Electronics > Beginners

Theory about NiMH battery recovering voltage in between discharges

(1/2) > >>

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:

--- Quote from: MosherIV on July 04, 2018, 08:27:56 pm ---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.

--- End quote ---
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

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod