Solid State Drive Control Questions and Answers

Answer: C

The machine does generate into the link, but the diodes cannot pass that power onward, so the link voltage rises until the drive trips or a braking chopper intervenes. An active front end is what makes return possible.

Answer: D

Ripple and audible noise improve while loss, device temperature and cable stress all get worse, so there is an optimum rather than a monotonic benefit. Where the loss can be afforded, silicon carbide shifts that optimum upward.

Answer: A

Correct answer: the junction temperature limit is reached at a lower loss

The junction sits at the ambient plus the loss times the thermal resistance, so a higher ambient leaves less of the temperature budget for the loss. Since the junction limit is fixed by the device, the permissible loss must fall, and with it the current the drive may deliver. The derating curve is that arithmetic expressed as a graph.

Answer: B

Correct answer: Carrying the load current to the machine

The load current is carried by the power stage - the semiconductors, the busbars and the output terminals - and the control electronics operate at signal levels and are galvanically isolated from it. The control functions are the other three: running the loops, generating the modulation, and detecting faults.

Answer: C

Correct answer: a power stage fault cannot reach the control wiring

The power circuit operates at hundreds of volts and can fail with the full link voltage appearing where it should not. Isolation confines that to the power stage, so a fault cannot put dangerous voltage on the control terminals or the fieldbus, and cannot damage the equipment connected to them.

Answer: D

Capacitive coupling from a fast transient on the collector can inject charge into the gate and turn the device partly on, which a few volts of negative bias prevents. It is standard practice on IGBT modules of any size.

Answer: A

A small resistor switches quickly and cheaply but generates a steep transient and a large overshoot; a large one is gentle but costly in loss. The value is one of the more consequential choices in a power-stage design.

Answer: B

Voltage overshoot at turn-off is that inductance times the rate of change of current, and it eats directly into the device's voltage margin. Laminated busbars and modules with integrated capacitors exist to minimise it.

Answer: C

The device with the lower on-state drop takes more current, gets hotter and, in some technologies, takes more still, which is a runaway. Symmetric busbars, matched modules and a positive temperature coefficient of on-state drop keep the sharing stable.

Answer: D

Leakage currents differ, so resistors are needed to share the steady voltage, and turn-off times differ, so capacitors are needed to share the transient. The difficulty of doing this reliably is what made multilevel topologies attractive.