Can you calculate lumen degration per 0.1 volt?

DIPSTIX

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I would like to know if there is a method to see how many lumens i am typically losing for every 0.1 volt my 18650 battery loses. If my flashlight has 3x XPL HI LEDS on a single 18650 on turbo mode. (Directly driven)
Also how would the lumen drop be affected switching to a 2x 18350 setup with extra voltage.
 

reppans

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You can make a BLF plumbing elbow/LX1330 light box for ~$45 - mine ties up well to HDS and ti-force (reviewer), the two sources I trust for true ANSI. Or just use the $5 plumbing elbow part and a free smartphone light meter app, or DSLR light meter, etc. (still far better than the human eye).

For V, just wire a multi-meter up.... or use amp current or runtime to match up with HKJs review graphs of your battery.
 
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jon_slider

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how many lumens i am typically losing for every 0.1 volt

You could test it and get some idea
1. How many lumens does it start with
2. How many volts does it start with
3. 0 lumens when the light turns off
4. How many volts when the light turns off

example data
1. 3000
2. 4.2
3. 0
4. 2.2

calculate
#1 / ((#2-#4)*10) =
3000 / 20 =
150

those data assumptions suggest you lose an average of

150 lumens per .1 volts
 

Lexel

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more important is how much voltage drop your cell gets
Samsung 30Q or similar types have very small drop between 2 and 7A

Easiest way to determine lumens drop is to measure with an external power supply the currents as you drop voltage slowly

Many things make the lumens drop at same current like heat inside the emitting area, so DTP MBPCBs will get more lumens at same current compared to cheap aluminium trays

also at lower currents the LED efficiency gets higher, so the same voltage drop will have more impact on output lumens

the most iomportant thing is when the voltage gets close to the LED junction voltage then the drop is exponential

in LED datasheets you can see the voltage vs current graph that gives you some point to start, but in praxis every LED has their own variations
 
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jorn

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I guess you will get a flash, then 0 lunems (broken light) with 2x18350 on a direct drive light.
 

StandardBattery

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more important is how much voltage drop your cell gets
Samsung 30Q or similar types have very small drop between 2 and 7A

Easiest way to determine lumens drop is to measure with an external power supply the currents as you drop voltage slowly

Many things make the lumens drop at same current like heat inside the emitting area, so DTP MBPCBs will get more lumens at same current compared to cheap aluminium trays

also at lower currents the LED efficiency gets higher, so the same voltage drop will have more impact on output lumens

the most iomportant thing is when the voltage gets close to the LED junction voltage then the drop is exponential

in LED datasheets you can see the voltage vs current graph that gives you some point to start, but in praxis every LED has their own variations

I guess you will get a flash, then 0 lunems (broken light) with 2x18350 on a direct drive light.
Maybe not if you use some cheap batteries :) Using external resistance only would still be considered direct drive, inefficient but...

The take away here is it makes no sense to try calculate lumens from an LED based on the measured voltage of a battery. Specifying that the light is direct drive does not change this. One could add all sorts of conditions and constraints on LEDs and batteries and still it really makes no sense what so ever outside of maybe a single specific use case which would not have any useful extension of the data.
 

jorn

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When he says "3x XPL HI LEDS on a single 18650 on turbo mode. (Directly driven)"
Mentioning "on turbo mode". I guess it's a fet driver like the blf x6 driver, and not a resistor. In that case, something will say :poof: if he tries to run it with 2x18350.
 

StandardBattery

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When he says "3x XPL HI LEDS on a single 18650 on turbo mode. (Directly driven)"
Mentioning "on turbo mode". I guess it's a fet driver like the blf x6 driver, and not a resistor. In that case, something will say :poof: if he tries to run it with 2x18350.
I was just joking... but if the batteries are really bad then their internal resistance would be high enough to have a large voltage drop and thus in theory no cause a poof. not to say in general you were 100% right which is why I quoted your answer, as I'm in agreement. This thread is like many on here these days that make no sense, but will probably drag on for 5 pages.
 

DIPSTIX

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Standard battery. You're post is like alot like others on here. Make no attempt to make forward progress on a topic and attempt to raise you're post count by stating that many members are speaking hooplah instead of assisting on some method that might work such as Lexel.
 

Lexel

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http://pct.cree.com/dt/index.html

here you can see all the Cree emitters on their typical specs

cree.png
 
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