2 x CR123 brighter than 2 x 18650?

I guess I don't know why, but for some reason I thought my Solarforce MPP-1 turbohead would be as bright or brighter running two 18650s instead of two CR123 primaries? I thought it would be as bright or brighter because it would be 8.4v instead of 6v. It pulls 1.15A at the tailcap with two CR123s, but only .7A at the tail cap with 18650s. It draws more current with a single 18650 than with two 18650s, but still not nearly as high as with two CR123s.

I guess I missed this flashlight knowledge somewhere along my journey, but could someone explain to me why this is the case?

Thanks.

There is a buck circuit in MPP1. That means it takes some higher voltage in and drop it to emitter voltage, something around 3.5-3.7v. So, to supply the same output current it needs less input current (the same power at different voltages) for 8.4v than 6v. The math seems to be ok for these two, and shouldn’t be much difference between these two configurations, but…
… Buck drivers usually need an input voltage above the output voltage. 1v difference is usually the best. Since a single li-ion can’t go that high the math no longer works and the circuit goes to direct drive or switch inefficiently.

So let me know if I understand this correct...

Lets assume the buck driver in the MPP-1 needs 1v more than the XR-E forward voltage (3.7v) to fully regulate the light. That would mean that you have to drive the MPP-1 with more than 3.7 + 1 = 4.7v. If the input volt is less than 4.7, the light will go into direct drive. One 18650 minus the driver, can supply between 3.3 and 3.5v to the led, which according to the datasheet will draw 350 - 700mA.

However, if we supply the MPP-1 with more than 4.7V we go into regulated mode. When coupled with 2 x CR123 we get a reading of 1.15A (according to motodeficient), which equates to 1.15A x 6V = 6.9W. The same happens when coupled with 2 x 18650 we get a reading of 0.8-0.9A (what I've observed), which equates to 0.8A x 8.4V = 6.72W

So this means the input watt is ~ 6.9W and if we assume a 90% efficiency, the output watt should be ~ 6.2W. Since the forward voltage of the XR-E R2 is 3.7v, the current going the XR-E is 6.2 / 3.7 = 1.675A And its at this point that I'm really confused, because everyone always say Solarforce underdrives the MPP-1 and that the Dereelight 3SM driver is better for XR-E R2's. Am I missing something? Even in this post https://budgetlightforum.com/t/-/4992#comment-94538 if you assume that the forward voltage is 3.8v at 6.2W, the current would still be 1.63A

Please correct me where I'm wrong, because I just ordered a 3 - 18V input and 1.6A output driver for my MPP-1 and after thinking a bit about it, there would be absolutely no gain