How to compute dq current in a delta-wound motor?

Hello,

I am currently using the Three-Phase PMSM Drive example to model a Field-Oriented Control(FOC) Delta-Wound PMSM. I looked under the PMSM FOC block mask, but I could not understand the method it use to compute dq current. Why should the line current be divided by √3 and why should I add a -20° offset to the electrical angle?

Could you please help me figure it out?

Thank you in advance for your support.

Best regards

Answer 1

[Sabin]

From the control design point of view there is no difference between wye or delta windings. However, some inputs required by the control algorithm are not directly available. In wye configuration we measure directly the phase current. In delta configuration we cannot measure the phase current, we can only measure the phase-to-phase current which we need to convert to phase value (we can do this by diving it by sqrt(3)). If we look at the current waveforms in wye and delta configuration, we can observe that there is also a phase shift between phase and line waveforms. We can correct this by adding an offset to the electrical angle. If we don’t add the offset to the electrical angle, we will not get the expected torque in delta configuration.

[子瑜]

Hi Sabin,

Thank you for your answer.

In my point of view, the offset should be -30° rather than -20°. Am I wrong? Or it doesn't matter?

Best regards

[Sabin]

It is correct that there is a 30 degrees phase shift between phase and line currents. However, this does not directly translate to 30 degrees offset in the electrical angle. The electrical angle has an offset of -120/N degrees where N is the number of pole pairs. In the Three-Phase PMSM Drive example N=6 and therefore the offset is -20 degrees. I expect -30 degrees will still somehow work for N=6 but will start deviating for higher values of N (e.g., something like N=32).

[子瑜]

It helps me. Thank you so much!

Answer 2

[TED MOSBY]

Hi,

In a 3-phase circuit, the relationship between a line current and phase current is:

iLine = (√3)*iPhase + 30° phase displacement

The magnitude factor √3 and the 30 ° phase displacement are the direct result of subtracting two equal 120 °‑spaced phasors (Kirchhoff’s current law at a delta junction).​ Also as each line current leads its corresponding phase current by 30 degrees thus the angle correction has to be done.

Hope this helps!

[子瑜]

It helps me. Thank you so much!