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.

  • 7
    Doesn't sound off-topic for EE.SE to me... May 10, 2014 at 17:20
  • 2
    A power source is electronics. May 11, 2014 at 0:42
  • 1
    You mentioned that you have several Arduino for your robot, hence I deduce you probably have several actuator devices (those that need high current), so why not use each battery as a distinct power source? If needed, you'd just have to connect all grounds, but they would not be used in parallel.
    – jfpoilpret
    May 11, 2014 at 7:33
  • 2
    Sometimes I think EE.SE is suffering from overzealous moderating.. I hope Arduino.SE remains as open as it has been. Aug 9, 2014 at 3:49
  • 1
    What's worse, is that when EE.SE flushes sound engineering-principles questions elsewhere, they often get answers of a lower quality. Aug 14, 2014 at 17:08

2 Answers 2


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:

      |+  diode
|     |        |
|              |
|     |+       |
+----||--->|---+----o V++
|     |        

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.

  • What happens if I have different voltages on the batteries? A short between the different potentials?
    – frodeborli
    Aug 12, 2014 at 13:23
  • To a first approximation, you only draw significant power from the highest voltage one, until it discharges to match another. Aug 14, 2014 at 17:07
  • @chris-stratton What you're saying sounds obvious when you say it :)
    – frodeborli
    Aug 15, 2014 at 11:34

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.

  • I have two numbers about the packs; voltage at full charge (16.5V for the good packs) and a remaining battery time as reported by the PC; most are around 3 hours. A few are really bad (50 minutes or less battery time left, at full charge).
    – frodeborli
    May 10, 2014 at 23:11
  • Since each battery pack has built in over- and undercharge protection, I assume any leakage between the packs will be no problem. One pack at 14.8 volts cannot overcharge another pack to more than 14.8 volts, and I will never connect an empty battery pack together with fully charged battery packs.
    – frodeborli
    May 13, 2014 at 18:54
  • 1
    @frodeborli I wouldn't assume that with lithiums as the charging cycle is a bit more complex. The protection circuitry stops under and over voltage on the battery cells, but it might expect a constant, high-current source for its input. You will need to find out what that input is as well. Aug 9, 2014 at 3:54

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