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Now, I know d and q axis current can be calculated by phase current ia, ib, ic in a PMSM, via clarke transform and park transform.

ialpha = sqrt(2/3) * (ia - ib / 2 - ic / 2)

ibeta = sqrt(1/2) * (ib - ic)

id = ialpha * cos(theta) + ibeta * sin(theta)

iq = ibeta * cos(theta) - ialpha * sin(theta)

Now, if motor running at a steady state, id and iq should be in DC form.

But how to calculate RMS of phase current ia,ib,ic by id,iq.

I think RMS(ia) = RMS(ib) = RMS(ic) = K * sqrt(id^2 + iq^2)

What is the value of K.

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  • \$\begingroup\$ You already have to measure them: ia, ib, ic prior to transform in d,q-coordinate system. ia, ib, ic are peak values - so a RMS value is Ia_rms = ia/sqrt(2) -> K=1/sqrt(2); \$\endgroup\$
    – Marko Buršič
    Commented Mar 3, 2019 at 9:56
  • \$\begingroup\$ If a motor is labeled : rated current 1A, peak current 3A. So, I should limit the value sqrt(id^2 + iq^2) <= 3 * sqrt(2) = 4.24A ? \$\endgroup\$
    – Hyz Yuzhou
    Commented Mar 3, 2019 at 11:58

1 Answer 1

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If you have your \$i_d\$ and \$i_q\$ currents, you can perform the inverse Parks-transformation (thus you obtain the \$i_\alpha\$ and \$i_\beta\$ currents) and subsequently the inverse Clarke-transformation. Then you have the three sinusoidal currents \$i_a\$, \$i_b\$, and \$i_c\$, which have the same amplitude and a phase difference of 120° to each other.

The RMS value of a sinusoidal signal is easy to derive, and it is given by \$\frac{A}{\sqrt{2}}\$, where \$A\$ is the amplitude of the signal.

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