DC Motor Drives Questions and Answers

Answer: D

Correct answer: 90 degrees

Since \( V_{dc} \propto \cos\alpha \), the mean output passes through zero where the cosine does, at 90 degrees. Below that angle the bridge rectifies and delivers power to the machine; above it the mean voltage is negative and, with the current still flowing in the same direction through the thyristors, power flows back to the supply. Ninety degrees is therefore the boundary between rectification and inversion.

Answer: A

Correct answer: inverting, returning power to the supply

Beyond 90 degrees the mean output voltage is negative while the current, constrained by the thyristors, still flows in the same direction, so the product of the two is negative and power flows from the dc side to the ac side. The machine must be generating for this to happen - it is being braked, and its kinetic energy is returned to the mains rather than dissipated. This is regenerative braking through a fully controlled bridge.

Answer: B

Correct answer: light load and large firing angle

Discontinuous conduction occurs when the armature current falls to zero before the next thyristor is fired. That is most likely when the current is small, that is at light load, and when the conduction interval is short and the applied voltage low, that is at a large firing angle. In the discontinuous region the output voltage no longer follows the simple cosine law and the drive’s gain changes, which upsets a controller tuned for continuous conduction - hence the smoothing inductor.

Answer: C

With the current interrupted, the mean voltage rises above the value the continuous-conduction formula predicts, so the machine runs faster than expected. The relation between firing angle and speed also becomes nonlinear.

Answer: D

Correct answer: reduce the current ripple and extend continuous conduction

The inductor opposes changes of current, so it smooths the ripple the rectified waveform would otherwise impose and keeps the current flowing through the gaps, extending continuous conduction down to lighter loads. That matters for three reasons: less ripple means less additional heating and torque pulsation, and staying continuous keeps the converter’s voltage-versus-firing-angle relation linear, so the control loop behaves predictably. It does not change the mean voltage at a given firing angle.

Answer: A

Delaying the firing displaces the current from the voltage, so the displacement factor falls as the cosine of the angle. A drive spending its life at low speed therefore loads the supply with reactive current as well as harmonics.

Answer: B

The distorted current interacts with the supply impedance to distort the voltage, and the harmonics cause extra loss in transformers, cables and capacitors. Line reactors, filters and multi-pulse arrangements are the usual mitigations.

Answer: C

The mean voltage is the supply voltage times the duty ratio, so the switching frequency can be far above the supply frequency and the ripple correspondingly small. Choppers are used on drives fed from batteries or a dc link.

Answer: D

Correct answer: \( DV_s \)

The chopper connects the supply to the load for a fraction \( D \) of each period and disconnects it for the rest, so the mean voltage is

Varying the duty ratio between 0 and 1 therefore gives any armature voltage from zero to the full supply, continuously and with very little loss - which is the whole advantage of chopper control over the resistance control it replaced.

Answer: A

Correct answer: one quadrant only

A class A chopper has a single switch and a freewheeling diode, so the armature voltage and the current can both be of one sign only. It can therefore motor in one direction and nothing else - no braking, no reversal. It is the simplest and cheapest arrangement and suits a fan or pump drive that never needs to be braked or reversed.