I'm doing some testing on a battery (MC002787). I've noticed that the voltage after a load takes a few minutes to go back to the open voltage (voltage without load for a long time). Can anyone explain why it takes so long for the voltage to stabilize?

The battery is in a battery holder soldered on a PCB with 2 wires connected to an electronic load (no circuit on the PCB/no other component assembled). The electronic load is an SDL1020X-E by Sigilent. The electronic load is controlled and monitored using NI-Visa and Python.

For the test, I simply apply a current load on the battery (100 mA) for 1 min or 10 min using the electronic load. The sample rate is 1 Hz (1 sample/second). I can see the voltage slowly go back to a stable value when there is no current load on the battery.

Side note: Is it possible to share an Excel file on Stack Exchange (to share the data collected during both tests)?

Graph of battery voltage showing it slowly recover after a load

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    \$\begingroup\$ It is a very odd Lithium battery producing 15 times less max current than a common 18650 Lithium battery. Aha, it is NOT rechargeable. \$\endgroup\$
    – Audioguru
    Sep 15, 2022 at 15:14
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    \$\begingroup\$ The phenomenon is called "relaxation" if you want to look it up. It's because batteries are not homogeneous. They are a mix of stuff inside and the electrodes can't be everywhere, reacting everything at the same time. \$\endgroup\$
    – DKNguyen
    Sep 15, 2022 at 15:18

2 Answers 2


The main reason is the time it takes for the active chemicals in the cell to diffuse into the electrodes and reaction products of the chemical reactions to disperse.

The active chemicals are consumed when the cell is being discharged and waste products produced.

This is closely related to the ability of the cell to provide high current discharge without excessive voltage drop.

When the cell is being discharged the chemical concentration within the electrode becomes lower as they are consumed and reaction products build up thAT slow further reaction. This causes a drop in the terminal voltage. Active ions from electrolyte at a distance from the electrode will then diffuse from more concentrated regions to the active region but that only occurs slowly within time scales of minutes. As the concentration returns to normal the open-circuit voltage recovers.

As Winny's answer describes this diffusion can be modeled as RC time constants when creating an electrical model of the cell for simulations. It may take multiple capacitors and resistors to create an accurate model.

The question doesn't state what type of cell is being referred to but all types suffer from the problem to a greater or lesser extent the details of the chemical reactions will vary. The rate of recovery will also vary with temperature and the mechanical construction of the cell.

In Leclanche type cells such as zinc-manganese dioxide (alkaline or zinc-carbon cells) the effect is known as polarization. Hydrogen gas is created by the chemical reaction that impedes further reaction. It takes time for the manganese dioxide to oxidize the hydrogen to water.

Lead-acid car starting batteries are another example; they are designed with thin electrodes to allow them to provide the high currents for starting without the voltage dropping too much because the sulphuric acid only has a short distance to diffuse. That does however result in a less robust battery with a shorter life; deep-cycle batteries have longer life but less capability to provide high-starting currents.

For lead-acid batteries the effects of diffision are expressed in the form of Peukert's Law (Wikipedia) that relates how the effective capacity of the cell varies with discharge current.


I can't answer what happens on the chemistry or molecular level here. When I designed battery chargers for a living, the chemists I worked with called it relaxation so that would be the search term I recommend if you want to look for deeper insights. If you take a look at an equivalent circuit for a battery, there are several things going on:


simulate this circuit – Schematic created using CircuitLab

There are many ways to model a battery and you tend to make one which fits your needs. This is just one of them. As @user253751 pointed out, modeling your battery to fit the observed behavior leads to circular reasoning, it doesn’t explain the chemistry behind it, it just helps to understand the behavior on macro level.

R1+R2 will form your DC ESR of your battery. For a step response, C1 will act like a short circuit. Current flowing through R1 (by charging or discharging your battery) will introduce a voltage drop which will charge C1. When you stop charge or discharge said current, an RC circuit formed by C1 and R1 will give you the time constant you see in your oscillogram.

Play around in your SPICE of choice and make it match your real life tests. Don’t be scared if your C1 ends up in the kF range while R1 is in the mΩ range to give you that minute long time constant, as pointed out by @mkeith.

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    \$\begingroup\$ I think this is circular reasoning. The equivalent circuit is designed to mimic the effects we see in real life, so you can't really point to it to explain those effects. \$\endgroup\$ Sep 16, 2022 at 1:20
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    \$\begingroup\$ It might be worth noting that C1 will likely be measured in Farads, and the time constant of C1 and R1 will be many minutes. \$\endgroup\$
    – user57037
    Sep 16, 2022 at 4:21
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    \$\begingroup\$ @user253751 It certainly is. I tried to make it clear in my first sentence. I’ll re-phrase it. \$\endgroup\$
    – winny
    Sep 16, 2022 at 6:52
  • \$\begingroup\$ Despite not explaining the chemistry, this diagram is useful if you need an accurate simulation of a battery in a circuit. \$\endgroup\$ Sep 18, 2022 at 15:54

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