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I have been trying for some time to solve what for me is a real puzzle. The internal resistance of a battery can be measured by a DC method. The open circuit voltage \$ V_{ocv} \$ is measured, then the voltage \$ V_{load} \$ and the current \$ I_{load} \$ are measured during DC discharge. The internal resistance is \$ R_{int} = (V_{ocv}-V_{load})/I_{load} \$, and it exhibits typically low values.

On the other hand, when performing electrochemical impedance spectroscopy (EIS), at low frequencies batteries are characterized by the so-called Warburg impedance \$ Z_W = \frac{A_W}{\sqrt{\omega}}(1-i)\$, with a constant \${45}^\circ\$ phase. It is evident that as the frequency tends to zero, both the real and the imaginary part of \$ Z_{W} \$ diverge.

I would assume that at very low frequencies the impedance of the battery should approach the DC resistance, but this is clearly not the case. Could anyone please explain to me the reason for this difference between the two methods?

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    \$\begingroup\$ For my own benefit and for others who are not familiar with the term, here is an article of explanation: palmsens.com/knowledgebase-topic/warburg-impedance \$\endgroup\$
    – PStechPaul
    Jul 27, 2022 at 23:39
  • \$\begingroup\$ The open circuit voltage seems meaningless. In my mind, I think of it as almost like static electricity, that with ultrahigh resistance between terminals, almost any chemical EMF at all presents a voltage close to when new, and so highly misleading. \$\endgroup\$ Aug 1, 2022 at 12:13
  • \$\begingroup\$ You would do better to calculate the internal resistance from two or three loads close to the load of interest, and average them. I consider that open-circuit voltage calculation useless and misleading. \$\endgroup\$ Aug 1, 2022 at 12:13

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""DC"" method!

At low enough frequencies, the battery wholly discharges (or overcharges -- or other, shall we say, nonlinear, consequences!). Whether the Warburg impedance is the furthest asymptotic case before that happens (and, at any given signal level), depends on battery chemistry. What's important is that it spans a wide enough frequency range to be a useful model. (Just as a resistor is only a resistor up to some frequency, or a capacitor or inductor over a certain range.)

Specifically, a Warburg impedance models ionic diffusion, or a number of other collective effects that manifest as generally fractional power-law responses. It might not be the totality of a device's impedance at all meaningful frequencies -- such is modeling.

The catch is, often they don't tell you that they're modeling, and present it as irrefutable fact and fundamental to the component...

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    \$\begingroup\$ @Hybridslinky I believe Tim’s point is that “DC for a short period” is in fact not DC but rather a very low frequency. \$\endgroup\$
    – winny
    Jul 27, 2022 at 22:05
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    \$\begingroup\$ @Hybridslinky No idea. I was just pointing out what I believe to be a misunderstanding between Tim and you. With pure DC, it would charge or discharge forever. I too use the term DC loosely. \$\endgroup\$
    – winny
    Jul 27, 2022 at 22:11
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    \$\begingroup\$ @Hybridslinky perhaps if you have some data to compare EIS with pulsed test, a more specific answer can be had? I think you will find, during the pulsed test, the voltage does indeed droop slowly, corresponding to an EIS response. \$\endgroup\$ Jul 27, 2022 at 23:36
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    \$\begingroup\$ The internal resistance of a battery can be measured by a DC method. There's your problem, 'DC' is not well defined. Warburg is. Once you've defined a DC method, or several, you can compare the answers, see whether they agree, see how they relate to Warburg conditions, and get a realtionship going between them. \$\endgroup\$
    – Neil_UK
    Jul 28, 2022 at 5:39
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    \$\begingroup\$ @TimWilliams You're right, the voltage has a sudden drop followed by a slow drop. \$\endgroup\$ Jul 28, 2022 at 16:19

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