[Solved] How to estimate the value of a chart? (Battery tester)


I found a post about battery tester, and I noticed that the battery charge has a power that needs to be estimated in a slightly more complex way.

Could someone tell me where I could find an example code that can estimate what a person could complete when using a chart?

Take my case, with a 18650 battery, with 4.13V maximum voltage (no load) and 12.1A maximum current, measured with a multimeter (Victor VC97), we could have a false calculated power of 49.973 Watts. But in fact this battery was being used in a notebook, and the notebook was showing a power of approximately 26W. As the notebook battery consisted of 6x 18650 type batteries, we can estimate that each battery could supply only about 4.33 Watts.

Going back to our case, when plotting a graph, and delimiting the shear voltage at 3.7V, we can find an intersection point where the lines intersect. At this point we can note that for 3.7V the battery could only present 1.2A. And if we calculate 3.7V multiplied by 1.2V, we get 4.44 Watts. Which seems much more consistent with the result that the notebook was presenting.

Thank you.

Test source: https://www.instructables.com/id/Digital-Battery-Load-Tester/

rtek1000: I found a post about battery tester, and I noticed that the battery charge has a power that needs to be estimated in a slightly more complex way.

I don't think that piece you have quoted is at all reliable.

AFAIK an 18650 battery uses Li-ion chemistry so I suspect it can produce a great deal more than 12 amps when short-circuited and allowed to self destruct (don't try this at home).

The watts output of a battery has to be a combination of volts and amps measured at the same time while powering a load.

There is a big conceptual difference between watts and watt-hrs. The latter is a measure of the capacity of a battery. The former is a measure of the instantaneous power being consumed.

What is it that YOU want to measure, or estimate?


Well, I ended up remembering how to do it, I need to find the equation of the line, so I'll be able to find the current value, relative to the value I'm looking for, in the case, 3.7V.

I found a site that handles this, I think I'll study this part again,


Anyway thank you!

Hello again, on the subject of electric current, yes, it is dangerous, it is subject only to highly trained technical personnel. Not recommended for beginners, not even amateurs.

I found a problem in a hand led flashlight that I have, it uses 3x18650, these batteries have the 6800mA label, of course it is fake. In two of them gave 5.8A, and one gave 0A. In other words, they are connected in parallel, and the charger can not recharge properly, as a result, the battery will degrade ahead of time without even realizing it. That is why the charge lasted very little, only two batteries were accepting charge.

Thank you, have a good weekend.