"Torque" constant for shunt-wound motor

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

Jamie

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Surely a shunt-wound motor would be the same as a permanent magnet motor where the torque is proportional to armature current since the magnetic field form the stator is essentially constant.

A series would motor would have a torque proportional to the product of the armature current and field current to give a squared term since they are the equal.

kevin

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

It's late in the day and I got my terms mixed. Usually I know perfectly well what "shunt" and "series" are. I certainly had the right picture in my head -- why didn't you look at that instead of my post?

At any rate, for a SERIES (dammit) wound motor, is there a symbol that's normally used for the constant of proportionality between current squared and torque?

Isn't it something like G*Ia*If, where, in series wound, Ia=If, then it's G*Ia^2, where "G" is the coupling coefficient between the windings... from ancient memory cells, my apologies if I'm wrong

A series-wound motor is weird. Ignoring copper loss, with a voltage applied to the terminals, the stall current is infinite, so the torque is infinite. The faster it goes, the more back EMF is generated by the armature, which then reduces the supply voltage available to the field. Less field current reduces the armature back EMF. The theoretical end point is zero current at infinite speed.

They usually wind up going really fast, limited by losses.

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

You can still saturate the iron, though, after which the torque increases linearly.

Cheers

Phil Hobbs

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.

armature current and field current to give a squared term since they are the equal.

I'm thinking it would be torque proportional to current-squared at low currents, proportional to current when one of the irons saturates (either rotor or stator), and constant if/when both the rotor and stator reach saturation.

Just to be picky, though, there's also frequency dependence, in the rotor and stator, due to core eddy currents (there's a phase shift). This is about a UNIVERSAL motor, after all, there's no reason to assume it has DC input, it could be AC. And even DC gets commutated to AC in angular-velocity-dependent fashion.

When spinning (for just the rotor) and when supplied by AC (for both stator and rotor). After all, the rotor sees a field that, though it may be relatively constant in the middle of the rotor or at any given point in time, its direction is constantly changing.

Tim