Originally Posted by XrayFK
It should hold the cells at 0.85v because of the transistors. As soon as the voltage creeps above 0.85v the transistor base/gate will be opened and the cell would be discharged back down to the cutoff. I'm sure Axxis can confirm this for sure, but I believe that is how it is designed.
Looks like you do know your electronics... (but in a transistor (NPN) the current flows from the collector to the emiter with the base controlling the flow. FET's have a SOURCE, GATE, and DRAIN
As we all know batteries start regaining their nominal voltage when a load is removed. As a semiconductor is FORWARD BIASED current will flow and will keep the cell at the set voltage (0.85 volts). Since the 030 pulls a very large amount of current (25 amps @ 0.85 volts) the cell will cut off immediately (since it is at a high current, it is difficult to get exactly 0.85 volts...). After several minutes the cell voltage will start rising and the semiconductor will be forward biased again, thus bringing the cell down to 0.85 volts. (This is a continuous cycle…)
If the current would be lower (< 4 amps) better equalization can be achieved when the cut off is reached. What we recommend with the 030 is to leave the cells in the tray for an additional 15 to 30 minutes in order to get better equalization. As more current is drained out of the cells at the preset voltage the closer the cells cutoff voltage will be to each other.
I still recommend EQUALIZING the cells at a low current setting...because then you get real equalization. The lower the current setting, the better the equalization.
For those that want to treat their cells at a lower current, I still recommend cutting off one of the resistors (in each cell... R=0.05Ω) in the 030. By cutting one resistor off you will still discharge at 20+ amps, get better equalization, and your batteries will not reach such high temperatures. Cells last longer if NOT discharged at high currents, temp is kept lower, and the inner cell pressure is not pushed to the limits.