Field Effect Transistors Questions and Answers

Answer: C

Each transition moves charge onto or off a capacitance, and that energy is dissipated whatever the resistance in the path, so the power is proportional to capacitance, to the square of the supply voltage and to the frequency. Both devices do conduct briefly during a transition, adding a smaller shoot-through term, but the capacitive term dominates. Thresholds and on-resistance do not depend on frequency. That the power law contains the supply voltage squared is why reducing supply voltages has been the single most effective way of controlling power in digital circuits for forty years.

Answer: D

Because the energy per transition contains the supply voltage squared, halving the supply quarters the switching energy, which is a far larger saving than any other single change offers. Thresholds have in fact been scaled down along with the supply, oxides have become thinner rather than thicker, and channel resistance rises rather than falls as the supply is reduced. When a quantity appears squared in a power law, it is the first place to look for a saving, and that has driven the whole trajectory of the industry from five volts to under one.

Answer: A

A lower threshold means the device is never quite off, and the sub-threshold leakage grows exponentially as the threshold falls - so static power returns, which was the very problem the family was chosen to avoid. Oxide thickness, transconductance and the body effect all impose their own constraints but none is the binding one here. The trade is genuinely awkward: a high threshold wastes switching speed and a low one wastes static power, and the optimum has moved as circuits have grown. A limit that reintroduces the problem a technology was adopted to solve is the most instructive kind.

Answer: B

A vertical device carries its current from a source on the top surface down through the substrate to a drain on the bottom, which gives an enormous effective channel width for a given die area and lets the drift region be made as thick as the voltage rating demands. Gate charge is actually larger in such a device, the substrate is the drain rather than a separate terminal so the body effect question does not arise, and the structure is emphatically not symmetrical. Rearranging a device in the third dimension to escape a two-dimensional constraint is what made power MOSFETs practical at all.

Answer: C

The body region and the drain form a junction that conducts whenever the source is taken above the drain, so every power MOSFET contains a diode across it whether the designer wants one or not. The gate is insulated from everything, so no gate junction exists. In a bridge circuit this body diode is genuinely useful, carrying the freewheeling current an inductive load demands, but its slow recovery can also cause trouble at high frequency. A parasitic that is sometimes a feature and sometimes a nuisance has to be understood in both roles before a bridge is designed.

Answer: D

When the opposite device turns on, the body diode's stored charge must be swept out, and that appears as a spike of current through both devices at once - a real dissipation, and a source of switching noise as well. The diode carries the freewheeling current perfectly well rather than preventing it, and it affects neither the channel resistance nor the threshold. Reverse recovery in a parasitic that cannot be removed is one reason synchronous converters need careful dead-time control, and it is why some designs add a faster external diode in parallel.

Answer: A

Dead time is a deliberate gap in which neither device is driven, inserted so that one has certainly turned off before the other turns on. Without it both conduct together for an instant and short the supply, which is called shoot-through and is usually destructive. The other options describe turn-on delay, recovery and load behaviour, all of which influence how much dead time is needed but none of which is its definition. A margin inserted to cover the spread in two opposing delays is a recurring pattern wherever two switches must never be on at once.

Answer: B

The channel is enhanced progressively, so driving the gate to ten volts rather than to the four-volt threshold gives a much lower on-resistance and far less conduction loss - data sheets quote on-resistance at a stated gate voltage for exactly this reason. Threshold spread and the body effect are real but secondary, and a higher drive voltage does not itself charge the gate faster, since the charge required rises too. Driving a switch hard rather than merely past its threshold is the difference between a device that runs cool and one that needs a heatsink.

Answer: C

Applying the resistive loss law:

\[ P = I^{2}R = 100 \times 0.02 = 2 \ \mathrm{W} \]

giving 2 watts, which needs a heatsink in most enclosures. Note that the loss is quadratic in current: at twenty amps it would be eight watts rather than four. A bipolar device with a one-volt saturation drop would dissipate ten watts at ten amps but only twenty at twenty, so its loss grows more slowly - which is why the two families cross over as the current rises, and why the crossing point has to be computed rather than assumed.

Answer: D

The loss goes as the square of the current:

\[ P = 400 \times 0.02 = 8 \ \mathrm{W} \]

so 8 watts, four times the original rather than twice. Four watts would be the answer if the loss were proportional to current, which is the bipolar saturation-voltage case rather than this one. The order of a loss law decides how a design scales, and a quadratic law means headroom disappears far faster than intuition suggests as the current rating is approached.