Solid State Drive Control Questions and Answers

Answer: A

Linear sinusoidal modulation can reach only \( \pi/4 \) of the six-step fundamental, which is why the last part of the voltage range is reached by over-modulating. Space vector modulation and third harmonic injection recover part of that shortfall.

Answer: B

Adding a common mode third harmonic flattens the reference peaks so a larger fundamental fits under the carrier. It cannot drive current in a star-connected machine with an isolated neutral, so the machine never sees it.

Answer: C

Treating the three phases together rather than separately gives about fifteen per cent more fundamental voltage and slightly lower distortion for the same switching. It maps naturally onto a digital implementation, which is why it is now the default.

Answer: D

Each device conducts for a hundred and eighty degrees, which extracts the most from the link but leaves fifth and seventh harmonics in the output. Drives use it at the top of the speed range where voltage matters more than waveform.

Answer: A

If the incoming device turned on before the outgoing one had turned off, the link would be short-circuited through both of them. A microsecond or two of overlap protection is enough, and it is set from the devices' turn-off times.

Answer: B

Correct answer: low output voltage and low speed

The volt-seconds lost to dead time are the same at every switching, so they are a fixed subtraction from whatever the inverter is trying to produce. At full output that error is a fraction of a per cent; at the low voltages of low-speed operation it becomes a large proportion, distorting the current and causing torque ripple.

Answer: C

Correct answer: both devices conducting and short-circuiting the link

If the upper and lower devices of a leg conduct together, even briefly, they place a short circuit across the dc link and the current is limited only by stray impedance. Destruction follows in microseconds. Dead time is inserted between turning one off and the other on precisely to make this impossible.

Answer: D

Smaller current excursions between switchings mean less ripple loss in the machine, but each transition costs energy in the device. The optimum depends on the devices, the machine and how much audible noise can be tolerated.

Answer: A

Magnetostriction in the machine's iron at the switching frequency is what produces the whine, so moving it above about fifteen kilohertz removes it. The penalty is switching loss, and random or spread-spectrum modulation is the compromise where the loss cannot be afforded.

Answer: B

A short circuit drives the device out of saturation and its collector voltage rises while it is supposed to be on, which the drive detects within a microsecond or two. It then turns the device off gently to avoid an overvoltage from the circuit inductance.