Hello I have a simple question, I was reading in my Airplane Maintence book that in AC motors are frequency sensitive (I understand that) frequency changes can cause speed changes of the armature. But what I don't understand is this line: "Some components can overheat if the frequency drops and the current increases" I understand more current overheating, I don't understand how lower frequency would increase current.

Edit: Again thanks for those answers to my above question, but I guess another question I have on this train of thought(confusion) is with inductive reactance, where it opposes current at higher freq, yet dosen't use power or convert it into heat, where does it go?


An AC motor stator is an inductance. In an inductance, the lower the frequency, the lower its impedance. So having constant voltage over it, the current will rise if the frequency gets lower.


First, we need to understand the components of AC motor speed control. Their are only three components: Voltage, Current, Frequency. We only have control over Voltage and Frequency through an AC Motor Driver. However, Current is controllable, but dependent on motor Load.

AC Motor Speed control requires a Voltage/Frequency input relationship to control motor speed. The V/F ratio is different for different motors and totally depends upon the motor's rated values. Let's say you are using 400V 50Hz motor. The voltage to frequency ratio for that motor is 400/50. This can be achieved by using AC Motor Drivers. When you need to run motor at 25Hz, than you must supply voltage near to the 200V.

We get \$\frac{400V}{2} = 200V\$ because \$25Hz = \frac{50 Hz}{2}\$.

For 12.5 Hz Speed, you have to supply 100 Volts, and so on.

All motors have a minimum frequency value. You must not go below that value because driving a motor at a frequency value lower than the minimum can cause excessive current flow at the Armature winding. This is called "Eddy Current". Excess eddy current produces heat, which may burn the windings of the motor.

To solve this problem, almost all AC Motor Drivers have a function called "Slip Frequency". Slip frequency is the lowest possible frequency a drive can give to motor. It is different for different rated motors and is related to the motor's slip. Some AC Motor Drives automatically adjust this function according to rated values of motors.

If we try to operate a motor at a frequency that is lower than Motor Driver's Slip Frequency, then the Motor Driver will use the Slip Frequency to operate motor instead.

Hope this will help you.

  • \$\begingroup\$ Thank you both. I did a little reading just now and read about how at lower frequency inductive reactance is low, so I understand how more current is getting through, but what confuses me is that inductive reactance does not use power or generate heat. So how would more or less of this inductive reactance generate any different heat amount. Also if this reactance lowers voltage and then hence current, where does it go if energy (P=I * E) can neither be created nor destroyed? And then you have reactive power which seems to contradict impedence not consuming power or generating heat. \$\endgroup\$ – GW2500 Jul 29 '12 at 18:26
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    \$\begingroup\$ The apparent power S = V * I, where S arises from both real and reactive power. So if we have increasing reactive power without changing real power, we have increasing apparent power. Keeping the voltage constant, that means we are sourcing more current to maintain apparent power S. If we are sourcing more current without yielding more real power, then it is being dissipated elsewhere. \$\endgroup\$ – Envidia Mar 10 '17 at 0:42
  • \$\begingroup\$ @GW2500 Ok, so if lower frequency increases current, this current has to flow through some wire, which always has some resistance, causing it to be heated up through dissipation. The more current, the more dissipation, the more heat and the wire can melt, or the insulation can burn. \$\endgroup\$ – HarryH Jul 2 '18 at 21:29

The inductive reactance (\$X_L\$) more or less acts as a resistance within a circuit. It means that it offers a specific resistance to the circuit to which it is connected.

This resistance is associated with the imaginary component of power (the reactive power in VAr). The reactive power originates due to phase difference between current and voltage phasors, which is quite a definite case in inductors (the current lags the voltage). So this is indeed the power which is of your concern, I suppose.

This is the power which is associated with an inductor. For a detailed explanation, you can certainly have a look at Inductive Reactance.


2nd question: Where does it go?

The energy sent to inductors (and to capacitors!) is temporarily stored. Then it flows backwards, going back into the power supply. During 1/2 of an AC cycle, ideally all the energy has returned, and the coil or capacitor has consumed zero.

So, if we connect an inductor to a battery, the coil current rises and energy gets stored in the magnetic field. Next, we suddenly reverse the coil connections. This reverses the current, which then falls smoothly to zero. The magnetic field collapses, and energy flows backwards across the circuit, to "recharge" the battery. Finally, just when the current hits zero, disconnect the battery. That's a simulation of one half-cycle of AC. Ideally, all the energy that went into the inductor, has now flowed backwards and returned again to the battery.

Real AC: hook an ideal, zero-ohm inductor to an AC generator, and the generator will send energy out to the inductor, then suck it all back again, twice per cycle. (It's twice per cycle because energy is sent out and back during the positive phase, and also sent out and back during the negative phase.)

Practical effects: the coil and the connecting wires will warm up because of their resistance. We'd only get 100% energy returned if the coil and wires were superconducting. Real coils also act like resistors, like electric heaters. Also, your AC dynamo rotor will vibrate at 120Hz when trying to drive a large inductor. Twice per cycle the generator sees a load, a large current, then sees an "anti-load" from reversed current, and its shaft gets a forward kick from the returning energy. The average energy flow is zero, yet significant energy is "sloshing" back and forth between the dynamo and the distant inductor.

To eliminate this effect, add a "tuning capacitor" across the inductor, and adjust its value for resonance at 60Hz, 400Hz, whatever was your AC system frequency. Now the "energy sloshing" takes place only between the inductor and capacitor, while the dynamo sees a constant AC load.


protected by W5VO May 21 at 14:51

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