I just bought a bright headlight for my bicycle: Dorcy 41-4001. It's fairly bright and beefy, which is good. It also runs on 3 AA alkaline batteries, which isn't so much. I want to convert this headlight to run on a Li-ion cell instead. This would give it a longer runtime and make it rechargeable.
Li-ion cells produce a higher voltage than alkaline ones (nominally 3.7V instead of 1.5V), so this isn't necessarily a drop-in replacement. The headlight uses the 3 AA cells in series to get 4.5V nominal. The headlight description says
LED: USA Made CREE XML (T6) High Power White LED Power Driven:5.0Watt (3.10~3.15V 1500~1600MA ± 5%)
The LED manufacturer description lists a typical forward voltage of 3.1V and a maximum current of 3A, so this is all consistent.
To replace the power source I need to know how the driving circuit works. I naively assumed that there's a switching regulator being used, feeding back on the current to drive the LED efficiently and to maintain a constant current (and thus constant light output) in a range of voltages. To test this theory I ran the light from my benchtop power supply instead of the AA batteries. I expected to see the input current track the input voltage inversely (keeping the input power roughly constant). Instead, I saw them moving together, with the light brightness varying noticeably with input voltage. This is what you'd see if there was a simple resistor used to limit the LED current. Seriously???
Headlight body characterization
The headlight is composed of
- the main housing (contains the LED and the battery pack)
- battery cap (contains ON switch)
The initial assumption was that everything we care about lives in the housing, so I characterized that first. I scanned a range of input voltages, taking note of the input currents drawn. Let me assume a trivial resistor-diode driver circuit and try to fit my observed input i-V data (source)
I didn't go to the nominal triple-alkaline voltage (4.5V) because my power supply can't produce more than 3A. The fit isn't perfect, so there's likely something else going on. The fitted resistor value is 0.28 Ohms! Could this really be true? The alligator clips could have resistances in this range. Furthermore, the LED datasheet says that it produces 280 lumens at 0.7A, but we get this at a very low voltage and the headlight is rated for 220 lumens. This is a highly inconsistent piece of data. Something else is going on.
Headlight body and battery cap characterization
What if the battery cap is more than a dumb switch? It's beefy and sealed, so there could be something in there. I hooked everything up cables, and scanned the voltages again. Once again, let me assume a trivial resistor-diode driver circuit and try to fit my observed input i-V data (source)
Much better! The inferred resistance is 1.9 Ohms. Low, but I can believe it. The body was just the LED, and the resistances were my (poor) connections. Alkalines generally start at around 1.6V, then decay to 1.0V or so as you use them. Ballpark average: 1.3V, so 3 of those are at ~4V. The data we just gathered puts this current draw at 600mA, which is a bit below 700mA; this makes sense since the rated light output is 220 lumens, a bit lower than the 280 lumens we're supposed get at 700mA.
I'm now confident I know what's going on. I'm satisfied that Li-ion won't break anything, but it will be dim. Li-ion batteries have varying discharge curves; generally they start at 4.2V or so, then drop to 3.0V-3.5V as you use them. This will result in a dramatic drop in the current and thus light intensity.
This is just lame. Without regulation the light gets dimmer as the batteries are used up and it's just inefficient. Taking their nominal operating point of 4V you get 3.1V in the LED and you throw away 0.9V in the resistor for an efficiency of ~75%. In this scheme the efficiency gets worse at higher voltages (brighter light). I guess you could do worse, but it's just dumb.
My particular rechargable plan won't produce great results. I guess I can lower the resistance by installing something in parallel. Maybe I will, but I will complain the whole time.