Field Effect Transistors Questions and Answers

Answer: A

The gate-source voltage, and hence the channel resistance, changes as the signal itself moves the source potential, so the on-resistance is signal-dependent and the switch distorts. There is no offset voltage, conduction is bidirectional, and the gate needs no current. The remedy is a complementary pair in parallel, whose resistances vary oppositely and largely cancel, which is why analogue switches are built that way. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

An n-channel and a p-channel device in parallel have on-resistances that rise and fall in opposite directions as the signal moves, so the parallel combination is far flatter. Halving the resistance is a side effect. Both devices conduct both polarities. Placing two components in parallel so that their opposite errors cancel is a general technique, and it is why a transmission gate is drawn with two devices rather than one.

Answer: C

A MOSFET is a majority-carrier device with no minority charge to sweep out, so it switches far faster and has no storage delay. Its conducting drop can exceed a bipolar device's saturation voltage at high current, since the channel is a resistance. It has no current gain in the bipolar sense, and its safe operating area is better in some respects rather than all. Speed rather than conduction loss is what won it the field.

Answer: D

A resistance dissipates current squared times resistance, so doubling the current quadruples the loss - unlike a bipolar device's roughly constant saturation voltage, whose loss is proportional to current. That difference decides which device wins at a given current: MOSFETs at moderate currents and low voltages, bipolar or insulated-gate devices where the current is very high. A loss law of a different order changes the answer as the operating point moves.

Answer: A

Gate charge is the charge a driver must deliver to turn the device on, and it decides the drive current needed at a given switching frequency. It is not leakage, not channel charge, and not depletion charge. A device with no steady gate current still needs substantial pulses of it, which is why gate drivers exist and why switching loss rises with frequency even in a device that draws nothing when static. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

Each transition moves the gate charge, and that energy is dissipated in the driver and in the device, so the total is proportional to frequency. On-resistance and threshold do not depend on frequency, and the channel current is set by the load. That a static device with no steady input current still costs power to switch is the essential economics of every switching converter and every logic gate. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

Keeping the drain voltage small holds the device in the ohmic region, where the channel is a resistance the gate voltage sets - the basis of automatic gain control and voltage-controlled attenuators. Beyond pinch-off it is a current source instead, and the remaining arrangements produce neither. The small drain voltage is essential, since the resistance becomes non-linear as the drain voltage grows towards pinch-off.

Answer: D

As the drain voltage grows, the channel is progressively depleted along its length and the resistance stops being constant, so the element distorts the signal it is attenuating. Heating is not the issue at these levels, the gate never conducts, and the transconductance is not what is being used. A resistance that is only a resistance for small excursions has to be kept within them, which limits the signal amplitude such attenuators can handle.

Answer: A

Field effect devices win decisively: a MOSFET's gate is insulated and draws nothing, and a junction device's gate leaks only nanoamps, against a bipolar base current of microamps or more. The configuration does affect the figure, but not enough to close a gap of several orders of magnitude. That one comparison decides most choices between the families where the source is delicate.

Answer: B

Bipolar devices give about forty millisiemens per milliamp against a few for a field effect device, because an exponential characteristic is far steeper than a square-law one. The load resistance multiplies both equally and so cannot change the ranking. The two families trade input impedance against transconductance, and knowing which way that trade runs settles most topology choices. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.