Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Battery self discharge rates

Status
Not open for further replies.

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!
 
It's genrally 10% of the charge remaining so it would be .9 x .9 x .9 ...
 
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:
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.
 
Status
Not open for further replies.

New Articles From Microcontroller Tips

Back
Top