Direct Draw & Voltage on p2d/L1d/p1d Fenix Question

Hello,

I'd like to understand, but I do not find the answer anywhere.
If the p2d is on direct drive, what would be the current draw? Would this be dependent on the battery's direct current output?
That is on the aw rcr it would be what? In essence depended on the battery?

I know that power = draw*voltage.

Please help me to understand.

Thanks

Yikes, no one knows? help!!

The best way to find out is by measuring it.

The current will be VERY depend on the voltage and the values for one P2D does not necessary match other P2D's.

HKJ said:

The best way to find out is by measuring it.

The current will be VERY depend on the voltage and the values for one P2D does not necessary match other P2D's.

Click to expand...

Thanks, that's great, but can you let me know how to do that?
I have a multimeter. Where are the points for me to measure it?

timcodes said:

Thanks, that's great, but can you let me know how to do that?
I have a multimeter. Where are the points for me to measure it?

Click to expand...

I would get a variable power supply and a 1 ohm resistor, then connect it this way:
power supply + ---- resistor ---- led/flashlight ---- power supply -

Start with the power supply at 3 volt, measure voltage over the resistor (and calculate current from this) and the led. Then increase the voltage in 0.1 volt steps and do it again, while drawing a current/voltage curve.

The stop point would be 1 volt over the resistor (i.e. 1 ampere), this is the rated maximum current for the led.

Then your know that a battery, just of the charger can be up to 4.2 volts, but working voltage is in the range of 3.6 to 3.7 volts.


Measuring direct between a battery and a flashlight is difficult, because the voltage drop over the meter would affect the result and give to low a reading, it is also depend on the battery charge state. But it is done this way:
battery + ---- meter in 10A range ---- led/flashlight ---- battery -