# Torque and Back EMF Waveforms of Brushless Motors

For simplicity, I'm going to be talking about a single stator phase.

I understand that torque is proportional to current, and that back EMF is proportional to angular velocity. However, after watching this video on brushless motors (only the first minute is relevant), it seems that the graph of torque as a function of rotor position is identical to the graph of back EMF as a function of rotor position. How is this so?

• What do you mean by identical? They obviously can't be literally identical since one is a voltage and one is a torque. – Eric Aug 16 '14 at 6:43
• I mean that the graph of torque vs rotor angle, and the graph of back EMF vs rotor angle have the exact same shape, and are in phase. – 1overcosine_c Aug 16 '14 at 9:13

Both torque and back-emf are produced by the magnetic field acting on the windings, so they are in phase because they are both proportional to magnetic field strength.

However that does not mean that they must be identical. Increased back-emf tends to reduce current flow due to the lower effective voltage across the windings, so unless phase current is held constant at every instant in time the torque waveform will not match the back-emf waveform. And since back-emf is produced by velocity, any change in rotor speed will affect it. In the extreme case of a stalled rotor, maximum torque is produced at the same time as zero back-emf!

All electrical machines can be transformed to a simplified two-axis machine. In this two-axis System you have one axis which represents magnetization (the $d$-axis), and one that represents torque generation ($q$-axis). In our case it is not important wether it is a ACSMPM(Synchronous PM) or a BLDC, as they just have different waveforms fed to them.

The torque can be represented with the help of transformed currents:

$$T=\frac{3}{2}\Psi_{PM}i_{q}$$ The back EMF is the induced voltage in the stator coming from the rotation of the rotor: $$\vec u_{EMF,dq}=\omega_{el}\Psi_{PM}\matrix[0 \quad1]^T$$ or $$u_{EMF,q}=\omega_{el}\Psi_{PM}$$ $$u_{EMF,d}=0$$

Okay I guess now you ask yourself a few questions:

• Where are the harmonic functions? The transformation is very nice, it makes the machine model look almost the same as for the standard DC machine.
• What is the $\frac{3}{2}$ in the torque equation? It is because of the transformation, it represents that you go there from an amplitude invariant transformation while calculating a power related quantity(there are power invariant versions of the transformation).
• Does this means that my the EMF is in the same ($q$) axis as the torque? Yes.
• What should I read up to understand this better? Vector control.

In an ideal BLDC motor each phase conducts for 120deg electrical. In any motor torque is directly proportional product of stator mmf, rotor mmf and sine of angle(load angle) between them. In BLDC motor rotor mmf is constant because pf permanent magnets however load angle is changing from 120 deg to 60 degree when a phase conducts. Hence you wont get the constant torque unless you control the stator mmf by controlling stator current. Thus you can say torque developed by a BLDC motor is dependent on rotor spacial angle.

If you see DC link current waveform on scope the torque wave shape will be exactly similar to that.

• I understand how three phases can be combined to produce constant torque overall, but I'm still unsure how for a single phase, the torque and back emf waveforms have the same shape in phase. – 1overcosine_c Aug 16 '14 at 9:15
• It is wrong that i would say (i have not seen the complete video link you have provided ) For getting torque of each phase just divide the product of two times per phase back emf and current by instantaneous speed over 120 deg interval. As i told you it will be same as DC bus current waveform. – chatty85 Aug 16 '14 at 18:19

The reason the graph appears to be identical for torque and backEMF is because for the ideal case $K_e = K_t$ & it is an over-simplified video, plus some external controller deals with the stator energising polarity.

In practice they are different as $K_e$ is defined as the open terminal voltage while $K_t$ is usually taken at maximum design current (and thus has some aspects of stator saturation)

To generate reasonable torque the controller MUST align the stator voltage to the rotor backEMF otherwise you end up with a phase-advance/retard which is essentially field weakening.

So the controller MUST energies the stator in-phase with the rotor (its a sync machine at the end of the day so the frequency must equally be equal).

Why are they the same profile? as stated $K_e \approx K_t$ and equally the controller firing the stator must provide a similar voltage profile to the BackEMF to maximise electrical --> mechanical conversion

The y-axis scaling however will be different.