When operating as a motor, the stator rotating field induces voltages (and hence currents) in the rotor windings and makes it spin (hence the rotor produces a torque).

When operating as a generator I rotate the rotor faster than the stator rotating field. this should induce in the stator windings some voltage, but how if it was connected to the voltages that are generating the rotating field? how do I ''get'' from the generator the output current/voltage?

I need the stator rotating field to have currents on the rotor, so I haven't turned it off... what happens? Do I have extra windings with no potential in the stator?

basic induction motor:

enter image description here

  • \$\begingroup\$ Same place you put the excitation voltage in. \$\endgroup\$ Jul 12 '16 at 21:55
  • \$\begingroup\$ Could you grace us with a photo or a diagram showing wire colors, etc. The basics are simple, but we need help from you for a usable answer. We do not like to guess. \$\endgroup\$
    – user105652
    Jul 12 '16 at 22:00
  • \$\begingroup\$ @BrianDrummond this sounds weird. how can I ''get'' a voltage from a port where I ''put'' another voltage? woudn't the output be equal to the input? \$\endgroup\$ Jul 12 '16 at 22:18
  • \$\begingroup\$ @Sparky256 done, not sure if it helps tho. \$\endgroup\$ Jul 12 '16 at 22:19
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    \$\begingroup\$ I agree it sounds weird... \$\endgroup\$ Jul 12 '16 at 22:24

With a squirrel-cage induction motor, you can feed energy into an active supply simply by driving the motor above the synchronous speed. The power flow will reverse at the stator terminals. You can not easily and reliably use an induction motor as a stand-alone generator.

You can find on the internet instructions for using capacitors with an induction motor to make a stand-alone generator system, but those types of systems are not entirely satisfactory. The capacitor value and the load both have an effect on the generated voltage. Every time you start the system, you need to perform a starting sequence such as: first charge the capacitors from some external source, then connect them to the motor with a switch, then connect the load. You may need to increase the driving speed after you establish operation. If you connect or disconnect load items, you may need to add or remove capacitors.

If the motor is energized through an inverter, it is possible to design a satisfactory stand-alone system, but you still need an external battery or other source to get it started. The electronic system will be fairly complex.

Additional information re real and reactive power

An induction generator needs magnetizing current to operate. The magnetizing current does not represent real power but current that is out of phase with the voltage and causes energy to circulate back and forth between the source and the motor. Capacitors can be used for that purpose, but they need to be "tuned" to the motor and load. If the induction generator is connected to an AC source, that source can supply the magnetizing current (reactive volt-amperes), while the generator supplies power (watts) to the load.

enter image description here

I don't know if I'm too tired to see what's wrong in my question, but I'm indeed asking for the generator. which is the same component as the motor, only put in motion by something external at a speed higher than the synchronous. The problem is that generators GENERATE power, all I'm asking is where this power is. on the stator winding? how can it be, if there's already the excitation there? – user3149593 2 hours ago

I think it should be safe to assume that we are talking about a squirrel-cage induction motor. Wound rotor motors are included in electric machinery texts and courses, but they are almost never used compared to the many millions of induction motors in use.

Power in AC circuits is all about the phase relationship between voltage and current. To make things simple, we can assume that the current coming from the AC source is 90 degrees out of phase with the voltage and thus no real power comes from the source. The current in the load can be assumed to be in phase with the voltage. The current in the induction machine is the sum of the two. That is the mechanism that causes the power to come totally from the induction machine even though it is receiving magnetizing current from the AC source. That assumes just the right generated voltage. With the right driving speed, you can assure that no real power is taken from the source.

enter image description here

  • \$\begingroup\$ qsl.net/ns8o/Induction_Generator.html claims that it's not necessary to "charge the capacitors from some external source"; the system works just fine with capacitors hard-wired to an AC induction generator without any other source of real electrical power or any other source of reactive electrical power. \$\endgroup\$
    – davidcary
    Jul 13 '16 at 0:20
  • \$\begingroup\$ Yes; but it also says that you may need to magnetize the motor when you start the first time and that it can lose magnetization if you shut the engine off with the load connected. I would expect that it can also become demagnetized if you don't use it for a while. You will find different claims and procedures from various sources. These schemes work, but they are not necessarily satisfactory. \$\endgroup\$ Jul 13 '16 at 0:31
  • \$\begingroup\$ so, in the most basic configuration I can implement... I drive the rotor above the synchronous speed and then turn off the AC source and I will get on the stator windings an output current. Is this what you mean with the power flow will reverse? Or should I continue to feed the machine leaving the AC supply on? very good answer tho, thanks a lot! \$\endgroup\$ Jul 13 '16 at 1:36
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    \$\begingroup\$ oh I think I got it: the AC power supply is left on and provides an excitation current (or magnetizing) that brings the reactive energy to the circuit. what I get to the load is whatever current comes out from the induction in the stator's winding minus that current...or I could say that the active power from the generator goes to the load. It's hard to figure in my mind, but you made things a lot clearer. \$\endgroup\$ Jul 13 '16 at 1:45
  • \$\begingroup\$ Worth pointing out that the reason you can't use an induction motor as a stand-alone generator is that you need to supply AC excitation. Which is where the question started. Having supplied excitation, the OP will find out that the more torque load you place on the shaft, the more current it will draw, and the greater its slip speed ( the more it falls below synchronous speed). Once he understands that, he can externally drive the shaft ... when it is faster than synch speed, it will start to return current to the supply via exactly the same means. \$\endgroup\$ Jul 13 '16 at 13:37

What the other answers aren't telling you is:

Connect the motor to the incoming AC power supply. The motor spins at its rated speed.

Now spin the motor faster than its rated speed using some form of external rotation power. This can be a water-wheel or windmill or whatever.

The motor now acts as a generator, feeding power back into the supply source.

Note that the motor speed won't increase significantly above its normal speed. But the faster that you try to spin the motor, the more energy this takes. It is that extra energy that is converted to electricity and fed back into the power source.

  • \$\begingroup\$ this is a nice answer. do I turn off the power supply when the machine is above the synchronous speed? I have a problem figuring out how can I feed and ''take power'' from the same winding. \$\endgroup\$ Jul 12 '16 at 23:02
  • \$\begingroup\$ If spinning faster or slower than its output is asynchronous to the power line. OP seems confused about the 'reactive' properties. The link the OP provided offers details. \$\endgroup\$
    – user105652
    Jul 12 '16 at 23:02
  • \$\begingroup\$ or maybe I leave it on and its voltage is subtracted from the induced one. then I would get a current flowing into my voltage source...which is weird. as far I know induction machines are always asynchronous (pm machines are synchronous) so I'm not really following sparky. \$\endgroup\$ Jul 12 '16 at 23:10

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