Testing the internal resistance of a li ion cell and making determinations

newb123

100 W
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Jun 7, 2019
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102
I tested the internal resistance of a 18650 sanyo cell using the following method

No load test = 4.01 V

2 ohm 5 watt resistor load test = 3.80 V

ohms law 3.80 V / 2 ohm = 1.9 Amps

Kirchhoff voltage law (sum all algebraic values of V drops = 0)

4.01 - 3.80 = .21 V (4.01 - 3.80 - .21 = 0)

.21 V / 1.9Amps = .110 ohm (internal resistance of cell)

Does this value (.110 ohms) indicate that this cell has been over cycled/ charged? The internal resistance from the spec sheet indicates less than 100 milli ohms is normal

I tested another cell from the same pack, same voltage drop (3.80) was measured


Tested another cell from another set, charged at 4.10 no load 4.03 2 ohm resistor load, came up with .065 milli ohm
 
You should prioritize determining residual capacity via precisely timed CC load testing, say at 1C.

 
The 0.11V drop you noted is not internal resistance. Most of it is voltage sag from the chemical process of producing current. If we had nearly that much resistive loss in our packs, e-bikes would be a very different game.

It's difficult to quantify cell resistance because it's inseparable from voltage sag. But voltage sag doesn't make heat in proportion to the square of the current. Voltage sag represents power that isn't available, not power that's being converted to heat.
 
Discharge rate and reduced voltage sag is indeed inversely proportional to IR.

But a go/no-go "cell condition" decision should not be based on IR measuremrnt alone.

Precisely controlled and accurate, discharge tests for capacity are really the first step, much more important than resistance.

Discard any below 80% SoH just for safety, match your groups based on Ah capacity.

Use periodic checks of cell IR to identify cells on the verge of failure, keep records of each cell's IR over time.
 
newb123 said:
I tested the internal resistance of a 18650 sanyo cell using the following method

No load test = 4.01 V

2 ohm 5 watt resistor load test = 3.80 V

ohms law 3.80 V / 2 ohm = 1.9 Amps

Kirchhoff voltage law (sum all algebraic values of V drops = 0)

4.01 - 3.80 = .21 V (4.01 - 3.80 - .21 = 0)

.21 V / 1.9Amps = .110 ohm (internal resistance of cell)

That first “no load” voltage reading is not representitive of the cell under load.
You should do the calculation , but use two different “loads”. IE,. 2 ohm and 10 ohm,..for two voltage rreadings under load .
Having said that.... depending on which Sanyo cell you tested, and its history, the 0.110 ohm IR could be a typical result anyway. !
 
ok so how about this capacity tester? will it give more accurate results?
Im using a 2P10S pack right now, perhaps the range of such a used pack is not that good, but the 1v drop per km seems very bad
tec 06 battery cap tester.jpg
 
Balmorhea said:
The 0.11V drop you noted is not internal resistance. Most of it is voltage sag from the chemical process of producing current. If we had nearly that much resistive loss in our packs, e-bikes would be a very different game.

It's difficult to quantify cell resistance because it's inseparable from voltage sag. But voltage sag doesn't make heat in proportion to the square of the current. Voltage sag represents power that isn't available, not power that's being converted to heat.

I can't find anything to support this theory - do you have a reference/source for it?
 
Punx0r said:
Balmorhea said:
The 0.11V drop you noted is not internal resistance. Most of it is voltage sag from the chemical process of producing current. If we had nearly that much resistive loss in our packs, e-bikes would be a very different game.

It's difficult to quantify cell resistance because it's inseparable from voltage sag. But voltage sag doesn't make heat in proportion to the square of the current. Voltage sag represents power that isn't available, not power that's being converted to heat.

I can't find anything to support this theory - do you have a reference/source for it?


