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Bringing NiCd's from the dead

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  • Nicolae Sfetcu
    Bringing NiCd s from the dead (From no-idea) The failures the article talks about occur in mutli-cell Ni-Cd battery packs, and are due to the voltage
    Message 1 of 1 , Nov 19, 2001
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      Bringing NiCd's from the dead

      (From no-idea)

      The failures the article talks about occur in mutli-cell Ni-Cd battery packs, and are due to the voltage differences between cells. Say you have four 1.25 V cells in a pack connected to a 200 ohm load. The load "sees" 5 volts and draws 25 mA. Since each cell must pass the entire 25 mA and each cell's potential is 1.25 volts, Ohm's Law tells us that each cell sees the equivalent load of 50 ohms.

      But in practice, no four cells in a battery ever exhibit exactly the same output voltage. Assume that one cell is delivering only 1.2 V, and the others are at 1.25 volts. Now, the 200 ohm load sees 4.95 volts and draws 24.75 mA. Since all four cells must pass the entire 24.75 mA, each of the strong cells at 1.25 volts sees an equivalent load of 50.5 ohms; the weak cell sees only 48.5 ohms. The weak cell works into the heaviest load and as a result will discharge more rapidly than the other cells. If the pack is charged for only a short period of time, the weak cell, which has been working the hardest, is also the one that receives the least charging power.

      This usually doesn't matter if you trickle charge after each day of flying. The inequality is small for any given charge or discharge cycle, due to the relatively flat output voltage NiCd cells exhibit over most of their range. But a combination of incomplete charges and deep discharges will exaggerate the energy difference between a weak cell and the other cells. Operated continually in this manner, the weak cell invariably reaches its "knee," the point at which its voltage decreases sharply, long before the other cells reach the same point.

      Now comes the problem! Suddenly, the weakest cell sees an increasingly heavy load, which causes its voltage to drop even faster. This avalanche continues until the cell is completely discharged, even as the other cells continue to force current to flow. The inevitable result is that the weak cell begins to charge in reverse, which eventually causes an internal short. Once an internal short develops, recharging the cell at the normal rate is futile. The short simply bypasses current around the cells active materials. (Even though the cell is apparently dead, most of its plate material is still intact.) If the small amount of material that forms the short could be removed, the cell would be restored to virtually its original capacity once again.

                  300 ohm              Charge
                     5W             /  Switch
      20-40 + O---/\/\/\----o------o  o------------o----------------o
      VDC                   |                      |                |
                            |      Zap             |                |
                            |      Switch          |               +|
                            |      ___|___         |           -----------
                            o------o     o---------o              -----
                            |                      | +    Shorted   |
            6000 micro-     | +                 -------    Cell     |
            Farad, 40V  _________               |     |             |
            Capacitor   ---------               |_____| Volt        |
                            |                      |    meter       |
                            |                      |                |
            - O-------------o----------------------o----------------o
      

      Using the circuit shown, the internal short can be burned away in a few seconds. In operation, energy stored in the capacitor is rapidly discharged through the dead cell to produce the high current necessary to clear the short. Current is then limited by the resistor to a safe charge rate for a small A cell.

      Several applications of discharge current are usually necessary to clear a cell. During the "zapping" process, it is a good idea to connect a voltmeter across the cell to monitor results. Momentarily close the normally open pushbutton switch several times to successively zap the cell, allowing sufficient time for the capacitor to charge up between zaps, until the voltage begins to rise. Then, with the toggle switch closed, watch as the potential across the cell climbs to 1.25 volts. If the potential stops before full voltage is reached, some residual short remains and another series of zaps is in order. If you observe no effect whatsoever after several zaps and shorting out the cell and taking an ohmmeter measurement indicates a dead short, the cell is beyond redemption and should be replaced.

      Once full cell potential is achieved, remove the charging current and monitor battery voltage. If the cell retains its charge, it can be returned to charge and eventually returned to service. But if the cell slowly discharges with no appreciable load, the residual slight short should be cleared. To do this, short circuit the cell for a few minutes to discharge it, zap again, and recharge it to full capacity.

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