Synchronous Motor Drives Questions and Answers

Answer: C

Correct answer: Hall effect sensors

Three Hall sensors spaced 120 electrical degrees apart in the stator detect the rotor magnet poles and give six distinct states per electrical cycle - exactly the information the inverter needs to decide which two of the three phases to energise. They are cheap, robust and adequate for trapezoidal commutation. Higher-performance drives use an encoder or resolver, or estimate position from the back emf to dispense with sensors altogether.

Answer: D

Correct answer: the stepwise nature of its commutation

A brushless dc drive energises two phases at a time and switches to the next pair every 60 electrical degrees, so the torque produced dips at each commutation instant, giving a ripple at six times the electrical frequency. The trapezoidal back emf is designed to match this square-wave excitation and reduce ripple, not to cause it. Slip does not arise in a synchronous machine, and position feedback is present by definition.

Answer: A

The rotor aligns its low-reluctance axis with the field, and the torque follows the difference between the direct and quadrature reactances. It needs no magnets and no rotor excitation, which makes it cheap and robust.

Answer: B

Because all the flux must be supplied from the stator, the magnetising current is larger and the power factor poorer than in a machine with magnets. Modern rotor designs with deep flux barriers have narrowed the gap considerably.

Answer: C

No rotor current means no rotor copper loss, so the machine is smaller and more efficient for the same output. The magnets cost money and lose strength if they get too hot, which is where the induction machine still wins.

Answer: D

Both heat and a strong opposing field from the stator can drive the magnet past the knee of its characteristic and cost it permanent flux. Current limits and temperature monitoring are therefore part of the drive's protection.

Answer: A

Correct answer: 1,000 rpm

The synchronous speed is

and the machine runs at exactly that figure whatever the load. The 960 rpm answer would be an induction motor of the same pole number at four per cent slip - a distinction worth keeping in view, since a synchronous machine never runs below its synchronous speed.

Answer: B

Correct answer: 30 Hz

Rearranging the same relation for the frequency,

The converter must reduce the voltage in proportion, holding \( V/f \) constant so that the flux stays at its design value. This is the standard way of obtaining variable speed from a synchronous machine, whose speed is otherwise rigidly fixed by the supply.

Answer: C

Correct answer: 0.8 per unit

The pull-out power is the maximum of \( (EV/X_s)\sin\delta \), which occurs at \( \delta = 90 \) degrees:

The machine can therefore carry only 80 per cent of its rated power at this excitation, which is why over-excitation is used not only for power factor but to enlarge the stability margin.

Answer: D

Correct answer: 30 degrees

Using the same expression with the pull-out power already found,

The machine is at half its pull-out power and a third of the way to the 90 degree limit. The margin is comfortable, which is the point of the calculation: the angle is a direct measure of how close the machine is to slipping poles.