Balmorhea has interesting opinion here, which is yet uknown in the world of Li-ion cells internal resistance measurement and methodology. :)

Here is short quote from such methodology study :

There are a number of phenomena contributing to the voltage drop, governed by their respective timescales: (i) the instantaneous voltage drop is due to the pure Ohmic resistance R 0 which comprises all electronic resistances and the bulk electrolyte ionic resistance of the battery 16 , (ii) the voltage drop within the first few seconds is due to the battery’s double layer capacitance and charge transfer resistance R CT which is attributed to the charge transfer reaction at the electrode/electrolyte interface 8 , and (iii) the shallow, linear (or close to linear) voltage drop is due to polarisation resistance R p which accounts for ionic diffusion in the solid phase and is usually considered to be the rate determining step for Li ion batteries. s 27 .

The contribution of these three parts can be calculated separately for an intuitive understanding of the complex electrochemical processes involved in the battery system. On the other hand, a bulk total cell resistance can be calculated from the total voltage drop for the pulse, as is often done in the literature 15,28 . The other drawback of using DC current to obtain resistances is that only the imposition of all the different contributions to resistance, hence there in an inability to completely separate the different resistance components.

End of quote


0,5C 10s.jpg

0,68-3,4 A   10-1 s.jpg

3A 1s.jpg
 
Hmm. Interesting.

The difference between DC and AC (@1kHz) impedence being due to cell capacitance is well established but this goes much further. Especially with the capactive capacity lasting as long as ~2 seconds and then a sudden, sharp drop in voltage - seems like such an obvious characteristic should be widely observed?

Li-ion cells are widely modelled as a cell in series with a resistor and the resulting IR blamed for ohmic heating and wasted capacity. This probably is simplified, with more really going on, but it would presumably be "close enough" to be useful in working out cell heating and usable capacity.
 
Again, since cell capacity is easily measured directly,

is most consistent and

a much simpler more straightforward measure of relative health

just do that.

https://www.skyrc.com/BD200

https://www.rcgroups.com/forums/showthread.php?3391743-Isdt-fd-200

Chargers: 4010-Duo, FMA PL8, Thunder 0620AC
Hyperion, 15A390W

past post

https://endless-sphere.com/forums/viewtopic.php?p=1488274#p1488274



 
Punx0r said:
Especially with the capactive capacity lasting as long as ~2 seconds and then a sudden, sharp drop in voltage - seems like such an obvious characteristic should be widely observed?

Explanation to the pictures : all three are recordings from DCIR measurement of LG MJ1 cell. Voltage drop is always result of the measuring constant current. At the first picture is current 0,5C = 1,7 A , at second recording is measurement according IEC 61690-3 standard 0,2C = 0,68A for 10 s then immediately increase to 1C = 3,4 A. Third picture is record of 2 A pulse DCIR measurement.

I agree with john61ct that in this case (newb123) is probably better to prefer capacity measurement. Internal resistance can serve as an accessory info, probably isn´t very accurate and you don´t know DCIR value of the new fresh cell. First you need to know exactly what Sanyo cell it is.
 
Punx0r said:
Balmorhea said:
Voltage sag represents power that isn't available, not power that's being converted to heat.

I can't find anything to support this theory - do you have a reference/source for it?

No. I have only my observations that very saggy batteries don't get as hot as P=I^2R would imply. Mostly they don't get noticeably hot at all, despite the chemical reaction in lithium cells being exothermic on discharge.

Consider a typical 18650 cell. At 5A, it should be making something like 3W of heat if newb123's measured voltage drop was purely ohmic resistance. That would mean 11W of heat per cell at 10A. In practice, we don't observe this degree of pack heating even when we drive our packs that hard.

Maybe other folks are getting a kW of heat from their batteries between resistive and chemical heating, and I simply haven't encountered this phenomenon?
 
thanks john; that skyrc model looks very nice, but the one I posted from ebay, is it about the same thing? Looks like both are from China

Voltage sag represents power that isn't available, not power that's being converted to heat.

To the other posters,thats a bit over my head, sounds like something similar to wasted volts reactive power in A/C theory, but for DC... :)

I have to read that over again many times to understand it, I thought voltage sag was the resting volts then the volts when the battery is under load. Like 36 no throttle, 33 full throttle...3V sag?
 
Dunno man at some point getta do your own research.

