Questions: Checking internal resistance of CR123

[scott.cr]

Hi group,

I've been reading a bit on how to test the internal resistance of a CR123 using a voltmeter & would like some feedback.

To test a single CR123 this is how I did it:
1) Check open circuit voltage
2) Check voltage while under load (approx. 1 amp using 3.1 ohm resistor)
3) Perform simple calculation:

Rb = R1 (Vo / Vl -1)

Where:
R1 = test resistance
Vo = open circuit voltage
Vl = voltage under load

On one of my CR123s the open circuit voltage was 2.986, voltage under load was 2.610 and the result of the calculation was 0.447.

-Is that result in Ohms?

-When testing voltage under load, the voltage continuously plummets as long as the test leads are connected. Is there a time cutoff that's common accepted practice?

-Is 3.1 Ohms an appropriate resistance to use for this type of test?

Also, I noticed that a new CR123 would have an open circuit voltage around 3.2, and after a short load test of around 10 seconds the open circuit voltage would be around 2.98. Is this normal?

 

[bcwang]

It's not too clear to me how one should go about checking internal resistance of a cell either. Voltage drops under load continuously, and doesn't spring back up to the same voltage after the load is removed. I've also tried doing calculations by using many different loads and noticed that calculations using the numbers gathered don't always end up with the same calculated value.

I wonder if this is why ac impedance testing is done instead of DC.

If anyone knows the proper way to do internal resistance testing without a dedicated internal resistance meter, please post.

 

[mudman cj]

I have tested my cells using a 2 Ohm load because I found that at the lower current levels with a 3 Ohm load I did not get as much differentiation between batteries. All I did was measure the voltage drop across the load resistors in order to determine the current and the voltage across the loaded battery and calculated the resistance from Rb=Vb/I. Of course, to get the current from the voltage across the resistors you use Ohms law again: I=V1/R1. In my case R1 was 2 Ohms since I measured across both resistors in series. This could be simplified by measuring the voltage drop across only one of the series resistors (while still using both in the circuit) so that I=V1/1 Ohm=V1. Then the substituted equation becomes Rb=Vb/V1

Even easier is to just measure the voltage across the resistors. A better battery will hold voltage better under load and therefore drive more current through the constant resistance test resistors. So, a higher measured voltage indicates a higher current and therefore a better battery. I just wrote the last two digits (after the decimal) of the voltage on each battery with a Sharpie marker and match them that way. There was no point in writing the first number in the voltage because it was the same for every battery anyway.

I don't see how the open circuit voltage of the battery has a place in the calculation at all. :thinking:

I went with the readings obtained within the first few seconds because I did not want to drain the cells very much in the process of testing.

Impedance analysis is used in cases such as with electrochemical cells when a resistance (impedance actually) builds over time at an electrode due to some rate limiting step(s) such as gaseous exchange, ion mobility or some chemical conversion process such as the plating of lithium. Impedance analysis can break down the resistance as a function of frequency to look at how each of the components to the total resistance stack up and contribute. In the case of testing CR123 primary lithium batteries I do not see how anything but the DC case (ie. low frequency regime) is of importance because what we really want to know is how they perform while driving an essentially resistive load without switching the direction of current flow and the internal resistance of the battery increases with discharge, so performance under load corresponds with its remaining capacity.

 

[mdocod]

The following post should be read as a hypothesis/theory, not fact.

The problem I see with any sort of testing for resistance is that I believe voltage loss under a load in cells comes from 2 phenomenon. There isn't just resistance, there is also a chemical function that produces it's own natural voltage drop when "loaded" that has nothing to do with resistance. However, like many things when dealing with electricity, a particular behavior can often be described as resistance because that's the way the circuit treats it, but the "increased resistance" is actually a change in voltage... As an example: an electric motor as it spins faster has an effectively higher resistance on the circuit, but that effective resistance has it's roots in something called back electromotive force, in effect, the coils of the motor always have a relatively static resistance (changing more with temperature than anything else), however, when the motor spins up, there is voltage produced on the coils as the magnets pass by (or vice-versa depending on the design) (a motor is a generator and a generator is a motor, the difference is pretty much just optimization of design for application). The only reason I bring this up is to point out that effective resistance can come from more than just resistive material on the path that the electrons are flowing.

