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I am using this BLDC motor [https://uav-en.tmotor.com/html/2018/navigato_0402/52.html] as a generator. The device providing the mechanical drive turns the motor about half a turn and not at a linear speed, thus my output is not a perfect sine wave. I am using 0.3R resistors as a load in a star configuration, and I believe, but not certain, that the motor is wired in a star too, so I think this is my schematic. Note, the motor just has 3 wires, no neutral and I checked for continuity with the motor case, in case that is the neutral.

EDIT 8th Apr - the schematic below is incorrect, I have just got a reply from the supplier of the motor, it is actually in a Delta configuration.

Schematic

Before I get into the actual output, can I just confirm a few things first. I have searched online for how to calculate 3 phase power (assuming a normal case of a perfect sine wave) and got this in a few places:

P = √3 × pf × I × V

However, I'm unsure if the voltage above is line to line, or line to neutral, which I don't have anyway. I am measuring the voltage across one of my three load resistors, so that's line to line, yes?

As I don't know my current, but know my load, can I rewrite this as such:

P = √3 × pf × (V^2 / R)

I am assuming that the voltage should be RMS, and that my load, being resistors, have a pf of 1.

Next is the issue of taking my imperfect sine wave output and trying to get an average power figure. So I need to get an RMS value. I did the following [https://www.techwalla.com/articles/how-to-get-the-rms-in-excel]:

Export the data from my 'scope and ran ABS on it, converting negative values to positive. Then squared each data point value, then averaged these values, and took the square root of the average.

This is the output of my motor/generator (just showing one phase here):

enter image description here

So using my rewriten 3 phase formula, and my RMS method, am I getting the correct average power output?

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    \$\begingroup\$ Most digital scopes can do RMS calculation for you. Just use that if available. If not, then you need to square the voltage, average it, then take a square root. That’s what root-mean-square is after all. \$\endgroup\$ Apr 5 at 16:38
  • \$\begingroup\$ I'm using a cheap PC 'scope and yes it does show RMS, but, I cropped the image, there is a lot of flat line either side, it just does RMS for the whole screen, so get a figure that's too low. Re your instructions, is that not what I did? \$\endgroup\$
    – KevInSol
    Apr 5 at 16:41
  • \$\begingroup\$ You may ground your R neutral or use 3ph Delta like the motor. If you need DC, then you need 6 diodes or FETs controlled with something Then sense load current on ground side with 50mV max shunt and multiply V*I and average to get Pout. Consider diode rating much higher than motor \$\endgroup\$ Apr 5 at 16:46
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    \$\begingroup\$ For some reason, they wouldn't give me free merchandise at Home Depot on my card. decades ago. \$\endgroup\$ Apr 6 at 13:21
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    \$\begingroup\$ Because of my name, my simulation runs much faster than the grid along with privileges from decades of racing experience in the GHz range. \$\endgroup\$ Apr 6 at 14:01

1 Answer 1

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With 3 wires the motors is most likely a Delta configuration. Although it could be a Y wired motor and logically are the same with 3 wires.

They create an artificial low impedance Neutral on the grid using delta-Y transformers.

Then Neutral is protective earth, PE grounded for safety.

Remember Ground is simply any "0 volt reference". It can even be floating relative to PE ground.

So if you create a high resistance Neutral with resistors, this can see how well balanced the current in each phase is measuring neutral current. But you don't need that unless you are planning to convert AC into DC and choose some 0V point and sense current thru it.

e.g If you ground any phase then that Neutral point you created becomes the phase you grounded. proof by simulation Yet this does not alter the differential output voltage.

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  • \$\begingroup\$ Many thanks for that. I have emailed the supplier asking how the motor is wired. By "Y" I assume you mean star? Your details of the high resistance Neutral are above my head TBH. Do I need to worry about this? I'm just doing an experiment (for an actual project) and am dealing with under 5V, so safety is no concern, if that's why you are discussing Neutral and ground. In this case I am just using AC, however I have also converted to DC using the same circuit in your simulation, but this is not relevant to this question. \$\endgroup\$
    – KevInSol
    Apr 6 at 16:37
  • \$\begingroup\$ For this question, I am just trying to determine if I have used to the correct methods to see how much power I have generated. (ran out of room in first comment) \$\endgroup\$
    – KevInSol
    Apr 6 at 16:38
  • \$\begingroup\$ The simulation ought to verify your experiment, Y=star same thing. Change to your imagination. Properties of resistor plot include power and Pavg, Vrms etc \$\endgroup\$ Apr 6 at 18:05
  • \$\begingroup\$ Thanks Tony, I'll have a play around with it. \$\endgroup\$
    – KevInSol
    Apr 7 at 13:46

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