# Acceptable resistance for good quality 18650?



## dazed1 (Jun 17, 2014)

What would be acceptable resistance for 1.5 yr old cells, Panasonic green 3400 mah protected? around 60-70 recharge cycles at most.
Also i have an ongoing debate with my friend, about charging current. He isisnt that charging at lests say 0.375ma would results in noticably higher charge capacity of the cell, compared to 1A.
Is this true? did anyone measured obtained capacity at different charging current? 0.3, 0.5 vs 1A? thanks.


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## RetroTechie (Jun 18, 2014)

dazed1 said:


> Also i have an ongoing debate with my friend, about charging current. He isisnt that charging at lests say 0.375ma would results in noticably higher charge capacity of the cell, compared to 1A.
> Is this true? did anyone measured obtained capacity at different charging current? 0.3, 0.5 vs 1A? thanks.


I think you meant 375 mA (or 0.375 A) there?  Anyway your friend is wrong. In the case of Li-ions, proper charging (CC-CV) involves constant current for some time, then constant voltage. The charge current just determines how fast the _bulk_ of the charge goes into the cell. That constant voltage (eg. 4.2V) determines how much charge goes in. The _cell_ determines how much current it needs to 'saturate' itself to reach that voltage. Compare with a balloon: higher pressure -> more air goes into the balloon. But up the pressure too much, and it pops.

When the charge current drops below some value (say, ~50 mA for 18650's?), charging is stopped. So obviously where that cutoff is, also determines how much charge goes in - _but only a little_.

With NiCd's or NiMH's it's more complex, but basically a (proper) charger will detect a voltage peak while charging. Stop charging before that, and the cell isn't charged completely (regardless of charge current). Stop after that, and charging efficiency drops quickly to the point where you're just converting energy into heat.

Your friend might still be right in his/her observations, though. Which could mean:
* He/she is using really, really poor cells. Or
* He/she is using a really dumb charger. Or
* He/she saw that behavior with NiCd / NiMH's, and thinks it also applies to Li-ions (NOT!  ).


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## HKJ (Jun 18, 2014)

dazed1 said:


> Is this true? did anyone measured obtained capacity at different charging current? 0.3, 0.5 vs 1A? thanks.



The charge does mainly depends on charge voltage and charge termination current, only to a little degree on charge current.
But according to LiIon manufacturer a lower charge current will charge more energy into the battery, as long as the voltage and termination current is the same.


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## dazed1 (Jun 18, 2014)

Thanks guys. To make it simple (for example only)

How much mah will each cell have after being fully charged, and will the volts drop under load be the same? assuming both are equal quality, and exactly the same internals and both charged on some ultra high end charger....

a) Charged @0.350A

b) Charged @1A

Thanks. I need generals figures, not scientific or something, my friend insists that the energy will be quite a bit higher on the slower charged cell...

Edit, one more thing, what is an general acceptable resistance? for NCR18650B protected after 1.5 years?


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## HKJ (Jun 18, 2014)

dazed1 said:


> Thanks guys. To make it simple (for example only)
> 
> How much mah will each cell have after being fully charged, and will the volts drop under load be the same? assuming both are equal quality, and exactly the same internals and both charged on some ultra high end charger....
> 
> ...



I do not have any numbers for different charge currents, only for different charge voltage and termination currents: http://lygte-info.dk/info/BatteryChargeVoltageCurrent UK.html 

I do not have the resistance either, but it will depends on battery usage and also on how you measure it.


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## dazed1 (Jun 18, 2014)

Thanks alot for your help!


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## LEDite (Jun 21, 2014)

dazed1 said:


> What would be acceptable resistance for 1.5 yr old cells, Panasonic green 3400 mah protected? around 60-70 recharge cycles at most.



I sell quite a few Panasonic 3400 mah unprotected cells.

They start at ~ 60 milliohms when new.

After 1.5 years of use, I would normally see a little over 100 milliohms.

Protected cells have more Internal Resistance due to the protection circuit.

LEDite


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## markr6 (Dec 17, 2015)

Bumping this old thread because I'm wondering as well. Most of my cells measure around 50-80 milliohms new. I have an Eagletac 3400 protected which is several years old, stored at 4.18V in my Jeep year-round -20°F-120°F or more...rarely used so it has taken a beating. I'm measuring around 210 milliohms.

I came across this online and wondered if it's a good estimate:


*Milli-Ohm**Battery Voltage**Ranking*75-150mOhm3.6VExcellent150-250mOhm3.6VGood250-350mOhm3.6VMarginal350-500mOhm3.6VPoorAbove 500mOhm3.6VFail


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## ChrisGarrett (Dec 17, 2015)

Isn't HKJ measuring in the .04 ohm range on some of the newer hybrids, for starters?

