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We have a project to build an RC car from scratch and are only allowed to use NiMH batteries, no LiPo. We're hoping to power the car with an EDF (https://hobbyking.com/en_us/durafly-vampire-v2-rcaf-edf-and-motor.html or similar) which runs at 4s or ~14.8V, 60A. Instead of using two heavy (1kg) 8.4V batteries in series, we would like to try and power the EDF with 6 x 2.7V 60F supercapacitors, charged before the race with the aforementioned heavy battery packs.

So my questions are:

  • Do capacitors in series need some sort of balance charging like LiPo's?
  • Do they all discharge at the same time or one at a time, i.e. will we be constrained by the rated 20A discharge rate or the low voltage cutoffs of the ESC?
  • Is this safe and doable?
trijoco
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1 Answers1

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Quickly simulating what you described, it looks like you will run out of power way before you cross the finish line. 6x 2.7V 60F capacitors will hold 1300 J assuming they're charged at their nominal voltage. The EDF you linked will draw way too much power for your capacitors to handle.

schematic

simulate this circuit – Schematic created using CircuitLab

They would indeed charge/discharge at the same time, so to supply 60A you would need 3 parallel lines of capacitors.

So, without even considering charge/discharge complications, this doesn't seem doable for a race lasting more than a second.

Harnex
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  • Interesting, I was under the impression that 60F was more than enough. Using this https://electronics.stackexchange.com/questions/73330/how-to-calculate-capacitor-required with our requirements gave (60A*10secs)/16.8 = ~36F – trijoco Aug 26 '19 at 14:11
  • @trijoco that answer presumed constant current, which capacitors really cannot supply. Use it for estimating. Whether 1300J is enough... How many Joules do you need? (That will be a function of weight, max speed, altitude changes, and various losses) – Chris K8NVH Aug 26 '19 at 14:33
  • @trijoco Even with this calculation, you're not taking into account that your capacitors are in series. 6x 2.7V 60F capacitors in series are equivalent to a single 16.2V 10F capacitor. Remember that the energy stored in a capacitor is proportionnal to the capacitor's voltage **squared**. First, you need to find how much power you need and for how much time, that'll give you the energy you need to supply. You also need to take into account that the power consumed by your fan isn't quite the same as the mechanical power it will push your contraption with. – Harnex Aug 26 '19 at 15:13
  • Thanks for the feedback guys, looks like we'll need to ditch the capacitors and deal with the extra weight of the batteries. 10F is obviously not enough for such a high current draw. – trijoco Aug 27 '19 at 00:56