From the practical viewpoint:

- As the load increases, the motor continues to rotate at the same speed, but the phase of the rotating magnetic field lags the applied voltage, so the current increases in an attempt to reduce the phase angle.
- Similarly, if the output shaft is driven from an external mechanical source the phase begins to lead, and the motor starts to generate power back into the supply.

I suspect that your circuit represents the motor under no-load conditions. If the motor was delivering say 100W of mechanical power to its output shaft, then there is no way your circuit can model this (unless R1 is a variable resistor).

I last did maths like this at university 40 years ago, so someone more academic will need to provide a more scholarly answer.

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.

## Best Answer

Fist make sure that the three-phase power voltages supplied to the motor are the proper voltage and balanced.

Check the resistance U1 to V1, V1 to W1 and W1 to U1. The resistances should be very close to being equal. Do the same for U2, V2 and W2. You should not find continuity between the groups (U1, V1 and W1) and (U2, V2 and W2).