A Tale of Two Batteries

This is concerning two different 6-volt lantern batteries that have behaved the same exact way (I believe.) Here is a datasheet for the type of battery, a Rayovac # 941 General Purpose Carbon Zinc , but I have seen this pattern in alkalines as well.

Discharge has been with a constant current circuit running a white LED at 1mA, starting with a full battery until the LED went out, so from 6V down to about 3V. (Not sure how many months this took.) Then I did it again with another battery of the same kind.

I expected the cells to have evenly discharged, meaning that we went from 6 volts down to 3 volts, so 3/4 = 0.75 volts is what I expected each cell to be showing.

The Surprise

I opened up each battery and was surprised to see that the discharge was highly asymmetric, leaving two cells at zero, and the other two cells mostly charged. What is the explanation for this? Does this effect have a name? The most surprising thing was when almost the same exact thing happened with a second battery.

Due Diligence

These are the questions that I have looked at before posting this:

Found somewhat relevant:

  1. Why do batteries in consumer electronics get used unevenly?
  2. Batteries in series discharging unevenly
  3. Battery Pack Management: need to balance during discharge?
  4. Inequal discharge across battery cell
  5. Odd failure mode for AAA battery

Not sure if relevant:

  1. Why is it only one of four batteries (Li Ion) is drained?
  2. How to maintain equal discharging on batteries
  3. Why is only one out of 2 batteries is discharging when connected in series?

Building a New Battery

As a result of this, which ends up being fortunate, I took the 4 good cells from the two batteries and made up another battery to run the LED for another few months. This has now been running for about a day now.


The constant current measures at 1.039mA

During the 1.039mA constant current discharge, the cells measure:

  • 1.541 V
  • 1.540 V
  • 1.367 V
  • 1.373 V


  • 5.820 V - Measured in series
  • 5.821 V - Mathematically summed via calculation


You can see that two of the cells were actually charged. This alone to me indicates that something out of the ordinary is occurring.

Circuit Details

Here is the model of the circuit that I am using: (I use the pulse voltage source to show how the circuit responds to voltage change)

Constant current 6v into 1mA for a White LED

It should not matter, but:

  • R1 measures 097.2 K-ohm.
  • R2 measures 975.0 K-ohm.

How many ways can current pass through a cell?

These are the possibilities in my mind:

  1. Normal (chemical reaction)
  2. Semiconductor conduction (like a bipolar transistor)
  3. Static electricity (enhanced, through electrolyte)

This is pure imagination and speculation on my part, but based upon what I see as strong evidence of something happening which I don't understand, but seems to me to be obviously present. Whatever is happening is happening inside the battery, along side what we normally experience when a cell discharges. Because it can't be seen, it's easy to speculate it away to non-existence with excuses.

(On a possibly related note, I just took a bunch of corroded cells out of a pack of alkaline batteries that were all new, not hooked up to anything, and yet corroded. Why? The dead cells I removed were also corroded. I have no idea if the corrosion is related to the uneven discharge in the two cases I observed, but I don't think it started out that way.)

If the current passing is low enough, my hypothesis is that for the stronger cells, the current passes through the cell causing normal cell chemical reactions, whatever that is (this effect seems to happen with carbon-zinc, alkaline, and lithium-ion, perhaps lead-acid). For the weaker cells, my hypothesis (continuing on with the same proposed example) is that the current gets through the cell by another mechanism whereby the chemical reaction is suppressed, but the electrolyte is conducting the current without the normal chemical reaction taking place. In other words, conduction occurs through all cells equally (for a 4s-1p for instance) but the chemical reactions occur preferentially, happening in the strongest cells, and suppressed in the weaker cells. What makes a cell stronger or weaker is the result of a wide variety of factors.


Does anybody know of any existing (or speculated) models of this?

How can any of this be proven or disproven?

You never know what the practical implications for a new understanding might be. We can certainly use better batteries. And I for one would appreciate my devices not being destroyed by corrosion. For either of those to be affected is a long shot, but you never know.

My new direction

I have decided not to use the cells as a night light, but rather to discharge them and record how much life each one of them have remaining at this point.

The reason this is important is because of speculation in one of the due-diligence links that even though the voltage shows as being high, there's really not much life left in the cells. So I want to measure the actual life left to prove whatever is the truth of the situation.

I will report back with the results. If you have a better idea, or any information to help, I would appreciate it if you would please let me know.


I know it's a lot of information, and I probably asked a lot of questions, but I am really only asking the one question:

How do I prove the existence of a novel conduction path in weaker battery cells?

I have taken numerous Li-ion battery packs apart, and they seem to exhibit a similar pathology. That is why I think that this is a multi-chemistry effect.

