I've come across a phenomena for the BLDC that I cannot explain only realize.

A BLDC motor is working as dynamic brake. At some point when all three phase are short circuited the back-emf will drop and the current in the coils will reach a steady-state value. This will lead to that the produced torque will drop to zero or close to zero.

I think this is a phenomenon only with PM motors. It seems like the short circuit current creates a flux that opposes the flux of the magnets.

Does anyone recognize this behaviour for BLDC motors?

 

Shorting together the leads of a BLDC motor for dynamic braking purposes is sometimes used to get a faster emergency stop. The motor's own phase resistance becomes the load for the motor acting as a generator. The electrical loading creates an opposing torque that decelerates the motor (which, since its leads are shorted together, has been disconnected from any drive power).

Basically, this produces an exponential decay in the motor velocity, as the loading decreases the speed, which decreases the back-emf, which decreases the current in the phases, which decreases the loading, and so on, until it is stopped.

Remember that you are dissipating all of the motor's (and load's) kinetic energy as heat inside the motor when you do this, so it is not anything to be done on a consistent basis -- do not use it to counteract a steadily applied mechanical torque on the motor. (Some people think you need external resistors between the phase leads even for emergency stop purposes.) But conceptually, if you weren't worried about the heat, it would reach an equilibrium against a steady applied torque with enough load torque at some velocity to get a steady-state condition.

You are correct that dynamic braking only works with PM motors, because you need the magnets to keep the rotor field alive when the leads are shorted together.

 

Thank you for your answer Curt.

I think I need to extend my inquiry.

If the load on the shaft of the shorted 3-phase motor is increasing rather than decreasing the motor will accelerate. It will accelerate constantly to a point when the load forces the motor acclerate very rapidly. The braking torque from the shorted motor will drop towards zero.

The torque-speed curve will be similar to that of an induction motor. The shorted motor can be seen as an inverted induction motor with the shorted stator as the induction motor rotor and the magnets working as the rotating flux produced by the stator.

 

I disagree with your analysis. If the motor accelerates, the back EMF due to the permanent magnet field will increase. If the motor leads were open-circuit, this would just appear as higher AC voltages between the terminals. With the leads shorted together, the higher voltage across the phase resistances means that more power is dissipated in the resistors, which means that more, not less, counteracting torque is generated.

This is completely different from the torque generation mechanism of an asynchronous induction motor. I don't see any mechanism here for inducing the kind of currents in the stator that you see in an induction motor rotor. The stator iron is designed specifically, through features like lamination stacks, to prevent the kind of axial-direction current flow that is the key to generating the induction rotor field.

And I repeat my concern from the previous response that the motor's thermal design will not be able to dissipate this type of heat on any type of continuous basis.

Curt Wilson
Delta Tau Data Systems

 

I'm sorry that I haven't been able to reply sooner.

Yes the counteracting torque will increase but not forever. There is a knee on the torque/speed curve where the counteracting torque will rapidly reduce.

I have actually done experiments that shows that this actually happens. A 57 mm motor couldn't hold 0.4 Nm at 1500 rpm.

This can be seen as a symmetrical 3-phase fault in the stator and I supposed that has been carefully studied earlier.

 

What is the inductance of the phases? To keep things simple in my earlier analysis, I ignored the effect of inductance. A purely resistive load would stay in phase with the rotor magnets, but inductance will introduce an L/R time constant that will cause the response to get out of phase at higher frequencies. To the extent that it is out of phase, it will not generate counteracting torque as well.

Curt

 

The motor have a 1.7 mH inductance and 0.58 ohm resistance.

I think it is definately a phase shift in the flux due to inductance that causes this. However it becomes really theoretical and I can´t find any example of this.

I do find reports on symmetrical short circuit faults and there they show how the current is asymptotically stable and the torque will drop to zero at high and moderate speed.

I just want to say that I really appreciate that you answer my thread.

Fredrik Strååt

 

With those R and L values, I calculate peak generated torque at about 1600 rpm for a 4-pole motor, declining slowly after that, but not toward zero.

Fundamentally, the applied voltage is:

Ke * w * sin(wt)

going into a series resistance and inductance. The torque is proportional to the resulting current magnitude and the cosine of the angle between voltage and current (the phase lag of the L/R circuit). It reaches a peak when the phase lag is 45 degrees, at an electrical frequency of 341 rad/sec, a little over 1600 rpm for a 4-pole motor.

 

Your calculations appear to be very accurate. Please tell me how you calculated it.

I don't think the generated torque will drop to zero but definately towards zero.

 

Does anyone know of a controller or suitable relay for this type of application - It's exactly what I need.

 

Fredrik:

The phase lag in your windings is equal to:

Lag = arctan (w*L/R)

where w (omega) is the electrical frequency in rad/sec.

The normalized magnitude of the windings in amps/volt is equal to:

AperV = 1 / sqrt(w^2*L^2 + R^2)

The torque is proportional to the frequency times the normalized magnitude times the cosine of the lag:

T = K * w * AperV * cos(Lag)

I made a table in a spreadsheet to confirm, but the math comes out like the torque/slip curve of an induction motor, even if the dynamics are a bit different, where the maximum torque is at 1 over the time constant, where there is a 45-degree lag. Like the induction motor torque/slip curve, as the frequency increases the torque tends to a pretty constant value, but well below the peak.

Remember that if you put external resistors in series, you will increase the frequency for peak torque, and move a lot of the dissipation out of the motor.

Curt Wilson
Delta Tau Data Systems

 

Fredrik:

The phase lag in your windings is equal to:

Lag = arctan (w*L/R)

where w (omega) is the electrical frequency in rad/sec.

The normalized magnitude of the windings in amps/volt is equal to:

AperV = 1 / sqrt(w^2*L^2 + R^2)

The torque is proportional to the frequency times the normalized magnitude times the cosine of the lag:

T = K * w * AperV * cos(Lag)

I made a table in a spreadsheet to confirm, but the math comes out like the torque/slip curve of an induction motor, even if the dynamics are a bit different, where the maximum torque is at 1 over the time constant, where there is a 45-degree lag. Like the induction motor torque/slip curve, as the frequency increases the torque tends to a pretty constant value, but well below the peak.

Remember that if you put external resistors in series, you will increase the frequency for peak torque, and move a lot of the dissipation out of the motor.

Curt Wilson
Delta Tau Data Systems

Hi Curt,
I'm having trouble with the unit analysis on the torque equation. What motor constant goes into that equation? Is it Km, Kt, or Ke, or possibly Kv(SI Units)?