AC Induction Motor works in low-speed area 2

[ecenus]

I am using a VFD to drive a 3-phase AC Induction motor. A v/f scalar control method is used.

Many textbooks define such motor working range is usually from 8 or 10hz to, say, 60hz. I think that is because the torque is insufficient to drive the motor at the low-speed range. So, I would like to use "torque boost" to make the motor run in the low-speed area. That means, I will control the voltage by adjusting the v/f profile when the motor is commanded to run at low-speed. But, people say, usually the v/f ratio is kept the same when the motor runs.

So, experts in this forum, please clarify my confusion: how do you make the motor run at the low-speed by keeping a constant v/f ratio?

Your advice and ideas are highly appreicated!

 

[itsmoked]

People are not always correct. Well the truth of the matter is that you don't keep the V/f ratio when using torque boost.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Skogsgurra]

There is the complex, correct explanation - and there is the shorter, simpler explanation.

The simpler one is not wrong, but it doesn't go into what happens in the rotor. It just covers stator current and stator field.

Simply put. The motor needs a magnetic field to work. This field is created by current in the stator winding. Current alone. Nothing else.

To drive an AC current through a pure reactance (inductor without any resistance) you need a voltage that is proportional to the frequency, hence the constant V/Hz in a scalar drive.

But, in real life, there is also resistance in a winding. This resistance needs to be overcome. So, at very low frequency, you need voltage to cover the resistive drop. That's why there is a residual voltage (the "boost" voltage) also at very low frequencies.

As frequency increases, the voltage needed to overcome the reactive part of the impedance gets higher and higher and the small portion needed to cover the resistive part gets more or less insignificant in comparison. The resisitive part also is at a right angle to the reactive part, so it doesn't contribute much to the total voltage (the "hypotenuse" in the voltage triangle).

That's the simple explanation. Much more can be said. But I leave it there. others will surely fill in what I left out.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[ecenus]

Skogsgurra-
Thank you for your detailed answers.

I am still trying to iron out a weird thing happening to my motor: When I try to start the motor, I have to give a higher v/f ratio than the ratio at the steady state so as to turn the motor; Otherwise, if I keep the steady state v/f ratio to start the motor, I not only failed in starting the motor, but the very high current makes my inverter trip out!

That totally confuses me as I know the motor might not turn at low v/f ratio initially but I can not figure out why the stator current goes so high to trip out the inverter so quickly. Please note that when I use higher v/f ratio, I can start the motor at the same acceleration rate as the scenario when the steady-state v/f ratio is used. But, with steady-state v/f ratio, I can not turn the motor at all although I give it at a very low acceleration rate.

Now, I used the method of start the motor with the high v/f ratio, then switch to steady-state v/f ratio after the motor turns...Now motor can run but I think I am shooting in dark. I can not explain why it is like that in the starting process...

 

[itsmoked]

I'm not understanding your conflict. A motor that doesn't turn is going to sit there drawing a huge current limited only by the motor's stator resistance. The drive will trip. As it should. Why does this confuse you?

The motor's static losses allow you to raise the V when the motor is at a very low/zero speed without rapidly overheating the motor as the motor is not going to be loaded like when running near full speed.

The increased V/f only occurs at the low speeds generally then goes away or becomes negligent.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[ecenus]

Thank you itsmoked.

Maybe I did not put it clearly.

The scenario is when I use the steady-state v/f ratio to start the motor, my inverter trips out directly(within 400ms, by then the motor did not turn at all); but, when set a higher v/f ratio, with the same acceleration rate, the motor can turn. Once it turns, I switch back to steady-state v/f ratio...

My confusion is why the higher v/f ratio, which enables higher voltage, does not trip out my inverter(?)

 

[itsmoked]

Hi ecenus.

Question: Does the motor, when a higher V/f ratio is selected, start turning immediately?

Your 400ms is almost 1/2 a second. For a motor this can be an eternity. I have a 5hp single phase motor that starts to put out gray smoke in less than a second if it doesn't start turning.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[ecenus]

Itsmoked-

Thank you for your prompt reply.

The motor does NOT start turning immediately when I use a higher v/f ratio. It takes about 2-3 seconds before the motor spins. The acceleration rate is 0.23Hz/sec. Werid...

BTW, my motor is 125hp 3-phase AC Induction Motor...

 

[itsmoked]

Interesting indeed. A huge amount of stuff goes on inside VFDs. I wonder if the VFD changes alarm limits when you change the V/f setup. Do you know what brand and model number it is?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[ecenus]

The inverter is from Semikron and the rest of board, i.e. DSP board is made by me and a friend of mine. The VFD's current limit is 450A which is the hardware limit that could not be changed by the code...BTW, I use TI TMS320L2407A chip and download the TI's code (with v/f profile) from their website as a reference, which I guess should not be a problem as well. I used that successfully driving a 20hp 3-phase induction motor...