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Calculation of a battery's energy loss using a voltage difference

Calculation of a battery's energy loss using a voltage difference

Calculation of a battery's energy loss using a voltage difference

Hi !

I im trying to collect data concerning a device's battery life(camera). I would want to calculate the energy my device consumed during a whole week without knowing exactly what current it would theorically draw. My idea was to use the difference between the initial and final battery voltage to extrapole the total energy; is it even possible to do that ?

Knowing that Ui = 8,28V and Uf=8,19V and that my Li-ion battery's nominal capacity is 2.0 Ah, I would have donne

E=(8,28V*2Ah)-(8,19V*2Ah) = 0,18 Wh

but I am pretty sure this is not very accurate, considering that the formula E= C*Uavg uses an average voltage and I am now using specific voltage.
I have done the same exact week test with 8 AA alkaline battery (1,5V 2,5 Ah, 4x2 battery in parallel which would give a 6V 5Ah battery pack). Ui =6,27V and Uf=6,19V, which would give

E= (6,27V*5Ah)-(6,19V*5Ah)=0,4 Wh ....

That's more that 2x the Li-ion's energy loss for the same use! That is the big why I've got doubts on my use of E=C*Uavg.
Do any of you have an idea of how I could know the loss of energy in my battery only using the battery spec and the initial and final voltage ?

RE: Calculation of a battery's energy loss using a voltage difference

No, it doesn't work like that. The ampacity curve is only barely linear with voltage. I suggest you search for "battery university" and/or look for "Li-ion discharge curve"

TTFN (ta ta for now)
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