Edit 1:

The best answer so far says "The voltage alone doesn’t tell the whole story" but the voltages that I reported were under 1mA constant current load.

To strengthen my argument, one multimeter that I have has a battery tester built in, and reports the mA resulting from what must be a 360 ohm resistor inside. For a 1.5mA cell 4.0mA represents fully-charged, and for a 9V battery, 25mA represents fully charged, according to the face plate label. I report the voltage for all the cells under this load:

With the 360-ohm resistor, the cells measure:

  • 4.2mA at 1.491V
  • 4.2mA at 1.488V
  • 3.7mA at 1.328V
  • 3.7mA at 1.340V

And altogether 15.6mA at 5.574V.

This should furnish a decent voltage under load. Also, it should be apparent that two of the cells were actually charged instead of discharged (or at the very least did not discharge), and the other two are close to fully charged. (These are the 4 good cells out of the 8 total cells).

I'm working on discharging the cells at a higher rate, to prove how many mAh they have. This may take a week or so.

Do our models have to change?

I don't think I've ever heard of a battery being discharged to fifty percent the original voltage, and have two completely dead cells, and two completely charged cells. Would somebody with the "tomes" about this, and the knowledge about this, tell me -- is this to be expected, or is this a new discovery? In particular, I want to know if our model(s) have to change.

Edit 2:

enter image description here

New evidence

I just found a multimeter after having misplaced it a while ago, and the 9v battery was dead. I just opened up the battery (copper-top) and it has 6 x AAAA cells in it, and there is no corrosion visible, but one cell is dead, and the others are 2.9, 3.3, 3.3, 3.4, and 3.7 mA (out of 4.0mA). One was zero, the others were 75% to 92% (tested with a 360 ohm resistor, 4mA = full). This is additional evidence that a leakage-level discharge causes highly asymmetric discharge.

My new hypothesis is this: Like the base-emitter current enables a larger current collector-to-emitter in a bipolar transistor, so a very small leakage current allows the cells to rebalance themselves, and that to be the primary dynamic.

I encourage you... Find an old device, long forgotten, with a dead 9V battery in it, and open up your own slowly-discharged 9V to see what I'm talking about. I'm sure this is widespread and not rare. But it is not widely known or widely understood. Why does it do this? So...

The title for this question used to be "How to prove novel conduction path in weaker battery cells in two 4s-1p carbon zinc batteries discharged at 1mA from 6V to 3V?".

  • \$\begingroup\$ @All - Stop the unfriendly comments and Be Nice. Comply with the site's Code of Conduct (CoC) in everything you write. Nothing snarky / sarcastic / condescending (or worse). Some comments here were deleted due to being unfriendly. Polite constructive criticism is allowed in comments - with the emphasis on polite & constructive. If your comment includes words like "you" or "your" then check that it still complies with the CoC, because you are at more risk of breaking it and possible consequences. Thanks for your understanding. \$\endgroup\$
    – SamGibson
    Commented Apr 6, 2022 at 15:52
  • \$\begingroup\$ The new data is insufficient to support any new claim. browse here wikiwand.com/en/Nine-volt_battery and a few References to get up to speed on proper battery tests also from batteryuniversity.com/search/results?q=test+method Anything can be modelled accurately with sufficient data like , constant current tests, pulse current tests with ESR, C and leakage all being dynamic variables with SoC age and brand. There are only two paths of charge flow conductors (ESR, Rleak) and insulators ( dielectrics, C) \$\endgroup\$ Commented Apr 7, 2022 at 6:20
  • \$\begingroup\$ I use a simple RC model or RC//RC/Rleak with a base potential voltage just below 0% SoC (state of charge) But when full discharged ESR rises several orders of magnitude so the model of a ideal low V cell in series with RC does not contribute charge to the C, rather models the voltage of the battery threshold for "dead" Batteries typically drain to -15% of initial value not -50% due to high ESR and low C near end of life \$\endgroup\$ Commented Apr 7, 2022 at 6:28

4 Answers 4


Thanks for all the useful information to tell me your assumptions, predictions and measurements. Many students can learn from you.

There is no novel conduction method but the model may be new to you. It is still an R1-C2//R2-C2 simplified model.

You may have heard how difficult it is to have electrolytic caps with tolerances < 5%. Batteries are about a thousand times the C-value in capacitance in size as an e-cap. Tolerance deviations are no easier even with tight material and process controls but better in the same batch.

The simple battery model should clear up all your assumptions and explain why matching is so critical. It is rare that any set of cells will all decay or age at the same rate. There is a quadratic effect where the weakest cell drains faster and ages more when under or over voltage in rechargeables. The C and also effective series resistance both affect the short circuit current and discharge times. The ESR creates self-heating which improves the capacity from expansion and gap shrinkage and also accelerates the aging in most types. The greater the differences, the more rapid the differences expand. So that at end of life the weakest will be more than dead while the better ones are much less discharged due to effective capacitance.

