> Hello again,
>
> Since you are so knowledgeable in these matters, I am wondering if you
> can give me some ideas on how linear (or not) the internal resistance curve
> of a lead acid battery is with respect to load current. Being partly ohmic
> and partly chemical in nature, I would guess it to be complex. I see some
> battery testers that only draw an amp or two and are able to extrapolate
> the performance of the battery at 100 amps. Does that work?
>
> Thanks,
>
> Alastair Couper


Thank you for the flowers, I am not sure how justified this is. As they say:
Every new discharge (or charge!) a new surprise.

The impedance of a battery is pretty complex because there are many different
processes taking place. In terms of the discharging voltage that exists as a
result of the discharging current, the answer can be simplified as this:

The current depends exponentially on the voltage: I ~ exp-k*(E-Eo) where Eo
is the equilibrium potential and E the electrode potential. Eo can be attempted
to be measured (long periods of rest) whereas E is basically outside the
measurement possibiliby (ohmic resistance and the ion distribution on the electrode
surface are the two factors that need to be "removed". However, for all
practical purposes, if you draw the current against the logarithm of the voltage
you get a straight line. This is because the other factors are negligible
compared to this exponential curve. At very high currents the ohmic resistance
becomes a major factor, and at voltage close to the equilibrium voltage, the
exponential dependence is no longer valid. However, there are time delays to the
voltage development which are associated with the change in the ionic distribution
on the electrodes surface ( a few milliseconds) and electrolyte diffusion
(hours to day). Measuring the straight line is therefore very time consuming,
even if points are taken that are close to each other because the SOC changes
during the time it takes for the time delays to vanish. For this reason, the
exponential curve is usually only measured for a fully charged battery during float
charging - no change of SOC here.

Now, if you take two current values (or voltage values) with a well defined time 
difference between the measurement points, write this down and measure the
same thing again sometime later you can trace changes. This are relevant and
show changes of the internal suface area (influences the time delay constant)
and degradation of catalytic behaviour (e.g. old batteries have a higher charging
current at the same float voltage).

If these devices go from 2 to 100 A and carefully analyse the voltage response
over time and then compare this to older measurements of the same (!) battery,
I am sure this can be made to provide a good indication as to whether the
battery should be tested more closely or not. - Comparing the measurements with
other batteries will reduce the information considerably and when you go to
different types, sizes and manufacturers/alloys I think that nothing will work
any more. Only measuring at one current means that different SOC, temperature, 
connections etc, will start to play a big role and I would be very sceptical if useful 
results could be provided. But perhaps they compare the difference between 0 A and
the test current! Doubtful if the test result would be able to discriminate between 
broken and aged, partly charged batteries or batteries at rest for a long time, etc.


Of course devices that use signal frequencies of more than 50 Hertz, better 1 kHz
range to measure the complex resistance only measure the ohmic resistnace of the
system, all capacitance effects will disappear and changes are signficant for SOC,
corrosion and drying (VRLA) because all these effects increase the ohmic resistance.

I hope that this helped.

Heinz Wenzl
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