When talking about closed-loop motor control, people usually refer to a rotational frame in DQ terms.
In motor specifications(see image below) the value of current is more likely to be specified in the phase or line terms.
In the control algorithm, is it safe to assume that the Iq should be no bigger than the rated current – 8.2A in this case?
Best Answer
There are two questions here
1) what does rated current mean
Does rated current represent the stall, the starting or the rated power point?. This all depends on the motor in question.
From this particular motor we know the rated power (2200W) rated velocity ( 4500rpm) and the flux linkage ( 0.0976Wb) From the rated power and rated velocity the torque at this operating point can be derived: \$ P = T\omega\$ thus T = 4.66Nm
Now the provided information also states rated torque is 4.7Nm and rated current is 8.2A so there is some correlation that the stated values are for operating power not startup/stall. To double-check the the torque constant can be derived from the flux linkage
\$ K_t = \frac{3}{2}\cdot P \cdot \lambda = 0.439Nm/A\$
How many amps to produce 4.66Nm? 10.6A This is higher than the "datasheet" but in the same ballpark, thus it is reasonable to state that for this particular machine the ratings are for operation and not startup/stall.
The stall currents will be a lot higher and based upon the voltage and phase resistance, it could be around 321A.
2) What does Iq represent with regards to peak current
In short Iq represent the peak of the 3 phase currents
This can also be seen from the the algebraic representation of Idq vs Iabc
\$ I_d = \left[ \frac{2}{3}I_a - \frac{1}{3}(I_b - I_c)\right]Cos\Theta - \left[ \frac{\sqrt{3}}{3}(I_b - I_c) \right]Sin\Theta \$
\$ I_q = \left[ \frac{2}{3}I_a - \frac{1}{3}(I_b - I_c)\right]Sin\Theta + \left[ \frac{\sqrt{3}}{3}(I_b - I_c) \right]Cos\Theta \$