# Runtime difference between Li Ion and NiMh batteries in same light on same modes?

#### HighlanderNorth

##### Flashlight Enthusiast
OK. If you have one of the many LED lights that are designed to run on either NiMh or Li Ion batteries, what will be the differences in runtime if using the light on the same modes when comparing the Li Ion vs. NiMh batteries. Lets say you have a Zebralight SC52. Its listed to use either an AA battery or a 14500 Li Ion. If you were to install an Eneloop 2000mah AA battery one day, then a 14500 Li Ion battery with 750mah capacity the next, and you experimented by running the light on the exact same modes one day til each battery was run down to its safe lower limit, how would the run times compare as long as you were comparing the low and medium modes, not the much higher turbo mode that you would likely see using the Li Ion battery?

The Eneloops are 1.2V nominal with 2000mah capacity and the 14500 is 3.7v nominally with just 750mah. But I assume the additional voltage would essentially make up for the lack of capacity if used on the same lower and medium modes. How does that work out? For instance, lets say the medium mode you were comparing was 100L and the low mode 25L approx. on both batteries....

Would the run times be fairly close when comparing these 2 types of batteries on the same modes?

Well... at least in my personal experience, very different battery voltages for the same physical size is the exception, rather than the rule. User error, bla bla...

I suspect most lights that take both NiMH and Li-ion will show somewhat different behavior with each voltage. But assuming all else equal, it becomes simply a matter of -useable- energy contents of a battery. Which is measured in Watt-hours, or [nominal voltage] x Ah. So...

The Eneloops are 1.2V nominal with 2000mah capacity and the 14500 is 3.7v nominally with just 750mah. But I assume the additional voltage would essentially make up for the lack of capacity if used on the same lower and medium modes. How does that work out?
That would be ~2.4 Wh for the Eneloop, vs. ~2.8 Wh for the Li-ion. Not that big a difference.

A remaining factor is efficiency of the driver circuit. Most likely it's designed to be most efficient at ~1.2, OR most efficient at ~3.7V, not maximum efficient at both voltages. I'd expect a DC-DC conversion easier to do when starting with a higher voltage, but who knows how good ~1.2V powered circuitry is these days.

All this of course assuming the battery does what it says on the box. Which is more likely for an AA NiMH than for a 14500 Li-ion.

The answer is going to depend upon the driver being used. If you have a buck driver with a linear regulator, then the main determination of run time is going to be Amp hour capacity of the battery system. A linear regulator simply converts the excess available voltage to heat in the device instead of light. So with a linear regulator in the circuit, the higher amp hour capacity in the NiMh cells may provide longer run time than the 14500 Li-Ion will.

If you are using a boost driver, or a better quality buck regulator, then the determination will be mostly a function of energy capacity of the battery system.

In general a 1.2 volt system will be less efficient than a 3.7 volt system because the higher current results in higher I^2*R losses, and there is generally a fixed loss of voltage across a semi conduction junction.
and at 1.2 volts that is large portion of the available voltage than at 3.7 volts.