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I am curious about the impedance of the three phase BLDC motor when it is in operation. If the speed of the rotor of a BLDC motor is proportional to the frequency of the three phase current, and the reactance of the stator windings is also proportional frequency of the three phase current, then wouldn’t the impedance be proportional to the speed of the rotor (as inductive reactance of the stator windings can be defined as 2*pifL)? In this case, wouldn’t the motor draw less current as the RPM increases, but only due to impedance (not considering back-emf or other effects).

Or is the increase in impedance (with RPM) balanced out by some other factor in the design of a BLDC motor? If so, what exactly balances it out?

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the speed of the rotor of a BLDC motor is proportional to the frequency of the three phase current

Beware of the above phrase: there are two frequencies involved. The most important is the synchronous one, i.e. the commutation of the phases carried by electronics in order to generate a rotating magnetic field. For example, a BLDC motor rotating at 1 turn/second could have a driving frequency of 6 Hz (six-step for one turn). For this frequency, I would not say that the speed is "proportional", I would say that it has to match exactly. Try to vary the frequency while the motor is running, and bad things will happen (big vibrations, big current consumption and spikes, heating). In reality, the frequency of the phases is dictated by the motor speed, not the contrary. To vary the speed of the motor you vary the voltage to it (using the duty cycle of the PWM).

The other frequency involved is normally around 20 kHz, it's the frequency of the PWM used by the output bridge, but it has nothing (or very little) to do with the speed of the motor. The inductance of the motor has a lot to do with this frequency.

When the speed of the motor increases, it draws less current because:

  1. The BEMF increases and contrasts the power supply.
  2. The commutation times decrease, so there is less time to pump current in the coil. As soon as the current start to rise, it's time to give up to pass to the next coil. A lower inductance lets the current rise more quickly than a higher inductance.
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Increasing the voltage in proportion to frequency increase balances out the increase in impedance.

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  • \$\begingroup\$ Thanks for your answer. If you increase voltage in proportion to the frequency, wouldn’t the power output then depend on the frequency (and therefore RPM of the rotor)? \$\endgroup\$ – Ani Vasudevan Nov 23 '17 at 7:00
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    \$\begingroup\$ power = rotational speed * torque. So for the same torque (i.e. the same current), the power output inevitably depends on frequency. \$\endgroup\$ – Brian Drummond Nov 23 '17 at 7:10
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Inductivity of the winding is proportional to the number of turns and so is the BEMF voltage, too. The voltage constant and torque constant are related in same manner. Larger is the voltage constant [Vs/rad] or [V/RPM], larger the torque constant [Nm/A].

As a conclusion we can say, compared motors with the same size: the one with higher inductivity, would have higher torque but lower maximum speed.

EDIT:

The BLDC is driven by six step pulse method, where the current is limited by a chopper. The resulting voltage will rise with speed.

The PMSM could be driven with FOC method, where the current is PWM controlled in closed loop.

With both methods the resulting voltage will follow the V/f ratio, so at higher frequencies also the voltage has to be higher.

$$I=\dfrac{V}{j\omega L}=\dfrac{K_{V/f}\cdot \omega}{j\omega L}$$

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  • \$\begingroup\$ Thanks for your answer. I see how the BEMF is related to the number of turns on the stator coils, but I am a little confused as to how that (the BEMF) relates to the impedance of the stator could while rotating. Won’t the BEMF simply reduce the current even further, on top of the increase in reactance? So considering that, how do engineers determine what inductance to have for their stator windings given their operational frequencies (of the three phase current)? \$\endgroup\$ – Ani Vasudevan Nov 23 '17 at 6:56
  • \$\begingroup\$ "compared motors with the same size: the one with higher inductivity". Actually not. More inductance means more turns, which means that you drive it at a higher voltage and lower current -- and the various increases and decreases all balance each other out, so the torque per watt dissipated in the windings, and the L/R ratio, remain -- at least roughly -- the same. \$\endgroup\$ – TimWescott Jan 10 at 1:15

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