Isn't CV/ CC for charging li-ion ....shouldn't the eneloops be charged at .5 amps the whole time?
I'm sorry to tell you. . . In order to get 1900 mAh out of a battery, you always have to put more than 1900 mAh in. We're bump up against the laws of thermodynamics here.
Thanks, I stand corrected. The energy efficiency must be less than 100% but in some cases as you say, the Coulombic (amount of charge) efficiency can be 100%. At 100% Coulombic efficiency, energy efficiency is less than 100% because the average charge voltage is greater than the discharge voltage.This is indeed true for NiMh chemistry, i.e. unlike Li-ion, its charge (Coulombic) efficiency is not 100%, so you need to put more charge in than you get out. Actually they are 100% charge efficient below 90% SOC, e.g. see AA Cycler's NiMH Charge and Energy efficiency graphs for Eneloops.
I figure this is one of those simple things where I am missing a fundamental understanding. I have a Nitecore D2 that charges 2 channels at 500mA each. When I put 2 AA 1900mAH Eneloops in it takes them 5.5 hours each to charge. Is it not as simple as dividing the capacity by the charging rate? I expected 4 hours max given the math. Is the extra hour and a half some ramping up and down thing that the charger doesn't display?
Isn't CV/ CC for charging li-ion ....shouldn't the eneloops be charged at .5 amps the whole time?