How to calculate the real charge and run time of a 18650 battery?

[aymannox]

Hi, I am making an outdoor weather station, with solar panel to charge two batteries in series (3.7v , 1600mAh) for each one, I have measured the capacity using a zb2l3 model.
By the way the most important is how to calculate the real run time, because as I know:
Runtime = Capacity/currentOfTheLoad= 1600mAh/12mA =133hours.

But the 18650 batteries use just the voltage between 4.2v to 3v is about 30% of the full voltage, so does we need consider just 30% of the capacity to calculate run time?Runtime=1600*30%/12=40hours.

And also for charging time , the current of charging is 70mAh , does i need just consider 30% of 1600mAh , and also which Efficiency Loss i need to choose below in the image?

[slipstick]

No. Battery capacity is specified as that available between full charge and the minimum allowable voltage. So you should get 1600mAh between 4.2V and 3V.

Therefore when charging you need to add the full 1600mAh if you have discharged all the way to 3V. Lithium batteries typically have very high charging efficiency so a simple "capacity/charging current" calculation works pretty well. However other battery types are very different.

Steve

[jremington]

Battery voltage is not a simple indicator of capacity. By the time the voltage of a lithium-based battery has dropped to about 3V, very little stored energy remains.

More here: BU-501a: Discharge Characteristics of Li-ion - Battery University