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

9 visualizaciones (últimos 30 días)
子瑜
子瑜 el 8 de Mayo de 2025
Comentada: Neil Bajaj el 28 de Jul. de 2025
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

Respuesta aceptada

Sabin
Sabin el 8 de Mayo de 2025
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.
  4 comentarios
子瑜
子瑜 el 9 de Mayo de 2025
It helps me. Thank you so much!
Neil Bajaj
Neil Bajaj el 28 de Jul. de 2025
This assumption only holds for steady state or sinusoidal commutation, correct? If truly doing FOC with dynamics, can you actually solve for phase currents (given only line currents) in a delta without making a sinusoidal assumption? I am doing FOC control on a BLDC motor with high bandwidth position control rather than speed control.

Iniciar sesión para comentar.

Más respuestas (1)

TED MOSBY
TED MOSBY el 8 de Mayo de 2025
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!

Productos


Versión

R2024a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by