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Field loss in a shunt wound dc motor

  1. Jun 27, 2010 #1
    In a field loss event, why does a shunt wound dc motor run away? From my understand of electric motor, the armature spins because it's magnetic field pushes/pulls against the field winding's magnetic field. If I remove power to the field winding, then shouldn't the motor slows down because the armature's magnetic field has nothing to pushes/pulls against? Isn't like a permanent magnet motor running without the magnets?
  2. jcsd
  3. Jun 27, 2010 #2
    In case the flux drops to "phi"_res the internal generated voltage dops with it.(EA=K *'phi'*w) And thus causes the armature current to increase, likewise the torque(K*'phi'*IA). And speed increases until it breakes or the generated torque reaches load torque.

    Due to hysteresis inn the core the flux("magnetic field") never reaches zero ("phi"_res), if the applied voltage is removed.
  4. Jun 28, 2010 #3
    Very good!,I enjoy from this conceptual Question and wonderful answer.

    Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com
  5. Jul 24, 2010 #4
    huh? I don't get it. Currently my understand of electric motor is very superficial so that's why I'm having a hard time understand your answer from a mathematical pov.
  6. Jul 24, 2010 #5
    Its actually not that hard to understand, my explenation may be a little diffuse.

    I dont know which notation you are used to, but;
    - E_a is the internal generated voltage.
    - K is contant
    - "phi" (greek letter, not pi) donates the magnetic flux
    - w (omega) is the rotational velocity

    The steel core (ferromagnetic material) of the stator have a remanent magnetic field after the current in the coils are switched off. Look up hysteresis. The core thus becomes an permanent magnet which "drives" the rotor. But the flux is weak and the equations yields that the motor runs away.
  7. Sep 5, 2010 #6
    o I see. In a shunt dc motor, what factors determine it's base speed?
  8. Apr 6, 2011 #7


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    You should think about in terms of how a shunt motor operates..

    As you know a shunt motor has its field winding connected in parallel w/ the armature winding. It is a high resistance winding (about 10 ohms) and as such accepts a lower amount of the applied current.

    The shunt winding does not saturate as quickly as a low resistance heavy gauge series winding.. as in a series motor. This results in good speed regulation from no load to full load, as well as good torque regulation although it is quite low (because not much current reaches the shunt, it has a high impedance).

    However with the shunt field open, There exists only the residual magnetism (not really like a permanent magnet motor without magnets as you said..) This residual magnetism is not enough field strength to oppose the armature magnetic field, which now has all the applied current passing through it (now that the shunt is open).

    w/ such a high application of current, we have a very powerful magnetic field w/ not enough opposition from the residual magnetic field to control it. The motors speed tends toward infinity as quick as it can until its housing cannot contain the force and it destroys itself.

    From a mathematical standpoint.. the motor speed (N) equals to the ratio of motor constant times current applied minus current lost to field flux. N = K(I-VR) / phi .. as the field is opened and the flux collapses the armature current is becoming increasingly high, and as such the motor speed N is increasingly proportionally.

    Hope I helped, K. Tucker
    Last edited: Apr 6, 2011
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