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Synchronous motor features and lagging/leading operation

Short answer

A synchronous motor runs at constant speed, is not self-starting, and its power factor is adjustable via field current. Under-excitation (Ef<V) makes armature current lagging; over-excitation (Ef>V) makes it leading. This occurs because Ia = (V - Ef)/jXs, so the resultant voltage E shifts, changing the current phase relative to terminal voltage.

State the characteristic features of synchronous motor and explain how a synchronous motor can be operated to draw lagging current as well as leading current.

A synchronous motor runs at constant synchronous speed Ns=120f/PN_s = 120f/P and, unlike an induction motor, its power factor can be set by the DC field current.

Characteristic features

  1. Speed is constant at NsN_s from no load to full load; it depends only on supply frequency and number of poles.
  2. Not self-starting; started by damper winding, pony motor or variable frequency.
  3. Operates at lagging, unity or leading pf depending on excitation.
  4. Over-excited motor on no load acts as a synchronous condenser.
  5. Breaks down (pulls out of step) if load exceeds the maximum (pull-out) torque.
  6. Requires a separate DC excitation source.
  7. Prone to hunting on sudden load changes.
  8. High efficiency, especially for large, low-speed drives.

How it draws lagging or leading current

For a motor (with RaR_a neglected): V⃗=E⃗f+jXsI⃗a\vec V = \vec E_f + jX_s\vec I_a, so

I⃗a=V⃗−E⃗fjXs=E⃗rjXs\vec I_a = \frac{\vec V - \vec E_f}{jX_s} = \frac{\vec E_r}{jX_s}

The current lags the resultant voltage ErE_r by 90∘90^\circ. Keeping the load (and so Iacos⁡ϕI_a\cos\phi and Efsin⁡δE_f\sin\delta) constant, only EfE_f is changed by the field current.

1. Under-excitation, Ef<VE_f < V - lagging current

   V ---------------------->
    \  Er (from Ef tip to V tip)
     \ delta
      '-----> Ef (short)
   Ia = Er/jXs lags V   -> lagging pf

The resultant ErE_r is nearly in phase with VV, so IaI_a (90° behind ErE_r) lags VV. The motor takes magnetising (lagging) reactive power from the supply to make up for its weak field.

2. Normal excitation - unity pf

EfE_f is such that ErE_r is 90∘90^\circ ahead of VV; then IaI_a is in phase with VV, and current is minimum.

3. Over-excitation, Ef>VE_f > V - leading current

   V ---------------------->
    \          Er points above V
     \ delta
      '-----------------> Ef (long)
   Ia = Er/jXs leads V  -> leading pf

ErE_r swings ahead of VV by more than 90∘90^\circ, so IaI_a leads VV. The excess field means the motor supplies lagging reactive power to the network, behaving like a capacitor.

ExcitationEfE_fCurrentpf
Under<V< Vlargelagging
Normalabout VVminimumunity
Over>V> Vlargeleading

This control is the basis of the V-curves and of using synchronous motors for power factor correction.

✓ Answer checked by Claude Sonnet 5.5

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