3
\$\begingroup\$

From Peukert's Law, we know that when discharging a lead-acid battery, if the discharge rate is high, the effective capacity of the battery will be reduced due to the battery reaching a minimum cutoff voltage earlier (after delivering less total energy) than if it were discharged at a lower rate. To use this remaining charge, the battery must be allowed to recover to a higher voltage before discharge can resume at a lower rate. This is why lead-acid batteries are rated at a 20-hour discharge rate - their capacity will be different at different rates.

My question is, does this apply to charging too? If charging a lead-acid battery to a given absorption voltage (say, 2.45V per cell), will a battery reach that voltage after consuming less total energy if charged at a higher C-rate? If this happens, can the battery be charged further (without overcharging the cell) by allowing the voltage to "recover" as can be done while discharging, and then continuing at a lower charge rate?

If this is possible, in a proper CC/CV/float charge process, will this ever happen, or will the reduced charge current during the CV and float phases ensure that any additional charge the battery can take at a lower current, it will?

\$\endgroup\$
3
  • \$\begingroup\$ I can’t say for certain whether the same factors apply to discharging as to charging, but yes it is observed that fast charging results in reduced capacity. Many chargers use multi-stage techniques where the current reduces as the state of charge increases, and this seems to mitigate this effect. \$\endgroup\$
    – Frog
    Commented Aug 7, 2021 at 2:20
  • \$\begingroup\$ Since internal resistance is the most important factor in Peukert's Law for discharging. It will impact charging (heat), but rate will be less because charging is just a few volts over battery voltage. \$\endgroup\$ Commented Aug 7, 2021 at 2:35
  • \$\begingroup\$ smartgauge.co.uk/peukert_depth.html \$\endgroup\$
    – D.A.S.
    Commented Aug 7, 2021 at 2:50

1 Answer 1

4
\$\begingroup\$

From Peukert's Law, we know that when discharging a lead-acid battery, if the discharge rate is high, the effective capacity of the battery will be reduced due to the battery reaching a minimum cutoff voltage earlier... the battery must be allowed to recover to a higher voltage before discharge can resume

It's a bit more than that. If terminal voltage was the only issue you could just specify a lower termination voltage under load, but at high current the available current becomes limited by charge mobility. Having a rest period allows the slow moving ions to 'catch up' and produce higher current again for a while longer.

This effect isn't normally seen during charging because the charge current is much lower (typically C/5 or less). During charging the main issue is keeping the voltage low enough to avoid gassing.

If charging a lead-acid battery to a given absorption voltage (say, 2.45V per cell), will a battery reach that voltage after consuming less total energy if charged at a higher C-rate?

It will store less energy due to reaching the maximum permitted terminal voltage earlier, as a result of higher voltage drop across the battery's internal resistance at higher charging current.

If this happens, can the battery be charged further (without overcharging the cell) by allowing the voltage to "recover" as can be done while discharging, and then continuing at a lower charge rate?

Yes, but having a 'recovery' period doesn't help much. Simply maintaining a fixed voltage and allowing the current to reduce by itself works just as well and is faster.

in a proper CC/CV/float charge process, will this ever happen, or will the reduced charge current during the CV and float phases ensure that any additional charge the battery can take at a lower current, it will?

In CC/CV float charging the voltage is held at a level which ensures the battery can't gas no matter how long it is charged for.

This is different from multistage charging which uses a higher voltage during the absorption stage, then drops back to float voltage when the battery is fully charged. During the absorption stage some gassing may occur. Sealed lead acid batteries can recombine the gasses provided the pressure doesn't get too high and cause venting.

Some chargers can also apply an even higher 'equalization' charge at the end of the absorption stage, to ensure that all series cells in a battery get fully charged, and/or 'refresh' charges to allow maintaining a lower float or storage voltage without risking sulfation.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.