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EMD: IPM hysteresis control for speed driven

Builder
Builder

Hi,

 

In EMD, there is a demo for electric machine Hysteresis control. For my case, i want to run my motor (IPM) with a given excitation at a constant speed. So i changed the demo model by setting a constant speed (k=3000rpm)  to the motor shaft as shown below. The given excitation is I=80A and Phase shift=45 deg, which  means Id=Iq. However, the simualtion results show that Id and Iq are far away from what i expected. In addition, the sqrt(Id^2+Iq^2) is larger than 80 A. Also, the rotor relatvie speed and electromagnetic torque are both negative. So i am wondering if the model setups are correct for my study purpose. 

 

Thanks,

Kan

 

2018-09-17 10_48_22-Plot - 1.png

2018-09-17 10_05_46-Simcenter Amesim - [C__Users_kachang_Desktop_Simcenter Amesim_MS Amesim_Hysteres.png

 

However

5 REPLIES 5

Re: EMD: IPM hysteresis control for speed driven

Siemens Valued Contributor Siemens Valued Contributor
Siemens Valued Contributor

Hi kan,

In this demo you have to enter the angle in rad. Could you try to enter pi/4 instead of 45°?

Best regards,

Florent

Re: EMD: IPM hysteresis control for speed driven

Builder
Builder

Hi fpasteur,

 

Thank you so much! It works now. But i am still confusing about the relationship between dq current and the phase current. I got Id=Iq=69.9A, the phase currrent is following the command that has Ipeak=80A. For 45degs, my understanding is  Id=Iq=80/sqrt(2)=56.6A. it seems that Id=Iq calculated in Amesim is sqrt(3/2) times what i expected. Could you please give some comment?

 

Another question,i set the speed value 3000 RPM to the machine shaft , with the given current, i got negative torque and negative torque. It means the machine works as motor but in the third quadrant.  I tried to change the speed to negative 3000 RPM, the rotor relatvie speed becomes positive, but the torque is still negative, which means it works as generator. So i am wondering how i should setup the model so that it works as motor in the first quadrant.

 

Best Regards,

 

Kan

 

 

Re: EMD: IPM hysteresis control for speed driven

Siemens Valued Contributor Siemens Valued Contributor
Siemens Valued Contributor

Hi kan,

yes you're completly right.

The sqrt(3/2) factor is coming from the Amesim convention of the park transformation. We are using the power balance convention and not the amplitude one. You can get more detail here:

qthelp://lmsimagine.lab/ame_dir/libemd/doc/html/utils/parkabc2dq0.html

Your misunderstanding is because the controler of the hysteresis control demo does not respect the Amesim convention. If you look at the formulas entered to control the current, it is:

Iamp*sin(thetae - theta)

with Iamp, the current amplitude, thetae the electric angle (we have spoken about it in your previous post) and theta the phase shift you can control.

The Amesim convention for rotating machine is:

Iamp*cos(thetae + theta)

If you use this formula for your machine current, theta is now becoming the control angle of your machine (ie the angle between the stator and rotor field). Now With theta = 90, you have the full positive torque, a 0 torque for theta = 0 and for theta = -90 the full negative torque.

I have modified the Amesim demo to respect this convention in the attached file. You just have to take care the angle are given in degrees and no more in radiant with that update.

I will modify this demo for a next Amesim release, it will be less confusing like this.

Best regards,

Florent

Re: EMD: IPM hysteresis control for speed driven

Builder
Builder

Hi Florent,

 

Thanks again. Since i always use the non power balance convention park transformation and i did not read the help doc carefully, i got the confusion. Now it is more clear.

 

About the command current signal formula, i noticed that and then added 90 degs to my command shift angle, now i got the expected results.

 

Still, the speed setting in my case, i should set -3000rpm, am i right?

 

Best Regards,

Kan 

Re: EMD: IPM hysteresis control for speed driven

Siemens Valued Contributor Siemens Valued Contributor
Siemens Valued Contributor

Yes indeed.

This is because of the orientation of the component that impose the speed:

RotatingSpeedSet.png

The Speed is considered positive when it enters the motor, whereas the rotor relative speed is considered as positive when it comes out of the motor.

Best regards,

Florent