Originally Posted By: Quattro Pete
Originally Posted By: JHZR2
There are now 2550 mAh Eneloop Pro cells. Im not sure if the impedance is similar or different, but remember that high power cells often don't store as much energy, so the 2000mAh may actually be better (assuming the impedance is lower and rate capability is higher) than the pro. Just speculating though...
My charger reports the internal resistance of regular Eneloops to be around 70 mOhm. The Eneloop Pro - around 50 mOhm.
Not sure how accurate this is.
Well if we go by the definition that:
I = C * dV/dt
Then through arrangement you can see that:
I/dV = C/dt
Since Resistance = V/I, then the above equality when integrated also relates to the reciprocal of resistance.
But long story short from above, one can generally see that how voltage drops when charge is removed should define impedance. So intrinsically, a higher capacity cell will reduce voltage less for a certain amount of charge removed, which is intrinsically sound given that higher energy cells can have more energy removed before voltage (which is an indicator of state of charge) is meaningfully decreased.
Which also should mean a lower voltage drop for a given load.
The question becomes if it is higher cost weight that is used to get more energy in the pro cell, or if some other way of reducing one of the many internal resistance factors means that more capacity can be extracted before a minimum voltage is reached, or some combo.