Understanding BLDC electric motor constants - The Kv torque fallacy

It is a common misconception that if you have two otherwise identical electric motors, one with a low Kv and one with a high Kv, the lower Kv motor will be capable of producing more torque with less waste heat.

This assumption is incorrect.

The specific torque density of an electric motor (torque per unit volume) is independent of its Kv. Similarly, the heat generated by an electric motor while producing a given torque value is also independent of Kv. Read on to see why.

Torque produced by a BLDC motor for a fixed current density

The torque capability of a  BLDC motor is determined by the average magnetic field strength produced by the stator which acts on the rotor, the average magnetic field strength produced by the rotor magnets which act on the stator and the dimensions of the rotor itself.

If we have two otherwise identical motors, one with a low Kv and one with a high Kv, then we can assume that the average magnetic field produced by the rotor magnets and the dimensions of the rotor itself (i.e. its radius and length) are the same. This leaves only the average magnetic field strength produced by the stator as a possible difference.

The average magnitude of flux provided by the stator which acts over the entire surface of the rotor is determined by many factors (flux gap size, stator core material, the geometry of the motor etc.) but we can once again assume that these are all the same between our high and low Kv motors. Therefore, the only possible difference between our two motors can come from the average current density in the stator windings.

Looking at a cross-sectional view of the stator we can see that there is only a fixed area available to place the copper windings.

Let's look at just a single stator tooth and the impact that a different turn number will have on the applied magnetic field strength when placed in the available winding area.

The motor with fewer turns of wire will have a lower induced voltage produced by the rotor magnets as they pass by the tooth, giving it its high Kv rating when compared to the motor with more turns.

The high Kv motor has 4 turns of wire each at 10 A for a combined total of 40 A/tooth. The low Kv motor has 10 turns of wire each at 4 A, for the same total of 40A/tooth. Therefore these two motors will provide the same magnetic field strength and have the same torque output. 

Yes, you could increase the current in the low Kv motor to be the same as the high Kv motor at 10A and produce more torque. However, this is fundamentally no different than increasing the current in the low Kv motor with the same end result. Therefore, rewinding a motor to increase its Kv only makes sense when you wish to match the motor current draw to the current limit of your existing motor controller (ESC). You could just as easily achieve a higher torque output by purchasing a new motor controller with a higher current limit and keeping your existing motor unchanged. Alternatively, if you have a motor with a very poor copper fill factor (area in the stator slot filled with copper vs empty air) then it may also make sense to rewind your motor.

Note that for the purposes of this argument we are ignoring the production of any useful reluctance torque (like that used by a reluctance motor) which will be true for almost all motor you encounter as a hobbyist.

Now let's consider waste heat generation for our high and low Kv motors.

Waste heat produced by a BLDC motor for a fixed torque

The power dissipated by a motor winding is given by:

`P=I^2R` 

where I is the current in the windings and R is the resistance of the windings. As the power dissipation in the motor scales with the square of the stator current, it feels only natural to assume that the low Kv motor, with its 4A current draw, will produce less heat than our high Kv motor with its 10A current draw. However, this assumption fails to take into consideration that the total area of the copper windings is fixed and therefore the current density remains the same.

The DC resistance of a wire is given by:

`R_{DC}=\frac{l\rho}{A}`

where l is the wire length, `\rho` is the conductivity of the conductor and A is the conductor area. In order to simplify this argument lets assume we are using square cross-section conductors.

In order to fit more turns into the same area, we had to reduce the cross-section of each individual conductor, which reduced its area and therefore increased its resistance. If we assume that each turn of wire has a length of 1 and that the total conductor cross-sectional area is also 1, then enter the current and turn numbers listed above we find:

`P=I^2R=I^2\frac{l\rho}{A}=10^2\frac{4\rho}{1/4} = 4^2\frac{10\rho}{1/10} = 1600\rho`

Therefore, for a given torque (fixed current density), copper fill factor and copper winding area, the power dissipation is not changed by altering the motor Kv. 

If you do wish to increase the specific torque density of an existing BLDC motor then you have a few options:

1. Rewinding the motor to increase the copper fill factor by more efficiently packing the conductors.
2. Replace the permanent magnets in the rotor with higher energy density magnets.
3. Reduce the flux gap distance between the rotor and the stator.

Increasing your peak current output of your motor controller so that the motor windings run with a higher current density will of course also increase your peak torque output. However, this will require additional cooling to handle the extra waste heat and you run the risk of saturating the core material

Note that in the above example we assumed a DC resistance. In reality, a motor will operate with an AC current. At very high frequencies it may make sense to rewind a motor to use many parallel small conductors rather than singular thick conductors in order to minimise the skin effect.

Conclusion

You will not improve the specific torque density or lower the power dissipation for a given torque output by rewinding a motor to have a lower Kv. However, it can make sense to rewind a motor so that its peak current draw will be better matched with an existing motor controller. 

Equations were produced in this post with the help of arachnoid.com and are based on those found in the book Electric Motors and Drives: Fundamentals, types and applications by Austin Hughes. If you have noticed any errors in the above article then please let me know.

January 2021 addendum

An anonymous commenter has pointed out that the above argument does not consider the impact that changing the wire diameter has on the lengths of wire between each wound tooth or to the ESC. See the comment below for more details. In short, if you were to decrease the Kv of a motor by doubling the number of turns and halving the conductor area you may think that the total length of wire from one phase terminal to the next is also doubled. However, this turns out not to be the case because the length of wire from the ESC to the motor and from one wound tooth to the next does not actually change. Therefore, the total increase in the length of wire is slightly less than double, making the lower Kv motor technically more efficient at producing the same amount of torque. However, this effect is likely to be very small in most scenarios. 

Thanks, Anonymous!