it clear to others. If what they mentioned was an option, I would certainly have bothered! \$\endgroup\$ – Arsalan Oct 24 '17 at 16:54

This depends on the kind of battery you are using.

If you are using lead acid batteries; this is all fine and will work the way you describe.

If you are using practically any other type of battery; you should not do this, the batteries could charge at unsafe rates. It would be safer to wire a "schottky power diode" in series with each battery so that they can both power the system but nether can charge the other. You would swap the batteries and charge them away from this system.

  • \$\begingroup\$ With this arrangement, whichever battery currently has the highest voltage will power the system; the others will be idle. If two are equal, they may end up sharing the load. \$\endgroup\$ – Simon B Oct 24 '17 at 15:20
  • \$\begingroup\$ Or use a charge controller circuit or IC. \$\endgroup\$ – Passerby Oct 24 '17 at 15:26
  • \$\begingroup\$ Thanks for your help so far. As I'm new to this, let me recap your solution, and please correct me if I got it wrong: We'll go with the setup from the diagram. In such a case, the battery with the higher voltage would power the device, and the battery with the lower voltage would be idle. Then, if we remove the battery with the higher voltage, the other battery will take over. But for safety reasons, we should add a charge controller or a "schottky power diode" between the two batteries. Can you give me a little bit more information why we need the charge controller/diode? \$\endgroup\$ – Arsalan Oct 24 '17 at 17:51

As Bracken has described, a diode bridge would work. Especially with the main power supply being a higher voltage. Keep in mind that the current draw may be an issue so the swap may cause a brown out, so a capacitor after the bridge would be recommended. You would need to test.

A better solution, would be to use an integrated charge controller. This would handle the charging for the smaller battery, and allow pass through power. Once the main battery is removed, the smaller battery should kick in through the charger circuit.


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