POH
Newly Enlightened
I'm thinking of using the MaxFlex v4 in a new light. The MaxFlex will drive 3 Luxeon K2 (200lm - 3.65Vf each) in series using a 4xAA lithium (3000 mAh) battery pack or 6xAA Ni-Mh battery pack (2650 mAh).
Part of my intent on going this route is I could in theory use the same light hooked up to a dynamo+bridge rectifier and run it that way if I end of going that route. I'm not keen on hooking up a side mount bottle generator to my nice carbon frame and while I look into a hub dynamo I wanted to have a new working light that I can use.
I plan on running the light usually at 350mAh/LED for approx 285lm which in most situations on the road will do. In odd cases I would want to turn that up if needed. I say this now but I doubt I would want to crank it up to 1000mA - don't see the need on the road.
If I run this at 350mAh is there a way to determine how much current draw there will be? My understanding of the law of energy conservation is poor. I'' suspecting that to boost the voltage to 10.95 Vf at 350mAh I would consume about 630mA (10.96V/6V=1.82 350mAhx1.82=637mAh draw from the battery).
Does anyone know if this sounds about right? If not do you by chance know what this would draw? Yellow replied in another post and I seem to think this is roughly what he noted.
I also looked at Martin's PCB and that would be an option though I thought a voltage double as included in that design was only for AC to DC current? I may be wrong though.
Part of my intent on going this route is I could in theory use the same light hooked up to a dynamo+bridge rectifier and run it that way if I end of going that route. I'm not keen on hooking up a side mount bottle generator to my nice carbon frame and while I look into a hub dynamo I wanted to have a new working light that I can use.
I plan on running the light usually at 350mAh/LED for approx 285lm which in most situations on the road will do. In odd cases I would want to turn that up if needed. I say this now but I doubt I would want to crank it up to 1000mA - don't see the need on the road.
If I run this at 350mAh is there a way to determine how much current draw there will be? My understanding of the law of energy conservation is poor. I'' suspecting that to boost the voltage to 10.95 Vf at 350mAh I would consume about 630mA (10.96V/6V=1.82 350mAhx1.82=637mAh draw from the battery).
Does anyone know if this sounds about right? If not do you by chance know what this would draw? Yellow replied in another post and I seem to think this is roughly what he noted.
I also looked at Martin's PCB and that would be an option though I thought a voltage double as included in that design was only for AC to DC current? I may be wrong though.
