need help dropping voltage...11.1 to 6???

[929freak]

o.k. heres the deal I have a little fuel caddy that uses a small electric pump to fill my bird....the pump is rated fo 6-12 volts,I am using a 11.1 v lipo I had laying around,everything works fine,except that it pumps a little too fast.What I'd like to do is drop the voltage to around 6 volts,is there an easy way to accomplish
this?? resistors? what watt and what resistor? I've seen plenty of discussions on here about electronics
by some of you edjumicated folks(ha ha) so if ya can,please help me out!! thanks in advance........
here is a pic of my litle fuel caddy,so ya know what I'm working with.

[devious17]

You could try a variable resistor connected in series of the +ve lead from the battery.

[kgfly]

How much current does your pump draw ? The obvious solution is a 6V BEC. Cheap ones that can do 1.5A continuous (claim 3A but they'll melt) are about $12, otherwise you can work your way up to 3A, 5A or 10A. A SportBEC will to 3A continuous. Hercules BEC 5A.

[JEEPWORLD2002]

u could always use the balance connector wiring and just use the neg and the cell 2 wire so ur out put will be 7.4v fully charged ? or step it down from there

[929freak]

jeepworld,good idea!! but if I want to use a resistor,how do I know what resistor??? I'm not a total buffoon,but I am a little unsure of how to go about it.......

[kgfly]

Using a series resistor is wasteful and inaccurate but if you are determined to do it that way then: V = IR so R = V/I

To know what size resistor to give say a 4V drop you need to know what current your pump draws. If you knew that you could just use a BEC which would be a lot more reliable and efficient. If your pump draws 1A then you would need a 4ohm resistor to drop 11.1V to 7.1V. Of course you would be dissipating 4W in the resistor so you would need a large 5W sandbar resistor and a way to mount it so that it doesn't melt a hole in your fuel bottle! Now if your pump draws 2A you need a 2ohm resistor to drop 4V and it will dissipate 8W so you need a 10W resistor.

Running off the balance lead is only viable if the amp draw is less than 4A, in which case a BEC running off the main power lead is a better solution by far. since tapping off to use 2 cells out of a 3 cell pack is going to make charging difficult. You would have to charge just the two drained cells as a 2s pack since the imbalance between the drained cells and the unused one would be too large for any decent 3S charger to allow you to charge the pack.

[929freak]

all of you gave valuble info,I thank you indeed...but after testing the pump I don't think I need to drop the voltage after all(sorry!) this new pump is slower than the old one,so even with the higher voltage it is still very managable,not to fast....so basically I wasted your time....beatings may commence at your will.
again thanks for the info,it will come in handy at some point,no doubt.