# Brushless motor specs to maximize stall torque

I'm interested in a brushless motor having a high stall torque (not much in RPM).

What specifications of a BL motor would give a fair indication of its stall torque, provided that

• voltage is fixed ( 12V )
• current is also fixed ( say 10A max )

I'd like to identify quickly from a list of motor specs (found on the Net) what to expect from a torque point of view.
For instance, number of poles, kv, weight, diameter, length...

## 4 Answers

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.

• Thank you for this clear, detailed and to-the-point "in a rush" answer. – Ring Ø Dec 5 '11 at 8:40
• Btw, where is R in the equation? – Ring Ø Nov 7 '13 at 7:44
• @ring0 That took a long time :-) - Hmm - R seems to be out to lunch. But, id and iq will relate almost wholly to Vin and R for stalled torque. The rest seems to apply OK. – Russell McMahon Nov 19 '13 at 6:39
• More turns: higher flux and therefore pulling strength at cost of higher back emf which will drown driving voltage once speed starts to rise. However, to avoid high I^2R heat losses, you may need to use thicker wire (higher volume/weight/cost). Basically, lower RPM means the stator coils will act more resistive than inductive.
• More rotor pole pairs: closer distance to "pull in". Consider a hybrid-type stepper motor as an extreme case.
• Larger diameter: take advantage of mechanical "leverage" of having the stator/rotor interaction occurring far from rotation axis. For a given motor volume, this video mentions that torque rises linearly with length, but quadratically with diameter. Therefore, for a given volume, you would want a "ring" motor, or at least a "pancake" motor.

See "torque motors" or "direct drive" motors, e.g. from Allied Motion, Kollmorgen, Moog, etc. They utilize all three of the above, resulting in a speed/torque curve with relatively flat, high torque in a low rpm region that quickly drops off as speed increases. "Hub motors" commonly used in e-bike applications are similar in design. From an Allied Motion torque motor: As far as voltage goes, it doesn't affect the efficiency at the motor level (though maybe at the supply level) assuming the copper volume is the same and the copper thickness is "right-sized" for the voltage/current. To prove this, compare I^2R (heat loss) and NI (proportional to stator flux) for a given (V, I, R) to (2V, .5I) applied to 2x the number of 1/2 cross-sectional area windings (resulting in same copper volume), which will result in 4x the resistance.

P.S. -- Feel free to correct me if applicable.. I am still learning about this motor stuff.

• Here's another example of a "ring" motor: thingap.com/high-power-density. Note they emphasize power density (high surface area/volume allows for high heat dissipation) and torque density (copper and magnet volume all concentrated in thin ring far from rotation axis). – abc May 14 at 22:08

Trying to guess stall torque from other parameters is not a good idea. Good specifications will tell you the stall torque at some fixed current. There are too many tradeoffs in motor design that you can't reasonably infer this parameter from other single operating point parameters.

• Thank you for a rational comment. However I must admit the other comments offer some interesting - while less formal - digressions. – Ring Ø Dec 6 '11 at 9:58

In your question you asked for a motor with a high stall torque. This property should be given by the manufacturer, i.e. it is a specification. However, diameter, the motor constant are generally proportional to torque capabilities in the same motor family. In addition, you get more torque with additional current.

In general however, when specifying a motor you want to know its operating point, i.e. its speed and torque. You generally get this from the motor's characteristic torque-speed curve. Torque and speed are linearly related in an electric motor. Generally this curve is defined by the stall torque and the "no load speed". The "stall torque" is the torque at zero speed. The "no load speed" is the speed with zero torque resisting the motor's rotation. These should be given in the motor specification and they define the torque-speed curve: The operating point is somewhere on the torque-speed curve. There are many ways to obtain the operating point. However, since you've given me the electrical power, it can be obtained from it, the motor's efficiency and the torque-speed curve. The mechanical power is the a quadratic function of the speed and the integral of the torque-speed curve. You've specified the electrical input power, i.e. 12V*10A = 120W. The motor turns this electrical power into mechanical shaft power with some power is lost to heat. Typical DC motors are more than 85% efficient in this energy conversion process, so as a rough approximation, say you get out 100W of mechanical power. The efficiency should given by the manufacturer.

Thus, there are two possible speeds for any one torque. But the torque or speed can be found using these equations. The power is these equations is the mechanical power, not the electrical power in.

All this only applies to a steady state of the system and doesn't consider the transient portion of the motor's motion.

You seem to require a lot of torque and not much speed, therefore, I would seriously consider a geared motor. If you gear down the output speed, you shift the torque speed curve by reducing the no load speed and increasing the stall torque. This can lower the cost of your overall system by reducing greatly the size of motor required. Its is not practical to get a lot of torque from an direct drive system. You know you've chosen a motor badly if it is not running near its peak power, i.e. at half the no load speed. Why buy a really big expensive powerful motor then run it at 1% power output!? This is silly!! Instead by a motor that will run near its maximum power output but be geared down to deliver the same torque. However, you have not specified what your mechanical requirements are. You should start there.

• Some interesting input. I'm indeed interesting in the zero-speed torque. – Ring Ø Dec 6 '11 at 10:24