If I turn one of my Fenix lights on and leave it running until it goes out, take the 18650 out and measure it with a multimeter it reads about 2.91v. So the light cuts out under 3v. If i then wait a couple of hours and measure again it reads 3.00v.
What dark magic is this? Why does the voltage creep back up to 3v? and how did the controller in the Fenix light know when to cut the power if the raw voltage reading of the battery at that point was 2.91v?
It’s normal for a healthy cell to creep back to its nominal voltage. The the 2.9v you see was taken right after it had undergone a “load” from the light.
Normally its 3.4 to 3.6v for a healthy cell to be in fully-discharged, after sitting for sometime.
If you recall, even your car battery will do this if the car won’t start and you grind it down to the solenoid “Click, Click”. After a few minutes it generally will regain enough voltage to try and turn the engine over, albeit somewhat slowly and not for long.
All battery voltage ‘sags’ under load. The higher the load, and the poorer the battery > the more pronounced the sag. Remove the load, which included the cut-off in a light > voltage rebound.
You can take a crap battery, completely full, put it under a high load in a demanding application, and have it fail/quit almost immediately. But, check it with a low load and almost all the capacity is still there.
Which is why some high output/lower capacity batteries, will do better in some high demand lights, than a low output/higher capacity battery.
The ability to bounce back after a deep discharge is one of the signs of a healthy cell. Cells that don’t bounce are culled out and rejected for use in some aerospace applications.
> how did the controller in the Fenix light know when to cut the power
from your comments it seems the Fenix has built in LVP (Low Voltage Protection)
you dont say which Fenix, so I cant look it up, but LVP should be listed as a feature
your comments also suggest the Battery Cell is UnProtected, otherwise when protection in a battery is tripped, it would read 0 Volts.
we dont know what the LVP is set to trigger at, but my guess is 2.7v (could also be 2.5v). It is better for the battery cycle life, to recharge before going below 3.4v.
the 2.91v you measured was without a load, and may have already had a slight rebound.
rebound is normal, but dont let it fool you, and dont repeatedly redrain a depleted cell, just because it rebounded (a Very temporary condition). It will trigger the LVP again, quickly, and this is Bad to do.
I see, thanks. Sounds like when it first steps down the output it would probably be sensible to switch off and remove it from the light then. I normally never get below 3.8v after evening walks so not really an issue for me, I only noticed this behaviour when deliberately running it down to see what happened.
Have there been any incidents of leakage or fires with batteries inside a well made aluminium torch? I would assume there isn’t enough oxygen in one for combustion and i doubt there’s enough energy to blow the metal apart… just curious if keeping batteries in lights is relatively safe or not. Obviously if it leaked it would still be bad
> Kinda feel like I want to get protected batteries now!
that is a reasonable approach, for lights that lack LVP
> Have there been any incidents of leakage or fires with batteries inside a well made aluminium torch?
yes, but that is not your primary risk. The main issue is related to recharging a cell that has been overdischarged.
also, the main risk is from ICR cells… the newer chemistries INR, and IMR, are less likely to get damaged by overdischarge (and are more likely to be sold unprotected).
maybe you can deduce what caused this: (Im not sure)
> oxygen
I dont think the issue is “normal” combustion… it seems to be more of an electrochemical heat discharge.
you could spend some quality time googling for more info… I started to search and found this (battery fire, but not a flashlight):
Pretty sure my light has LVP as evidenced by it cutting out when the cell read 3v. Not sure id gain much by using a protected cell, i’d presumably still have to monitor it while charging, the safety mechanisms are still there with an unprotected cell they’re just in the light and the charger rather than on the cell itself.