When the chemical reaction in the cell is "balanced" (IE: no load condition, open circuit), then there is a maximum buildup of potential ready to be on the move, once electrons start flowing, the chemical reaction is in a constant state of trying to restore that balance. There has to be some pressure "drop" (voltage drop) to induce the reaction.

I think we have to think of voltage loss of loaded cells in 2 separate ways: Here's how I would split it up...
1. Voltage Sag: A chemical function
2. Voltage Drop: A function of resistance.
Combine the 2 and call the total "voltage loss from all factors"

In most cells, there is a natural rise in resistance as the cell is drained, but I would hypothesize that most of the voltage loss towards the end of a discharge is caused by voltage sag rather than voltage drop. As a cell heats up, resistance would increase slightly, however, the chemical reaction is excited, which tends to over-come the higher resistance and produce higher effective voltage under a load.

Of course, we have to ask ourselves, what percentage of voltage loss comes from sag and what from drop. I'd hypothesize that one could estimate this effect by comparing a cell at various loads, from light to extreme (like near dead short)... Some of the tests would not be entirely safe to get a very clear picture on CR123s.

I really like mudman's approach using 2 separate loads. This would help eliminate some of the initial voltage sag needed to induce the reaction from the results, (taking that open circuit voltage out of the picture). While there is still going to be a variance in voltage sag between each load induced chemically, a large chunk of that effect is removed from the comparison on a cell like a CR123 where the maximum current that the cell can flow into a dead short is much greater than most normal applications.

-Eric

 

[mudman cj]

In essence, I agree with your understanding of battery resistance, but if I may I would like to comment on it in the hope that I can help others to come to a better understanding of batteries and how to test them.

The voltage drop which is a function of resistance will not be constant as current is varied as you know because a fixed-value resistor always drops more voltage as the current flowing through it is increased. Now, the voltage drop of a battery at a fixed current level will also change as the battery is depleted because the internal resistance of the battery changes as well. This is essentially what you are calling voltage sag. For some battery chemistries such as Nimh, the internal resistance (voltage sag) of the battery actually decreases into the discharge cycle of the battery before increasing again towards the end. For this reason, the load test can be misleading for some battery chemistries and should not be considered to be the ultimate test for remaining life across the board of all battery types.

The reason that the internal resistance of the battery changes as it becomes further depleted is because of the rate-limiting chemical processes that occur at both electrodes and in the electrolyte as well. The exact chemical changes depend upon the specific battery chemistry, and may include interactions with a gaseous phase such as in the case of zinc-air batteries or lead-acid batteries, but always include the transport of some ion through the electrolyte (lithium ions in the case of lithium batteries) and the reactions at the electrodes that liberate ions at one electrode and then accept them at the other electrode.

The method I outlined in my post was not actually at two different loads. I was trying to make my method as simple as possible, hence my suggestion to use a 2 Ohm load but only measure the voltage drop across one of the 1 Ohm resistors to simplify the calculation further. However, you bring up a good point. A more accurate determination of the batteries' state of charge can be determined by measuring its performance under two different loads. But, unlike the case presented by the OP, neither load should be open circuit. For this case, the internal resistance of the battery is calculated as follows:Rb=(V2-V3)/(I2-I3). I chose to use subscripts of 2 and 3 rather than 1 and 2 in order to make the next paragraph easier to follow.

If I were to implement this method for testing primary lithium batteries I would use resistances of 2 Ohms and 3 Ohms. V2 and V3 are the voltages across the battery while discharging through 2 Ohms and 3 Ohms respectively. I2 and I3 are calculated by measuring the voltage drops across the resistors and dividing by the resistances. As before, you can simplify the current measurements by measuring across only one of the series resistors in each case so that the resistance is 1 Ohm and then the calculations reduce to I=V/1=V. In that case, the equation can be rewritten as Rb=(Vb2-Vb3)/(Vr1-Vr2), where Vb means voltage across the battery and Vr means voltage across one of the 1 Ohm resistors.