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I was thinking about nameplate parameters of BLDC motors/PMSM and wondered what happens to those values if I operate the motor at a lower voltage as rated.

Are my following conclusions correct?

  • If one reduces the voltage to the half of the nameplate value, Mr. Ohm tells us that the starting current is also halved. So I expect the starting torque to be halved as well. The same can be expected for the pull-out torque.
  • If the voltage is halved I only can reach half of the rated speed without field weakening.
  • The maximum power output is quartered because the voltage and the current are halved or the maximum speed and torque are halved.

Thanks for your help.

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  • \$\begingroup\$ @BrianDrummond If you supply the motor with max available current, that is limited only by the staror resistance in real life on few k€ PMSM you get the magnets ot of it. \$\endgroup\$ Commented Feb 5, 2016 at 17:24

2 Answers 2

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  1. No, as long the driver can supply the rated current, the torque is the same. The resistance is pure loss, the current regulator has to apply more voltage, since the back EMF voltage at zero speed is zero, then it should be no problem to supply the rated current at start (n=0).
  2. Yes, as the V_BEMF rises when motor is turning there is no longer possible to inject the current, when V_BEMF = Vcc
  3. P = M * omega, torque remains as is, the speed (omega) is halved, so the power is halved.
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  • \$\begingroup\$ But the current is limited by the resistance isn't it? so the maximum current at half of the voltage should be halved as well? \$\endgroup\$
    – c-a
    Commented Feb 5, 2016 at 16:09
  • \$\begingroup\$ @c-a No, the current is not limited by resistance, the current is fed by motor driver which is a current source. The resistance smaller is better the motor is. V_driver = V_motor_bemf + I_set * R_winding; V_motor_bemf = k * N_motor; you find k parameter in motor data. \$\endgroup\$ Commented Feb 5, 2016 at 16:43
  • \$\begingroup\$ @c-a No, it isn't. The winding resistance is purely a parasitic effect that reduces efficiency, it isn't "required" for limiting current. If you were to make a motor from superconducting wire it still wouldn't pull infinite current. What limits the current is the motor controller: it tries to maintain the current within a specified value by varying the effective voltage. The voltage is varied with pulse width modulation. \$\endgroup\$
    – jms
    Commented Feb 5, 2016 at 16:44
  • \$\begingroup\$ ok, but there is still something I don't get: if I feed a sinusoidal signal (generated by a PWM) with the respective maximum amplitude into the motor, there must be a difference in the starting behaviour if I supply different voltages since the maximum amplitude is different, isn't it? \$\endgroup\$
    – c-a
    Commented Feb 5, 2016 at 16:53
  • \$\begingroup\$ @c-a Assuming that the motor controller can measure the supply voltage and/or motor current, it can compensate for a lower supply voltage by raising the duty cycle (so that the motor current remains the same). So no, there wouldn't be a difference. \$\endgroup\$
    – jms
    Commented Feb 5, 2016 at 16:56
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  1. Yes, but only if the controller allows it. Stall current is usually much higher than normal operating current, so the controller may limit startup current to avoid blowing things up.

  2. Yes. Speed is directly proportional to voltage.

  3. Yes (if you ignore 'iron' losses, which are usually a small proportion of total power consumption). Maximum power output is achieved when the motor is loaded down to 50% rpm, then half the input voltage and power is lost in the winding resistance.

enter image description here

enter image description here

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  • \$\begingroup\$ Sorry to say, but I think you are not an expert in BLDC/PMSM motors. \$\endgroup\$ Commented Feb 5, 2016 at 17:19
  • \$\begingroup\$ I'm a bit confused by that graphs. They look more like those for "real" brushed DC motors. \$\endgroup\$
    – c-a
    Commented Feb 6, 2016 at 23:06
  • \$\begingroup\$ Primary characteristics of a PMDC motor are the same whether it is commutated mechanically or electronically. electronics.stackexchange.com/questions/210471/… \$\endgroup\$ Commented Feb 7, 2016 at 2:09
  • \$\begingroup\$ ok sounds reasonable :) thank you for your effort! \$\endgroup\$
    – c-a
    Commented Feb 8, 2016 at 10:44

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