# li-ion discharge limit - 2.5v/2.7v/2.8v/3v???



## Huz (Nov 30, 2009)

I am looking on info on safe lower limit of discharging li-ion. I have seen some protected batteries w/ 2.5v over discharge protection and I am not sure that is safe. Are voltage supervisors best way to implement voltage cutoff in LED circuits? Thanks for your help.


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## csshih (Dec 1, 2009)

2.5V I believe would be the Li-Ion in a fully discharged state under load.
when the load is relieved, the cell should recover back to 3.6V


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## TranceAddict (Dec 1, 2009)

practice to measure battery voltage after a long run, and if they fall around 3.6 to 3.7 OC, recharge them. after you got familiar, your instinct will tell you when to recharge without to measuring anymore.


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## DM51 (Dec 1, 2009)

For an average light (and yes, I know there's no such thing) if you discharge it under a typical load to the 2.5V-2.7V low-voltage cut-off, the cell will usually recover to >3.0V after a few minutes at rest.

If it does _*not*_ recover to >3.0V, it will have suffered some damage, depending on how low its voltage is at rest. The key will be to recharge it as soon as possible, as the longer it spends in a fully-discharged state, the more it will deteriorate.

Cells that are below 2.5V at rest should be treated with great care when recharging. There are threads on how to deal with charging such cells. It sometimes works, but sometimes not.

The best policy of all is *never* to let Li-Ion cells get that low. You can top them up as often as you like.


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## TooManyGizmos (Dec 1, 2009)

Best not to discharge Li-Ion below 3.55 volts ... IMO .

"voltage supervisors best way to implement voltage cutoff in LED circuits?" ..... NO ..... you are .
.


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## Huz (Dec 2, 2009)

I want to drive a 2.1v red XPE w/10280 -how am I supposed to know when to stop?? particularly while hiking where I dont carry voltmeter?


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## BillyNoMates (Dec 2, 2009)

Simplest way is to use a cell with built-in protection, it will cut-off to prevent over-discharge.


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## buickid (Dec 2, 2009)

BillyNoMates said:


> Simplest way is to use a cell with built-in protection, it will cut-off to prevent over-discharge.



Incorrect, the built-in protection is a fail-safe designed to prevent the battery from creating a dangerous situation (fire!). It is not meant to cut-off the every time you discharge the cell. It is likely to fail if used in such a manner.

Read: Think of the protection circuit as the air-bags in your car. You hope you never have to use it, but its good to have when the SHTF.


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## Huz (Dec 2, 2009)

That is why I am asking if voltage supervisors are the way to go? 10280 doesnt come w/ built in protection?


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## DM51 (Dec 2, 2009)

AFAIK there is no 10280 made with protection. Protected 10440s have only just started appearing - they're new on the market, and the 10280 is not a popular enough item to make it worth anyone's while to fit it with a protection circuit (which in any case would add ~3 mm to the length).

I don't know what light you are running, but if it is a Draco or similar bright mini-light, you will see a very noticeable dimming as the cell nears empty. At that point, change the cell straight away and recharge the empty one as soon as you are able to.

Aside from that, the only other reliable method of avoiding over-discharge is to time the discharge and change cells well before you think you need to.


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## Huz (Dec 2, 2009)

DM51 said:


> I don't know what light you are running, but if it is a Draco or similar bright mini-light, you will see a very noticeable dimming as the cell nears empty. At that point, change the cell straight away and recharge the empty one as soon as you are able to.



It dims because in direct drive battery voltage drops below Vf of the emitter which is normally 3v+ . Red XPE has Vf 2.1 to 2.3V. So it wont dim until it way over discharged.

Since this will be a mod I am thinking of using buck converter that will efficiently lower the output voltage to 2.2V. It needs min 2.7V input. So it may give some warning or stop working. The purpose of the thread was to find whether discharge to 2.5V or 2.7V is safe or if I need to add a voltage supervisor.


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## VidPro (Dec 2, 2009)

the start of the dive on the discharge curve 
when it starts to tank in output , different cell types will take that dive earlier or later than thier differentally build counterparts.
a battery with heavy plates and fast discharge capability will tank sharper on a graph than a slower battery. so high discharge batts will be more depleated when they (finnaly) droop in voltages. (higher discharge batteries dont droop in voltage as much with similar loads on both)

any high drain devices will slam a smaller battery so hard the voltage will tank when it reaches this location, so the protection kicks in earlier.

a Very slow drain on same said battery (like say low PWM rates or direct drive slowdowns) by the time the battery reaches any protection level of 2.5 the battery will be discharged MUCH further. 

so your underLOAD voltage will depend on the battery and the load, and all the HUMAN has to remember is "Before the cells capacity is gone" so to speak. 

