Motors and back EMF
Motors and back EMF
(OP)
I’m really trying to understand the concepts for AC induction motors but still struggling with how back EMF is produced. This is what I’ve got so far:
As the voltage is applied to the motor windings of an Induction motor the windings create a rotating magnetic field this magnetic field cuts the rotor bars this change in magnetic field therefore produces a current in the rotor bars which you in turn create a magnetic Field in the opposite direction to the magnetic field that created it.
Now this is the bit that confuses me is this second induced emf in the rotor bars what causes ‘back EMF’ and is it due to this emf cutting the windings as it rotates producing the back emf. Therefore as more slip occurs the rate at which it cuts the windings is less so hence more current drawn when there’s more slip?
As the voltage is applied to the motor windings of an Induction motor the windings create a rotating magnetic field this magnetic field cuts the rotor bars this change in magnetic field therefore produces a current in the rotor bars which you in turn create a magnetic Field in the opposite direction to the magnetic field that created it.
Now this is the bit that confuses me is this second induced emf in the rotor bars what causes ‘back EMF’ and is it due to this emf cutting the windings as it rotates producing the back emf. Therefore as more slip occurs the rate at which it cuts the windings is less so hence more current drawn when there’s more slip?
RE: Motors and back EMF
Let's look at this in terms of how the requirements for motor action and the requirements for generator action are met. Every rotating electromagnetic machine has both motor action and generator action occurring when operating, but it isn't always easy to see how they are caused.
Requirements for motor action to produce a force or torque (Lorentz's force law):
- A current-carrying conductor
- A magnetic field
Requirements for generator action to induce a voltage (Faraday's induction law):- A conductor
- A magnetic field
- Relative motion of the conductor with respect to the magnetic field
As the voltage is applied to the motor windings of an Induction motor the current in the stator windings create a rotating magnetic field this magnetic field cuts the rotor bars this change in magnetic field thereforeproduces a currentinduces unequal voltages in the rotor bars, which, since the rotor bars are shorted, cause currents to flow in the rotor bars, meeting the requirements for motor action (a current-carrying conductor in a magnetic field), and the rotor spins. Now, the stator conductors are exposed to the magnetic field created by the current in the rotor, and since there is relative motion between the rotating rotor field and the stator windings, the requirements for generator action are met (stator conductors, rotor's magnetic field, relative motion), whichyouin turncreate a magnetic Fieldinduce a voltage in the opposite direction, (due to Lenz's law), i.e., the back emf to themagnetic field that created itvoltage applied to the stator windings.Therefore as more slip occurs, i.e., the rotor slows down further below synchronous speed, the rate at which its magnetic field cuts the stator windings is less so less back emf is produced hence more current drawn when there’s more slip
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RE: Motors and back EMF
In the above thought experiment, the rotor is acting as a generator. You are putting torque in at a certain speed and your power input is creating the amount of power associated with rotor I^2*R (assuming steady state).
Everything we just described is a valid description of a motor operating in motor mode at low slip, except that we were viewing the situation from the synchronous reference frame (the speed we observed in that synch ref frame is what we call slip speed in our normal ref frame). The rotor currents that we computed were correct. The torque that we computed was correct. The associated power flows associated with that torque change when we change our definitions of speed but they can be reconciled by careful analysis.
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(2B)+(2B)' ?