There is a surface charge effect, which is capacitive, and a chemical effect which is much slower.
I would suggest switching a resistive load at, say, 1kHz, and measuring the amount of ripple.
This is similar to your "point 3 and point 4" method but it does not give it time to progress to point 5 (and back to point 2)

Ok, I could apply a ~1 kHz signal driving the load for some time (what time?). I would get peak-peak voltage ripple values- I could capture them and average them maybe. But still how would the IR calculation look like exactly?

 

Ok, this part is clear to me now. But how to calculate the IR? Assuming that point 4 amplitude will be a lot higher than in point 3, it will mean "good" battery, so should yield low IR. But in the equation it will be the contrary, as higher 4-3 voltage, will yield higher IR: (V_4 - V_3) / (I_3 - I_4). So the equation surelly has to be formed in a different way, but what?

On a good battery V4 and V3 will be close together. On a poor battery there will be a big difference, so the equation is perfectly correct. Isn't I4 zero?

 

On a good battery V4 and V3 will be close together. On a poor battery there will be a big difference, so the equation is perfectly correct. Isn't I4 zero?

I_4 is few mA, as the measurement circuitry is powered from the battery itself. Your logic is correct, but there is also that effect I described- the speed of going back to point 4. The faster, the better battery, but higher IR in the calculation... Your logic is based on the fact that point 3 (its amplitude) will decide on whatvthe IR is really, not the ability of coming back to 4. Im not saying its wrong, there are just 2 sides to it.

 

But in the equation it will be the contrary, as higher 4-3 voltage, will yield higher IR: (V_4 - V_3) / (I_3 - I_4).

Yes, a higher voltage change means a higher internal resistance, which is correct.
Why do you think that's wrong?

 

Yes, a higher voltage change means a higher internal resistance, which is correct.
Why do you think that's wrong?

Because in this context, the higher the "return" voltage, the less resistance it should be, as the battery is able to regenerate faster.

 

Yes, a higher voltage change means a higher internal resistance, which is correct.
Why do you think that's wrong?

For better explanation, please consider this plot:



At V1 the first measurement is taken (under load).
Now there are 2 cases. Lets say the first battery after releasing the load stops at V2 and the measurement is taken there. The second battery stops at V3 when the load is turned off. The second battery is clearly better, as the regeneration voltage was raised at the given time to a level that is higher than in case of the first battery. But with the calculation logic proposed by you, the second battery would have higher IR, as the potential difference between V3 and V1 is higher than between V2 and V1.

Do you see what I mean? This is why I am really not sure how to tackle the regeneration approach. The only reasonable way to do this IMO is to also measure OCV and subtract the V3/V2 from it and only then divide the result by the current. But I havent seen this method anywhere, its just how I think it could make sense...

 

Because in this context, the higher the "return" voltage, the less resistance it should be, as the battery is able to regenerate faster.

That may be a good criteria for the condition of the battery, but it has nothing to do with the internal resistance, which is what I thought your were trying to measure.

 

That may be a good criteria for the condition of the battery, but it has nothing to do with the internal resistance, which is what I thought your were trying to measure.

Condition of the battery is directly related to the IR. You must agree its difficult to imagine that a battery with a higher IR has better SoH than the one with lower IR (assuming same type, production batch etc).

So I have stated my point. In case I havent explained yet why It is that this meyhod has a flaw yet, maybe I still dont understand your point? Could you try to explain in more depth, on an example maybe?

 

Condition of the battery is directly related to the IR. You must agree its difficult to imagine that a battery with a higher IR has better SoH than the one with lower IR (assuming same type, production batch etc).

And if the battery has a higher internal resistance, the voltage will rise more when the load is removed.

 

Could you try to explain in more depth, on an example maybe?

There are two battery conditions factors being discussed.
One is the open-circuit voltage recovery time after the battery load is removed.
The other is the internal battery resistance.
You can measure one or the other but not both with the same two measurements.

So you need to decide which one you want to measure, or if you want to measure both.
I don't know which one is more important but, if I were only doing one measurement, it would be the internal resistance.

