# Minimum battery capacity

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 !

• That makes perfect sense. Think about how much energy is available from one battery level to another. – winny Feb 22 '17 at 15:21
• Discharging to 80% of its capacity implies that you're only using 20% of the capacity. That's what the "1-0.8" is doing in the equation. – pjc50 Feb 22 '17 at 15:22
• I don't know if I understand what you said @pjc50 ... Does that mean if I take a 1.46Ah battery my system will stop to work at 50% ? Therefore at 4.5V ? Oo I think I'm missing something... – Tagadac Feb 22 '17 at 15:36
• Also you are assuming perfect voltage regulation. – Passerby Feb 22 '17 at 15:42
• 50% is a number you pick for how much depth of discharge you'll accept. It absolutely does not correspond to battery voltage, which declines on a curve then drops steeply when almost empty. – pjc50 Feb 22 '17 at 15:47

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.

• I think I understand a little bit more.. But I din't understand what are those 50% so ? Doe's that mean if I take a 3V battery with the same capacity it's not going to change anything about the total duration of my system? – Tagadac Feb 22 '17 at 16:04
• It is going to change because you're changing the supply voltage but your energy consumed in Wh is not going to change, so if your voltage is smaller then your DC DC converter is going to draw more current from it. – Marcelo Espinoza Vargas Feb 22 '17 at 16:17
• Your system is consuming 0.3mA at 2.5V as you stated, so it's around 0.75mW of power consumption, if you're using a 9V battery you must use a DC DC converter to turn those 9V into 2.5V, then your battery is not going to drain 0.3mA, it's going to drain 2.5V*(0.3mA/9V)=0.083mA ideally, if you use a smaller battery like 3V its going to drain 2.5V*(0.3mA/3V)=0.25mA, the change in voltage will change the current drawn but not the power indeed , it'll be always 0.75mW – Marcelo Espinoza Vargas Feb 22 '17 at 16:31
• Exactly, in that case you must check the drop-out voltage of your DC DC converter and see if it can manage to convert 3V into 2.5V, once you pick your DC DC converter include its efficiency on the equation and the self discharge rate of your selected battery – Marcelo Espinoza Vargas Feb 22 '17 at 16:42
• At this low of a current draw, you need to add in the Iq of the device, of the regulator and the self discharge of the battery. And frankly, you may want to use 3x AA and an efficient ldo. – Passerby Feb 22 '17 at 18:02