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Battery self discharge rates

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Speakerguy

Active Member
If a battery is rated to have a self discharge of 10%/month, over the course of several months is that:

Vbat = V(initial) * .9 * .9 * .9 .......
(aka 90%, 81%, 73%, etc)

or

Vbat = V(initial)) - 0.1*(V(initial)) - 0.1*(V(initial)) - 0.1*(V(initial)) ........
(aka 100%, 90%, 80%, 70%, etc).

Simple question, but it makes a big difference when my product has to sit on the shelf for a long time before use. Thanks!
 

crutschow

Well-Known Member
Most Helpful Member
It's genrally 10% of the charge remaining so it would be .9 x .9 x .9 ...
 

Speakerguy

Active Member
Yes, of course. I knew that, but must have had a brain fart when I typed V(initial). Should be Capacity(initial) and Capacity(bat) in mAH.
 
Last edited:

audioguru

Well-Known Member
Most Helpful Member
The voltage of a self-discharging battery drops only when it has a load.
It is the capacity of a battery that discharges, not its voltage.

The 9V battery in my smoke detectors is almost 9V when it is old and dead. But it cannot provide enough current to sound the beeper. When the smoke detector tests the battery then it provides a load.
 
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