# Internal resistance measurement algorithm using DC method

#### bremenpl

Joined Oct 30, 2013
21
I have a circuit that allows me to add a load to a battery and measure the voltage, current and temperature of the battery. I am trying to derive what is the best method for measuring the internal battery resistance using the simplest DC load method. In theory that method is not complicated, but the problem is that the results I am obtaining are far from the battery resistance specified in the datasheet and measured by Hioki meter for the same battery. So far, I tried the following method:

1. I measure open circuit voltage
3. I wait for 1 second.
4. I measure the closed circuit voltage.
6. Calculate the IR with formula: (V_open - V_closed) / (I_closed - I_open).

The result will highly depend on the load application time, as it can be seen in the scope capture graph. Loading the battery for a longer period of time, will make the voltage drop even more. I tried a different method as well in which I take under consideration not the open voltage but the regeneration voltage:

The calculation results are similar and still far from the reference.

I am fully aware of the fact that the plain DC method is not very precise, but based on the fact that my results are ~70 % off (IR is too high compared to the reference), there has to be something I am doing wrong.

I would appreciate all help!

#### sparky 1

Joined Nov 3, 2018
544
The question would require extensive discussion beyond scope of the forum. The state of health expressed as % full charge.
Improved accuracy of State of Charge (SOC) estimation. The cell voltage, which is one of the main parameters from which the SOC is derived, can be measured when the cell is in the open circuit condition and the chemical transformation in the cell has stabilised at the end of the rest period. This results in a more accurate measurement.

Having derived an accurate profile for a specific cell type can also be helpful in tailoring an SOC.

#### Ian0

Joined Aug 7, 2020
2,289
What type of battery is this? Is it lead-acid?

#### crutschow

Joined Mar 14, 2008
27,243
Likely better would be measurements at two different loads, with the load current not changing more than perhaps 10-20%.

#### bremenpl

Joined Oct 30, 2013
21
The battery type is lead-acid. I know that there is a method which utilizes 2 different loads (light and high), but the hardware does not allow for that.

#### bremenpl

Joined Oct 30, 2013
21
The question would require extensive discussion beyond scope of the forum. The state of health expressed as % full charge.
Improved accuracy of State of Charge (SOC) estimation. The cell voltage, which is one of the main parameters from which the SOC is derived, can be measured when the cell is in the open circuit condition and the chemical transformation in the cell has stabilised at the end of the rest period. This results in a more accurate measurement.

Having derived an accurate profile for a specific cell type can also be helpful in tailoring an SOC.
Thanks for the answer. The geberic IR meters available on the market do not know the battery profilr and still provide quite accurate measurements. I am only inyerested in IR, not full SoC or SoH

#### bremenpl

Joined Oct 30, 2013
21
What type of battery is this? Is it lead-acid?
This is lead acid battery type, yes.

#### bremenpl

Joined Oct 30, 2013
21
Likely better would be measurements at two different loads, with the load current not changing more than perhaps 10-20%.
In practice, the open voltage measurement is done at some small load, as the circuitry doing the measurement is drawing current as well from the battery (around 10 mA). The applied load is around 2A (depends on exact battery voltage).

#### Ian0

Joined Aug 7, 2020
2,289
It's very difficult to do - lead-acid batteries exhibit a phenomenon called coup-de-fouet, where the voltage rebounds after the dip due to the load being applied, after that happens the terminal voltage reduces steadily. You can get any reading you fancy depending on exactly when you read it.
The best estimation of state of charge is open circuit voltage, but the measurement is only valid if the battery has been left disconnected for about 24 hours, and frankly, that's really not very helpful.
If you have a load which generates a decent amount of ripple current (for instance something connected via an inverter) then you can divide the RMS voltage ripple by the RMS current ripple to get the impedance.

#### bremenpl

Joined Oct 30, 2013
21
It's very difficult to do - lead-acid batteries exhibit a phenomenon called coup-de-fouet, where the voltage rebounds after the dip due to the load being applied, after that happens the terminal voltage reduces steadily. You can get any reading you fancy depending on exactly when you read it.
The best estimation of state of charge is open circuit voltage, but the measurement is only valid if the battery has been left disconnected for about 24 hours, and frankly, that's really not very helpful.
If you have a load which generates a decent amount of ripple current (for instance something connected via an inverter) then you can divide the RMS voltage ripple by the RMS current ripple to get the impedance.
Thanks for the reply. Thing is, there are devices that calculate IR only by aplying resistive load to the lead acif battery, thats why I am sure its possible. I just dont kbow how they do it...

#### Ian0

Joined Aug 7, 2020
2,289
Thanks for the reply. Thing is, there are devices that calculate IR only by aplying resistive load to the lead acif battery, thats why I am sure its possible. I just dont kbow how they do it...
The commercial conductance meters generally use a high frequency (1kHz). There are plenty of battery-measuring devices on the market with dubious credentials, and fancy looking displays.

#### bremenpl

Joined Oct 30, 2013
21
The commercial conductance meters generally use a high frequency (1kHz). There are plenty of battery-measuring devices on the market with dubious credentials, and fancy looking displays.
For the better quality ones thats true- they use the AC based method. But there are some that only consist a dc load to calculate the resistance.

#### crutschow

Joined Mar 14, 2008
27,243
The most accurate would likely be to measure the voltage under load, and then immediately after you remove the load (<1s between measurements).

#### bremenpl

Joined Oct 30, 2013
21
The most accurate would likely be to measure the voltage under load, and then immediately after you remove the load (<1s between measurements).
But what is the algorithm here exactly? Because assuming healthy battery, the voltage after releasing load would jump right back to its open circuit state- the faster it goes back, the higher is the voltage drop, and the higher resistance. So this kind of stands in contradiction.

#### bremenpl

Joined Oct 30, 2013
21
Based on what I have observed during the load application on the scope, the battery voltage goes in different "stages"...

1. First stop in voltage drop,
2. second stop in voltage drop,
3. Getting to the third stop take longer period of time. The instantaneous voltage drop would stop at this point if the load were to be applied longer.
4. After the release- first voltage rise stop,
5. second stop- getting to the ~ovc voltage would take some time now.

The main question I have (and for which the answer is fuzzy depending on who you ask) is: at what time after applying load should the voltage measurement happen? Some say les than a second, some say few, some say sub second times... Cannot find any fact based description for this.

#### crutschow

Joined Mar 14, 2008
27,243
after releasing load would jump right back to its open circuit state- the faster it goes back, the higher is the voltage drop, and the higher resistance.
True.
That's why you don't want to wait until the battery voltage settles back to it's open circuit voltage.
You want to measure at point 3, release the load, and then as rapidly as possible, measure at point 4.
That should make that voltage change largely due to the internal resistance, not the normal change in battery voltage as it settles back to the long-term open circuit voltage (point 4 to point 5).

#### Ian0

Joined Aug 7, 2020
2,289
The most accurate would likely be to measure the voltage under load, and then immediately after you remove the load (<1s between measurements).
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)

#### bremenpl

Joined Oct 30, 2013
21
True.
That's why you don't want to wait until the battery voltage settles back to it's open circuit voltage.
You want to measure at point 3, release the load, and then as rapidly as possible, measure at point 4.
That should make that voltage change largely due to the internal resistance, not the normal change in battery voltage as it settles back to the long-term open circuit voltage (point 4 to point 5).
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?