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In Field oriented control of induction motor it's necessary to do:

  1. Clark Transformation
  2. Park Transformation

According to the note (page 3): http://ww1.microchip.com/downloads/en/AppNotes/ACIM%20Vector%20Control%2000908a.pdf To calculate Park trans. (flux and torque stator current components /Id, Iq/) rotor flux angle is needed - it's described at page 4 how to calculate it.

My misunderstanding is how to do it if at these equations (EQ 1: Magnetizing current, EQ 2: flux speed) there is a necessity to have Id, Iq current which we haven't (as I mentioned to calculate it rotor flux is needed)?

What are the steps to do it exactly?

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  • \$\begingroup\$ Not clear what is the question but: You get Id and Iq through Clarke->Park transformation. The flux can be estimated with use of estimator. \$\endgroup\$ Commented Sep 3, 2017 at 18:24
  • \$\begingroup\$ Thanks for reply. My question was how to calculate rotor flux angle which is needed to calc. Clarke->Park transformation. \$\endgroup\$
    – Fasset
    Commented Sep 3, 2017 at 18:27
  • \$\begingroup\$ You get it from the encoder. Actually you need to know the slip frequency, which is \$\omega_S - \omega_R \$ \$\endgroup\$ Commented Sep 3, 2017 at 19:28
  • \$\begingroup\$ To be more precise I'm trying to write my own soft to FOC. \$\endgroup\$
    – Fasset
    Commented Sep 3, 2017 at 19:54
  • \$\begingroup\$ Too complicated, good luck. \$\endgroup\$ Commented Sep 3, 2017 at 19:57

2 Answers 2

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Start with a flux angle of 0. As there is no current in the machine at the beginning, this is correct. Then you can do the Park Transformation and calculate Id and Iq. With these you can calculate the flux angle for your next iteration. Texas Instruments described it under 3.9.1 here: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=2ahUKEwiZuuCU_YfdAhUi_SoKHTN0CewQFjADegQIBxAC&url=https%3A%2F%2Fwww.researchgate.net%2Fprofile%2FSai_Sudheer_Reddy_Bonthu%2Fpost%2FWhat_are_the_methods_to_implement_PARK_transformation_on_C2000_DSP_micro_controller%2Fattachment%2F59d64d7079197b80779a7006%2FAS%253A488146115796992%25401493394511043%2Fdownload%2FFOC.pdf&usg=AOvVaw2TEZzD_6dVsD1KEAMmm-IQ

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  • \$\begingroup\$ ehm...the texas link is broken, can you fix it? \$\endgroup\$
    – Deep
    Commented Jul 21, 2018 at 6:43
  • \$\begingroup\$ Seems as TI removed the article from their website. ST has a similar application note, the rotor flux angle estimation by the current model is descibed in 2.4: st.com/content/ccc/resource/technical/document/application_note/… \$\endgroup\$
    – Tim
    Commented Aug 15, 2018 at 11:44
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My understanding is that at the beginning the induction motor is deexcited. There is no magnetic flux in its magnetic circuit. In these circumstances you have these initial conditions: \$I_d(0)=0\$, \$I_q(0)=0\$ and \$\theta(0)=0\$. Then in case some torque is requested it is necessary to build the magnetic flux at first. During this "excitation" period the reference value of the quadrature component of the stator current is \$I^*_q=0\$. As soon as the magnetic flux in the machine achieves its nominal value the machine can produce torque i.e. you can start requesting non-zero reference value of the quadrature component of the stator current.

As far as the magnetizing current usage I would say that it is only for sake of simplicity. In case I write so called "current-speed" model of the induction motor in rotating coordinates coupled with rotor magnetic flux I have following differential equations

\$\frac{d\psi_{2d}}{dt}=L_h\frac{R_2}{L_2}i_{1d}-\frac{R_2}{L_2}\psi_{2d}+(\omega_s-\omega)\psi_{2q}\$

\$\frac{d\psi_{2q}}{dt}=L_h\frac{R_2}{L_2}i_{1q}-\frac{R_2}{L_2}\psi_{2q}-(\omega_s-\omega)\psi_{2d}\$

In case the (d,q) reference frame is coupled with the \$\hat{\psi}_2\$ we can write \$\psi_{2q}=0\$. The \$(\omega_s-\omega)\$ coefficient is the slip speed or rotor speed \$\omega_2\$. In case we take into account the aforementioned conditions the equations simplify in following manner

\$\frac{d\psi_{2d}}{dt}=L_h\frac{R_2}{L_2}i_{1d}-\frac{R_2}{L_2}\psi_{2d}\$

\$\omega_2\psi_{2d}=L_h\frac{R_2}{L_2}i_{1q}\$

We can define rotor magnetizing current in following manner \$i_{m2}=\frac{\psi_{2d}}{L_h}\$ and then we can rewrite the above mentioned equations into following form

\$\frac{di_{m2}}{dt}=\frac{i_{1d}-i_{m2}}{T_2}\$

\$\omega_2=\frac{1}{T_2}\frac{i_{1q}}{i_{m2}}\$

where \$T_2=\frac{L_2}{R_2}\$ is the rotor time constant. The last set of equations is the base from which the Microchip difference equations were derived from.

In case we can calculate the rotor speed \$\omega_2=s\cdot\omega_1\$ and measure the mechanical speed \$\omega_m\$ we can sum them and we obtain synchronous speed which is magnetic flux speed of rotation \$\omega_1=\omega_2+\omega_m\$. Integration of the \$\omega_1\$ gives us the transform angle for the Park and inverse Park transformation.

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