Understanding motor constants Kt and Kemf for comparing brushless DC motors

So for a practical definition, the torque constant Kt specifies how many Nm of torque you'll get from the motor at a certain current? Is a bigger Kt always better?

I am more confused about the backemf constant Kemf. This is the amount of back EMF produced by the motor running at a certain speed. Does the back EMF limit the voltage the motor can accept and hence it's speed?

Would an ideal motor have a very high Kt and very low Kemf?

What happens if the back emf voltage reaches the input voltage, does the motor stop running?

Thank you for any answers, Fred

Answer

Does the back EMF limit the voltage the motor can accept and hence it's speed?

Hmm. I think you're a little confused. Back-emf limits the motor speed because it dictates how much voltage you need to achieve a given speed.

Would an ideal motor have a very high Kt and very low Kemf?

No. (The symbol \$K_e\$ is usually used for the back-emf constant, by the way.)

If you use SI units (Nm/A for \$K_T\$, V/(rad/s) for \$K_e\$), then \$K_T = K_e\$ for DC motors and permanent-magnet synchronous motors (aka "brushless DC"), and depending on the type of motor and how you define \$K_T\$ and \$K_e\$, the ratio of the two should be a fixed proportionality constant.

Proof of why this is true for DC motors:

At constant operating point (constant speed, voltage, current, torque):

• \$V_T = K_e \omega_m + IR\$


• \$T_m = K_T I\$

(\$V_T\$ = terminal voltage, \$\omega_m\$ = motor angular velocity, \$I\$ = motor current, \$R\$ = motor resistance, \$T_m\$ = torque, including frictional losses)

Electrical power in = \$V_T I = K_e \omega_m I + I^2 R\$

mechanical power out = \$T_m \omega_m = K_T I \omega_m \$

Losses = \$I^2 R\$

Conservation of energy means electrical power in = mechanical power out + losses

This is true if and only if \$K_e = K_T\$.

What happens if the back emf voltage reaches the input voltage, does the motor stop running?

No -- what happens is that the ability of the motor to produce torque decreases with speed. Back-emf voltage "uses up" voltage from the electrical power source; the remaining voltage available to the motor is what's left for IR drop in the motor and the inductance drop \$L \frac{dI}{dt}\$, and since torque is proportional to current I, the available torque decreases. The system reaches equilibrium at some point where the electromagnetic torque matches the motor's mechanical load. If you increase the mechanical load, current will increase to match that torque as speed slows down, making more IR drop voltage available.

Is a bigger Kt always better?

No. Rule of thumb with DC motor selection (also true for brushless DC motors to a large extent) -- pick a motor with a \$K_T\$ and back-emf constant such that the supply voltage you have available is well-matched with the back-emf at your maximum speed. You usually want back-emf voltage to be 80-95% of the supply voltage, but the exact number depends on the load torque and the IR drop in the motor at that operating point.

If you pick a \$K_T = K_e\$ too high, you'll run out of voltage and won't be able to achieve the speed you need. If you pick a \$K_T = K_e\$ too low, the current needed to achieve the torque you need will be higher than necessary.