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Unread 20-05-2014, 04:17
Rauhul Varma Rauhul Varma is offline
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by DampRobot View Post
Could you (or someone else) give an example of how Ke and Kt are used?
This is how I understand it (disclaimer I am just starting to learn this stuff).

Ke is the constant that relates the back emf generated by the motor to the angular velocity, Eb=Ke*omega. Kt is the constant that relates the torque output of the motor to the current drawn, Tau = Kt*I. In an ideal motor Ke and Kt are equal.

You can use these numbers to calculate the current drawn by the motor under a given load (tau) or speed (omega). Using the earlier definition of Kt, calculating the current under any torque is simple.

I = Tau/Kt

Solving for a given speed is slightly more complicated. Say you had the following circuit:


We are assuming the motor has been running long enough such that the inductive effects of the motor windings are ~0. (This is called steady state operation)

E is the applied voltage to the circuit, Ra is the resistance of the motor, Eb is the back emf generated by the motor and Ia is the current through the circuit (what we are trying to solve for).

Using Kirchhoff's loop rule we know that ΔV of a loop is 0. The voltage drop across the resistor is I*R (from ohm's law) and Eb = Ke*omega; from here we can solve for Ia.

ΔV = 0
E-Ia*Ra-Eb = 0
E-Eb = Ia*Ra
(E-Eb)/Ra = Ia
(E-Ke*omega)/Ra = Ia

If Kt = Ke the two currents should be the same

(sorry if this is poorly written, its late)