Duo conversion using 18500

Fjell

Well-known member
I imagine this has been done before, but in our lockdown interlude I was inspired by various posts to have a go at a Duo conversion with an old one I had. We were already using several Duo inserts on Explorer helmets, but I am very reluctant to mess with them, so are using AA?s. Unless you commit to changing out cells on a longer trip, it?s a little difficult to use the full capability of the Custom Duo V42 - specifically running on full flood all the time.

I gutted it and put on another helmet. As you can see I moved the cable entry to one side. It is driving the warm-white version of the V42, as I also do in my Scurion. Being more used to carbide I suppose.

I started by making a twin 18650 (6800mA.hr), which had a few trips. However, it?s rather cruel to the 20 year old plastic around the lid and seal as it is too tall. Probably it would be OK, but it offends my engineering purity.

I then made a triple 18500 (6000mA.hr) as shown. This works pretty well and fits nicely. I also made a single 18650 as a spare because that can lie diagonally and not interfere with the lid.

I reckon you can make the whole thing including helmet and insert for about ?140 with some judicious Ebaying, and you end up with a very decent light for the money.

 

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Ian Ball

Well-known member
What did you use to waterproof the holes?

and how did you make 3 cells fit vertically if 2 cells were a bad fit in the same orientation?
 

Fjell

Well-known member
It?s sealed with glue (poured into corner). You can make a pool with a glue gun (you could get this out again), or go for epoxy (not coming out). The cable is clamped both sides prior to that. Feeding through the helmet makes it a bit more robust (you could also seize the helmet entry). The cable now being near the corner of the box helps with glueing.

You need to be careful about the screw at the bottom depending on how enthusiastically you gutted the box, and whether you intend to use the bottom screw. Given the extra space with the 18500 maybe leave it alone.

The battery is double-wrapped, which makes it snug. It is a rattle fit without the wrap.

I did originally look at triple 18650, but it is very tricky to fit and probably even more unwise in terms of stressing the lid nearer the end.
 

Ian Ball

Well-known member
Aaah pay attention Ian.

Please could you explain something to me energy density wise and the like, my maths and physics or maybe chemistry is not strong enough.

Would a lamp being powered by a LiIon cell of 2000mAh offering 3.7V achieve a longer run time than a NiMH 4.8V pack rated at 2000mAh? If the output was maintained at the 350mA mark (such as in Custom Duo setups).

If the current draw is 350mAs regardless of voltage then what effect does the lower voltage have in the system, reduced visible output, lower heat generation (wasted energy).

The reason I ask is I have a fair few older LiIon cells that are recycled out of laptop batteries and using an analyser I see they hold 2000mAh. 

However the four Ikea Ladda AA I use hold an average of 2300mAh.  And so would a single 18650 LiIon cell running more efficiently in terms of self discharge, loss of capacity in cold temperatures, efficiency at high drain make the switch to a single Li-Ion holding 2000mAh still be worth it over the current AA setup?  the benefit of reduced weight isn't something I see as an important benefit whereas the likelihood I bugger it up certainly is.

Obviously moving to a dual liion cell in parallel pack holding 4000mAps would be a better bet but the struggle to fit them in is off putting.


 

andrewmcleod

Well-known member
OK... to first approximation (i.e. ignoring efficiencies that may depend on current etc):

Total energy is power (watts) times time (seconds). The more energy, the longer you can have a constant power drain (like a light on a constant setting) running before you run out of juice. The quantity that you really care about in a battery (for most purposes) is the total energy 'in' the battery (note the quotes for the pedants).

Power is voltage/potential difference (volts) times current (amps).

The confusion starts when people start using mixed units. For example, mAh is milliamps (current) times time, which isn't directly comparable except for batteries of the same voltage. So you can compare two lithium 3.7 V batteries and say that a 2000 mAh battery has twice the 'energy' available of a 1000 mAh battery.

More useful is Wh, which is watts (power) times time, which is *actually* a unit of energy (because power x time = energy). What it means is that 1 Wh is enough to run a 1 W load for 1 hour, but it is the same as 1 W * 3600 secs = 3,600 Joules (Joules being the SI unit of energy).

If you have a 3.7 V lithium battery with 1000 mAh (or 1 Ah), that is 3.7 V * 1 A * 1 hr = 3.7 Wh.
If you have a 7.4 V lithium battery with 500 mAh (or 0.5 Ah), that is 7.4 V * 0.5 A * 1 hr = 3.7 Wh.
So those two batteries have the same amount of energy available.

So to answer your original question, there is more energy in a 4.8 V 2000 mAh battery than a 3.7 V 2000 mAh battery. There are of course differences, and the effective capacity of batteries does depend to some degree on the current draw (you get less useful energy out if you draw high currents due to heating in some batteries).

