Help with the spec's from a light guru

[gcbryan]

I would like to understand the spec's as it relates to this light:

Diving Cree MC-E (K-WC) 3-Mode 640-Lumen LED Flashlight Kit (2*18650/4*CR123A/4*16340)

So 640 lumens from 2 18650 batteries and this light is Digital Regulated 1500mA Current.

If the 4 leds are wired in parallel then the voltage requirement would be 3.6 volts I believe and the minimum mA would be 1400mA (4*350mA)

If that is how it's done how do you get 670 lumen as the Cree spec says 370 lm at that minimum current?

If it's wired in series the voltage drop would be 3.6 x 4 = 14.4 V so that's not it.

It seems to me that the mA would have to be higher than 1400 for this chip to reach 640 lumen.

What am I missing?

Also, there are some divers locally who do have this light and have tested it out underwater and it is noticeably brighter than a 10 watt HID (Light Cannon) which is rated at 450 lumens. So, it does seem to produce 640 lumens.

I'd just like to understand how.

Any help would be appreciated.

 

[gcbryan]

Bad form though it may be, I'll answer my own question with the info I got elsewhere in case it helps someone else.

The leds in this case can be wired independently so with 2 in series and 2 in parallel you have 7.4 V at 1400mA. This is with each die seeing 3.7 V and 700mA (the two in series would be 7.4 V at 700ma and the two in parallel would be 3.7 V at 1400mA).

If you get 370 lumens at 350mA then it's reasonable that you could get 640 lumens at 750mA.

A couple of more points (that I just learned) for anyone interesting in these kinds of things...the batteries that can be used differ in voltage with 4 CR123a in series producing (3V*4) 12 volts and the rechargeable 18650's with a working voltage of 3.7V * 2 or 7.4 volts.

This choice of differing battery voltages is possible due to a buck converter which is a type of voltage regulator which "pushes down" the voltage to within specs.

Now perhaps you know a little more about led lighting terminology...I know I do!