I'm trying to control the speed of a PMSM in the simulation software PSIM.

Now, I have learned that in a Synchronous Machine the steady-state speed is directly proportional to the frequency of the source, independently of the load (assuming it is in a supported range).

Therefore, I was hoping to be able to control the speed in a open-loop, by just changing the frequency using a inverter (and the voltage accordingly, to avoid flux saturation). But I have not been succesful and every article I've found so far uses Current Control to achieve speed control, which increases the complexity.

Is it not possible to control the speed of a PMSM without caring about the current?


1 Answer 1


You have described specifically how you have been unsuccessful. I suspect that the motor is drawing more current than you source can supply. The only way you can successfully operate open-loop is to start the motor at a very low voltage and frequency, 1 or 2 Hz. Once the motor is operating at a low speed, increase the voltage and frequency together at a controlled rate. All speed changes must be made at a controlled rate. The motor will draw a higher current just accelerating its own inertia if the rate of frequency increase is too great. If the frequency is reduced too quickly, the motor generate current back into the supply in braking the motor to a lower speed, so the rated of frequency reduction must also be limited.

  • \$\begingroup\$ Thanks! I have now gone slowly to 30Hz and, since my machine has 4 poles, the expected speed is 900 rpm. It is indeed around 900 rpm, but it's fluctuating at a high frequency from around 700 to 1000 rpm. Do you know what is causing this? \$\endgroup\$ Commented Jun 16, 2017 at 22:45
  • \$\begingroup\$ I recall that stability can be a problem with a PMSM, but I don't remember any of the details. Stable operation may not be possible with a load inertia exceeding some multiple of motor rotor inertia. If you look for papers with earlier dates, you may find information about open-loop control of PMSM. \$\endgroup\$
    – user80875
    Commented Jun 17, 2017 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.