 

There are two battery conditions factors being discussed.
One is the open-circuit voltage recovery time after the battery load is removed.
The other is the internal battery resistance.
You can measure one or the other but not both with the same two measurements.

So you need to decide which one you want to measure, or if you want to measure both.
I don't know which one is more important but, if I were only doing one measurement, it would be the internal resistance.

What I want to measure is the IR, this we can set as fixed.

 

And if the battery has a higher internal resistance, the voltage will rise more when the load is removed.

Yes, this kind of make sense, but based on graph I have drawn, It is clearly a contradiction case.

 

Yes, this kind of make sense, but based on graph I have drawn, It is clearly a contradiction case.

Sorry, I see no contradiction, clear or otherwise.

I don't believe your graph shows what a real battery would do.

 

Sorry, I see no contradiction, clear or otherwise.

I don't believe your graph shows what a real battery would do.

Ok, I see your point. Maybe the issue is indeed that the shown by me graph does not express a real situation.
In this case I started to wonder when should one measure during loading and when during relaxation in order to get proper IR calculation. On the battery loading graph I have have noticed 4 characteristic phases:


A- Initial voltage drop,
B- secondary drop,
C- Initial voltage rise,
D- secondary voltage rise.

I have noticed that phases A and C have similar amplitude (but not time). The same applies for phases B and D (similar amplitude, different time). So this looks like a kind of hysteresis. Also based on this, it seems indeed that there is no big difference whether I measure during loading or relaxation. Thing is, how to take into account this kind of behavior of the battery in calculations? What I have noticed as already stated is that the IR calculated by me is way off (too high) compared to the one from the battery DS. Assuming I would only take into account the potential difference in areas B or D, then the value in the numerator is lower, which divided by the current would yield lower IR, maybe closer to the right one...

 

Yes, I would think the jump a A or C would be due to the battery internal resistance.

 

Yes, I would think the jump a A or C would be due to the battery internal resistance.

That makes sense for me too. The problem is that if that voltage (around 0.5V) is divided by the load current, the obtained IR is still far from the one specified in the DS of the battery (still too high). If instead of using A or C, I woild use B or D, the IR would get quite close to the DS value. But in your opinion these regions dont correspond to the IR?

 

The problem here is measuring the non-linear ionic resistance of the battery. How do you measure the opposition to current flow due to internal factors such as separators, electrode surface area and electrolyte conductivity that don't directly translate to easily measurable electrical parameters? The computed resistance is strongly dependent on the timescales of the technique employed because the ionic resistance is a non-ohmic time-based electrochemical reaction. This means there is usually a requirement to linearize measurement data when making operating point calculations. The standard cheap way to do it with a nonlinear element is to have a linear elements in the circuit.
https://mathscinotes.wordpress.com/2011/07/22/thermistor-mathematics/

The two-step resistor loading method of measuring the ionic resistance of the battery is similar to the method used for linearizing thermistor resistance. IMO your single load measurements will always be less accurate because of simple mathematics.

 

The problem here is measuring the non-linear ionic resistance of the battery. How do you measure the opposition to current flow due to internal factors such as separators, electrode surface area and electrolyte conductivity that don't directly translate to easily measurable electrical parameters? The computed resistance is strongly dependent on the timescales of the technique employed because the ionic resistance is a non-ohmic time-based electrochemical reaction. This means there is usually a requirement to linearize measurement data when making operating point calculations. The standard cheap way to do it with a nonlinear element is to have a linear elements in the circuit.
https://mathscinotes.wordpress.com/2011/07/22/thermistor-mathematics/

The two-step resistor loading method of measuring the ionic resistance of the battery is similar to the method used for linearizing thermistor resistance. IMO your single load measurements will always be less accurate because of simple mathematics.

Thanks for the answer!
I am actually ok with the measurement being less accurate, but what I have observed is thatvthe measurement is pretty muchbthe same from run to run, but consists of a fixed "offset" towards the proper, theoretical value.

In general, I know that with only one load resistor it is dofficult to achieve the proper IR, but I am insisting on doing this because I know that there are devices working based on this principle. The measurement time and points I can afjust freely. What I cannot do is add more loads...