"Torque" constant for shunt-wound motor

WHen you say DC motor (not shunt), are you referring to a PM or series?

Jamie

PM. And I just realized that I've been saying shunt wound when I meant series. D'oh, d'oh d'oh.

Series motors know no limit. The factor I^2 as you have noted, however, the limit comes from winding resistance and heating for torque. This information is normally supplied per motor.

HP rating on a motor can generally give you an idea of max current, take that and you can work back towards your torque..

As a rule of thumb, a 600V DC motor at 1 amp is 1 HP, 300V DC motor at

A series motor can generate lots of HP in a small package with lack of self speed control. This is due to the field never being matched with the armature to balance the incoming driving source. as current drops, so does the current in the series field winding, weakening it even and the armature tries to spin faster. Of course you lose your torque and at some point the load will level the RPM's.

And thinking back, if the current drops 50% in a series motor I think the RPM's increases 4 times, or something in that neighborhood, I would have to look that up. You also need to factor in DC resistance, too.

Jamie

If you neglect losses, then the motor elec/mech powser balance is

Power_Elec = Power_Mech , so V * i = T * w

The PM DC motor elec/mech power balance is then

V = Ke * w and T = Kt * i

Ke * w * i = Kt * i * w , giving Kt = Ke

Now if T = Kt * i^2, then the series DC motor power balance would be

Ke * w * i = Kt * i^2 * w , giving Kt = Ke / i

dave y.