To directly compare batteries of different voltages, compare the Watt-hour (or Joule) ratings of the batteries. You can convert Ah to Wh by multiplying by the voltage. You can convert from mAh to Ah by dividing by 1,000.

Caveats: you normally have to multiply by the average voltage, as it varies over discharge (hence typically using 3.7 V for lithium ion batteries which go from 4.2 V to 3 V or less).

Your efficiency questions are more complicated and I don't know.

Final note: LED drivers are a bit weird. They are generally designed to provide a constant current _across the LEDs_. They keep the potential difference/voltage across those LEDs at about 3 V per LED (regardless of input voltage). The current draw may vary depend on the source voltage. An LED driver designed to run on a particular voltage will probably work across a range of source voltages, and should draw a constant power (not current) - so when being supplied with a lower voltage will draw more current, and vice versa. However, it may not work if the input voltage is too low and may go boom if the input voltage is too high. Some drivers are designed to drop a higher voltage to a lower voltage (e.g. run 3 V LEDs on 7.4 V input); typically there is some overhead (some fraction of a volt) and they won't work below 3 V plus that overhead (so may not work on 3.7 V input). Other drivers work the other way round (e.g. running 3 x 3 V LEDs in series which is therefore 9 V off a 3.7 V supply) and typically won't work if the input voltage is close to or above the output voltage. Some drivers can do either and will be fairly flexible?
 

maxf

New member
The closer your input voltage is to the output voltage the more efficient the driver will be
 

Fjell

Well-known member
I have always assumed these drivers are DC-DC converters with constant current set points. Need the Man for gory details.

The lower the current you pick, the closer to 3V you can get before it starts to die on you. I have one on at the moment that was discharging at 700mA. It has dropped off to about 500mA at 3.1V odd. The units usually go phut at around 3.0V, before the BMS kicks in.

On the Scurion the BMS kicks in first at just under 6V to protect the battery, because the Scurion can run on 4V at low power.

With Ikea selling 4 AA?s at ?4.50 at the moment, it is hard to beat that price point. And NiMH?s are more robust and less likely to be nasty to you. Four of those is about equivalent to a single 3000mA.hr 18650.
 

Ian Ball

Well-known member
A round of applause thank you all  :clap:

thanks Andrew, you managed to explain all of that in text form as well!  (y)

https://batteryuniversity.com/learn/article/discharging_at_high_and_low_temperatures

The temperature of a cave being around 10 degrees in the uk would have an effect on performance I am sure, and I can't seem to find the document from Energiser I think which discussed the effect across different chemistrys, Alkaline Manganese, Li-Ion, NiMH, NiCD I think.  It showed a 10% loss of performance for NiMH at 10 degrees celsius but only a negligible loss for LiIon.  So the loss of performance effecting NiMH, plus the loss of performance for using a high current which effects NiMH more (where is that high current? 500mA  Peukert's boggles my mind).


Aah, found something

https://data.energizer.com/pdfs/nickelmetalhydride_appman.pdf

"Use of nickel metal hydride batteries in cold environments may
force significant capacity derating from room-temperature values"


Although the accompanying graph shows the effect to be less pronounced than I remembered.

 

royfellows

Well-known member
NiMh not my thing, but a few bits of info you may find useful.

First the cells. I am not the expert on NiMh but they are cells that can be fully discharged. I would therefore expect that a lamp specifically designed for these cells to use some form of boost driver so as to get the full capacity of the cells.
LiIon cells have a discharge range from 4.2V fresh off charge down to (ones I use, refer to data sheets) 2,7V when the management circuit will disconnect them. 3.7V is the nominal voltage, a sort of 'stab in the middle'. If over discharged they can become dangerous so a quality charger will fail to charge a cell below as certain safe voltage limit.
Protected cells have a management circuit mounted in the base and therefore cannot be over discharged, so safe to use in a lamp that utilises a boost circuit.

Driver efficiency is a funny one as there are many different types of driver. Linear drivers generally, current in equals current out, the excess voltage being burned off as heat, and some need a certain level of voltage overhead to function correctly, a sods law.

Switch mode drivers much better, but best efficiency on those I have tested is only at full power where the max output is set by a current sense resistor. I have seen 95%. Lower modes using pulse width modulation, (PWM), takes a dive, I have seen 65% on lowest, but over 90% on max.

I cannot second guess design of someone else's lamp design, so above comments general, but hope its helpful.
 

royfellows

Well-known member
I have just written a paper on the subject of "Lithium Ion batteries, what cavers need to know". I have uploaded it to aditnow and am happy for it to be available on this site.
I cannot see how to do this, so anyone who wants a copy can download it from aditnow, or admins can make available here.

