Figuring out theoretical runtime for 2x18650 and upcoming emitter

I'm trying figure out the theoretical runtime for different emitter and battery combinations (neglecting inherent circuit resistance and driver inefficiency) and I'm not sure I'm doing it right. Here is the information I'm starting with (This is for an Osram LEW E3B Ostar emitter):

Theoretical Batteries: 2200mAh Protected Lithium 18650s x 2 (3.7V each)
Datasheet Emitter Stats by If:

If = Current Forward
Vf = Voltage Forward (given in datasheet as Uf)
Φv = Luminous Flux (total output in lumens)

If............350mA ....700mA ....1000mA
VfTyp .....19.5V .....22V .........24.5V
VfMax......22V ........25.8V ......29.8V
Φv .........260lm .....420lm ......504lm

The datasheet and other information can be found here.

I wanted to figure out shortest battery life running at full power with maximum voltage forward... so I figured out the approximate Watt*hours of the batteries:

Power = Voltage * Current
Watts = Volts * Amps
...so why can't I just drop in an hours unit?
Wh = V * Ah
So if that's correct, that would give us:
2.200Ah x 3.7V = 8.14Wh
Twice that would give our theoretical battery capacity of 16.28Wh... yes?

Next the LED:
Power consumption I would guess would be the way to start...
P = IV
P = 1000mA * 29.8V = 29.8W

So that would mean 16.28Wh/29.8W = 0.546 hours runtime?

If I'm doing this wrong somebody please let me know. However, if it's correct, that would mean it would be possible to build a single-emitter light (THROW!) running on two 18650s with emitter lumens and runtime as follows:

With Vfmax:
260 lumens for 2.11 hours
420 lumens for 0.90 hours
504 lumens for 0.54 hours

With VfTyp:
260 lumens for 2.39 hours
420 lumens for 1.06 hours
504 lumens for 0.66 hours

I think this will be a good candidate for the new "C" li-ion batteries as well. I have a feeling I've left something out, but can't think of what it might be. Thanks!

Note: The above lumenous flux figures for a given If seem to be "average" - a closer look at the LEW E3A datasheet, (emitter without optics as opposed to with optics as on the E3B, for which the datasheet is as yet unreleased) reveals that the bins will vary quite a bit... I read it as anywhere from 240 to 820 lumens, but I could be wrong. Hopefully somebody with a bit more experience can take a look... that datasheet is here.

I did not do the math, but you are starting with 2.200Ah at 7.4 volts (if in series) and need 1 amp at 30 volts.

That's a 4 amp load on the batteries if the converter is 100% efficient. You need a converter and batteries that can handle that.

Frequently, the higher the load (current) the lower the capacity of the battery, so your 2.2ah may drop significantly.

Daniel

Can somebody elaborate on what gadget lover said? I understand that under higher load, battery voltage and capacity dip, as indicated by the battery tests in the stickied part of this forum, but.. what did he come up with the 4 amp load on the batteries? I mean, calculation-wise....?

Try using the LED Pro Modders Program Great program mad by fellow CPF'ers.

what did he come up with the 4 amp load on the batteries? I mean, calculation-wise....?

Click to expand...

The batteries are rated at 3.7v each so in series gives supply voltage of 7.4v.
For a 1A current (for your emitter) you need a voltage of approx 30V.
Assuming 100% efficiency in your voltage convertor (which will 'boost' 7.4V to 30V).
Power out = power in
Vout*Iout=Vin*Iin
Therefore Iin = (Vout*Iout)/Vin
Iin=(30*1)/7.4=4A
Current into convertor from batteries = 4A

Ah, that makes sense now, thank you! Boy... that was kind of an important detail to leave out wasn't it?