Be sure that the batteries aren't getting ho

July 12 [Fri], 2013, 11:22
It is common knowledge that Nickel Cadmium like full discharge/recharge cycles. The reason is that Nickel Cadmiums store their energy in crystal form. When it is charged, crystals are created. As the battery is discharged, the crystals shrink in size. When you charge the battery before most of the crystals are gone, it both creates small crystals and enlarges the old crystals. After a few half-dischargea href="http://www.batteryoffice.com/sony-vgp-bps13s-battery.htm">12 cells VGP-BPS13S cycles like this, you can end up with many larger crystals and not very many small ones. The result is that the energy is still there, but there is less surface area (many small crystals have more surface area than a few large ones of the same mass), resulting in decreased output. You'll need to completely wear-down the larger crystals.

The problem is that modern electronics can't operate on just a few crystals that are slowly giving their power up, so the oversized crystals never go away.
Read and open the case: Nickel Cadmium batteries don't need protection circuits, so you might be able to do this without opening the case. It's not very likely to be a problem, but I would recommend opening the case anyways. Worry not, if your batteries are shot, what good is the case anyways? If you do fix it, a little glue (I recommend cyanoacrylate) ought to put it back together. Look on your case and determine the total capacity (mAh) and voltage (V). My laptop battery shows 11.1V, 3800mAh. If you don't see it written anywhere on the case, you may be able to calculate it.

Determine your resistor: You'll want to discharge the battery over about a 1 hour period. Since the capacity (3.8A, or 3800mAh for me) is expressed as amp-hours, it will be our draw-amperage. Calculate the size of your resistor with this formula: ohms = volts / amps. Go to Radio Shack or Digikey, or make your own and find the closest resistor to this value. It needs to handle a power (watts) of at LEAST battery volts (11.1V for me) multiplied by draw-amperage (3.8A for me) (11.1V * 3.8A = 42.18 watts). This is going to be a big resistor, and it will get hot when you're using it. Be sure that the batteries aren't getting hot. If you're comfortable dissipating more energy, you could discharge the battery much faster (at 4C, for example), or if you can't find a resistor that can handle the watts, you could try lowering your draw amperage and finding an appropriate resistor.

Connect the resistor: Connect the resistor across your battery pack (the pack itself; not the + and - output on the protection circuit), and a voltmeter (in parallel). Because of various voltage drops and anomalies I'm12 cells VGP-BPS14B not going to go into, you may need to leave the battery discharging for more than an hour! You're not done until the voltage is very low (such as .1 volt per cell or lower) If you're also using an ammeter (in series), your measured amperage should drop as the voltage drops. If your resistor is adjustable, you can decrease the ohmage in order to keep the amperage the same, but you don't need to.
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