Your load tends to be far greater than ESR so the lifespan of batteries depends on usually a 15% voltage range discharge rather than an RC=T 63% asymptotic mostly linear decay towards the target voltage of a capacitor after which it becomes exponential.

Read here for more details. ( sorry longish rambling, not edited yet)


This will show the memory model for a battery with 2 or more caps in parallel. How determinate parameters for thermal-model battery (Advanced question about battery modelling)

  • \$\begingroup\$ As a software developer, I understand all too well the limits of abstraction. Finding that reality isn't covered by the model in a corner case shows the need for a new model -- possibly. Can you explain my observation using your current model? \$\endgroup\$ Commented Apr 5, 2022 at 18:32
  • \$\begingroup\$ You're probably right, though. I just can't see it yet. I find we struggle the most in matters of magnitude. In other words, I didn't really understand the math, and particularly, how big or small something was. \$\endgroup\$ Commented Apr 5, 2022 at 18:59
  • \$\begingroup\$ The battery or capacitor rate of change simplified is Ic=C*dV/dt so the rate of change depends on C for cells in series. Thats a simple formula which explains it for say a constant current. Smaller the C the bigger the rate of change dV/dt like mV/second or V/s depending on I. So the weaker batter drains faster even fully charged in terms of C in Farads, yet we do not rate batteries in C rather use Amp-hours and I * dt = C * dV \$\endgroup\$ Commented Apr 5, 2022 at 22:43
  • \$\begingroup\$ Derivatives can be measured in small units of time dt, when there is a change in voltage dV thus rate is dV/dt \$\endgroup\$ Commented Apr 5, 2022 at 22:48
  • \$\begingroup\$ Thank you for your answer. (I'm thinking about it). Also, thank you for the recognition that I tried to do a good job on the question. \$\endgroup\$ Commented Apr 6, 2022 at 22:02

I think you're overthinking the problem.

You started with 4 cells in series giving 6 volts (more or less). You continued until the total battery voltage dropped to 3 volts.

So, what would you expect? Let's say, just to put numbers on it, your 4 cells had capacities of 1.1 A-hr, 1.1 A-hr, 1.05 A-hr, and 1.0 A-hr.

You discharged at 1 mA. After 1000 hrs, the weakest cell was discharged and effectively became inoperative. The battery voltage was now 4.5 volts

After another 50 hours, the second cell was discharged and became inoperative. The battery voltage now dropped to 3 volts, the LED went out, and you stopped the run.

Now, when you look at the cells, two show about 1.5 and the other two show about 1.4. This is not a voltage under load. If you had checked the cells when the LED went out but was still drawing current, I expect you'd have seen something like 1.5, 1.5, 0 and 0. The 1.4 volt levels show what happens when a discharged cell is allowed to relax with no current being drawn. For instance, if you try loading each cell with 1.5 kohm, you'll probably see the two good cells providing 1.5 volts, while the other two show much lower voltages.

Don't let the behavior of unloaded cells fool you about how they behave under load.

  • \$\begingroup\$ Sorry, I must have been unclear in my question, but you're close. I started out with one 6V battery, 4 cells @ 1.5V ~1Ah each, & discharged as you described. At the end, I had two cells with high voltages, and two completely dead cells. Then I repeated the experiment with another completely different 6V battery and got the same exact results, except this time the the two high voltage cells actually seem to have been charged. Whereupon I disassembled both batteries and created a new one out of the good 4 cells remaining out of the 8 total cells involved in the complete experiment. \$\endgroup\$ Commented Apr 5, 2022 at 18:55
  • \$\begingroup\$ Better yet, those cells you call zero still have current being forced through them, and will likely show a negative voltage when you actually measure them. \$\endgroup\$
    – gbarry
    Commented Apr 6, 2022 at 3:31
  • \$\begingroup\$ @gbarry -- I would normally agree with you. They did measure zero. My hypothesis for the reason for that is that the corrosion was so bad -- I don't think it was possible to make a connection. Unfortunately, I threw those cells out. I didn't want to mess with the corrosion. \$\endgroup\$ Commented Apr 6, 2022 at 4:03
  • \$\begingroup\$ This took me months, to discharge these lantern batteries into a white LED at 1mA constant current. So, what I'm observing happens under "current leakage" dynamics, which seems different from regular discharge. \$\endgroup\$ Commented Apr 6, 2022 at 4:14
  • \$\begingroup\$ You said, "Don't let the behavior of unloaded cells fool you about how they behave under load." Please see at the bottom of my question, a new section added under "Edit 1:" where I give voltages under load, and an estimate of cell life using a multimeter battery tester. \$\endgroup\$ Commented Apr 6, 2022 at 5:26

You are talking about corrosion.

