- 2075 Chaitra · 7 marks
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 and, unlike an induction motor, its power factor can be set by the DC field current.
Characteristic features
- Speed is constant at from no load to full load; it depends only on supply frequency and number of poles.
- Not self-starting; started by damper winding, pony motor or variable frequency.
- Operates at lagging, unity or leading pf depending on excitation.
- Over-excited motor on no load acts as a synchronous condenser.
- Breaks down (pulls out of step) if load exceeds the maximum (pull-out) torque.
- Requires a separate DC excitation source.
- Prone to hunting on sudden load changes.
- High efficiency, especially for large, low-speed drives.
How it draws lagging or leading current
For a motor (with neglected): , so
The current lags the resultant voltage by . Keeping the load (and so and ) constant, only is changed by the field current.
1. Under-excitation, - lagging current
V ---------------------->
\ Er (from Ef tip to V tip)
\ delta
'-----> Ef (short)
Ia = Er/jXs lags V -> lagging pf
The resultant is nearly in phase with , so (90° behind ) lags . The motor takes magnetising (lagging) reactive power from the supply to make up for its weak field.
2. Normal excitation - unity pf
is such that is ahead of ; then is in phase with , and current is minimum.
3. Over-excitation, - leading current
V ---------------------->
\ Er points above V
\ delta
'-----------------> Ef (long)
Ia = Er/jXs leads V -> leading pf
swings ahead of by more than , so leads . The excess field means the motor supplies lagging reactive power to the network, behaving like a capacitor.
| Excitation | Current | pf | |
|---|---|---|---|
| Under | large | lagging | |
| Normal | about | minimum | unity |
| Over | large | leading |
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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