EX: a 5mm white led put on a li-ion and allowed to droop to nothing will have completly drained the cell (~2.3-2.4v) with no bounceback. 
a 1c load from a Boosting driver pushing the battery current up as the battery is depleated in voltage will reach a low voltage point much faster.

if your talking the RESTING voltage, then ALL of those would be concidered to low and in dire need of recharging.
To put ONE number on a the Voltage location for stopping a load, would not take Actual remaining capacity into account.
a perfect stop alogrythm in, would be at a Percentage of *actual *remaining capacity instead (that being neer impossible). 

i point that out only because people want some NUMBER a definable location, and a specific voltage number is not what is needed, plate seed, or chemicals in place or whatever they want to call it is what is needed, and no specific voltage number can define that for every occurance.
if there was a visual voltage representation it is when the voltage starts to take a dive, aka some parts of the chemical conversion processes are completed.

so in general flashlight use with protected cells, when your draining the cell rapidly the average 2.4 2.5 2.7v protection will work ok, but if your doing the slow low drain thing, then YOU Certannly NEED to stop before you drain the thing so far it cant re-charge fully later.

if your making a driver or protection with a cutoff built in, then more than ONE voltage cut-off point could be usefull, but then tiny little brains only have so many lines of code , leaving big human brains available to compensate 

if you had more than one voltage point in a driver cut-off , with normal discharge batts ,it would be:
higher when the load is smaller (like cutting off at 3.2v or higher even)
Lower when the load is larger (like cutting off at 2.5v when load is 1c)
so your devices cuts off when it reaches the "about empty" location.


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## B12 (Dec 2, 2009)

buickid said:


> Incorrect, the built-in protection is a fail-safe designed to prevent the battery from creating a dangerous situation (fire!). It is not meant to cut-off the every time you discharge the cell. It is likely to fail if used in such a manner.



What makes you think so? It is solid state electronics without moving or wearing parts.


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## MWClint (Dec 3, 2009)

Get one of these tiny voltmeters for $3. they are not much bigger than your 10280 cell itself. i keep one in my laptop bag..they weigh nothing.

solder two short wires onto the first two pins..and then trim all the pins down.
I've got a few of them from this ebay seller and they have all been accurate.


http://cgi.ebay.com/NEW-RC-Lipo-bat...Control_Parts_Accessories?hash=item3efabdbb40

Weight: 4.6g
Size: 23mm x 25mm x 9mm
Input voltage: 0.01V - 4.50V Per cell
Precision : 0.01V
they can do up to 6 cells, and total pack voltage..but ive just been using them to test single cell li-ions.
i dont let cells go under 3.6v..so testing voltage at rest is good enough.


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## TooManyGizmos (Dec 3, 2009)

Does it put the battery under any type of load for accuracy ?


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## MWClint (Dec 3, 2009)

TooManyGizmos said:


> Does it put the battery under any type of load for accuracy ?



nope it's a straight up dvm. load is not 100% necessary in the field. if your resting voltage is 3.6..it's a good idea to swap to a fresh cell, and 
if you choose to continue using the battery..then at least you know it could suffer some damage. 
so you'd keep this in mind the next time you go charge this cell. 

_if the charge is above 3.6v then you know it has some life left. exactly how 
much is unknown_, unless you keep track of cell charging habits/cell health..if 
cell condition is known, then you will have a decent idea of capacity left vs 
resting voltage. you get used to it after a while...and you do end up
noticing runtime differences on older/abused cells..and thus retiring them.

it's also a good idea to mark cells once they being to degrade, so you dont
accidentally expect too much runtime out of them during use. it helps
to have a decent hobby charger. 


you could easily make a small load device for use with that dvm..simple
would be a 6D maglite bulb connected in series to this tiny dvm.
but again not necessary when you know the health of your cells and just
need an idea of when to stop using it.


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## TooManyGizmos (Dec 3, 2009)

?
Can't a weak batt. read 3.6v at rest .....

But then drop a lot when put into the light under load ?


I'm not sure I follow ?
.


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## DM51 (Dec 3, 2009)

3.6V at rest is pretty much empty.

Here is a useful table which gives a rough idea of state of charge of LiCo (Li-Ion) cells. Bear in mind the figures below are *resting* voltages, where the cell has been resting (not under load) for ~15 minutes. The figures are approximate, but they are a good guide.
 
4.2 volts 100%
4.1 about 90%
4.0 about 80%
3.9 about 60%
3.8 about 40%
3.7 about 20%
3.6 empty for practical purposes
<3.5 = over-discharged


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## TooManyGizmos (Dec 3, 2009)

I guess I said that wrong ...

Can't a batt. read 4.0 volts at rest , under no load ....

But then drop to much lower (such as 2 volts) when used under load ?

I'm just saying , can't a batt. tested with no load , read good , but be totally dead when put into use under a load ?
.


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## Illum (Dec 3, 2009)

pardon the pun but discharging to 2.5V is yanking it pretty hard, I can see this sort of things happening for direct drive lights but for most Buck regulators for LEDs they usually drop out before that, where you would notice it before it goes deeper.

I've never completed a full cycle using AW17670s but I'm pretty sure the protection circuit cuts out around 3.6V/OCV

When my cells discharge to 3.8V/OCV its back on the charger for them.

