The driver isn't seeing 39A though.
Can you explain that to me. I’m still on a learning curve
I’m glad I don’t have to take any exams.
Ohms law says:
Volts = Current (Amps) x Resistance (Ohms); or as more commonly expressed: V = IR. So, we can see that there is a direct relationship between voltage, resistance, and amperage.
You are measuring 39A at I'm guessing around 4 volts at the emitter (it is probably lower than that, but we'll just use it because it's easier math!) We know that you are putting about 12 volts through the driver and the emitters are wired 3S, so they each see about 1/3rd of the series voltage (4V.) But, what matters is the driver, switch, and wires are only seeing 13A. If you tried to 1S6P wire those emitters, the wires, switch, and driver would be seeing the full 39A. Resistive loss (heat) increases as amperage increases, which is why 3S makes much more sense here. You can carry twice as much current in the same size wire if you double the voltage. Of course, as the voltage gets higher there can be other issues (arcing, etc.) but as long as you remain 'in spec' then that won't be an issue.
Okay. Thanks.
I’m going to copy and paste that into my Evernotes. I was really starting to get scared.
thats crazy. how many watts are you running?how many lumens is this light?
What i have undstood:
with cells in series the voltage sums up, but share the Amps reading at the tailcap!
in paralell they ad their amps but share the voltage
each cell 4V / 3 Amps
3S0P: 12V / 3A
0S3P: 4V / 9A
volts add up, divide the total current measured to determine how many amps each “cell” is providing
If he is reading 13A at the tailcap…it’s pushing 13A total…but divided by 6 cells, each cell is pushing a little over 2A each
No, if he's pushing 13A in series then each cell is pushing 13A each.
So does this mean in a buck driver configuration with two Li-ions, if the tailcap reading is 3A than each cell is pushing 3A?
Also when in series configuration does the capacity of each Li-ion add up? for example two 26650’s are in series, does that mean they have a total of ~8000mah? Or does this only apply to batteries in parallel?
Never was any good at series and parallel circuits and principles :~
what counts is the Watts a cell delivers…
V x A = W
if the delivered Voltage is below Vf you need a boostdriver otherwise a buckdriver
What you pointed out is why the mAh rating can be deceptive. You would think that a little 2000mAh AA Eneloop was coming close to the amount of energy that is stored in a 2000mAh 20R...when the 20R really has about three times as much stored energy, even though the mAh is the same.
mAh in series stays the same, but that doesn't mean the total energy changed, it stayed the same but is just measured differently. The law of conservation of energy applies to batteries too! 
ok have to say, true powerful light you have moded there Ouchyfoot 
regarding laws of Ohms I yet remember one thing my father teaches me when I was young :
and if you want little more information than …
Damn… your father taught you that when you were young? sheesh… my father only taught me how to ride a bike :_(
Okay guys. This light really is putting out 4.3A per emitter on high. I had it out last night and it’s just crazy bright. People’s jaws were dropping.
I was illuminating a small forested patch out back, and it lit it up so bright I was seeing purple spots in my eyes afterwards.
I think that AR lens I installed is helping to squeeze every available lumen out of the reflector.
Thanks Ervin.
I have zero background in the physics of electricity, but I think I can make heads and tails out of your charts. I’m going to sit down with my multi-meter and calculator today and play with some of these calculations. I think the exercise will help me get a better grip on things.
Yes - if 2 cells in series and bucking down to 1 LED, if tailcap is 3A, each cell is pushing 3A with total output of the cells ~8V, 24 watts of power, ~6A to the single LED.
2 4000 26650 in series will have 4000 mAh for providing 8V. When bucked down to ~4v though for one LED, effectively they have 8000 mAh -- hope I'm explaining this right, maybe RMM can confirm. Certaining in practice, you should get about twice the runtime from 2 cells in series driving a LED vs. 1 cell driving a LED, assuming at the same amps to the LED.
I’ve been playing around with Ervin’s charts (just doing math calculations with a C8) but I don’t know how to check resistance. I calculated the resistance using my known factors, but don’t know how to confirm the results. I bridged the battery and tube with my ohm meter, but get no reading.
Checking for small levels of resistance with a multimeter can be tricky. Maybe it's because I only have cheap DMMs but it has always been hard for me to get consistent results. Another easy way is to measure the voltage drop across a conductor. If you're reading 4.0v at one end under a 3 amp load, then 3.9v at the other end, you can calculate the resistance from that drop.
I got the light today, and it has XM-L, not XM-L2. I didn’t really expect it to have XM-L2, so I’m not heartbroken. Probably XM-L U2. Never trust DX.
The light says Kinfire on the side, but the box has a picture of the same light that says COURUI.
:/
Bad DX! Never trust DX. I would try to get a partial refund.
Mine came with XM-L emitters too. I expected that since the aliexpress seller did not mention XM-L2s. I had a little hope mine were XM-L2s though. Not a big deal. I got lots of spare XM-L2s from other lights.
At least mine is OEM. Which I like. :)
I really like the size of the light. :) Its pictured in OP here beside TK61 and BTU shocker.