This is a very recurrent question, about calculating the battery life given certain circuit current usage.

This example will be similiar to this one in terms of the battery used and the consideration of the load affecting the battery capacity.

Goal is to know if having batteries in parallel affects the way the load decreases capacity

So, let's set this example, with as much detail as I can.

An RPI 3B that's using 500 mA (assumed is steady, which it won't be, but let's say), that is based on this estimation and the official consumptions, seems like a good number.

Now let's assume that we have these batteries, which state that at 500 mA consumption, they will have 1500 mAh approx.

So, now the battery connection, let's say we have 2 packs of 4 AA batteries each, each pack is connecting the batteries in series, so you would have 6 V / 1500 mAh on each pack.

Here's where the question comes: in plain numbers, if you connect those packs in parallel, you would end up having an output of 6 V / 3000 mAh.

Therefore run time would be 1500 mAh / 500 mA = 3 hours, right?

But, wouldn't the load (in terms of stress that affects capacity) be less for each battery on each pack, if we're connecting those things in parallel, so the cell capacity drop is less?

My gut says this would be negligible, and therefore this is overthinking, but I'm curious.

  • 1
    \$\begingroup\$ It's not negligible since internal resistance of the batteries is not constant. If you look at a typical discharge curve usually a battery discharged at C/5 lasts more than 5 times the discharge at C. It really depends on the chemistry. But consider that some kind of batteries don't really like to be connected in parallel (that's why we see lots of series with maybe cell balancers) \$\endgroup\$ Commented Dec 1, 2021 at 7:53
  • \$\begingroup\$ Your "therefore run time" sentence may include accidentally wrong figures. One pack gives nominally ,3 hours and two packs nominally double that. In practice you will get somewhat more than double as effective mAh rises as load drops. | Note that 5:cells is about 6b when new and 5 Vat end of life. \$\endgroup\$
    – Russell McMahon
    Commented Dec 1, 2021 at 9:09

1 Answer 1


First of all, there will need to be a voltage regulator involved, which you haven't mentioned, since the RPI can't tolerate 6V.

But to answer your main question: yes using 2 packs in parallel cuts the current that each pack must supply in half. So (using your numbers), each cell would only be supplying 250mA, instead of 500mA. This will increase the number of mAh that each pack can supply. Thus the run time will be more than double. The effect may or may not be negligible. The battery datasheet should be able to tell you. 500mA is quite a lot for a AA, so it probably will make a difference.

One other thing you will need to consider to get the best battery life is the dropout voltage of your regulator. To get all the capacity out of a AA, you need to run it down to about 1V (4V for 4 in series). That means you either need a buck-boost regulator, or you need a buck only regulator with 1-2 additional batteries in series, to get a minimum voltage of 5-6V. If you only run the batteries down to 1.25V you're wasting a lot of capacity (again the datasheet can tell you exactly how much).


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.