I have three AA 1,5V@2950mAh Alkaline batteries in series with a discharge rate of 3% per year. My circuit has a 3V3 LDO that will work until 3V6 is reached, so each battery will need to have at least 1V2.

In order to calculate the battery life for my circuit and check if it is able to last for 2 complete years I have the following conditions:

  • The circuit consumes 380mA during 4 seconds 8 times a day.
  • The circuit will then remain at low power mode with a consumption of 1uA.

I have done the following estimation:

  • Operating duty cycle (per day) = 32 /86400 = 0.00037037
  • Average current during operation = 380mA x 0.00037037 = 140.740 uA
  • I consider Low power mode as 1uA 100% of the time
  • Total average current = 141.740 uA
  • Operating time = 2950 mAh / 0.141740 mA = 20812.64 h = 867.193 days
  • Adding discharge rate I'd have --> 867.193 - 867.193 * 0.06 (2 years @ 3%) = 815.162 days

So it seems the batteries will be able to last for the twop required years.

Are these previous calculations correct? For sure I´m missing something...

  • \$\begingroup\$ Looks correct to me! \$\endgroup\$ Commented May 13, 2020 at 7:37
  • \$\begingroup\$ Is the 3.3V regulators Iq less than that 1uA? equally is the leakage of the capacitors in your circuit less than that 1uA?, this stuff has caught me out so many times in the past. and as stupid as it sounds, keep silkscreen away from pads that could leak, contaminants build up along there ridges if cleaned poorly. \$\endgroup\$
    – Reroute
    Commented May 13, 2020 at 12:47
  • \$\begingroup\$ The LDO has a typical Iq of 2uA \$\endgroup\$
    – LazyTurtle
    Commented May 13, 2020 at 13:11

1 Answer 1


The first thing you need to check is what the battery manufacturer considers to be "flat". You may find that this is less than 1.2V. If that's the case, you can't use the 2950mAh figure to calculate the life.

The other thing to check is that the battery is still capable of delivering 380mA when almost flat. It's no use if the thing sits happily in standby for 3 hours, but the moment it powers up, the voltage drops below your minimum.

The better battery manufacturers will publish curves showing the expected life of the batteries at different current draws. You would have to check those to see how long it's capable of supplying the current you need at the voltage you need. That may turn out to be pessimistic, since batteries tend to recover when left un-used for a while. So 380mA constantly is a lot more work for a battery than 380mA bursts.

  • \$\begingroup\$ Wouldn´t the discharge rate per year include this? \$\endgroup\$
    – LazyTurtle
    Commented May 13, 2020 at 8:00
  • \$\begingroup\$ @LazyTurtle The discharge rate is how much power the battery loses if you leave it sitting idle. That's different to the discharge curves as you actually use the battery. Also, 380mA is quite a lot for an AA cell, so you may get a shorter life than you were expecting. \$\endgroup\$
    – Simon B
    Commented May 13, 2020 at 8:23
  • \$\begingroup\$ Thanks for your answer. OK, I´ll check if I can find this curves and come back with this info and see how to use it to estimate a more realistic value \$\endgroup\$
    – LazyTurtle
    Commented May 13, 2020 at 9:23

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