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  1. Banned
    Join Date
    Oct 2006
    Location
    london
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    1)Reason:
    a.The battery has no voltage.
    b.There is problem with charger, as it has no output electric current.
    c.Low charging efficiency due to external cause.
    2)Solutions:
    a.Test whether the charger has voltage and output electric current or not.
    b.Check if the charger is well contacted with battery.
    c.Activate the battery with voltage and current 1.5 time higher then the highest ones of battery. (Use this method only when the battery can't be charged.)

    SPAM team

    You are in breach of the forum rules and are being banned for spamming.
    / Moderator mats.hogberg
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  2. Member Krispy Kritter's Avatar
    Join Date
    Jul 2003
    Location
    St Louis, MO USA
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    Aside from the fact you are banned...this post makes no sense.
    Google is your Friend
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  3. Member
    Join Date
    Dec 2005
    Location
    United States
    Search Comp PM
    Very Informative!!!

    So, I have a battery pack, 9.6 VDC @ 900 mAH - it's an 8 NiCAD cell pack wired n series - each cell 1.2 VDC/Cell At 900 mAh.

    According to your formula (Activate the battery and current 1.5 time higher then the highest ones of battery).

    1.2 VDC x 1.5 = 1.8 VDC
    900 mAH x 1.5 = 1.35 Amps

    So, I should charge this sucker at 1.8 VDC @ 1.35 amps.

    You sir, are a ******* genious!
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