Higher KV means lower torque for DC motors?

Question

I'm in the process of trying to build an electric skateboard. On the websites I visited I read that torque and KV are inversely proportional which does seem counter intuitive. So I'm wondering why that is the case.

I think some beginning of answer can be found in this paragraph:

In summary, a low KV motor has more winds of thinner wire - it will carry more volts at less amps, produce higher torque and swing a bigger prop.

So yeah, lower kv motors have thinner wires. Thinner wires mean more resistance and less current for the same voltage. I don't really understand the rest though. I don't see the relation with more torque..

Answer 1

Motors can be difficult to reason about, as there are so many variables. So let's analyse a motor and keep the speed, torque and power output constant.

Let's say it has two seperate identical windings. Each winding, for the sake of argument, carries 1A and 10v. Now if we connect these in series, the electrical supply to the motor will be 20v 1A. If we connect them in parallel, the supply will be 10v 2A.

Both windings are receiving exactly the same power, and the motor is providing exactly the same speed and torque, yet the voltage and current at the terminals are different. In both cases the electrical power to the motor is the same.

We could regard the two parallel windings as a single winding with twice the area. In fact, some motors are wound with multiple parallel strands, as it's easier to do than using a single thick wire.

We can see that the speed constant and the torque constant change in exactly the same way, as we fill the winding space available in the motor with either many turns of fine wire or a few turns of thick wire.

Notice that the heat lost in our two windings is exactly the same, whether we connect them in series or parallel. So the efficiency of the motor is also the same whether wound with many thin or few thick turns. All that matters is the size of the motor to receive the wire, and the power you want to run it at.

Most people will match the speed, power and weight of the motor to their vehicle, and then choose a KV to get a terminal voltage at that speed to match a suitable controller and battery, not the other way round.

Answer 2

That site is misleading, a motor wound for low rpm/V (also known as KV) doesn't have greater or lesser torque handling than the same motor wound for high rpm/V. The real difference is that the low rpm/V motor requires more voltage at proportionally less current (same power) to spin a given load at a given speed. The efficiency, maximum torque and maximum power are still the same for both motors.
Replacing a high rpm/V motor with an otherwise identical low rpm/V motor only increases torque if the system is bottlenecked by the maximum current that the motor driver can deliver to the high rpm/V motor.

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Thinner windings in motor windings do mean higher resistance, but this balances out with the lower current draw. For example, think of a motor wound with 8 turns of 2 mm^2 wire for each stator pole. The same motor, but wound to half of the rpm/V value, would have 16 turns of 1mm^2 wire instead. Yes, the wire area is cut in half, so you get twice the resistance per turn, but you also have more turns, with the wire being twice as long (16 turns vs 8). Thus, the 16 turn motor would have half the rpm/V and 4 times the resistance.

This is not as bad as it seems. The 8 turn motor will have to draw twice the current of the half rpm/V 16 turn motor to produce the same amount of torque. Doubling the current passing through the winding resistance will increase resistive power losses 4 times (P = I^2 *R), so while the ½ rpm/V motor will have 4 times the winding resistance, only having to supply half of the current means that both motors have the same amount of resistive power loss. Both motors will also consume the same amount of electrical power to spin a load at a given speed: As electrical power is simply the product of voltage and current, the 8 turn motor will have to draw twice the current, but at half of the voltage, resulting in the same power consumption.

You do get a slight efficiency or weight gain from using a higher voltage system, but not thanks to motors. By doubling the voltage, you will halve the current, so you can reduce the cross sectional area of all wiring between components by three quarters while maintaining the same amount of resistive losses in said wiring.

Answer 3

The velocity constant, \$K_V\$, and the torque constant, \$K_T\$, are the same - you can see this if you examine the motor and generator equations for, say, a PM DC motor:

$$T=(Blr)i= K_T i$$ $$E=(Blr)\omega=K_V\omega$$

where: \$T\$ is motor generated torque; \$E\$ is 'back-emf' \$B\$ is flux density; \$l\$ is the length of conductor in the magnetic field; \$r\$ is the radius of the armature; \$i\$ is armature current; and \$\omega\$ is angular velocity in \$rad\:s^{-1}\$