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After researching a while on the subject, I noticed that several sources state that the back-emf constant of a BLDC motor satisfies the following equation:

$$ e = K_e.\omega_m $$

where:

  • \$e\$ - Back-emf
  • \$K_e\$ - Back-emf constant
  • \$\omega_m\$ - Angular velocity (in mechanical degrees, not electrical)

This makes perfect sense to me in the case of a DC motor, where we only have one phase. However, things get confusing to me when trying to understand the above equation for a three-phase BLDC.

When applying this equation in a BLDC context (as I've seen a couple of sources do), \$e\$ is the back-emf in which phase? Or is it an average value, or something else?


I'm trying to get answers from the book Analysis of Electric Machinery and Drive Systems but I have little background in this area, therefore hindering my reading substantially.

Any help would be appreciated, thank you.

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    \$\begingroup\$ Have a read about the six step commutation pattern for a BLDC motor. Basically only two phases contribute torque/produce a back EMF at any time. The last phase is disconnected by the motor controller (or brushes) during this period. Measuring the peak EMF between any two of the motor wires will give you the motor back EMF for a given speed. \$\endgroup\$
    – Jon
    Jun 22, 2015 at 21:39
  • \$\begingroup\$ Ok, things are clearing up for me. But then is this expression invalid for other types of commutation? (for instance, if I use FOC there will be moments when all of the phases are connected). \$\endgroup\$
    – JLagana
    Jun 22, 2015 at 21:48

1 Answer 1

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When talking about BLDC motors (and by BLDC I assume you mean a brushless, synchronous motor with a trapezoidal phase back-emf), there is no standard in the motor world as to whether \$ e\$ refers to line to line back-emf or phase back-emf, or whether it refers to peak to peak or RMS values. Similarly, there is no standard for what \$ K_e\$ means exactly. Often \$ K_e\$ means peak line-to-line back-emf constant. Good datasheets will explicitly tell you but many won't.

Your confusion between the back-emf constant for brushed DC motors and BLDC motors is common. Very little is published about the difference and many, many people just assume that what works for brushed DC motors also works for BLDC motors. I can explain the difference to you but first we need to understand the back-emf constant for brushed DC motors.

Consider your equation \$e = K_e * \omega_m\$. In a brushed DC motor there are a number of coils that produce close to a triangular-shaped back emf. But the commutator effectively acts as a rectifier to create a DC back-emf. So in your equation, \$ e\$ is a DC value.

In a brushed DC motor you can also say that \$T * \omega_m = e * i\$, where \$ i\$ is the DC current. If you substitute your equation into this equation you get \$ T = K_t * i\$, where \$K_t = K_e\$. All this tells us is that the back-emf constant and the torque constant of an ideal DC motor are equal.

When you start looking at brushless motors (generally with 3 phases), you have to take a couple of things into consideration. First, we don't have a commutator that serves to rectify the back-emf to DC. A brushless motor could have a trapezoidal back-emf or a sinusoidal back-emf or something in between. Second, our current isn't DC into a brushless motor. Typically a brushless motor will either be sinusoidal or it will be more of a square-wave that is positive for 120 electrical degrees, off for 60 electrical degrees, negative for 120 electrical degrees, and then off for another 60 electrical degrees.

As it turns out, if you look at a BLDC motor that has a trapezoidal back-emf and is controlled with a square-wave current as described in the previous paragraph, the equations for the back-emf constant and the torque constant are exactly the same as for an ideal brushed DC motor. The only caveat is that \$ K_e\$, \$ K_t\$, and \$ e\$ must be defined as in terms of their peak line-to-line values and \$ i \$ as a peak line current.

These equations do not necessarily hold for other back-emf and current waveforms. For example, you cannot say that \$K_t = K_e\$ if you have a sinusoidal back-emf and sinusoidal currents, assuming we are still talking about peak line-to-line back-emf and peak line current.

You can read more about these relationships in James Meavey's thesis found here or in Hendershot and Miller's book Design of Brushless Permanent-Magnet Machines.

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  • \$\begingroup\$ Thank you so much for your time and effort. So, when applying this equation to the trapezoidal, six-step commutation, BLDC context, \$e\$ means the peak line-to-line value of the back-emf? (I believe this is usually the setting they use when creating the datasheet, correct?) I ask this because I'm trying to get estimates for the linked magnetic fluxes between the rotor magnets and the stator windings. One more question, if you have the time: I've read in "MORETON, P. Industrial brushless servomotors" that if you want the per phase back-emf then you should divide \$K_e\$ by 2. Is this correct? \$\endgroup\$
    – JLagana
    Jun 23, 2015 at 22:12
  • \$\begingroup\$ @JLagana you are correct. The per phase back-emf can be found from the line-to-line back-emf by dividing by 2. This assumes the motor is Wye connected. If Delta connected, then line-to-line and phase are the same. \$\endgroup\$
    – Eric
    Jun 24, 2015 at 12:51
  • \$\begingroup\$ Got it, thanks! And what about the other question in my comment? \$\endgroup\$
    – JLagana
    Jun 24, 2015 at 14:27
  • \$\begingroup\$ Oh, sorry, missed that one. I would say it is pretty typical for e to mean peak line-to-line back-emf but I've seen it mean RMS as well. \$\endgroup\$
    – Eric
    Jun 24, 2015 at 16:50
  • \$\begingroup\$ @Eric If E is peak line-to-line, then how come Gieras (eq 6.2 on book page 228) defines the motor model as $$V_a = 2I_aR + E_{ll}$$. Using Va (per-phase) and E (line-to-line) seems incongruous \$\endgroup\$
    – techSultan
    Feb 11, 2018 at 5:07

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