# Over correction

1. Feb 17, 2014

### QwertyXP

I am looking for an explanation of why over correction of inductive loads can result in large circulating currents. Many sites give some form of vague verbal description but I have not been able to comprehend this phenomenon.

Can anyone explain it mathematically or provide a link that describes this concept in a mathematical manner?

2. Feb 17, 2014

### jim hardy

Hmmm. Intuitively,
If you drive power factor to unity, have you not made XC = XL, which is a resonant circuit?

Here's as good an explanation as i could find with a cursory search....

http://www.lmphotonics.com/pwrfact.htm
Any help ?

Sounds like "over-correction" means using a capacitor that draws close to a motor's magnetizing current, which is roughly the same as the motor's no load current . That's to stay away from resonance.

When disconnected from the source, the induction motor is an induction generator until its field dies away. If its inductance resonates with the capacitor recall that current and voltage depend on Q of the circuit, which is X/R and can be quite high.

Is that on the right track?

old jim

3. Feb 18, 2014

### QwertyXP

I don't remember much about Q.

Yes, if power factor is unity and the load is purely inductive, we have a resonant circuit. I had the understanding that at resonance, capacitor provides only as much reactive power as is required by the inductor. But now, looking at it from a different perspective, net impedance of the resonant circuit is zero so there should be an infinite current circulating in the circuit (I=V/Z)! Did I go wrong somewhere?

If, with a purely inductive load, no problem occurs in a normal situation at pf=1, why should any problem occur when the motor has been switched off and resonates at a lower frequency because of overcorrection?

4. Feb 18, 2014

### Baluncore

The difficulty here comes from the misleading term “circulating current”. Unfortunately, a mathematical explanation is more difficult than a literal description.

An induction motor can be operated as a generator by driving it faster than synchronous speed. When generating power to the grid, the grid will provide the synchronous excitation needed. Without a grid connection, capacitors can be used to self excite the motor as a generator. The frequency of the power generated when using capacitive “self excitation” will be determined by the resonant frequency of the motor inductance with the parallel capacitance. In order to change from motor to generator, the motor must be driven faster than the synchronous speed of the excitation LC resonance.

A reactive current has a phase of 90° to the supply voltage. The sign of that current is unimportant, it is still a reactive current. The inductive current “circulating” between the power station and the motor can be minimised by using parallel compensation capacitors to “neutralise” the inductive current. Zero circulating reactive current on the supply lines can be achieved when the inductive and capacitive currents are equal and opposite. That occurs when the LC circuit is resonant at the supply frequency.

When a compensated induction motor is connected to the supply, there are two separate reactive currents flowing. Those two currents are determined by the supply voltage divided by the inductive reactance of the motor, and the supply voltage divided by the capacitive reactance of the compensation capacitors. Those currents tend to cancel on the supply lines.

Overcompensation of an inductive current results in a capacitive current, which is not a big problem while the supply is connected, as the overcompensation of one machine will cancel part of the undercompensation on another. By overcompensating, the capacitance will be greater than required, and so the resonant frequency of the motor inductance with the compensation capacitance will therefore be lower than the supply frequency.

When the supply is disconnected however, there will be a short period of self generation while the motor decelerates from supply synchronous to below the LC synchronous speed. That deceleration will generate power at the resonant frequency of the motor inductance with the capacitive compensation. The self generating resonance can have a very high Q which may result in very high current in, and voltage across, the LC tuned circuit. That is not a reactive “circulating current” between the power station and the motor. It is a “circulating current” in the unloaded LC resonant circuit.

The mechanical energy available while the motor and mechanical load decelerate from mains synchronous to LC synchronous speed will be dumped into the unloaded LC circuit beginning at the moment of disconnection. The voltage and current can be very high which can damage both the motor insulation and the compensation capacitors.

That is the reason why overcompensation of an induction motor must be avoided.

5. Feb 18, 2014

### jim hardy

No, you're doing just fine.

Let's try a thought experiment to clarify the concepts, and see if somebody whose math is fresher than mine will elucidate further.

Initial condition:
Motor running with zero mechanical load so that its drawing basically just its magnetizing current,
meaning the motor looks like an inductor with not much resistance .
Capacitor in parallel with motor sized so pf= 1.0, ie Xc = XL.
Motor's counter EMF is in parallel also, and very nearly equal to supply voltage.

The supply lines set the voltage across all parallel elements.

