Battery question

Now if you could only track down the doof who figured that little plastic positive terminal cover was worthless and threw it away, instead of placing it on the battery as intended...

My car, there would have been a big sheet of Plexi or Lexan over the whole battery, or it would have been moved - Inside the passenger compartment is NOT a smart place to put a hydrogen gas generator with a half gallon of HCL acid inside.

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Reply to
Bruce L. Bergman
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Reply to
Jerry Wass

No HCl in any car battery I ever owned. Gerry :-)} London, Canada

Reply to
Gerald Miller

While there is no question that high voltage will damage a battery, in real life most battery chargers will be perfectly suitable since a battery's internal resistance increases as the battery is charged making the system is pretty much self limiting.

Try connecting a volt meter and an amp meter in parallel with the battery charger. Turn the charger up to about a 10 amps charging current and watch the volt meter. As voltage climbs the amperage decreases. With about a 10 amp initial charging current by the time the battery voltage reaches about 13.5 volts the current will have decreased to about 1 or 2 amps. Hardly a battery killing combination.

To put the battery charging cycle in prospective, A "3 stage smart charger" charges the battery to about 14.2 volts (depends on type of battery) before dropping back to about 13.4 for the bulk charge.

Bruce-in-Bangkok (correct email address for reply)

Reply to
Bruce in Bangkok

On Tue, 18 Mar 2008 10:07:45 +0700, Bruce in Bangkok wrote: ...

Internal resistance vs. charge depends a bit on battery type and history. [1][2] For lead-acid batteries, internal resistance decreases (not increases) as charge goes up. For typical car battery chargers, current drops as charge approaches 100% because the difference between charger voltage and battery voltage drops toward zero. Open-circuit charger voltage and charger resistance will be the main limiting factors.

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[1]
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section "Lead-acid Internal Resistance and SOC" at bottom of 4th page says "A fully discharged lead-acid cell has virtually no sulfuric acid in its almost pure water electrolyte" [so would have high resistance] [2] Near end of
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, the paragraph before Figure 6 (resistance vs. voltage) says "The resistance of lead acid goes up with discharge", etc.

-jiw

Reply to
James Waldby

Come on Bruce! I's well know that the Germans were still sore over losing WW-II, and they were just trying to get even with cheap imported rolling firebombs.

Reply to
Michael A. Terrell

Of course you are correct and electricity flows from negative to positive and transistors work because of "electronic holes".

But the average user has a bit of trouble visualizing that so the explanation that the internal resistance increases and thus the charging current decreases (which it does) is easily understood. It may not be correct but it does explain the action of the usual transformer battery charger in terms that the average user can relate to.

Bruce-in-Bangkok (correct email address for reply)

Reply to
Bruce in Bangkok

I would be very careful about connecting an amp meter in parallel with any battery charger. At the least, you will have to replace a fuse right away. At worst, all of the magic smoke will escape from either the meter or the charger (or both), or you may find yourself rushing towards the nearest fire extinguisher.

Joe

Reply to
Joe

Bruce in Bangkok wrote: (...)

It *is* Ohm's Law but you are changing the wrong variable. Current equals voltage divided by resistance. The difference in voltage between charger and battery drives current way more than does the change in the batterys internal resistance.

A given SLA might have an internal resistance that halves when the battery is taken from 'discharged' to 'fully charged' state. However that same battery is likely to experience a 15:1 decrease in current flow at the same time, because of the voltage difference between the charger and battery.

I think it would be as easy for the average user to visualize (a 'decrease in voltage difference' causing a charge current decrease) as it would be for them to visualize (a 'decrease in internal resistance' causing a decrease in charge current). That is fortunate because the first cause really happens and the second cause is completely backwards from reality.

--Winston

Reply to
Winston

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