Charging a 12V 7Ah Battery Charger with 12V 500mA?
Frage von Composer: Mai 2007
Hello, I have a 12V/7Ah NiCd Battery for my head light purchased. Now I wonder how long it with a batch with 12V / 500mA Output lasts until the battery is fully charged. Unfortunately, neither s.Battery still s.Charger a display of the charge. Can you just say it with a 500mA Charger will take 14 hours to 7Ah Battery to load, or do I have as a total mistake? Is there a way to calculate the loading or appreciate? Meters, Battery s.die I could connect to the test, I have unfortunately not. If someone could help me that would be really super.
Antwort von Markus:
"A / e Nameless / r wrote:
Can you just say it with a 500mA Charger will take 14 hours to 7Ah Battery to load ...?
That you have already correctly calculated (500 mA × 14 h = 7 Ah), but would in this case a small reserve to expect, because not all of the current 100% of stored energy is implemented.
Such a timer-controlled Ladeprozedur is the worst way, what do you agree with an Battery can. Condition is, inter alia, that actually 7 Ah missing, until the battery is fully charged. An overloading and damaging the batteries so that is actually occurring.
Cross-reference:
Antwort von Duisburger:
Although the Charger a maximum output current of 500mA has does not mean that with the maximum current load. The charging time is significantly higher.
Antwort von Markus73:
"Anonymous" wrote:
Can you just say it with a 500mA Charger will take 14 hours to 7Ah Battery to load, or do I have as a total mistake?
That would be under ideal conditions, the case that - The battery is completely empty, - Yet its full rated capacity, - The charging really exactly the specified value is reached and stops, and - The whole thing without losses expire.
With somewhere in the network is even the rule of thumb
Recharge = (capacity / charging) * 1.4
paths crossed. Source eg http://www.akkukonfigurator.de/akku-konfigurator_accu-rechner.aspx This should come closer to reality. But as already said, there are much better solutions than the whole thing with the clock to regulate.
Gruß, Markus
Antwort von hannes:
> = Loading (capacity / charging) * 1.4
basically correct, but the battery was empty? I would prefer a voltmeter dranhängen and 14.2V in the draw times.
Antwort von rtzbild:
You can use the Battery synonymous within half an hour with 14 A load, but it is believed that an ideal battery with 10% of its Entladekapazität loaded.
So 0.7 A * 10 h.
BTW: Was it the battery of LK-electronics Bischofsheim from ebay?
