7

I've had a lot of electrical problems with my car, and a couple of times, the car battery voltage got drained absurdly low (like below 2V) for short periods of time. I gather that fast charging completely flat batteries is bad for them, and that slow charging is better (to remove the sulfation as I understand it), so when this has happened I've revived the battery slowly with a solar panel inputting about 1-2 amps (not jump starting) and it has seemed to work. While the panel is connected in good light, the voltage reading exceeds 11-12 volts, which I think means there isn't a dead cell in the battery, and can charge like this back up to 12.6 volts (eventually!). Sometimes while charging (while the car's not doing anything) the measured voltage swings by +/- 3 volts, not sure if that's significant.

I've seen a lot of comments in passing many places saying that if a car battery voltage ever drops below about 10 volts it gets damaged, but little detail on how damaged how quickly and how to test how damaged one is.

The battery is new (6 months or thereabouts). It's a sealed "no maintainence" 60 Ah 12v lead-acid battery.

It recently went flat unexpectedly, which might be because of underlying electrical issues and damp draining it, possibly also because I hadn't recharged it as completely as I'd thought. I'm wondering how to judge whether I should just write it off and replace it, then look to see if there are any corrosion, bad cables etc causing a drain, or try to coax it back to health one more time (this will be difficult, it's rainy season so charging with the solar panel is a struggle, I might need to jump-start), make sure it's fully charged, then go looking thoroughly for possible drains.

What is the permanent damage that is caused by going completely flat, and how can I test the extent of such damage?

6

The bottom line is that if you venture below 10 volts, the battery is dead. Vampire dead. Done. Over. Kaput.

There were some magic solutions back in the day to try and "restore" such a battery, but again I refer to my bottom line.

You need a new battery. If you were ever down to 2 volts, the chemistry has shifted in such a way that there is no easy recovery. Sulphation forms on the lead plates, and six distinct "cells" kind of blend into a soup, which prevents the ordered electron flow that creates voltage potential. The acid loses it's acidity and approaches water. You can see this by checking specific gravity of the liquid electrolyte (on a non-sealed battery) with one of those clear turkey baster things with the various colored balls or weighted dial. While you can "charge" the battery to what seems a reasonable open circuit voltage, this charge is superficial and does not represent the SOC (State of Charge) that would exist on a non-damaged battery. The reliability and capacity will be diminished, as well as its maximum amperage flow. It's like pouring gin over ice cubes, letting the ice melt, and attempting to separate the gin and water and restore the cubes by putting the glass in the freezer.

In any case, the cost to "recover" would far exceed the cost of a new battery.

Sorry for the bad news, but unfortunately I speak the truth. I'm a cheap old coot and I've tried many things to voodoo lead-acid batteries back from this state, but I have to admit nothing works.

Get a new battery. Good luck!

  • I've read many comments like this but they never explain in what way it's dead - I have successfully revived this battery relatively easily to 12.6V and kept it there for a while, so it's not that. Do you mean, for example, it discharges faster than it should? Or stores fewer Ah at any given voltage? In what way is it dead? – user568458 Jul 30 '16 at 12:19
  • 1
    @user568458 I edited the answer adding information about why I feel such a battery would be dead beyond recovery. And I shoehorned an analogy using gin. – SteveRacer Jul 30 '16 at 19:43
  • @user568458 specifically the part about SOC mentioned here. A damaged battery may "appear" to charge up to 12.6 volts, but when actually put under regular load it will drain quickly again, due to the internal damage mentioned. It may "still kinda work", but if this is an auto battery for example, you will become the person that needs to get their car jumped every couple of days. – Christopher Hunter Sep 12 at 23:17
4

Surprised no-one mentioned this. I discovered it by accident the hard way, while researching the problem that knowing about this earlier could have avoided.

Internally damaged "weak" batteries can cause observable voltage fluctuations while charging and extra current draw that can cause alternators to develop a fault and break.

Damaged batteries including formerly completely flattened batteries can to get into a state where it seems to just about work, but is actually quietly damaging the alternator. They don't necessarily appear "dead", they can be made to seemingly work, but it's a potentially damaging undead zombie-like state where treating them like the living, allowing them to shuffle around, results in problems.

The danger signs that I'd observed but hadn't known how to interpret were:

  • Voltage fluctuations while charging. I mentioned in the question that I'd noticed these while connected to a battery charger. When driving, the alternator would respond to each such fluctuation by doing more work for longer, suffering excesive wear and (potentially) heat damage. Look for these while charging with a charger, not the alternator with the engine idling, since the alternator might be correcting and therefore masking fluctuations with the engine running, so no fluctuations with the alternator active might not mean no problem.
  • The battery voltage never quite getting as high as might be expected. I mentioned in the question that the battery could often (but not always) just about reach 12.6V (but never higher). I took this to mean it wasn't dead, since that was easily enough to start the car, but actually, it meant the battery was always hungry for current and therefore the alternator was always doing extra work, not just increasing wear but also increasing the likelihood of system voltage fluctuation because it's never not feeding current to the battery.
  • A third warning sign appears to be if the circuit voltage drops substantially while cranking the car to a start (or on other heavy load conditions). As far as I can tell from reading around (please correct me if I'm wrong), maintaining 11V or more while cranking is good, dropping below 9V while cranking is a warning sign of a weak battery.

I'd been using a relatively new (~6 months) but formerly completely flattened 'zombie' battery, that could be recharged and could start the car albeit temperamentally. My thinking at the time was, "well, it's clearly not dead like people say, since it seems to work". My alternator's voltage regulator wore out, and the alternator diodes and brushes got damaged.

The alternator eventually packed in and stopped working completely; ironically, shortly after I finally bought a new battery... Don't do what I did!


In particular, don't let all the talk of such batteries being "dead" fool you into thinking that they simply harmlessly fail. You also need to watch out for a battery harmfully appearing to work.


From the top-voted answer to the above linked question:

...when a car has a weak battery that battery will want lots of current. The current demand of the battery lowers the system voltage so the voltage regulator compensates by sending more current through the rotor. At idle, the alternator is incapable of making the needed current. Because of this the system voltage drops even more and the voltage regulator sends the maximum current through the rotor.

In this maximum load minimum speed conditions is where the wear happens. At the minimum speed the minimum amount of cooling is available from the built in fan. At the maximum load the voltage regulator will push the maximum amount of current through the rotor and through the brushes and slip rings. The brushes and slip rings get hot and with no additional cooling available from the fan they will wear faster.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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