how to drain most of battery

Discussion in 'General Electronics Chat' started by namara, Nov 15, 2010.

  1. namara

    Thread Starter New Member

    Oct 2, 2010
    im looking for help on how to drain 2 AA continous charge
    rechargable batterys without damaging them,
    i want to see how well they do charged by a solar panel,but first --to lose the batterys charge,
    is their any quick way to do this,
    i know this can be done quickly
    by connecting the 2 terminals ,
    but does this damage the battery,
    and how to avoid the batterys from being completely drained,
    thanks for any help,
  2. Audioguru


    Dec 20, 2007
    Connect a suitable resistor across each battery cell. A modern AA Ni-MH cell has a capacity of about 2300mAh and voltage of about 1.2V so use a 0.56 ohm/5W resistor connected for 1 hour to completely drain each battery cell. Don't drain both cells with only one resistor if they are in series.

    A Ni-MH battery cell is not damaged if it is completely drained but a Lithium battery is severely damaged.
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  3. iONic

    AAC Fanatic!

    Nov 16, 2007
    What is the battery chemistry? NiCd, NiMH, Lithium...?
    I don't think I'd discharge them below 1.0V per cell. Since a .56 Ohm 5W resistor is not a common resistor in home I might siggest draining them as follows:

    If you don't have the 1 Ohm resistors, use 10 ohm, or no resistors.
    Check the battery voltage every hour. It may take three hours.
    Last edited: Nov 15, 2010
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  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010

    The <snip> page tells you in the first sentence "USE THIS CIRCUIT"
    copyright my <snip>!
    Last edited by a moderator: Nov 15, 2010
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  5. Tahmid

    Active Member

    Jul 2, 2008
    Since it's just for experimental purpose, and not something you're going to be doing all the time, just connect a low-value resistor across each cell. 0.56ohms is good, but if you can't find it, just use 1 ohm. Will do, but discharging time will come down to about half as that with a 0.56 ohm one.

    Hope this helps.
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  6. maxpower097

    Well-Known Member

    Feb 20, 2009
    Back in the RC car days we made light bars to plug them into. As the bulbs got dimmers your packs drained. This was to completely drain a pack to charge for the next race. Back in the ole days a complete discharge was better then a medium charge.
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  7. iONic

    AAC Fanatic!

    Nov 16, 2007
    If you do not have .56 Ohms, or even 1 Ohm you can parallel multiple 2.2 or 3.3's, increasing the wattage 2x to 4x.

    Thus: (4) 2.2 ohm, 1/4W resistors in parallel will give you .55 ohms and 1W power disipation. (4) 3.3 ohm 1/4W resistors in parallel will give you a .83 ohm and 1W power disipation. Even (2) 1 ohm resistors in parallel will give you .50 ohms and 1/2W power disipation. And if you burn up a couple, who cares!
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  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    Light bulbs work well.

    If you are working with single cells, you can go down to 0.9v without hurting a rechargeable NiCd or NiMH.

    If the cells are in a pack, don't go below 1V/cell to call the cells "dead".

    The reason for this is when a pack of batteries, like 6 or more, are in series, some will drain faster than others. Charging to zero terminal volts usually means one of the batteries has actually reversed polarity rather than all of them have no charge. After this point, the charge life of the pack goes down quickly, and charges don't lat as long.
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