Looking forward to the Convoy L6 Mk2 then 
With all this extra output, manufacturers need to concentrate on better heat sinking. I find high output lights rather pointless if they have to stepped down quickly to avoid over heating.
+1
Hmmm…
At some point some professional markets will want to have modern LEDs that can be used in pure throwers, like search lights, don’t you think so?
I find it quite strange actually, that the developments in LEDs are apparently going in a different direction.
Matter of time maybe.
Momentary turbo button.
You’ll let go when it’s too hot and it can’t accidentally be left in turbo mode.
just an idea…
Interestingly, Led Lenser has this feature. Press and hold for 130% output. It’s barely visible compared to 100%, though I do like the idea of this feature for more powerful lights.
:+1: Looking forward to the highly efficient, higher output street lights! :person_facepalming:
lol 
Convoy L8 right around the corner!!!… oh geez just take my $$ now!!
I was using the ledlenser P7.2 momentary button last night for creating lots of starburst effects in a light painting photo.
I want this on an Acebeam K60
I extrapolated the two V vs I curves and for the 6v emitters, reaching 12 amps will take approx .75 to 1 volt less on the 70.2
Running 2S direct drive won’t be possible anymore due to extremely high currents at 7.4v that would kill the LED instantly.
Although 3S on a 12V emitter will be excellent for direct drive.
I think cree will go in the direction of factory de-domeding more LEDs and split the market up.
Unfortunately I don’t think running 3S on a 12V emitter will work. The XHP50.2 forward voltage is consistent with it being four XPG3 dies, so I’ll assume the XHP70.2 is four XPL2 dies.
From Texas_ace’s test of the XPL2 the forward voltage at 6A is 3.56V. In a 12V configuration that is 14.24V. 3S voltage is just too low to get good high current.
3s ia 12.6v max.
Go look at the graph, that would be 4A+ at full charge.
http://www.cree.com/~/media/Files/Cree/LED-Components-and-Modules/XLamp/Data-and-Binning/ds—XHP702.pdf
All i know is there are HI versions of XQ-C XP-L and XHP35, but that’s it.
You would think though that shaving / slicing / wet sanding the dome down would be the same result, but apparently it is not, or not always…
Based on Texas_Ace’s measurements of the XPL2 the forward voltage at 4A for the 12V emitter is 3.34V*4=13.36V. This is different from your extrapolation of the curve in the datasheet. From what I’ve seen this sort of data in the Cree datasheets has been not completely accurate. For example the cree data says the XPL2 voltage at 3A is 3.125V while Texas_ace measured 3.20V. Also it doesn’t look like you took into account the voltage drop for the batteries at 4A, which would be around 0.4V. I think you will get significantly less than 4A on a full charge, probably more like 2A.
3.2v is really darn close to 3.125v.
Also, if your batteries are dropping .4v at 4A then sorry but they are horrible batteries.
Well, 0.075V times 4 can make a significant difference in the resulting current. And 3 high drain 18650 cells in series plus some wire resistance does lead to about 0.1 ohms resistance. I guess we’ll just wait for you to test it and report.
I don’t have enough experience running LED tests and making graphs like djozz or texas ace do.
We’ll see how it performs when they do their tests.
I’m going to be running mine at 12A constant current so this isn’t really relevant to myself.
I tested a XHP50.2 two weeks ago in 6V set-up, that should give you some information too for the 3-battery discussion. Based on those numbers:
Say you want 2A at 12V, that is 4A at 6V for 3000 lumen. You need 6.3V for that in 6V situation, that is 12.6V in 12V situation, so 4.2V per battery under 2A load. So no, 2A is not feasable.
Try 1A, =2A at 6V, so 1800 lumen and 5.95V, that is 11.9V in 12V configuration, almost 4V per battery. With high drain cells fully loaded providing 1A, that should work for a very short period if there are no voltage losses in the flashlight.My conclusion again: does not work.
Or did I make mistakes in the above numbers?