AW protected 17670 with Malkoff M60L ?

[Niconical]

Hello.

I am trying to work out expected performance and runtime using an AW protected 17670 battery with a Malkoff M60L. There is info available on the forum on that subject, but only for the M60, and I'm specifically interested in the lower current draw of the M60L.

I have my own thought process for it, but considering my level of knowledge, or lack of it, my thought process would probably be better described as my semi-educated guess process, so I apologize in advance if any of my 'facts' below are off the mark.

So, here goes.

The AW 17670 has a listed capacity of 1600 mah.
The current draw of the Malkoff M60L is 350 ma at 6v.
The malkoff will run regulated at or above 3.8v.

I can work out my own (rough) figures based on that to tell me how much runtime to expect, however, what throws it off for me is "350 ma at 6v". I have no idea how to factor that in to the equation seeing as it will be a single cell, at a max of 4.2v, lower under load. The other thing I can't allow for with my level of knowledge is what voltage the cell will be at when under load, but with the relatively low load of the M60L compared to the 'full power' types like the M60.

So, if anyone can either tell me expected total runtime, expected total 'safe' runtime before getting too close to low voltage territory, and at what point during that cycle that it will start to run dimmer, I would be very grateful.

Alternatively, any ideas to point me in the direction of being able to work it out myself would also be appreciated.

Thank you

 

[Niconical]

 

[Al Combs]

I didn't bother to post at first because I'm not sure what I have to say is all that useful. But I'll give it a try. I have seen other threads about Malkoffs running on 17670's. Basically you're better off with a pair of AW 16340's in a 6P or a C2. A pair of 17500's in a 9P or a C3 would be even better. Well I guess that's a little after the fact and not much help here.

I don't think the Malkoff has an under-voltage cutoff (not sure). It simply falls out of regulation and goes into direct drive below 3.8 volts. Essentially it behaves as though the regulator was not there below 3.8 volts. That could be very bad for the battery. Let's say the protection circuit kicks in at 2.5 volts. At 2.5 volts, a Cree Q5 is only drawing about 30 milliamps. That's estimated from a graph titled, "Electrical Characteristics (T=25˚C)" in Cree's PDF on the XR-E. That's not enough of a load to cause the battery voltage to drop. The same 17670 would still be drawing considerable power with a xenon bulb. If you let your 6P run until the protection circuit kicked in, the battery's voltage would be below the 2.7 volt threshold any good charger would refuse to charge a Li-Ion. It would be basically the same as having no protection circuit in the battery at all. If you let it sit for a few hours it might come back. But it's never a good idea to discharge a battery to that level. There's the risk of plating metallic Lithium and subsequent venting with flames, very bad.

Then again, with only 30 ma going to the LED, you would seriously be into "Moon mode". You could never get to the protection circuit threshold without knowing it. Unless you accidentally put it down somewhere, left it on and forgot about it. Because of the nonlinear current drain as the voltage drops, the only way to know the runtime of a 17670 powering a Malkoff M60L would be to time it with a stopwatch. Maybe someone that already has can chime in here. If you hook up the Malkoff with clip leads and magnets to a DMM, measure the time it takes to reach 3.0 volts. That's a much safer runtime/voltage limit.

It would also give you an idea what it looks like at 3 volt output. Gene's page says the M60L puts out about 140 lumens. That's about 550 ma at the LED. Estimated from Cree's PDF in graph "Relative Flux vs. Current (TJ=25˚C)." The 350 ma is how much current the regulator needs to drive it to that level at 6 volts. Between the 2 graphs you can estimate the output to be about 70 lumens at 3 volts. That's roughly half the initial output. That should be easy to safely estimate in use. BTW, estimating from the same 2 graphs of the 2.5 volt "Moon mode" is about 20 lumens output.

Or you could go out and buy a 9P and 2 of AW's 17500's.

 

[Niconical]

Thanks for that, it's definately useful.

I think the end result is I won't use the 17670 for the M60L.
If I had no other option, then yes, it will work, but considering my other lights and batteries, it's not really worth it.

My main reason for wanting to was to try to standardize between the 2 lights I use most, a 3.7v lamp in a 6P, and a malkoff LED. 2 lights carried, but just 1 spare battery needed.

Instead I might focus more on these new cells from AW, 'LIMN'.

2 of those (RCR123A) in each, with just 1 spare set, and I could have the M60 running well, and a 9v incan lamp from Lumens Factory.

So, what I'm really saying is, I've hit a minor bump in a possible plan, and rather than solve it by A. just using 1 light or B. carrying 2 spare batteries, instead I'm going to solve it by buying more batteries and more lamps, and most likely at some point another light to use the 17670. Do I get a CPF T-shirt?

 

[thermal guy]

This thread might help http://candlepowerforums.com/vb/showthread.php?t=194879