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I am trying to build a drone generator with a gasoline engine and a 14-pole,170KV, 2KW BLDC skateboard motor (this one). Before designing the system and running it, I wanted to clear out a fundamental doubt that has been troubling me. Details are as follows:

At a given voltage, how is the maximum output current that my generator can deliver related to the input torque at the generator shaft? Would increasing the input torque on the generator increase the current delivering capacity of my generator?

I am aware about the fact that at constant supply voltage (hence speed), a BLDC will draw different currents for different propellers due to difference in aerodynamic loading.

Also,in brushed permanent magnet dc motors, at a constant voltage, the motor will draw whatever current it needs from the source (with a little voltage drop) OR if I use a brushed dc generator, I will be able to generate larger currents (at the risk of burning my coils) at constant voltage (const. shaft speed).

Will the same hold true in case of a BLDC?

Please share your views.

Cheers.

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  • \$\begingroup\$ What is a "drone generator"? \$\endgroup\$ – Transistor Nov 18 at 11:11
  • \$\begingroup\$ A Generator for powering the drone, something which can feed my drone with electrical energy at the required voltage. \$\endgroup\$ – Curious_Techie Nov 18 at 11:17
  • \$\begingroup\$ Is it an electrical generator powered by a gasoline engine? \$\endgroup\$ – Jakob Halskov Nov 18 at 11:20
  • \$\begingroup\$ Yes Jakob Halskov. I apologize if it was unclear \$\endgroup\$ – Curious_Techie Nov 18 at 11:22
  • \$\begingroup\$ With about 2% error - mechanical power into an alternator (or any shaft). Watts = kg.m torque x RPM. [That's a rehash of Power = work per second = Force x distance per second. For a rotating shaft distance = radius x 2 x Pi x RPM/60. Force = torque/radius. Cancel and you get Watts = 2 x Pi x RPM/60 x torque_Nm. = Torque kgm x g x RPM x 2Pi/60 ~~= RPM x kg.m torque :-) \$\endgroup\$ – Russell McMahon Nov 18 at 12:06
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A useful expression.
With about 2% error -
Mechanical power into an alternator (or any shaft).
Watts = kg.m torque x RPM.

  • That's a rehash of Power = work per second
    = Force x distance per second.
    For a rotating shaft distance = radius x 2 x Pi x RPM/60.
    Force = torque/radius.
    Cancel and you get Watts = 2 x Pi x RPM/60 x torque_Nm.
    = Torque kgm x g x RPM x 2Pi/60 ~~= RPM x kg.m torque :-)

Your BLDCM used as an alternator will make somewhat LESS volts per PPM ("KV") when used as an alternator than its motor spec.

Motors and alternators BOTH produce a voltage - in a motor we call it back emf and the motor speed stabilises when back emf + resistive drop = applied voltage. At the same speed with the same emf produced the alternator output is the generated emf LESS the IR drop. Murphy winds both ways :-).

Ignoring mechanical losses:

Motor: - Vapplied = Vbackemf + IR
Alternator:- Vout = Valternator - IR

ie with motor the terminal voltage is higher than the generated AC backemf and for an alternator the Vout is lower than the generated AC.

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  • \$\begingroup\$ where does the 2% error come from? Neglecting losses, it's surely 0%. With losses, 2% would be correct only for a huge machine, 2kW is small enough to easily see losses in the 10s of percent. \$\endgroup\$ – Neil_UK Nov 18 at 12:31
  • \$\begingroup\$ @Neil_UK That's just the rounding error in the equation for converting to different units, not the mechanical-electrical conversion efficiency. g x 2Pi/60 = 1.027 \$\endgroup\$ – Phil G Nov 18 at 14:47
  • \$\begingroup\$ @PhilG It's not rounding error in the normal sense - it's the result of mixing units which don't match with magnitudes that happen to cancel. g ~= 10 m/s/s (actually closer to 9.8) 2 x Pi ~= 6 ... whoops ... must be still waking up ... you've made that point. But, I suggest not rounding error as usually meant - rather, a convenient coincidence. \$\endgroup\$ – Russell McMahon Nov 18 at 18:34
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Assuming the load is able to draw some value of current at the terminal voltage delivered by the generator, then there are two limitations from the generator, whether it's brushed or brushless

1) The engine driving the generator has to supply enough torque. If it doesn't, the generator will slow down, the terminal voltage will fall, until the load is drawing a lower current so throwing a low enough torque demand on the engine. This limitation is instant and you'll notice the engine slow down.

2) The generator has to be able to supply that current without overheating. Improving the generator cooling can help with this. This limitation may take minutes or tens of minutes to become apparent, and if you ignore the smell and then smoke from the generator, it may destroy it.

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