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let's refer to the following (vector) control schemes:

abc

qd0

Qd0 is the synchronous reference frame, with d alligned to the pm magnetic field. the system is balanced and symmetric (so no 0 component).

the motor is an internal mounted permanent magnet ac motor.

for what I understood the main advantage is having vq and vd DC instead of va,vb,vc AC.

this means that at steady state I have a zero error (PI doesn't zero the error for AC? I'm not so sure about it). Qd0 scheme is more expensive due to the computational cost associated with the trasformation matrices.

Now I have this sentence on my book: ''Qd0 control is more performant since the current bandwidth does not depend on the speed''.

Why is that? I don't see why the current bandwidth is affected by the speed in the a,b,c reference frame, so I don't know wwhy this is not the case for q,d,0.

Is it because the speed (in a synchronous) pm machine is linked with the frequency of va,vb,vc (and so ia,ib,ic frequencies) ? I miss the actual connection to the bandwidth tho.

any advantage in using a,b,c? for now there doesn't seem any.

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  • \$\begingroup\$ If by "current bandwidth" the book is referring to the current in the motor, I agree with you - they would be identical in both cases. But if it's referring to the current computations to the left, then the QD0 has less computation there (assuming it is computations, and not analog summing junctions), but that is far outweighed by the Park and Clark transforms. \$\endgroup\$ – Mark Jul 11 '16 at 2:35
  • \$\begingroup\$ My understanding (which may be wrong) is that the QD0 system has its real advantage when you replace the permanent-magnet motor with an induction motor. By controlling Q and D independently, you can control both the motor torque and the rotor's field strength. \$\endgroup\$ – Mark Jul 11 '16 at 2:38
  • \$\begingroup\$ @Mark thank you mark. I was thinking: in abc the current is AC, so the overall current loop should have a bandwidth ''big enough'' to be able to control it (zero attenuation and phase lag). on the other hand the frequency of the current is the same as the speed of the shaft (it's a syncronous machine), so in a certain sense the speed of the machine influences the current bandwidth (if I want to go at x speed, the bandwidth should be more than x). in qd0 I don't really care about the bandwidth of the current loop, since it deals with DC currents. what do you think? \$\endgroup\$ – user3149593 Jul 11 '16 at 11:23
  • \$\begingroup\$ In both cases, the necessary current bandwidth is determined by speed. In either topology, the values of Va*, Vb* and Vc* would be identical for a given speed and torque. The only difference between the two is how Va*, Vb* and Vc* are calculated. So if the current bandwidth that is being referred to is in the amplifier, there would be no difference. \$\endgroup\$ – Mark Jul 11 '16 at 19:46
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    \$\begingroup\$ OK. I misunderstood. And now I know why that statement is true. With the first topology, the PID loops are trying to control the current in each phase. They are closing three loops that are oscillating at the commutation frequency. The faster the motor turns, the higher the frequency, and the faster the PID needs to react. With the second topology, the PID loop is only controlling the torque, and the commutation is hidden behind the transforms. In steady state, the second topology is basically sitting still, while the first topology is oscillating with the speed of the motor. \$\endgroup\$ – Mark Jul 12 '16 at 8:32

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