With the linear drivers we are all using brightness falls as voltage falls below whats needed to maintain full current to the LED due to battery charge depletion, and since the XM-L2 has a higher Vf, i want to know if as the voltage falls as the battery is getting depleted does the XM-L2 still have an advantage over the XM-L when running below full voltage/current

I’m not sure if this is making sense, so an example using guestimated numbers

At 3A the XM-L may need 3.4V to the emitter and the XM-L2 may need 3.5V. The XM-L2 should give higher lumens because both are running at the full 3A and the XM-L2 can give more lm/w

When the battery minus driver losses can only give 3.4V, the XM-L should still be running at 3A, but the XM-L2 may only be drawing 2.8A, so which emitter is giving more lumens now at the same Vf, the higher lm/w XM-L2 or the XM-L?

Lets say the battery is running low on juice and down to 3.3V and the XM-L is drawing 2.5A and the XM-L2 is drawing 2.2A, which is giving the higher lumen output?

Also is there a crossing over point, or is the XM-L2 always putting out more lumens when run at the same voltage as an XM-L?