# "Torque" constant for shunt-wound motor

A regular DC motor has a torque that is proportional to the current:
T = k_t * i
A shunt-wound motor has a torque that is proportional to the current
squared:
T = (something) * i^2
Most of the literature and data sheets that I see use k_t for the torque constant of a DC motor.
But what do they use for a shunt-wound motor?
TIA
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
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Tim Wescott wrote:

WHen you say DC motor (not shunt), are you referring to a PM or series?
Jamie
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On Wed, 17 Oct 2012 18:16:25 -0400, Jamie wrote:

PM. And I just realized that I've been saying shunt wound when I meant series. D'oh, d'oh d'oh.
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
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Tim Wescott wrote:

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 1 amp is .5 HP etc. At least that works in most cases.
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
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wrote:

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.
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