3.4V cut off voltage?

[seaside]

Normal protected Li-ion cuts off at arround 2.7V.
14500 Batteries I recently purchased cut off at about 3.4V.
It happened before, that was just 1 battery out of bunch of 16340 packs, so I just made a mark on that one, put it aside, and use other batteries.

But all four 14500 batteries I purchased... hmmm... what do you think? Should I ask for replacement?

Thanks.

 

[GrAndAG]

Which current did you use when testing for cut off?

 

[seaside]

I can't be sure for that because I was not able to measure that.
It was Q5 LED with DX 4735 driver. Good working driver is supposed to pull less than 1000mA out of battery.
Good one has 22uH inductor and faulty one which gets hot quick has 3.3uH on it and known to pull over 1.8A out of battery.
Mine has mystery inductor. Marking wiped out. Thus I have no idea about its value. But it acts like a faulty one.
It pulls initial 2A with fresh recharged battery. Then start pulling little less current. My 900mA 14500 last less than 30 min with that driver.

So.... My best guess is about 1200~1500mA range at the time of cut off.

 

[DM51]

Did you measure the cell's voltage immediately after cut-off, or was it some time later?

 

[mdocod]

So.... My best guess is about 1200~1500mA range at the time of cut off.

The PCB looks at voltage AT THE TIME it activates a protective mode. Not when you test it after the fact.

Under a strenuous load, a LiCo cell can be expected to suffer from significant voltage sag. The 2.5V bottom end limit of the PCB was reached, the load was removed, and the cells resting voltage rebounded to 3.4V....

FYI: anything between 2.5V and 3.5V resting is basically dead, the difference is, at 2.5V it's dead as a freaking doornail and ideally should be charged to a higher level ASAP to reduce wear-and-tear on the cell, at 3.5V you can let the cell sit around for awhile before you get to charging it and not take much penalty on cell capacity/life.

It doesn't sound like there is anything wrong with your cells.

What you'll tend to find is this:

When you have a pair of li-ion cells driving a buck regulated LED, or a single cell driving a boost and or boost/bucked LED, the current draw from the battery will rise towards the end of the discharge, which causes there to be a large gap between resting voltage and running voltage at the cell, especially towards the bottom end of the cycle.

When you have a single li-ion cell driving a LED in a buck regulated or direct-drive configuration, the current drops lower and lower towards the end of the discharge, which causes the gap between resting voltage and running voltage to stay very narrow right down to the very end of the discharge.

Some 1xli-ion "drivers" for LEDs will also start to really wane off on the boosting towards the end of the discharge, which offers some extended running at a lower output, but will also cause deeper slower discharges with lower final resting voltages at the cell.

-Eric

 

[seaside]

Thanks for the help.

When battery cuts off, naturally I took that out of the flashlight. It was 0V as protection kicked in. So I put it into charger for few second since this seems to close protection circuit. Then took it out, measure the voltage. It comes out to be arround 3.4V.

I am not sure if I understood it correctly or not. What I am getting after reading is that, it really is 3.4V but the battery protection circuit see it as 2.7V because of the voltage sagging at the end of battery life cycle due to the way the driver works?

If this is what happened, I think I should try this battery with my AMC7135 driver at low mode and see what's happening. Because I know other batteries cut off at arround 2.7V with that.

 

[DM51]

I am not sure if I understood it correctly or not. What I am getting after reading is that, it really is 3.4V but the battery protection circuit see it as 2.7V because of the voltage sagging at the end of battery life cycle due to the way the driver works?

That was why I asked when it was that you measured the voltage.

Under load, the voltage sagged to ~2.7V, then the protection shut it off. After a few minutes at rest, it rebounded back to 3.4V.

Sometimes cells need a quick jolt from a charger to reset the circuit so you can get a voltage reading. It's quite normal, and your cell is fine, as mdocod said.

 

[mdocod]

Even a few seconds on the charger would be enough to raise the cell voltage many tenths down in that range, like I said there is very little difference in stored energy between 2.5V and 3.5V.

 

[seaside]

Even a few seconds on the charger would be enough to raise the cell voltage many tenths down in that range, like I said there is very little difference in stored energy between 2.5V and 3.5V.

Thanks for the explanation. This eliminates my worries right away.

The batteries won't come back to life after few min, even in a day, once after they cuts themselves off. They still cut off at arround 3.4V with other driver circuit that draw less than 350mA. BTW, my other batteries except 1 16340 cut off at arround 2.7V with this driver. (That 16340 cuts off at 3.4V too, just like those 14500s I was questioning about) Quick jolt with charger brings them back to 3.3~3.4 in no time.

So I tend to believe these particular batteries and the other 16340 of mine have protection circuit that prematurely cuts off at around 3.4V rather than 2.7V. I found couple of people at DX had reported this behavior quite few times with cheap batteries, and it seems not mean any other problem rather than itself. I guess that's manufacturer's lack of quality control with protection circuit parts.

Since stored energy b/w 2.5V and 3.5V doesn't mean much, and I know those batteries can't be over-discharged, I think I don't have to worry about them. I will just use them.

Thanks guys for help me understand this.