The basic motor voltage equation (ignoring short term inducive effects) is:
V = Ke*w + R*I
where:
V is the voltage across the leads (constant applied in your case);
Ke is the back-EMF constant of the motor (volts per unit speed);
w (omega) is the angular velocity;
R is the resistance;
I is the current
The higher the motor speed, the more of the supply voltage is "used up" by the back-EMF term, and the less is available to drive current through the motor. Since torque is proportional to current, as your speed increases, your torque generation capability decreases linearly.
Depends on the type of motor. For a series motor the torque is related to the square of the current. For a shunt type motor, the torque is linear to the motor armature current. This is with a fixed field current. For the other posters, the heading is torque not voltage.
and to add to Curt's voltage makes speed equation is the other one, current makes torque:
T=Kt*I where Kt is same thing as Ke but in different units (torque/amp)
So studying them together one can see as the voltage is used up to make speed, less is left over to make I (V/R).
So as Curt showed, the faster you go with a constant voltage applied, the less is left over to make current in the motor resistance, hence less torque as u go faster.
So you see you stated it backwards by now; torque does NOT go down as speed goes down, but rather as it goes up.
Yea your right mike I did my question wrong, I meant why does torque decrease as rpm increases. I guess the correct way of stating the relationship is that torque and rpm of a DC motor are inversely proportional?
So how about brushless DC or PM AC motors? Is it the same deal?
