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I'm working on a medium sized robot project with multiple arduinos to control stuff. I've been planning on using a lead-acid battery pack, but then I was given 10 working and identical Asus laptop battery packs rated at 14.8 V 4400 mAh, and one old Asus laptop.

I opened up one of the battery packs, because it was bad and it has a 4S2P battery configuration and a protection board.

Can I hook all of these battery packs in parallel, or will power leak between the battery packs, and is that be a problem?

Is there alternatively some 4S20P or 3S25P battery pack with charger I can purchase to put these batteries in?

(I do have a step-down converter that will provide the Arduinos with the right voltage, regardless of input voltage up to 36 V)

I tried to ask a similar question on electrical engineering, but ot was deemed off topic for the site.

frodeborli
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2 Answers2

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I'd be cautious about hooking them in parallel. According to ADAfruit power leakage between batteries is a problem.

[multipack parallel] ... batteries are assembled by a company that is experienced and certified to test and assemble battery packs. The individual batteries are tested and sorted by machine so that each pack has matching batteries with the same capacity and internal resistance. Individuals do not have this equipment, which is why you should not try to make your own packs.

Are you certain all your batteries are the same age and have the same usage history? A brand-spanking new one will try to charge an old worn battery to its own fresh voltage.

Wayfaring Stranger
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In case you didn't know it already you can hook any batteries in parallel by adding a diode on every battery to avoid current from any other battery flowing back to it. Like this:

 battery
      |+  diode
+----||--->|---+
|     |        |
|              |
|     |+       |
+----||--->|---+----o V++
|     |        
|
_
V GND

The drawback is that you have a small drop of voltage on the diodes and of course a small power loss too. If for example you're drawing 0.5 Amp from the batteries and you're using an 1N4001 diode which has a voltage drop of about 0.7 Volts then expect 0.350 Watts of power loss (0.5A x 0.7V). So choose the diode with the lowest voltage drop that can handle the current you need. There are solutions to lower the power that is wasted but they're more complex.

ndemou
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