That's two 16340s in series. BAT+ to LED+. BAT- to 105c ground plane. LED- to the 105c's normal LED- pad. To keep the MCU alive, the positive to power the MCU is taken between the two 16340s, so it only sees half the total input voltage, or a single LiIon, just as it normally would. Input voltage is ~7.2v under load (cells pre-discharged for this test, see below), voltage across the LED is 6.23-6.28v on HI mode. All modes work as normal using the reverse clickie glued to the side of the battery box (even though the MCU remains powered at all times; I don't understand how it changes modes like this, but it does).
Running on LO mode:
Ah, now those caveats, damn it. First, these cells were discharged down to 7.68v (open circuit) before testing, with a hobby charger. Will it still work when they're fully charged? I suspect so, but no guarantees. At this input voltage the 7135s don't even get mildly warm.
Second, with a single pole switch, the power to the MCU part of the board stays on all the time. To do this correctly you'd need a double pole switch, with one pole on the BAT- and the other on the lead that powers the MCU. The LED shuts off completely as normal with the switch as that removes the ground path from the 7135s, but the + and - remain intact for the MCU part of the circuit. Just switching the power to the MCU instead and leaving the main ground circuit connected doesn't turn off the LED... it goes dim, but that's leakage through the 7135s when there's nothing controlling the Vdd pins. So both the full voltage and half voltage parts really need to be switched together to make it real-world usable.
I don't think I've come up with anything original here, all credit goes to Rufusbduck, DrJones, and PilotPTK, who in various parts have explained how it works in a way that simple-minded parts assemblers like me can understand. :p
Thanks for the experimenting. You had me scratching my head for awhile and read your post three times as I could only see one battery. The other is in camouflage blending in with the background. I believe you it works but if I was a liar I would tell you why.
With charged 750mAh 16340s, starting voltage 8.23v (did a rather fast charge, this Turnigy MAX80W seems to shut off early if the charge rate is manually set high-ish, it will fully top them up if I set it to something ridiculously low like 0.2A). I played around with it for a while before taking measurements, the driver has no problems with heat whatsoever. On HI the 7135s may be getting just slightly warm, but it's so slight it could be nothing more than my brain imagining the temp change.
Open circuit input voltage by the time I started taking measurements was down to 7.94v. These aren't particularly hot cells, so it still sags quite a bit under load, but from how the setup has behaved so far I just can't see there being an issue with stronger cells.
This driver is a "3.04A" 105c from Illumination Supply, just a traditional unit with the stars still used for group selection. Ammeter placed between the LED+ and BAT+ leads.
LOW: .15A, 4.95v at LED
MID: .91A, 5.30v
HI: 2.90A, 6.14v
On HI, input voltage falls to ~6.6v after around 15 seconds.
There is no measurable current flowing on the connection to the MCU with the switch off. There is some voltage present at the LED pads with the switch off, though if there is any current flowing (leaking thru the 7135s) my meter isn't sensitive enough to measure it.
This looks to be completely feasible to use in an actual light, the only hurdle being a pass-thru for the BAT+/LED+ that doesn't make a connection to the driver board, and a way to reliably make the MCU's connection to the half-voltage spot between the two cells.
Sounds very interesting! Of course it's above my head. If you do get to where you have enough time, would you put up a simple diagram please, as I am floundering through the text, (not your fault, my fault, can't grasp that stuff any more).
I think that, if I’m understanding what comfy did, he took 2 16340 batteries in series to power the MT-G2, but he connected the processor/MCU on the driver board to only “the middle” of the series, so that the MCU would only get 3.7 - 4.2V, but the emitter would see 7.4 - 8.4V.
If he didn’t do something like that, but fed the output of the 2 batteries in series directly to the driver, the driver would be seeing 7.4 - 8.4V, which would be too much, I guess?
On the bench it's simple. In a flashlight, the tricky part will be that the front battery's + cannot contact the normal spot on the driver board, it needs to jump around/through it instead, to connect direct to the LED+ (or there may be a trick to isolate the MCU's power feed from the center pad passthrough, haven't looked at it that closely yet). The rear battery's + needs a way to get around the front cell to power only the MCU portion of the board. The negative connections are exactly the same as a normal single cell/105c setup.
No, really. It really is that simple (which probably explains why the description didn't make any sense). You're running the MCU part of the board with one Li-Ion (3-4.2v), and the LED with two (6-8.4v).
Yep, just about anything that can get by with not using the driver as the contact plate will make it a lot easier. A side-by-side cell layout instead of inline would make it easier still.
I wonder if... it might not make more sense to locate the driver between the cells (in an inline cell light), instead of the traditional spot in the front? You'd still need to run a wire or something to carry the LED- around the front cell to get to the driver. Maybe have the driver on a fixed partition inside the tube, load one cell from the front and one from the rear.
> with a single pole switch, the power to the MCU part of the board stays on all the time.
Actually no, since the switch cuts the GND connection of the driver, the MCU doesn't get powered.
There are a few ways to get a suitable supply voltage VDD for the driver:
a) Pick up VDD from in between the batteries. (like shown above)
b) Pick up VDD from in between the LEDs in case of 2 LEDs in series (not possible with MT-G2 of course).
c) Use a simple resistor/Z-diode voltage regulator: (B+)—-(R)—+—(Z)—-(B-)
d) Use a 78L05 voltage regulator.
Might be easy to do with the Small sun z08 and the 2 18650 cells side by side. You’d have to alter the current path from parallel to series for the emitter, but should be easier in that case than one where the cells are stacked.
nice work there comfychair! Great to see something simple but useful demonstrated. It’d be tricky to implement for a bike light with a separate battery pack as you’d need a 3 pin plug and a separate charging jack (unless you find some way of wiring up the MCU to the battery balance tap), but options b) to d) that DrJones suggested would work just fine.
Very interesting, I’ll have to think about this. Now it just comes down to parallel LEDs + cells + stacked chips vs. series LEDs + cells + balance tap.
true, I hadn’t thought of that as I’m used to putting the driver in the light head for temp sensing, which the 105C doesn’t have. Your approach would work fine though, especially if the battery was permanently wired to the light. Twin LED all-in-one lights could also be wired up light this too.
Just ran it with topped-up 18650s, and heat might be a problem after all. The driver gets uncomfortably hot and after about 45 seconds on HI current to the LED starts falling off a cliff even while the cell voltage stays relatively flat. Let it cool down and switch on again and the current is back to ~2.90A. This may not be an issue with the driver mounted in a pill as at least some of the heat would transfer into the light compared to how it is now, just hanging in free air.