I guess if and when I get a li-ion charger that gives me an indication of this characteristic, I'll start worrying about it. I actually put more value in the I.R. number that my Maha gives me, than I do the capacity values!

Ignorance is bliss, perhaps?

Chris


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## markr6 (Dec 17, 2015)

ChrisGarrett said:


> Ignorance is bliss, perhaps?



You bet! I wish I knew nothing about IR and storing at high voltages. It's a lot more fun to just use 'em and abuse 'em. That's what I'm trying to do, but I'll run a refresh and test resistance now and then.


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## SilverFox (Dec 17, 2015)

Hello Markr6,

High internal resistance results in lower voltage under load. If you cell started having a mid point voltage of say 3.7 volts under a "normal" load (I use a 1 amp load) and the internal resistance of the cell goes up, the voltage will drop.

At some point the voltage will drop far enough that the cell will no longer be useful.

Tom


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## m4a1usr (Dec 18, 2015)

Here is pretty much one of the most simplest methods of diagnosing cells is question. You can find a few variations off this measurement method but it's still pretty sound with even todays cells. But cell IR has improved over the years so use this as a reference and not a rule set in stone.

Measuring a battery’s internal resistance is simple. All you need is a 4 ohm 5W power resistor or similar and a multimeter. Four common 1W 1 Ohm resistors in series would work. If you aren’t measuring an 18650, choose a load resistance that will load the cell but not overload it. Capacity/3 should be ok.
1) First, measure the battery’s voltage when fully charged and not hot. This is known as V1
2) Next, connect your multimeter leads to each side of the resistor and briefly connect the resistor across the battery. Make note of the voltage reading, this is V2.
3) Measure your resistor to get its precise resistance, this is R
4) The battery’s internal resistance (Ri) is calculated with the formula:
Ri = (V1-V2)*R/V2

Example: cell measures 4.2V unloaded at rest, 4.0V when connected to a 4 ohm resistor and has an internal resistance of 200 mOhms (0.2 ohms)
A new high quality 18650 battery will have an internal resistance under 100mOhms. 
A used up 18650 battery will have a resistance 400mOhms or more.


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## Gauss163 (Dec 20, 2015)

m4a1usr said:


> Here is pretty much one of the most simplest methods of diagnosing cells is question. You can find a few variations off this measurement method but it's still pretty sound with even todays cells. But cell IR has improved over the years so use this as a reference and not a rule set in stone.
> 
> Measuring a battery’s internal resistance is simple. All you need is a 4 ohm 5W power resistor or similar and a multimeter. Four common 1W 1 Ohm resistors in series would work. If you aren’t measuring an 18650, choose a load resistance that will load the cell but not overload it. Capacity/3 should be ok.
> 1) First, measure the battery’s voltage when fully charged and not hot. This is known as V1
> 2) Next, connect your multimeter leads to each side of the resistor and briefly connect the resistor across the battery. Make note of the voltage reading, this is V2.



A couple remarks worth mentioning.

1) IR is highly dependent on temperature and (less-so) on SOC (state-of-charge = %capacity). So do the IR test at temp closest to that of the target app, and do it above 20% SOC (= 80% DOD) since IR typically increases sharply below that (depending on temp), e.g. see the graph below.

2) If the load-test is too brief then the measurement will not be accurate because it takes some time for the IR from internal diffusion processes to build up to steady-state (esp. for less healthy cells). Ideally one should test while simulating the type of load in your target app. A too brief test would only be relevant for high-frequency (pulsed) loads. Look up "Nyquist plot" to better understand the full impedance spectrum.







Graph excerpted from p. 1-4 of Challenges and Solutions in Battery Fuel Gauging, by Yevgen Barsukov.


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## Benny Boy (Dec 21, 2015)

Your friends correct in a way. Lower amperage will generate less heat and get a bit higher capacity charge. It will also give more cycles over the life of the cell. If you don't mind waiting 6~10 hours per charge.



dazed1 said:


> What would be acceptable resistance for 1.5 yr old cells, Panasonic green 3400 mah protected? around 60-70 recharge cycles at most.
> Also i have an ongoing debate with my friend, about charging current. He isisnt that charging at lests say 0.375ma would results in noticably higher charge capacity of the cell, compared to 1A.
> Is this true? did anyone measured obtained capacity at different charging current? 0.3, 0.5 vs 1A? thanks.


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## Gauss163 (Dec 27, 2015)

Benny Boy said:


> Lower amperage will ... get a bit higher capacity charge.



Not necessarily. It depends on the charging algorithm, e.g. is the termination current fixed or a percentage of the charge current or some more complicated function? Is the charge algorithm pure CC/CV, or PWM or pulsed, etc? It's not easy to give a definitive answer without knowing such details.


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