Technically, 4 AA batteries will give out ~5V, which is equivalent to USB. Can you use 4 AA batteries to charge an iPhone, without messing up the Li-Ion cells used on iPhone?

So my question really has 3 parts:

  • Do the Amps add up with each battery? So if one battery is 2000 mAh, will 4 batteries = 8000 mAh?
  • Won't the AA batteries mess up the Li-Ion cells used on iPhones due to different chemical composition?
  • Is there ANYTHING wrong with charging Li-Ion batteries with regular AA batteries?


  • 2
    $\begingroup$ Hi NoobDev4iPhone, and welcome to Physics Stack Exchange! I think only the first of your three questions is actually on topic here. $\endgroup$
    – David Z
    Feb 25 '13 at 0:13

I am going to assume what you are thinking about is some sort of power pack that uses AA batteries like this one. There is no physical reason that you can not use AA batteries to charge a cell phone, as long as you have the correct adapters. What might be confusing the issue is that Li-ion batteries have definite charging issues. If overcharged, they can begin to break down and produce CO2, so there are definite safety considerations associated with Li-ion cells, however most of the products using them are designed with those issues in mind, so in general you shouldn't have any problems. However, I would always recommend checking the manufacturers warnings in order to prevent unsafe use of any product, and I would follow those recommendations rather than seeing what random people on Q&A sites tell you.

P.S. The summing of amperage will depend on whether the circuit is parallel or series. In general, if it is in series, the current will remain constant and the voltage will increase, if it is in parallel the voltage is constant, and the current scales with the number of sources. In your example, if the circuit is a parallel circuit, then you can add the currents together, however, since AA's only have 1.5V, your reference to 5V indicates a series hookup, in that case, the current will be constant.

A good tutorial can be found here.


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