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OK, so maybe it's not fair asking YOU why WE test something the way we do. ;)

Here's the thing -- Our company uses AGM Lead acid batteries in some products. The testing methods are ancient (but so is lead-acid technology). There's nobody left in the company to ask these questions of. We have "what" well documented, but rarely is "why" written down.

Our application is most similar to a golf cart or fork lift. "Traction batteries" I believe these would be called.

I've asked the battery manufacturers but I can't get a straight answer. I suspect this may be a case of Llama Dung

For a 12V battery (example) we do 150 cycles of the following:

  • Discharge battery at 13A until voltage reaches 8V
  • Rest 3 minutes (during this time, the battery voltage will recover)
  • Discharge at 5A until voltage reaches 10V
  • Perform slow charge (about 8 hours) until battery is fully charged
  • Repeat ad nauseum

The part nobody can explain is why we do a 2nd, lower current drain of the battery before recharging.

This may be a chemistry question (in which case I'm not qualified to really understand the answer I'm sure).

Anybody know something about this that I've failed to be able to Google?? (Believe me, I tried!!!)

Heck I even read the relevant sections of Lindens "Handbook of Batteries". All I could find was this one sentence..

" Intermittent discharge, which allows time for the electrolyte to recirculate, or forced circulation of the electrolyte will improve high-rate performance. "

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    \$\begingroup\$ You left out "install the spent batteries in the car of the VP of HR"... \$\endgroup\$ – TimWescott Dec 11 '20 at 1:10
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    \$\begingroup\$ I'd assume and hope that the testing is of a selected sample that is not going to be used subsequently for anything useful - ie to test a sample from a batch of batteroies. Yes? || Discharging a 12V LA battery below 10V is usually a bad idea [tm] unless you are intending to destroy it rapidly. I understand that the discharge to 8V is a one off per cycle load test. Aim is not certain. If you want long life from LA then discharge to a small fraction of full capacity is desirable. Maybe from 100% charged to 70% charged,. Maybe down to 60%. Lower encourages short lifetimes. \$\endgroup\$ – Russell McMahon Dec 11 '20 at 6:25
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    \$\begingroup\$ Much too much unknown here: Battery capacity. designed for deep cycle. Expected in use actual discharge profiles. Expected limit of DOD in normal use. Desired cycle or other life. Frequency of use. ... . | The question is a good start but knowing the real question will help real answers. || See batteryuniversity.com for lots on LA. \$\endgroup\$ – Russell McMahon Dec 11 '20 at 6:31
  • \$\begingroup\$ @RussellMcMahon Of course. These are disposed of properly. The cells I would consume in a test (about 100) is a drop in the bucket compared to our production quantities. \$\endgroup\$ – Kyle B Dec 11 '20 at 6:31
  • \$\begingroup\$ @RussellMcMahon The answers I got below are good enough for me. I'm not looking for help in dialing in the batteries... That's long ago been done and set in stone. For me, it's a curiosity - I wanna now why, not just what of these things. It helps me to better understand & deal with future issues. \$\endgroup\$ – Kyle B Dec 11 '20 at 6:33
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You could do the entire discharge at the lower rate but then it would take a lot more time.

When discharging at a high rate the voltage may collapse prematurely so the battery seems discharged before it really is.

By combining the two discharge rates you can discharge the battery in a reasonable time but be confident that it is fully discharged.

The lead-acid chemistry is somewhat unusual among rechargeable batteries in that the electrolyte is consumed in the discharge process. Being converted to water with the sulfate ions combining with the lead plate to form lead sulfate.

As a result at high rates of discharge, the electrolyte within the plates becomes progressively more dilute, the discharge for that portion of the plate slows down and the cell voltage drops. At lower rates of discharge, fresh electrolyte can diffuse into the plate fast enough to support the chemical reactions and the discharge can continue further. This is the meaning of the last sentence of your question.

Peukerts's Law

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  • \$\begingroup\$ Fantastic - Thanks for answering! Love it. My problem now is I can't decide whether I should select yours or Tims answer! They're both really good. \$\endgroup\$ – Kyle B Dec 11 '20 at 4:23
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A reason to do this would be to get a quicker assessment of the charge capacity at a constant 5A drain. Another may be because that more closely simulated the actual battery drain on the Llama farm 75 years ago.

All batteries tend to have a characteristic open-circuit voltage to which they'll recover under no load. For some chemistries (NiCd, dry cells), this no-load voltage is pretty constant. For some chemistries -- notably lead-acid -- this no-load voltage diminishes with diminishing charge (I'm looking at a chart* that's claiming 2.2V at full charge down to about 1.6V at nearly dead).

Beyond that ultimate no-load current, most cells have an equivalent circuit that looks something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

For quick discharges, you "see" the last resistor in the chain, for longer discharges you "see" resistors further back, etc.

For all cells, as the cell gets discharged, the apparent resistances of all the resistors goes up -- each time you discharge the cell, the voltage will go lower and it'll take longer to recover to the open-circuit voltage.

The reason for the above behavior is that as the battery discharges, the ions that conduct electricity inside of it either have more "dead battery" to go through (in the case of NiCd cells), or the electrolyte gets depleted and becomes less conductive (in the case of lead-acid cells).

So whatever they're doing with that test, it has something to do with that behavior of the cells.

* "Rechargeable Batteries Applications Handbook", Technical Staff at Gates Energy Products, Butterworth-Heinemann 1992.

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  • \$\begingroup\$ Great reply! Thanks so much for spending time to answer! Combined with Kevin's reply below I've gained excellent insight. \$\endgroup\$ – Kyle B Dec 11 '20 at 4:22
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    \$\begingroup\$ re "Rechargeable Batteries Applications Handbook", Technical Staff at Gates Energy Products, Butterworth-Heinemann 1992. -> used copies here \$\endgroup\$ – Russell McMahon Dec 11 '20 at 6:39

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