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How is back EMF generated in a pmsm motor and how does it vary.? Which parameters of the motor change due to the back EMF development.?

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Neethuraj
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2 Answers2

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As with all electromagnetic motors, the back-EMF is generated by Faraday's law of induction as the rotor and stator fields rotate relative to each other. The magnitude of the back-EMF \$\mathcal{E}\$ is a function of the rate of change of magnetic flux \$\Phi_B\$:

$$ \mathcal{E} = -{{d\Phi_B} \over dt} \ $$

Any loop of conductor (like the windings in the stator) will experience this EMF whenever the magnetic flux (such as that from the permanent magnets in the rotor) through that loop of conductor changes.

Back-EMF does not, in any sense I can imagine, change the parameters of the motor. It does, however, have a significant effect on the electrical load presented to the rest of the circuit. To change the current in the windings, one must apply a voltage to the windings in excess of the back-EMF. If the back-EMF is higher than the applied voltage, then the winding current will decrease, and you no longer have a motor, but a generator.

The back-EMF is also responsible for the easily observed phenomenon that a fast-spinning motor draws less current than a slowly-spinning motor when driven with the same voltage.

Further reading:

Community
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Phil Frost
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"Back-EMF does not, in any sense I can imagine, change the parameters of the motor."

This is a bit of an awkward statement.

Mathematically, the mechanical power ( minus all parasitic loses, e.g. friction, joule heating, flux leakage, etc. ) is the current * BEMF which, incidentally is rpm * torque. ( However, only the current that is generating the flux in the proper space and time relative to the conductor with the induced BEMF, often called Id).

From that you can deduce that BEMF is a function of flux ( i.e. the current which generates it) and the rpm. The faster the rotor spins the more BEMF. A motor will accelerate to an rpm such the BEMF is equal to supply voltage.

nik
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