I'm busy modifying a big 3xD cell maglite to use a luxeon III lambertian emitter (PW09) and thought I had everything working great until I measured the current. I want to run the LED at ~800mA or so @Vf = ~3.7V. I had some 1ohm 1W resistors on hand, so I wired used one to provide current limiting. Unfortunatly with the 1ohm resistor, I'm only measuring ~200mA or so through the LED. WTH? Also, I should be getting 4.5V from the batteries, but I'm only getting 4V with the LED on. So my thinking is that either the batteries are not fresh (they should be, they came right out of the package dammit) or that they are not able to supply the amount of current that I want. I thought that D-calls should be able to provide >2A? The Energizer spec sheet shows V/I graphs at 1.5A.... Did I just get bum batteries?
Unfortunately, with that much current drawn, small changes in resistance make a big difference. You calculated for the ideal conditions (batteries @ 4.5v, LED at 3.7v, .8A current), but, as you noted, the batteries only put out 4v instead of 4.5v. Calculating for a 4v supply, the correct amount of resistance is .375 ohms, almost 1/3 of the resistance you're using now. One thing you can do is stick 3 of those 1ohm resistors in parallel to get .333... ohms total. That should get the current drawn by the LED back up to .8A. A neater solution might be to find a silicon diode which has just the right amount of voltage drop at .8A. I can't tell you of any, mainly because I'm at work right now.
D cells should be capable of supplying a lot more than ~200ma without a serious voltage drop shouldn't they?
What is the actual voltage drop of your luxeon at the desired current? The Vf of the luxeon III devices can be up to 4.4 V. Check out this datasheet for more detailed information about luxeon binning. Most primary batteries can have significant internal resistance. A good quality alkaline D cell may lose up to 0.1 V at 500 mA. Lower quality or zinc carbon batteries may lose considerably more.
Spec'd Vf = 3.7V @ ~800mA Luxeon Datasheet Soooo, if the max Vf is up to 4.4V, and I'll definitly lose more than .1V from the batteries... could I just run the LED with no limiting resistor??
Yes, you could omit the current limiting resistor. But, this is not a recommended option. The problem is that you are relying on the Vf of the LED and the internal resistance of the batteries to limit the current. If you replaced the LED with the same type (but from a different batch) then you could burn it due to differences in Vf. I built a number of LED lamps which needed very low resistances and I bought several boxes of LEDs from lumileds. Leds in each box were matched, but the boxes themselves were not. I ended up having to empirically determine the appropriate resistance for each box of LEDs.