LiCoO2 Discharge curve

```
OCV                    CCV [Iload = 550ma]
100%      4.20V          100%      4.20V 
90%       4.06V          90%       3.97V
80%       3.98V          80%       3.87V
70%       3.92V          70%       3.79V
60%       3.87V          60%       3.73V
50%       3.82V          50%       3.68V
40%       3.79V          40%       3.65V
30%       3.77V          30%       3.62V
20%       3.74V          20%       3.58V
10%       3.68V          10%       3.51V
5%        3.45V          5%        3.42V
0%        3.00V          0%        3.00V
```

it is not recommended to drain it to 3V, but it will hit 3V if you let it drain


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## marschw (Dec 3, 2009)

DM51 said:


> AFAIK there is no 10280 made with protection. Protected 10440s have only just started appearing - they're new on the market, and the 10280 is not a popular enough item to make it worth anyone's while to fit it with a protection circuit (which in any case would add ~3 mm to the length).


I suppose you could pull the protection circuits off of a 10440 and attach it to a 10280. It would be ugly, add a lot of length, cost more than it should, and not have completely correct parameters for when to cut off, but it would be better than nothing I suppose. Probably.


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## Mr Happy (Dec 3, 2009)

Huz said:


> It dims because in direct drive battery voltage drops below Vf of the emitter which is normally 3v+ . Red XPE has Vf 2.1 to 2.3V. So it wont dim until it way over discharged.


This is not wholly or necessarily true. The discharge curve of a lithium ion cell is not linear. Take a look at a typical discharge graph like this one:








The top curve represents low rate discharge. You can see that once the voltage gets below about 3.5 V there is essentially no power left in the cell. Regardless of how low the Vf of the LED is, once the cell discharges to this point the LED will rapidly dim. The cell has no power left to sustain the voltage or deliver any current beyond that point. So it's really not important whether the Vf is 3.7 V or 2.1 V. As soon as the light dims, you should stop and change the battery.


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## MWClint (Dec 4, 2009)

TooManyGizmos said:


> I guess I said that wrong ...
> 
> Can't a batt. read 4.0 volts at rest , under no load ....
> 
> ...



not without you noticing that your flashlight doesnt turn on or is pretty dim. 
It would be instant indication that something is wrong and you should stop
using that light until you investigate it further. abused or old battery, short in 
the flashlight electronics, damaged led..etc.

so in reality, if your light+battery have been operating like _normal_, and you just need to see if you need to swap batteries...testing the resting voltage
and seeing if it's above or below the 3.6v threshold is good enough.

Hope that clears it up.

DM51's voltage/capacity chart in the posts above is a nice quick reference.


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## mdocod (Dec 4, 2009)

The lower Vf will result in a deeper discharge when dimming occurs compared to a "regular" Vf LED.

Given a scenario where a small unprotected cell must be used, my suggestion would be to just make a best attempt to prevent over-discharge by monitoring usage as best as possible. Just check the cell voltage before charging when you get back from your outing and see how you did. Keep track of the number of times that a particular cell is discharged to a less than desirable level (I'm going to call this less than 3.0V resting, between 3.0 and 3.5V is less than ideal but isn't the end of the world). Provided the cell is recharged within a day or two, I'd maybe put a limit of a dozen or less accidental over-discharges per cell then replace. Combine this with some voltage testing after coming off the charger and resting for a day or so to insure that the cell will hold charge voltage effectively. (ideally above 4.10V, when the cell will no longer hold above 4.0V it is effectively worn out). With some due diligence like this, I don't see a major problem. Alternatively, you might just make a habit of replacing the cells on a yearly basis or something to that effect (depending on how frequently the flashlight is used).

Considering the small stature of a 10280, the risk factor of having an undesirable "venting" event isn't much different than any other cell, but the intensity of said event on such a small cell would pose less threat. The difference between a 10280 and an 18650 is like comparing a 22 to a 30-06. Both weapons can kill and cause damage, both should be respected as dangerous, but we all know the underlying differences.

-Eric


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## TooManyGizmos (Dec 4, 2009)

MWClint ,

Thanks for your answer , it was a good one. It made me stop and think , why did I ask my silly question to start with. I was totally off in left field. My mind was in another place. I wasn't thinking about your original topic. 

When you mentioned buying your digital meter , it got me thinking about just testing batteries laying around in a box. Not about testing one's you just took out of your light. In that case I can see your point , and the immediate reading just after use would be accurate . And useful to determine the amount of charge left .

It was not intentional , but my points were way off topic. I just wasn't on the same page as you . I'll bet you were wondering why I just didn't understand your question. Thanx for explaining and waking me up.

Next time I'll try to pay closer attention and not get side-tracked with my own thoughts.
.


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## leukos (Dec 5, 2009)

VidPro said:


> if you had more than one voltage point in a driver cut-off , with normal discharge batts ,it would be:
> higher when the load is smaller (like cutting off at 3.2v or higher even)
> Lower when the load is larger (like cutting off at 2.5v when load is 1c)
> so your devices cuts off when it reaches the "about empty" location.


 
Good point, VidPro. High drain and low drain applications are often left out of the discussion when talking about li-ion cut off voltage even though this has a huge effect on the state of the battery and how effective protection circuits are. I would hazard to say that most protections circuits are designed for high drain applications. Running your protected li-ion down to cutoff with the low mode on your LED light is a recipe for battery death.


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