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If you have a three-phase generator (or motor run as a generator), what part of it governs the output voltage? For example the speed at which you rotate the shaft governs the output frequency, but what governs the output voltage?

The reason I ask is that I'm looking for cheap ways to produce three-phase power in an off-grid scenario, and electric motors from damaged EVs are relatively cheap. Many of these seem to be three-phase permanent magnet motors, but finding out what voltages they produce when driven as generators is very hard to come by.

I am wondering whether it's a function of some part of the motor design (e.g. the strength of the permanent magnets) or whether it's something you can control as part of driving the generator, as you do with frequency by varying the shaft speed as required. For example does the output voltage go quite high if there is no load, and enough of a load must be placed on the generator output to drag the voltage down to the desired level? I'm not sure if that's right because I thought the load was related to the frequency (and how hard it is to turn the shaft at the correct speed in order to maintain the desired frequency).

In other words, if you are designing a three-phase generator from scratch, what (if anything) do you change in the design to adjust the output voltage it produces?

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    \$\begingroup\$ If you are talking about synchronous generators, it's the excitation current. \$\endgroup\$ – Bart Jul 6 at 12:31
  • \$\begingroup\$ Same relation that governs their speed when driven from a voltage ... Kv the speed constant. (But losses subtract from the voltage generated instead of adding to the drive voltage required.) \$\endgroup\$ – Brian Drummond Jul 6 at 12:32
  • \$\begingroup\$ @Bart: Please excuse my ignorance - I am not sure what role an excitation current plays in a permanent magnet motor? I thought it was only applicable to induction motors. \$\endgroup\$ – Malvineous Jul 7 at 13:24
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The output voltage from an unloaded generator is the product of its speed, and the magnetic field. The voltage drops a little when loaded due to the winding resistance.

For a fixed field machine, that means the output voltage varies linearly with speed. Variable field machines, usually by using a wound field, can control the output voltage at a fixed speed by varying the field in the generator.

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  • \$\begingroup\$ So does this mean that if you take a bunch of different motors and run them all to produce a 50 Hz output, the differences in voltage between them is down to the magnetic field strength? \$\endgroup\$ – Malvineous Jul 6 at 12:40
  • \$\begingroup\$ For a permanent magnet synchronous motor, voltage depends on RPM, field strength, the number of turns per pole in the windings and the geometry of the construction. \$\endgroup\$ – Charles Cowie Jul 6 at 13:31
  • \$\begingroup\$ @Malvineous sorry, I over-simplified it. I took number of turns, and field conversion to total flux via the pole area, both as constants because they don't tend to change, whereas the magnetic field can change, deliberately as in a wound field machine or accidentally when a permanent magnet generator gets demagnetised by excessive current, shock or temperature. \$\endgroup\$ – Neil_UK Jul 6 at 13:43
  • \$\begingroup\$ Simple is good! So if I understand right, it is the magnetic field that governs the voltage, but it's the way that field interacts with the surface area of the windings, so things like the number of turns, number of poles, etc. all influence the final result, not just the strength of the magnets alone? \$\endgroup\$ – Malvineous Jul 7 at 13:22

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