Theory about NiMH battery recovering voltage in between discharges

[atmfjstc]

Hi everyone! I'm trying to find some detailed theory / information regarding a specific phenomenon that occurs with NiMH (and likely other) batteries.

I imagine many of us are familiar with the typical "S"-shaped discharge curve that is mentioned in basic battery theory and battery datasheets: as we start discharging a fresh battery, the voltage starts off high (around 1.3-1.4V for NiMH), then drops rapidly until we reach the nominal voltage of about 1.2V, then stays quite flat throughout the middle of the discharge, and it starts falling off more and more rapidly towards the end, culminating in a sudden crash (around 1.0V) when the battery is depleted.

The thing is, the battery only behaves this way if the discharge is uninterrupted. I've noticed that if I stop the discharge partway through, let the battery rest for a few hours, then start the discharge again, the battery will have recovered some of the "lost" voltage. In effect, instead of continuing from its last reported voltage, the battery will recapitulate the start of the discharge curve, except much faster. About 15 minutes into the discharge, the voltage will return to its normal trajectory.

Clearly the battery is not actually recharging during the pause - it must be then that the voltage of the battery is not a monotonic function of the state-of-charge (however nonlinear) but also depends on some other phenomena that can temporarily boost it.

Does anybody know of some resources with in-depth theory or explanation of the latter part, i.e. why does the battery appear to recover voltage during breaks, and maybe a model predicting how this occurs? I'm not having luck searching for "NiMH voltage recovery", ther results are mostly about resurrecting broken NiMH cells.

[The Soulman]

I've read about it in some datasheet, from what I remember it will take up to 16 hours for the voltage to settle
after a charge or (partial) discharge.
I can't remember what is the cause for this, maybe thermal effects? 

[JS]

It's about the chemicak process, I'll try to explain in simple terms.


The available ions depend on the charge, the ones in solution show the voltage, the equilibrium when the reaction is stopped you see about 1.4V. The speed of the reaction is much slower than the discharge to keep up with the consumed ions, so the concentration in the solution drops down till a point the ions jumps faster to solution and reach an equilibrium, that 1.2V level. When the ions are in this concentration the reaction is faster and the voltage is stabilized.

Once you are at that 1.2V and stop the discharge thebreaction continues till gets to the steady voltage, when you start the discharge again, the total available ions are less so the reaction is slower than at full charge so it reaches equilibrium faster.

Something similar is why at higher discharge rate usable charge is less, that and the ESR.

JS

[MosherIV]

From a purely electrical understanding, it is to do with the internal resistance of the battery.

When charge is drawn, some of the voltage is drppoed across the internal resistance.
When there is no current flow there is no V drop across the internal resistance.

When the cell run out of charge the internal resistance rises so there is a greater voltage drop.

Note: this is purely a simplified electrical way of understanding it.
It does not explain why it takes time for the voltage to recover.

[JS]

From a purely electrical understanding, it is to do with the internal resistance of the battery.

When charge is drawn, some of the voltage is drppoed across the internal resistance.
When there is no current flow there is no V drop across the internal resistance.

When the cell run out of charge the internal resistance rises so there is a greater voltage drop.

Note: this is purely a simplified electrical way of understanding it.
It does not explain why it takes time for the voltage to recover.

A next correction to the electrical model would be to add a cap in the output of the series resistor circuit, so at the begining starts the discharge of the cap and later from the battery, when no load the cap charges up again.

JS

[atmfjstc]

Hey there, sorry for resurrecting this, I had some issues and got distracted.

JS's answer makes a lot of sense. Is this somewhat similar to the phenomenon of 'surface charge' in lead-acid batteries by any chance?

Can you point me to some specific books and papers showing a model of this phenomenon in action?

[KL27x]

I remember reading everything I could to understand NiMh, back in the day. The understanding I end up with is that they are different than all of the other popular battery chemistries.

In every other battery type, you will get some "recovery" for the reason that JS stated. Give the battery a rest, and the float voltage goes up. NiMh has some other effect going on in addition to that.

My memory is notoriously bad, but what I inferred and internalized from all my research was that in NiMh, the voltage is not a function of charge, the way it is in other batteries. As the cell gets more charge, there are places in this process where the voltage will go down. So you can't even charge these batteries consistently when hooking them up in parallel. It is preferred to charge them in series. Even within one cell, there are some parts of the cell that will charge to a higher energy state than other parts, and they will not equalize. These two different energy states in the cell are both happy at the same voltage.

[KL27x]

I remember reading what appeared to be official NASA documentation for maintenance procedures for NiMh batteries, including use in satellites and whatnot. The thing read like a voodoo handbook. It appeared to be as much superstition as science.

[digsys]

The available ions depend on the charge, the ones in solution show the voltage, the equilibrium when the reaction is stopped you see about 1.4V. The speed of the reaction is much slower than the discharge to keep up with the consumed ions, so the concentration in the solution drops down till a point the ions jumps faster to solution and reach an equilibrium, that 1.2V level. When the ions are in this concentration the reaction is faster and the voltage is stabilized.

Once you are at that 1.2V and stop the discharge the reaction continues till gets to the steady voltage, when you start the discharge again, the total available ions are less so the reaction is slower than at full charge so it reaches equilibrium faster.

Something similar is why at higher discharge rate usable charge is less, that and the ESR.

Well written explanation ... As also noted "surface charge" , "electron migration" and others also have an influence. Throw in some "deposition" etc and it
really becomes fun :-)
I make up large LiIon packs for EVs and those bastids are just as difficult to quantify !! I run a few load lines / settling times etc, plot curves .. yadda yadda.
In saying all that, if you don't do all that, you may only be 5-10% out in capacity. What it is VERY useful for is - predicting EOL / intervening for better EOL.

[KL27x]

I make up large LiIon packs for EVs and those bastids are just as difficult to quantify !!

This would not even be practical with NiMh. Until li ion is developed, we are still using lead acid powered golf carts.