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02-08-2008 02:40 PM  10 years agoPost 1
snobdrs

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coatesville,pa-usa

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Ive read a few posts in which they stated that so much watts can charge a 6s pac at so and so amps. Now my question is P=I*E correct?
So lets say a charger is 150W. does this equate to a 6s pack or 22.2VDC which would charge at around 6.5amps? Or am i missing something in the translation? 22.2VDC(E) x 6.5Amps(I) = 144.3W(p)

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02-08-2008 03:26 PM  10 years agoPost 2
Invrted1

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Cincinnati, Ohio

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Keep in mind the capacity of the pack. You would only want to charge at that rate if you had a 6500ma pack. Charge at 1C, which is the capacity times 1. A 6S 3700 will charge at 3.7 amps.

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02-08-2008 04:00 PM  10 years agoPost 3
snobdrs

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coatesville,pa-usa

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Yes i know the one c rule. I was just confirming the output wattage of a charger and how that determins charge rate limit of higher cell count pacs. IE dont want a charger that can charge a 6 or 8 cell pack but only at like half c.

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02-08-2008 05:32 PM  10 years agoPost 4
JKos

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Redondo Beach, CA

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snobdrs,
Yes, you have the right idea. Just divide the output power rating by the pack voltage and you will get the max charge current at that voltage assuming the charger allows you to specify a charge current that high.

I would recommend using the maximum charge voltage of the pack when doing the math. So you would use 4.2*6 or 25.2 V for a 6S LiPo pack.

Every charger has three limits (voltage, current, and power) which can be reached independently or in combination depending on the situation.

- John

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