I have had to tag it to a mine at aditnow, but picked the first "A".

Its currently at https://www.aditnow.co.uk/documents/AARON-Mine/Lithium-Ion-and-Battery-Safety.pdf
 

royfellows

Well-known member
I have moved it on aditnow to personal files, had forgotten correct procedure. Now at https://www.aditnow.co.uk/documents/Personal-Album-128/Lithium-Ion-and-Battery-Safety.pdf
 

Cantclimbtom

Well-known member
Bumping an old post....
Thanks for the writeup Fjell And thanks also to you/Roy for the battery explanations
Thinking of doing the same and converting my Duo (Simple r02) to 3 x 18500s

In the past my Duo has staunchly defended its ground against Pixa, Pixa2 and "Chinese bicycle lights" as I believe Roy calls them. However after proximity in a confined space to a Dragon last week, I contracted a case of lamp-envy. I'd like to partially treat that by being able to run my Duo on the high setting for more than briefly looking up a stope/aven without fear the batteries will drain too fast, or to run it on medium power without counting the hours. (I don't currently get out enough to justify ordering a Dragon, but if I did... I would!!)

Also reading explanations here about drivers I have a better understanding of why my additional light will not work (for more than 40-45 minutes) with 1.2v rechargeables whereas they would always run like the Duracell bunny with 1.5v Alkalines

Hoping my rough estimated maths is OK, based on nominal voltages and ignoring driver (etc) cutoffs I'm reckoning it to be ~2.5 times longer light with the 18500s

Four standard/white Enloop AA
4 * 1.9Ah * 1.2v =  ~9Wh

Three 16500 @ 2Ah
3 * 2Ah * 3.7v =  ~22Wh

What I'm wondering those is a bit more mundane...
In your pics of the battery box, which connectors are you using?
What shrink material are you using for the 3 batteries?
 

wormster

Active member
Are you using an omirat insert from Biff in Somerset? I switched over from the standard inset and got better run time on my cells (helps if you splurge on eneloop cells (best C.O.T.S. AA recharable cell IMO))
 

PeteHall

Moderator
Only moderately off-topic, but another factor affecting the performance of a Duo is that when the plastic lens becomes scratched, the light output really drops.

I'd always advocated replacing the lens (for about ?2.50) on a fairly regular basis, until I realised that an old lens can be rejuvenated with t-cut. Or if it's really scratched, a polish with some 1200 grade wet & dry first, then t-cut.

Whatever upgraded battery or LED module you are using, a clear lens will still make a massive difference  (y)
 

Cantclimbtom

Well-known member
wormster said:
Are you using an omirat insert from Biff in Somerset? I switched over from the standard inset and got better run time on my cells (helps if you splurge on eneloop cells (best C.O.T.S. AA recharable cell IMO))
Yes it has one of Biff's Customduo inserts: Simple r02 (which replaced the Simple, which replaced the Duo Omni) and I do use Enloop in it (although the white 1900mAh not the "pro"). Biff's LEDs are incomparable to the original Petzl Duo and Zoom bulbs. I still fondly remember the warm yellowish glow you got putting a new battery in and setting off pre-dawn, up a route with a Zoom. Not so great for below ground though!


Thanks Pete, my lens is still very clear, got the thing on ebay before lockdowns hit, and I think it had lived in a drawer much of its life.
Noted: to replace that when it gets beaten up.

Edit: seplling correction ;)
Edit2: Just seen that you can get shrink plastic specifically for batteries, so please ignore that question in previous post
 

yuvals

Member
As part of 2200 lumens conversion of a DUO, I am converting the battery box to use 2*18650 cells (standard cells not a pack).
 

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Cantclimbtom

Well-known member
Oddly enough today I decided to do something very similar today. here's my attempt

I thought I could use 3 x 18500 (all the same orientation) and the 4th "battery" would be a thin wooden dowel that has a hole drilled top to bottom with copper wire and top/bottom caps so it is an electrical connector. The dream looked like this

6anenvy.png


This "cunning" plan was so that I could use the existing battery holder, although some extensive trimming to remove the dividers between the batteries and drop a thin metal plate into the top and bottom. That way I didn't have to gut the battery box and it could still work with normal AAs

I did manage to get it to sort of work, but it was awful, I had to trim so much plastic as the 18500 are too big for that the holder is ruined + the connections weren't great so it was flickering.

In my opinion the way to go... is to either not convert it at all, or to gut the lot and shrink batteries together using connectors, to make power packs and not use the battery holder. The half way approach is good for nothing, well my attempt wasn't good for much anyway. Thankfully I had an extra battery holder which had very wobbly connectors and that was ruined not my good one. I can now lick my wounds and decide if I am going to gut it and do it properly, or retreat
 
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