motor Ke and Kt Quiz

So, for most small, high Flux DC motors there is this phenomenon called magnetic field saturation. It happens when the permanent magnet doesn’t have enough field strength to drive the high amount of Flux through the moving parts of the motor. I am simplifying here to try to keep it short.

The bottom line is that at high current (aka, near stall), the magnetic field saturates producing less actual torque than ideal and this is one of the reasons for the difference between Kt and Ke.

However, it is not the only difference. This can really only be found with motor testing.

From Ether’s 3rd attachment, Ke = Kt + (other losses /Iw).
If there are “other losses” these will account for the difference.

Of course, there’s a lifetime of study in “other losses”…

Thanks Paul. I think that completes the explanation.

“Kt equals Ke” is a theoretical result which is only approximately true for real-world motors.

Firstly, the “linearity” (straight lines) assumption (notice the asterisks) used to compute Kt would not be valid when the magnetic field saturates.

Second, “other_losses” such as bearing friction, windage & viscous damping, eddy currents, and hysteresis in the power equation account for the difference between Ke and Kt.

If Ke = Kt (which would be a perfect world scenario)
Then the max torque the motor could output would be:
Tstall = Ke*Istall

Which for a bag motor comes out to:
Tstall =~ .32 Nm

This is less than the .4 Nm in the motor specs, and given we have come to the conclusion that the real world Kt is always less than Ke, wouldn’t the true stall torque of the bag motor be even less than .32 Nm?

(I also don’t understand why our calculated Kt is larger than Ke or is that a result of the curve actually being nonlinear?)

In the real world, Kt and Ke are not constant. So without enough data to compute them both at the same operating point, comparing them is like comparing apples and oranges.