Maybe start new threads on just that unit in the relevant forums
 
docware said:
Explanation to the pictures : all three are recordings from DCIR measurement of LG MJ1 cell. Voltage drop is always result of the measuring constant current. At the first picture is current 0,5C = 1,7 A , at second recording is measurement according IEC 61690-3 standard 0,2C = 0,68A for 10 s then immediately increase to 1C = 3,4 A. Third picture is record of 2 A pulse DCIR measurement.
So, in simple terms, the voltage change (SAG) is always in proportion to the current, by a constant factor...
....which we call “Internal Resistance “ ?
No mystery there , and we also know the cells heat in proportion to the voltage change.
I thought the relationship between deltaV, I, and heat, was well understood ?
 
Balmorhea said:
No. I have only my observations that very saggy batteries don't get as hot as P=I^2R would imply. Mostly they don't get noticeably hot at all, despite the chemical reaction in lithium cells being exothermic on discharge.
.
Well , my 18650 packs do get warm, even with relatively light loading of.1.5C. Max intermittent on 3C rated cells.
A “saggy pack” can be very different causes to measured voltage sag on a single cell test.
Maybe your pack “sag” is due to other causes... a few bad cells, poor connections, bad BMS, etc etc,
.. Also all controlled cell tests at significant discharge (1.0C++) show very noticeable heating.
 
Hillhater said:
.. Also all controlled cell tests at significant discharge (1.0C++) show very noticeable heating.

I don't dispute that. But I doubt that your 1.5C discharge generates a steady 3W of heat per cell from internal resistance alone. All sources of heat-- exothermic reaction, resistance in the interconnections and terminals, waste heat from the BMS, plus internal resistance-- don't usually amount to anything like the hundreds of watts of heat that you'd see if the entire voltage drop across each cell were solely a result of ohmic resistance.
 
It makes sense as a cell that sags under load will slowly recover its terminal votlage when that load is removed, presumably due to recharging of capacitance and the diffusion of ions "catching up". However, rather than voltage sag, lost capacity due to discharge at higher rates can only be due to heat-forming processes as the energy has to go somewhere.

To be fair, 100mohms for an 18650 cell is pretty terrible. That's probably a cell intended for a low discharge application at 0.5C.
 
Balmorhea said:
Hillhater said:
.. Also all controlled cell tests at significant discharge (1.0C++) show very noticeable heating.

I don't dispute that. But I doubt that your 1.5C discharge generates a steady 3W of heat per cell from internal resistance alone. All sources of heat-- exothermic reaction, resistance in the interconnections and terminals, waste heat from the BMS, plus internal resistance-- don't usually amount to anything like the hundreds of watts of heat that you'd see if the entire voltage drop across each cell were solely a result of ohmic resistance.
? Where do you get that 3 W per cell from ?
At 1.5C (5.0 A) a GA cell (0.04 ohm IR) only generates 1.0 watt of heat from internal ohmic resistance
In my 10s, 4p, 40 cell pack, that totals 40.0 watts.
I typically see a 1.5-2.0 volt drop on the pack at 20 amps draw.
So how do you figure that “hundreds of watts” could be generated ?
 
Balmorhea said:
Hillhater said:
.. Also all controlled cell tests at significant discharge (1.0C++) show very noticeable heating.

I don't dispute that. But I doubt that your 1.5C discharge generates a steady 3W of heat per cell from internal resistance alone. All sources of heat-- exothermic reaction, resistance in the interconnections and terminals, waste heat from the BMS, plus internal resistance-- don't usually amount to anything like the hundreds of watts of heat that you'd see if the entire voltage drop across each cell were solely a result of ohmic resistance.

Balmorhea, when the cells are at the phase of the exothermic reaction, the next phase is usually fire and destruction of the battery (and EV vehicle). :flame: :(
 
As you can see from these pictures, warming of the 18650 cells at 5 A discharge is quite significant :

M36 polystyren 2.jpg

M36 polystyren 4.jpg

LG M36 2A v 5A.jpg

LG HG2 2A v 5A.jpg

Samsung warming 30Q 2A v 5A.jpg
 
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