Yeh, that’s par for the course with those boost converters. It’s unreal how they need to be “fine-tuned” for a particular output. Coil inductance and resistance and size are huge factors. Beyond a certain total current in a specific component, they start to saturate the core. Lotsa heat and current, but no extra output. Like overcooking an LED ’til it turns blue with too much current.
I’d be curious if and where the bottom falls out on the low current end. Either in stability, ripple, or efficiency. 1-5 lumens might be something to shoot for… For data gathering purposes. 
Before putting together the graphs, here are a few numbers using NiMH running to 10%:
- Stock with 0.43 Ohm: 93 “maukka lumens” for 77 minutes
- Replaced resistor with 1 Ohm: 42 “maukka lumens” for 210 minutes!
Considering Panasonic/Sanyo makes 100% of the best NiMh batteries in the world (I’ve experienced this in Prius battery world and cordless tool battery world as well), I think somewhere at the company they definitely know what they are doing. Whether the engineers ChibiM spoke to in Belgium have the data at hand (and/or can share it) is another matter.
I don’t have sources to cite, but to my knowledge: reversed-polarity NiMh are considered failed. You can sometimes recover them but you’ll have noticeable capacity losses after even just one occurrence. At worst you’re risking them leaking. Leaking NiCd/NiMh cells is hard to do but when it happens the products are unmistakably toxic - I have dealt with them in the battery recycling world.
If you do get any that leak, please dispose of them safely. Probably best not to vacuum the powder as a household vacuum would likely distribute the dust throughout your home.
Very nice! 30-40 lumens is probably my sweet spot for around the house use. Will need to see what resistors I have lying around.
I recently bought a rainbow of C01s to use as reading lights / for younglings but I’d be also be curious how low it could go while maintaining a stable output.
Thanks for the information and the runtime test. I also did it with a 0603 resistor and changed the lens to a slightly narrower angle. I am very satisfied with the end result.
Alright, the low output runtime finally finished so here are the results:
- The only AAA alkaline I had sitting around was used, but measured at 1.55 volts so fairly full. That said, the runtime is understandably shorter than expected. You can see it held regulation for a little bit, then ramped down out of regulation. Output started out at 95 maukka-calibrated lumens.
- Using a NiMH (“1100 mAh” EBL LSD NiMH which is actually a 960 mAh cell but still very nice). Output started out at 95 maukka-calibrated lumens and stayed in regulation for 65 minutes.
- Also on NiMH, but with the output bumped by adding a 1 Ohm resistor in parallel with the stock 0.43 resistor. Output started at 133 lumens and didn’t regulate very well. It dropped hard at 35 minutes. This configuration is not recommended.
- Also on NiMH, but now using only the 1 Ohm resistor. Output started at 43 lumens and stay in regulation for 185 minutes. This is a good option if you like extended runtime.
- Also on NiMH, but now using only a 20 Ohm resistor. Not pictured. Output was around 2.3 lumens and ran for 69 hours (3 hours shy of 3 full days)! Efficiency was very good with amazing runtime.
- On all tests besides the last, I stopped the test when it dropped to around 2 lumens. For all of those, the battery at that point was around 1.1 volts. When I did the 2.3 lumens test, the light actually started blinking (probably the regulator struggling to enable) and voltage was 1.0 volts.
1 Thank
Fantastic, thank you!
Amazing work gchart, thanks for your time and efforts, I’m almost certainly going to get some to mod now.
No problem! I’m sure it’s been said somewhere, but just as a reminder: the stock current sense resistor is 0603 sized. I was able to carefully fit a 0805 on there, but if you have 0603 available I’d recommend those.
The question is: Can you replace it with a 1Ω resistor and a second 1Ω in parallel, but switchable somehow? 
Thank you, gchart!
Coming from a non-technical background it’s a bit hard to understand where all the energy went to… In case of alkalines, Duracell Coppertop AAA can deliver 0.5 W constant power for over an hour. Practically every LED light bulb manufacturer reported that their fixtures passed 200 lm/W threshold. Bare LEDs passed that like 8 years ago. So, one hour of regulated 100 lm on one AAA alkaline element should be common place by now, right? From the chart it seems like we get a quarter of that. Where did all the energy go to?
Yup, saw your comment about resistor size earlier in this thread. I only have 0805 so will try that first otherwise looks like I’ll be ordering more parts 
Not without replacing the driver. This is single mode only with no potential for having two modes.
The higher the current draw, the less apparent capacity. Especially for alkalines. NiMH have better discharge characteristics for higher current draw but there’s still a limit.
In looking at this chart in the Duracell AAA datasheet you linked to…
You can see that as draw goes up, capacity goes down drastically. In stock form, this driver is dialed in for 221 mA current to the LED. With an LED vF around 3.0 volts, that’s 663 mW. There’s some inefficiency… let’s say the boost driver is 85% efficient. That’s a total power draw of 759 mW. According to that chart, it shows to expect around 1/2 hour of service life at those kinds of power draw.
Resistors are small. Solder one in place, attach a second one in parallel with thin enameled wire, with a contact that can be switched somehow. Must be integrated in the plastic holder.
There are lots of 80%s in there.
The ’351 is rated about 170lm/W. Optics might have roughly an 80% efficiency (ie, 100lm from the LED = 80lm OTF [out the front]), the boost driver might be 80% efficient, etc. Losses add up.
Also, internal resistance goes up as a cell drains. Terminal voltage drops, to the point where the driver can’t keep supplying current, hence it dims and eventually turns off. The more current you suck out, the lower the terminal voltage.
I just burned down an alkaleak last night supplementing some light when reading. My E03 still had all 3 modes, but M was almost indistinguishable from H. Turned it off to flip through pages. Next time I tried turning it on to read something of interest, it wouldn’t turn on. Left on, it could’ve burned lots longer.
Point being, alkaleaks are pure evil. They do things just to screw with us and cause us misery.
Thank you for the explanation, that’s about the level I can still understand 
The LED has efficiency of about 100 lm/W then, right? That accounts for one half of my inflated expectations. So, we should expect 0.5 hour of fully regulated output. We are getting about half of that, if I eyeball it correctly. And we have already factored in driver inefficiencies (85) and the actual power draw of the LED. What does that mean? The driver is actually 43 efficient? Or the element is not quite Coppertop quality? It feels like we are being shortchanged at every turn
This is another important piece to keep in mind… I didn’t have any brand-new batteries to test. My alkaline was partially used.
Here’s Skilhunt’s chart. It probably pretty close to reality.
And yes… there are inefficiencies everywhere. Increased power draw decreases battery efficiency. The driver has some inefficiency (roughly 85% or so). You’ll lose some through the optic (~15% loss). It all adds up unfortunately.
The constant power discharge graph already accounts for that. Right? Falling voltage, rising current - it’s already there. When all is said and done Duracell promises you 750 mW for 0.5 hour = 375 mWh. Or, if you can lower you consumption rate, 500 mW for 1.1 hour = 550 mWh
It doesn’t look like like it’s 85. The LED consumes 221 mA at 3 V = 663 mW while in regulation. Once the light output drops I assume the power used drops proportionately, so I completely unscientifically integrate total power (663 mW * 5 min + 330 mW * 20 min) = 165 mWh. If we assume driver efficiency at 85 it’s 195 mWh out of the battery. So either your battery was half discharged (but then it wouldn’t measure 1.55 V) or the driver efficiency is about 50%.
I know even less about optics than I do about electricity, if that’s possible Let’s ignore that for now.