electric flight uk - British Electric Flight Association - Jan Bassett's

Method 1
Three FET devices are connected in series with 3 LiPo cells as shown below.
Note the 3 FET leads, SOURCE, DRAIN and GATE. The Charger is connected
negative to Lead 1, positive to Lead 4 and set up to charge 3 LiPo cells at 1C.
The Red part of the FET is totally resistive at all times so does not conduct
but the Black “conductor channel” conducts, so current (the same current)
passes right through the FETs and charges up the cells. However, as a cell
charges up its +ve voltage becomes greater of course and something called a
“field effect” puts a +ve charge on the Gate.
Now this increase in the “positiveness” of the Gate effectively “squeezes up” the
“conductor channel” and when the +ve reaches its “switching” limit (4.2 volts
for LiPo cells) the conductor channel actually becomes effectively “blocked”
and hence a total resistor and so it stops any more current flowing-hence
there is no more charging of anything.
So when the first cell to reach full charge “blocks” the conductor it blocks all
current flow so all charging ceases. No cell can become over-charged, but of
course two cells will remain under-charged. Probably not by much, but not
acceptable. The charger should indicate which cell is fully charged, and let us
say it is Cell 2. So now we have to “top up” Cells 1 &3.
So we firstly put the charger negative to Lead 1 and positive to Lead 2 and set
the charger to charge 1 LiPo cell at 1C. Again, when the +ve of Cell 1 reaches
4.2v the field effect from the positive of Cell 1 puts 4.2v on Gate 1 and FET 1
stops conducting. So Cell 1 is now fully charged but Cell 3 still needs a bit
more so we repeat as before, put the charger –ve to Lead 3 and positive to
Lead 4 and set the charger for 1 cell 1C. These “topping up” operations should
only take a few minutes but if it takes much longer it means that cell was very
out of balance and it is just as well you discovered it!
charged, and it would continue for the remaining 2 cells and then finally the
remaining 1 cell. There would be no need to “top-up” any cells separately as
would be needed with my Method 1. I show all this graphically in my Method
2 schematic, but I must emphasise that although I am conversant with FETs
of Method 1 I have no actual knowledge of Method 2 and all that I have
written is pure speculation. Be good if it was correct!
E.F.-U.K. 19