This implies moisture. It may be from the environment and/or from the battery electrolyte.

The moisture can create parasitic current paths inside the battery.

The most probable parasitic current path (giving the assembly geometry) is between the middle point and one of the terminals.

There are a lot of factors (both random and deterministic) that may make one of the parasitic currents much more than the other. E.g. the positive terminal builds up oxide layer, it soaks up and retains more moisture. The electrolyte has different capilary behavior depending on the potential, etc, etc...

In short, the battery self-discharges unequally between its elements. The process may as well have started long before you installed the battery and may have unequally depleted the elements.

  • \$\begingroup\$ I know what you are talking about. Yes, it is widespread. It is called low quality. Primary cells of high quality with reproducible parameters over the guaranteed shelf-life are a niche product and are expensive. Yours are of the other kind. What I am trying to tell you is that the leakage works for the whole shelf-life of the battery and what you start your experiment with is a profoundly unbalanced battery. You drain ~30% of the expected capacity and one of the cells is dead because the self-discharge removed the first ~70%. Other cells self-discharged less and still have something. \$\endgroup\$
    – fraxinus
    Commented Apr 6, 2022 at 21:46
  • 1
    \$\begingroup\$ p.s. If you are interested in the practical result more than the investigation, here is the recipe: use low self-discharge NiMH rechargeables (or better yet, Li-Ion rechargeables if your safety profile is compatible with Li-Ion). They, too, have somewhat different capacity and self-discharge from cell to cell, but (1) they are manufactured to somewhat higher standard and (2) you can at least top all of them up right before use, eliminating the most of the self-discharge issue. \$\endgroup\$
    – fraxinus
    Commented Apr 6, 2022 at 21:52
  • \$\begingroup\$ I don't think that a profound state of initial unbalance is necessary -- just a really, really slow leakage that allows, as you said, "self-discharge from cell to cell", to not only occur, but be the primary dynamic (or only dynamic). \$\endgroup\$ Commented Apr 6, 2022 at 22:06
  • 1
    \$\begingroup\$ When we are dealing with a primary cell, we cannot really know how much of the imbalance is from the factory and how much is because of the storage. In your case of very small discharge current, there is an additional failure mode specific to the zinc-carbon cells - their container is made of zinc that is used by the current-producing chemical reaction. At some point the container may develop small holes that enable electrolyte leaks and promote the self-discharge. Electrolyte leaks are slow and not important when you use the battery up with high current, but your case is different. \$\endgroup\$
    – fraxinus
    Commented Apr 6, 2022 at 22:18
  • 1
    \$\begingroup\$ You can see this everywhere, this is why rechargeable batteries usually employ equalizing circuits. But I am not aware of other chemistries (other than zinc-carbon) where some kind of discharge also promotes self-discharge. It is usually the other way round. \$\endgroup\$
    – fraxinus
    Commented Apr 6, 2022 at 23:06

Carbon-zinc batteries are a very old technology that are cheap with poor performance and no guarantee. In China they are called Super Heavy Duty.

Energizer brand alkaline batteries are made in USA and have a 10 years shelf life and are guaranteed not to leak even if they have been fully used for 2 years. They have excellent performance.

I have some Copper top Duracell batteries leaking, rusting and dead in in a short time. Some Duracell batteries are made in China and have very poor specs.

I do not use Rayovac batteries.

  • \$\begingroup\$ Thanks for the brand quality info. I agree with your quality assessment, though did not know the Energizer info. But since the batteries came free with flashlights... And as a night-light in the bathroom at 1mA into a single 5mm LED, they last for months, even though running continuously. Lately, I'm using Energizer Ultimate for my important electronics, because they don't leak or corrode. (yet!) \$\endgroup\$ Commented Apr 7, 2022 at 1:50
  • \$\begingroup\$ Energizer Ultimate Lithium 1.5V batteries are their most expensive ones. I was given free demos when they were announced but I never bought any. \$\endgroup\$
    – Audioguru
    Commented Apr 7, 2022 at 16:17
  • \$\begingroup\$ I've had good flashlights and good meters killed by corroded AA's, so the Energizer Ultimate AA cells are worth it to me. By the way, I opened up an Energizer 9V, but it has rectangular cells in plastic, all welded together. I can't get them apart (at least easily). This improves my confidence in your report that they're guaranteed not to leak. \$\endgroup\$ Commented Apr 7, 2022 at 23:43

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