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Now remove the supply line. Voltage is no longer restrained, is it?
Energy is available from motor's rotating inertia, and as it's slowing down it acts as an induction generator.
And since the capacitor provides the magnetizing current, the voltage can rise to whatever is the motor's counter-emf capability.

I think that's why they want the capacitor sized slightly too small to provide full magnetizing current. That assures voltage decays rather than builds up.

And i think "resonance" might be a red-herring here.

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In AC circuits we're taught that Q = X/R, also that it's the ratio of energy stored to energy dissipated per cycle.
It's tempting to think, as i did at first, that the circulating current goes high simply because of a high-Q resonant circuit.

But upon working this in my alleged mind i changed my thoughts----
If the "induction generator" cannot find enough exciting current to maintain counter-emf, voltage will decay to zero from lack of excitation.
That's why the notes cited earlier have such strong precautions against overcorrecting either an induction generator or a motor that has an over-running load. If the machine is spun by its load, it becomes an induction generator, and if it can find enough excitation, you'll have a positive feedback mechanism: more voltage = more exciting current = more voltage yet... just like any other self-excited generator

So right now i'm thinking it's more a matter of feedback math than of resonance math, and the point at which you reach instability from feedback theory just happens to be the same point at which it's electrically resonant. [insert red herring icon]

I may experiment with a pair of washing machine motors and some run capacitors when get back home.....

Meantime, perhaps a genuine 'power systems' PF member could correct me.

old jim

6. Feb 18, 2014

### jim hardy

@ Baluncore:

I understand the statement, but i'm having a difficult time with the "very high voltage" as one would expect in a linear circuit.
Will the motor not saturate, taking us away from sinewaves ?

Is that the source of the high voltage and current, huge peaks as motor switches between peak values of flux as opposed to ordinary current and voltage gain of a linear resonant circuit ??

Or is it more of a feedback phenomenon as i proposed ?

Thanks for your insight -

old jim

7. Feb 18, 2014

### Baluncore

I probably know less about this than you. I would like to do the disconnection experiment with an overcompensated three phase motor.

Firstly, nit picking, Xc = XL is not possible. I believe you mean Xc + XL = 0
Secondly, with a pf = 1, Xc = – XL , if the supply is removed there will be no generation. When an induction generator is at or below synchronous speed, generation is not possible. It effectively short circuits it's self excitation.

Induction generator current does not begin to rise until a speed greater than synchronous is reached. The generator speed is then limited by the available mechanical power input. However, when the supply is removed from an overcompensated induction motor, it immediately switches to become an induction generator with a very significant over-synchronous speed. I think the problem arises from that sudden transition to an extreme generation mode that is not normally attainable with an induction generator.

Conceptually, the available rotational energy is dumped into the resistive component of the LCR circuit. There is also circulating energy in the LC component. I suspect that the waveform will be distorted and that saturation will result in a loss of voltage control and so high voltages.

Without a better understanding of the behaviour of over-driven synchronous generators I find it difficult to predict exactly what will happen.

8. Feb 18, 2014

### jim hardy

Well you're not nit-picking , just pointing out my sloppy wording and inattention to details like signs. Thank you.
And i think you're ahead of me on this subject.

That's logical, for the oversize capacitor would seem to place resonant frequency below line frequency.

We're in agreement there .

This is an interesting phenomenon and i haven't yet wrapped my pea-brain around it.
From http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/content/ct_168468.pdf
Clearly i have some reading to do on "self excited induction generator"

Did you meany asynchronous generators?

A synchronous generator that's heavily overexcited offline makes a highly distorted waveform and the induction heating melts holes in the stator iron.
Picture courtesy these guys: http://www.exponent.com/Generator-Core-Failure/

We wrecked a big one ca 1981.

I don't know yet whether a self excited induction generator can do the same.

Cheers, and thanks !

old jim

Last edited by a moderator: Apr 18, 2017
9. Feb 21, 2014

### jim hardy

Did quite a bit of searching on this subject, and information is sparse.

In principle one could make an induction machine be self exciting by attaching capacitors that would "overcompensate" a motor. I actually found a paper where students did exactly that with some small machines built for early wind turbines. The machine walks up its saturation curve just like any other self excited machine, and saturation is what stops the increase making the machine stable..

If i can find that link again, will post it.

A manufacturer of induction generators has this on his site:
http://www.usmotors.com/TechDocs/ProFacts/Induction-Generator.aspx

So what i'm taking away is that one can design a stand alone induction machine that's self excited through external capacitors
but garden variety motors are not built for that service, they need somewhat different magnetics.

old jim