I am working on a project that it has to work at 2.5V-0.3mA, 12h a day during almost 2 years.
The probleme is that I don't understand something.
I did some calculation of the minimum capacity of the battery and I ended-up with some weird results.
Imagine we want a 9V battery.
We have 2.5V-0.3mA during 12h a day, it's 0.009Wh/day.
During 365 days*2 years, it's 730 days.
So finally my battery has to have a capacity of :
If I want an admissible discharge of :
– 50%, I have : 0.009*730/9/0.5 = 1.46Ah
– 80%, I have : 0.009*730/9/(1-0.8) = 3.65Ah
Wy does the battery has to have a higher capacity if I want to discharge my battery to 80% of its capacity than 50% ?
EDIT :
Does that mean if I take a 1.46Ah battery my system will stop to work at 50% ? Therefore at 4.5V ? I think I'm missing something…
Thank you !
Best Answer
The charge of a battery is not from 0V to xV, it depends on the chemistry of the battery, for example, a 3.7V LiIon battery will be 100% at 4.2V and 0% at 3.3V, so you assumption about being 50% of charge at 4.5V is not correct.
For your calculation you should just use:
Ah=(Wh*tHours)/(BatteryVoltage)
This will give you around 730mAh, then you must calculate the efficiency of your DC DC converter for an input of 9V to an output of 2.5V.
RealAh=Ah/Efficiency
Another thing to get in mind is that for such a great time the battery might self discharge, so you should know it's self discharging rate and add it into your Ah calculation.