Electronic – What happens to the torque/speed curve of a brushless motor when I increase the voltage

Rushing.
More later maybe ...

Note that "stall torque" is often used to mean locked rotor 0 RPM torque BUT you use it in the sense "dropout torque at a given speed". That's entirely fine as long as you note that some references will mean the former and not the latter.

Criticism (kind / constructive) welcome.
Written at a rush and unchecked. Better is possible.

See writer "Toper925" comment here

He notes:

• There really is no single equation that fits all states of a PMSM but this one works in general:

  • Te = 1.5p[λiq + (Ld - Lq)idiq]

Where:

• p is the number of pole pairs

• λ is the amplitude of the flux induced by the PMs in the stator phase

• Lq and Ld are the q and d axis inductances

• R is the resistance in the stator winding

• iq and id are the q and d axis currents

I'd need to read more on what he said to make total sense.

Stall torque is when torque is not sufficient to "pull in" the next rotor pole piece using the available magnetic field.

SO, I'd expect

• More pole pairs better. I'd expect better than linear gain as distance halves with doubled pairs BUT magnetic force at worst falls as distance cubed (only a considerable number of magnetic pole diameters away so not in most sensible motors), comes closer to falling with distance squared as gap falls to near pole width and at best can only approach linear at close proximity. SO more ples should give less interpole distance so ... (but pole sizes are down so ...).

• Torque = power per rev. If the power falls faster than RPM your margin is dropping until you reach the point of no pull in. At a quick glance I think that this is what this man here is alluding to about half way down below the graph. Leading to ...

• If you have a power curve you also have a torque curve as the two are related by motor rpm. (Torque = k x Power / RPM). If you have a speed-power plot for you load you should be able to overlay this on torque curve and see where load torque is > generated torque. This will be better than real world (probably).

• Lowest R should help as it allows greatest I but this is really a secondary effect for two motors with the same power at the same RPM.

• Induced flux should play an immense part. I'd expect non saturating magnetic (eg steel) cores to provide superior results EXCEPT if you can get all gaps so small that field is well maintained by magnet. Rule of thumb is you can get about 0.5 Tesla at an airgap of 1/2 a magnet diameter using a top class NdFeB magnet. Say N52? N45 won't be too bad.

• Note that the US process NdFeB magnets are cast but ground and sintered subsequently and are inferior in max possible flux to the Japanese versions. This should all be covered in the flux spec.

First, the torque constant \$k_t\$ for a DC machine is derived as follows. If you assume a constant speed and you neglect any losses or saturation, then the power into a motor equals the power out of the motor, or \$E*I = T*\omega\$, where \$E\$ is the line to line EMF, \$I\$ is the DC input current, and \$T\$ is the torque at speed \$\omega\$. From this we can say that \$k = \frac{E}{\omega} = \frac{T}{I}\$. Let me repeat the assumptions that were made for this equality to be true:

1. Constant speed
2. Lossless energy conversion
3. DC input current

If those 3 assumptions hold, then \$k\$ is a constant of proportionality. Traditionally for a brushed DC motor, we've given \$k\$ two different names, \$k_E = \frac{E}{\omega}\$ and \$k_T = \frac{T}{I}\$, where \$k = k_E = k_T\$ (assuming your units are \$\frac{Volts}{rad/sec}\$ and \$\frac{N*m}{Amp}\$, respectively).

Second, assumption #3 above poses a problem when we switch from DC motors to brushless motors because brushless motors are typically driven with either square (trapezoidal) currents or sinusoidal currents. Another issue that arises is that due to the commutator in an ideal DC machine, the EMF \$E\$ is a mean rectified EMF over all the coils in the machine. In a brushless machine, we aren't dealing with a mean EMF but rather an EMF with a waveform that depends on how the motor is built. The two ideal cases are a trapezoidal EMF and a sinusoidal EMF. Another issue that arises that in the ideal DC motor above the line to line EMF is just 1 phase while in a brushless motor, the line to line EMF be 2 phases in the case of a Wye connected brushless motor. In some cases (for example, a motor with trapezoidal EMF and driven by square wave currents), the \$k = \frac{E}{\omega} = \frac{T}{I}\$ equality still holds for a brushless motor. In other cases (for example, a motor with sinusoidal EMF and driven by sinusoidal currents), the equality does not hold.

Third, you can't calculate \$k_T\$ based on 1 data point. Typically a motor manufacturer would calculate \$k_T\$ by hooking the motor to a dynamometer and then measuring voltage, current, speed and torque while increasing the torque. They would then take a best fit line of the torque vs. current curve and the slope of that line would be \$k_T\$. This line will not go through the origin because of friction (that is, the motor requires a certain minimum amount of current to get the motor started) and many motors will not have a linear torque vs. current curve for high values of current (due to saturation). Also, generally this test is done at room temperature and it is done quickly to keep the temperature of the motor as close to room temperature as possible. The ratings in the chart above would have been performed with the motor windings at a high temperature (at the rated temperature of the insulation).

Fourth, why did I tell you all of that? Because most people when dealing with brushless motors assume things about \$k_T\$ that aren't true. Most often they treat it as the same as a DC brushed motor with a commutator. They also neglect losses due to friction and saturation. The other issue is that there really is no standard definition of \$k_T\$ in the industry. It could refer to the line to line or the line to neutral value. It could refer to the RMS or the peak value.

Fifth, if you forget about \$k_T\$ for a moment (since you aren't given enough information in your chart to determine \$k_T\$ anyway) your question boils down to "If 2 motors have the same rated torque but different input currents, what determines that input current?" Your intuition that it has to do with voltage is correct. If you increase the current from 48 V to 85 V, then in order to maintain the same input power, your current will decrease by the ratio \$\frac{85 V}{48 V} = 1.7\$. You'll see that that the current does indeed decrease by that amount \$\frac{2.4 A}{1.7} = 1.4 A\$. Motor designers have a rule of thumb where if have a motor design and you want to increase the voltage by a certain ratio, then all you need to do is increase the number of turns by that ratio and decrease the wire area by that ratio. Doing that changes the resistance by the correct amount but keeps the flux in the motor the same.