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How does back emf make energy conversion possible

  1. Aug 3, 2012 #1
    I've come across one thing about energy conversions: "To make an energy conversion possible, it is important that the supplied form of energy must be opposed by another force, otherwise this energy cannot be converted." Agreed!

    Once the motor starts generating torque, a back emf is produced in it, which in accordance with the Lenz law opposes the supplied voltage. So a supply of voltage must be maintained so that the motor keeps running. But even otherwise, just in case there were no back emf, supply voltage "has" to be there to keep the current flowing through the armature and thus rotate it in the presence of a magnetic field. So how does back emf, in this case, make the energy conversion possible, when supply voltage is anyhow a necessity, whether the back emf is generated or not?
     
  2. jcsd
  3. Aug 4, 2012 #2
    Are you familiar with the DC motor model shown here:
    CTMS Example: DC Motor Speed Modeling in Simulink

    If there's no back-emf, i.e. the rotor is locked, there will be no electrical to mechanical power conversion. The motor might be producing torque due to the armature current, but it won't be able to turn the rotor and thus it won't be doing any work on the rotor.

    If you apply a force to an object without moving it, you're not doing any work. Just like you're not doing any work on the earth by standing on it.
     
  4. Aug 4, 2012 #3

    jim hardy

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  5. Aug 4, 2012 #4
    Referring to the DC model I posted a link to (it's the same principle for any other electrical machine, circuit is perhaps just more complicated):

    When the supply voltage is larger than the back-emf, current will be flowing into the positive terminal of the back-emf and thus supplying power to whatever is producing the back-emf. This power is exactly equal to the mechanical power delivered to the rotor. If there's no back-emf, no power conversion can take place - it's the thing you "push" against in a mechanical analogy. The rotor has to have nonzero angular velocity, and thus back-emf, for the motor to output mechanical power.

    I'm not sure what you mean with regards to force pairs. If I apply a force to an object, sure, the object will apply a force of equal magnitude and opposite in direction to me, but that can't change anything with regards to the energy being transferred.

    - I apply force in same direction as displacement -> I do positive work on the object-system.
    - Object-system applies force equal in magnitude and opposite in direction to my body-system and does negative work equal in magnitude to the positive work done by my body-system on the object-system.
    - Net result -> Energy transfer from my body-system to the object-system.
     
    Last edited: Aug 5, 2012
  6. Aug 4, 2012 #5

    jim hardy

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    wow there's sure more than one way to think about things.

    by force pairs i meant
    conductors in the rotor experience a force because the moving charges in them experience F = QVcrossB.
    conductors in the stator experience force for same reason.

    there's the force pair acting on two different objects
    but no power is realized until displacement ( rotation ) begins, nor is counter EMF.
    so to my simple thinking counter EMF is a companion to power not cause of it.

    your link is an elegant explanation from a controls perspective.
    i bookmarked it....

    thanks !
     
  7. Aug 5, 2012 #6
    Right the supply voltage does push against the back emf. But at the same time, there will be no current in the rotor, if there's no voltage supplied. Unlike generators, where once a force applied to the armature would set it rotating owing to inertia, even after the force is removed.
     
  8. Aug 5, 2012 #7
    I see what you mean Jim, there's certainly no shortage when it comes to ways of understanding power conversion in electrical machines.

    My background is in control so I've always liked equivalent circuits and such, so it just seemed to me like an easy way to link electrical to mechanical power.
     
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