If I have a Brushless 3-phase motor that spins at 3000rpm when given 6.0A @ 20v (no load) will it produce 6.0A and 20v (or anywhere close) when used as a generator and spun at 3000rpm??

And i suppose this comes into it too... Will a 3-phase bridge rectifier make DC more or less as efficiently as an ESC turns DC into a 3-phase signal??

....UPDATE: So this is the motor I am thinking of using now:

MOTOR: 5065 KV: 270 MAX POWER: 2200W WIRE WINDS: 9 MAX AMP: 60A ESC: 80/120A MAX VOLT: 8S RESISTANCE (Ohm): .42 NO LOAD CURRENT: 1,5 SIZE: 50 x 65 ( without shaft ) WEIGHT (g): 0,380 SHAFT: 8mm with 3mm keyway

I will use a pulley to spin it at 18000rpm hopefully resulting in about 60v..

Does anyone know what current I can expect?

And being a 2200w motor (more than the engine that is powering it (700w)) ,when i connect it to the mppt charge controller will the engine struggle to turn as the load creates effort? Or can I expect that it will be creating far fewer amps than it draws when used as a motor at this voltage and under load?...


Similar but not equal. All the losses that subtract from 100% efficiency will work in the opposite direction as a dynamo.

So if the motor is 75% efficient under those conditions, then its 100% efficiency speed (unloaded) might be 4000rpm and it would draw 0A. You can roughly crosscheck by monitoring its stall current at the same voltage : in this case, it may draw 24A at 0 rpm. (The motor regulation would be 1000 rpm for 6A, or 167rpm/A, or 24A for 4000 rpm)

If that's the case, then you have motor constants of 4000/20 = 200RPM/volt, and a winding resistance of 20/24 ohms = 0.8333 ohms.

Given these values - and they are based on my guess of 75% efficiency for your motor - the open circuit voltage would be 3000rpm /(200rpm/v) = 15V.

And if you drew 6A from it, you would drop 5V across the winding resistance, so you would see 15-5 = 10V across your load, instead of 20V.

The bridge rectifier would then drop 2* the diode drop (say 1.4V for silicon diodes) giving 8.6V DC. Schottky diodes or synchronous rectifiers can improve on this, but probably at more cost than is justified for such low power.

Measure the motor regulation by running your motor at different loads and measuring both speed and current, and cross-check by measuring stall current, and you can arrive at the likely dynamo performance for your actual motor.

  • \$\begingroup\$ Wow, thanks!! I'm working my way through your answer and learning fast as i go! I haven't chosen a motor yet.. Would there be a smart way to guess at suitability for the job? My engine puts out about 700w at 6000rpm. Now obviously I can't create energy from nowhere so the 3kw brushless motors I've been looking at will only ever give me somewhere on the way to 700w. But will the bigger low kv motors find it easier (be more efficient) to make my goal of 60v and 10A? Should i be looking for motors that are rated way over 700w or is the rated power a 75% indication of suitability?? \$\endgroup\$ – Trewin Apr 2 '15 at 16:20
  • \$\begingroup\$ Find the relevant numbers for those motors - Kv is speed/volt (should be the 100% efficiency speed/volt), actual no-load speed and current at rated volts, stall current, and you should be able to work it out yourself. If you have a datasheet for a specific motor, link it in the question. \$\endgroup\$ – Brian Drummond Apr 2 '15 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.