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I'm trying to reconcile the physical steps that go into the operation of an AC induction motor. Looking online, I either find broad and general explanations that I already know and understand, or very specific and math heavy explanations that get a bit overly detailed.

I understand synchronous motor operation and describe it in my head like this: "From standstill, voltage is applied to the rotor. This creates a current in the rotor which generates a magnetic field which starts to turn the rotor due to attraction to the magnets of the stator. At the start, the rotor has little movement relative to the stator's field, so little back-emf is created, allowing for maximum current to flow, and so the machine has max torque at the start. As it gets up to speed, the rotor's quick movement cuts through more of the stator's flux, and generates more intense back-emf, which fights the input voltage, limiting current into the rotor, and torque drops off until some balance of speed/torque is achieved with the load.

But when I try to create the same "story" of what happens in an induction motor, I must have some fundamental misunderstanding (probably about back-emf), because I run into this: "At standstill, an AC voltage is applied to the stator windings, quickly manifesting a rotating magnetic field. This induces a voltage in the rotor coils, creating a current that manifests its own magnetic field such that the induced voltage/field opposes what created it. Now the relative motion of the rotating stator field and rotor coil would be maximum at startup, creating maximum possible back-emf.... and that's where my thinking fails. If the back-emf is biggest at the start, it'll fight the input voltage the most at the start, which will result in the least current and least torque at the start... and the thing will never get up to speed."

Obviously that's wrong, but I don't see where. In a DC motor, it makes sense easily because input voltage, back-emf, and rotation, are all in the rotor. So when the rotor speed is low, it cuts less flux, and creates low back-emf. When the rotor speed is high, it cuts more flux, and creates higher back-emf. A nice feedback loop.

But it seems reversed for the induction case the way I'm thinking of it. At a low rotor speed, the rotor would cut through more flux, because the stator field is quickly rotating around it. This would create a higher back-emf in the rotor, fighting the input voltage, limiting the input current, reducing torque, and slowing the machine or keeping it from moving.

Then at a high rotor speed, the rotor would cut through less flux, because it's catching up to the rotational speed of the stator's field. This would generate less back-emf in the rotor, which would mean the input voltage isn't being fought as much, increasing current, increasing torque, speeding the device up.

So my thinking ends up with the opposite of a self stabilizing device. If it slowed down, it'd want to keep slowing to standstill, and if it sped up, it'd want to keep speeding up...

None of that is right, so I must be messing something up about how back-emf operates in an induction motor. And I think the inverse nature of how the rotor speeding up leads to amount of flux being cut to be lower is also tripping me up...

Thank you so much for any help with a layman's quest for understanding!

amazonprime
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  • When you apply a DC voltage across the terminals of an inductor current begins to flow despite the back emf equaling the applied voltage. Does that not cause you the same concern? – Andy aka May 29 '20 at 18:26
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    You already asked this question yesterday at [Electrical Motors and Frequency](https://electronics.stackexchange.com/questions/502487/electrical-motors-and-frequency) - that you insisted on mislabelling an AC machine as DC does not change the fact that your question of yesteday was intimately specific to the behavior of an AC machine and covered the same issue you have posted again about today. – Chris Stratton May 29 '20 at 22:33
  • Note that only a 3-phase induction motor has a rotating magnetic field. A single-phase motor has only a _pulsating_ magnetic field. This is why a single-phase motor needs a starting capacitor or other mechanism for reliable starting. Your question doesn't specify 3-phase and I don't think it changes the answer to your question, but good to have in one's "mental model" :) – scanny May 30 '20 at 22:37

2 Answers2

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I must have some fundamental misunderstanding (probably about back-emf)

Yes, the concept of back emf doesn't work very well for induction motors.

The diagram below is used to explain the operation of a three-phase induction motor. It represents one of three phases of a wye-connected motor.

enter image description here Image from Fitzgerald, Kingsley, Umans, Electric Machinery, 4th ed

R1, X1, R2 and X2 represent the Stator and rotor resistance and reactance. Bm and Gc represent magnetizing reactance and the mechanism of power loss in the iron core. R2(1-s)/s is a variable resistance that represents the mechanism of converting electrical power to mechanical power. The slip, s is the slip / synchronous RPM. Note that the rotor circuit, is shown as being directly connected to the stator. An induction motor is like a transformer. Here it is assumed to be a 1:1 transformer and the secondary winding has been omitted.

When the motor rotor is not turning, the instant after power is applied to start the motor s = 1 and R2(1-s)/s = 0. The current is determined by R1 plus X1 plus the parallel combination of Gc, Bm and X2 + R2. The current is quite high, typically on the order of six times the motor’s full-load running current. Since the current in the magnetizing branch Im is considerably less than the motor’s full-load current most of the stator current also flows in the rotor. In actuality the effective stator to rotor turns ratio causes the rotor current to be perhaps 10 to 100 times the stator current.

The current in the rotor causes torque even though R2(1-s)/s = 0 and the electrical power into the motor is all converted to heat when the rotor is not turning. Remember mechanical power require both force and motion. No mechanical power is produced until the rotor begins to turn.

When the rotor begins to turn, slip decreases and R2(1-s)/s increases. The current decreases, but since R2(1-s)/s is no longer zero electrical power is converted to mechanical power. When the motor approaches the normal operating speed, s is about 0.03 and R2(1-s)/s is about 32 x R2. At that point most of the electrical power going to the rotor is converted to mechanical power and only about 3% is lost in R2.

See also my answers to the following:

Why does an induction motor draw more current when the load is increased?

When load increases in rotor of induction motor how does stator draws more current?

Torque-Speed Equation for Induction Motor

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What you're missing is that the rotor has inductance and resistance. So the initial rotor current is low, the rotor sees a high frequency field that can't excite much current against the inductance). Also, the initial phase relationship between rotor and stator is such that the lower the rotor resistance, the less torque you have at startup. At the same time, the lower the rotor resistance the less losses you have, and the closer the run speed is to the motor's synchronous speed.

Here's a picture (from this page) that shows the torque vs. speed for an induction machine.

enter image description here

TimWescott
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    I find this answer problematic. particularly in the assertion that the initial rotor current is low. –  May 29 '20 at 21:16