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    by jbkiwi ๐Ÿ‡ณ๐Ÿ‡ฟ ( Vice Admiral)

    Make Revision
    I got sick of charging LiPo batteries up for boats and planes then the weather packs up just as you have finished. You then have to storage charge your batteries unless you are using them fairly soon, which can take forever doing it with your charger (especially with bigger batteries). I decided that rather than have to drag out a plane to run the batteries down quickly, I would make a simple discharger with multiple discharge rates. I'd had some car bulbs lying around for years so decided to use those, plus a marine riding light I had spare for the loads. I now have 10w, 5w. 22w and 52w loads to suit different sized batteries,- larger loads for larger batteries, smaller loads for smaller, (depends on your batterys' output capability also, ie 20c, 40c etc).You could make a switched array of bulbs if you want to be fancy so you can adjust the Amps/load for each battery.

    I put the power analyser /balancer in between the battery and load and set it to battery check/balance mode (plug battery balance lead in for this method) so you can watch the individual cell voltage (you have to be there for this), or plug a voltage alarm into the battery balance lead (as in photo) and set it to 3.6v which reminds you to check it early.

    As it is with 12v bulbs, it will work for 2-3s batteries, for larger batteries you could try 24v truck bulbs or make resistance wire loads etc. There are dischargers available, but this is an easy method, and bulbs are cheap. You can just solder wires to the terminals (power) and body (earth). Brake / tail lights are handy as you have 2 wattages to choose from, ( I used the XT60 plugs (both positives to bulb terminals, ie 1 to each) and linked the 2 negative plug terminals and soldered that negative wire to the bulb body. The large load Halogen bulb is using the low wattage (55w) filament terminal, and the earth terminal (just an old bulb with a blown high beam).

    DC watts to amps calculation
    The current I in amps (A) is equal to the power P in watts (W), divided by the voltage V in volts (V):

    I(A) = P(W) / V(V) ie 10W / 8.2v (2s) =1.2A

    Information provided is user generated.
    This website accepts no responsibility for any inaccuracies.
    If you believe you have sufficient knowledge to improve the quality of this article, please click the "Make Revision" button at the top of the page.

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