I was shorting a Duracell AAA battery in a SAFE area with proper precautions. After 2 minutes I disconnected the circuit and measured the voltage, it dropped from 2.8V to 2.6V and the battery was hot. When I reconnected the circuit for another 2 minutes the voltage didn't decrease and the battery wasn't hot. I tried a few more times (after waiting 30 minutes) and the same pattern occurred: the battery drained a bit, but after some time it seemed as if the current stopped during a short. I used the battery to power up a led, it still works.

So I'm wondering if there are internal short circuit protections. If so what are they and can I get a few references?

  • 3
    \$\begingroup\$ What kind of AAA battery has a terminal voltage of 2.8V? \$\endgroup\$
    – Hearth
    Jul 30, 2019 at 1:34
  • 2
    \$\begingroup\$ I think you meant two AAA's. Measure each cell. WHen unbalanced, one cell will rapidly become the weak link. \$\endgroup\$ Jul 30, 2019 at 2:00
  • \$\begingroup\$ You most likely drained the battery, or heated it internally enough to damage the chemical structure and effectively ruined it. \$\endgroup\$
    – MadHatter
    Jul 30, 2019 at 2:23
  • \$\begingroup\$ There is no internal protection. The fact that it's not falling below 2.6v also depends in what you're connecting to. Please do tell. Was it an actual short? \$\endgroup\$ Aug 2, 2019 at 19:08

2 Answers 2


Alkaline batteries have an internal resistance. There isn't a current limiter as such, that's just the physical properties of alkaline chemistry and the limits of the relatively-small AAA battery.

The thermal vs. voltage effects you see relate to the the ionic properties of the battery, as it recovers from a high discharge. Multiple high discharges will deplete the battery faster than a continuous, moderate discharge.

More info here: http://data.energizer.com/pdfs/alkaline_appman.pdf

(Don't try that with a Li-ion, please.)

  • 1
    \$\begingroup\$ Don't try it with NiCd or NiMH, either. It probably won't be as spectacular as LiIon, but it won't be pretty. \$\endgroup\$
    – TimWescott
    Jul 30, 2019 at 2:29
  • \$\begingroup\$ Thanks for the info, I gave you best answer :) \$\endgroup\$
    – h4grid
    Jul 30, 2019 at 5:57

I think you meant two AAA's. Measure each cell.

When unbalanced, one cell will rapidly become the weak link.

ALL batteries have memory or dual-layer capacitance, each with different series resistance, ESR.

The ESR is the part that gets hot, capacitance is like "Super-Ultracap" or is that an "ultra-Supercap" with a much higher density.

But the 2nd cap also has a higher ESR resistance and higher capacitance enough to restore the battery voltage from short term short circuits or heavy loads.

But it cannot deliver the same short circuit current and takes a longer time to recover while it can usually power a small LED as can a thin lithium coin cell which has a very high ESR so it naturally limits the current.

When a primary or secondary cell depletes most of its charge or becomes "dead" as they say, the chemistry raises the ESR significantly and also reduces the capacitance significantly.

This is why good Lithium smart chargers will start charging slowly until some 5 or 10% State of Charge (SOC) and then the impedance drops so a rise in current does not result in a measured significant rise in voltage and the ESR is thus that ratio \$\Delta V / \Delta I= ESR\$

So Alkaline cells do not have a built-in fuse but the depleted charge off the low ESR side builds up some oxide layer that raises its ESR so that it cannot accept as high a charge or deliver the same as when it was "fresh" or new. This is also the same effect on e-Caps at end of life unless damaged by over-current or reverse polarity charge.


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