Field Effect Transistors Questions and Answers

Answer: C

Drain current through the source resistor raises the source above ground while the gate sits at ground through its bias resistor, so the gate is negative with respect to the source - exactly the reverse bias a depletion device needs. One resistor produces the required polarity without any second supply, which is why self bias is the standard arrangement for these devices and has no bipolar equivalent. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

Self bias makes the gate negative of the source, but an enhancement device needs the gate positive of the source to form a channel at all, so the arrangement cannot start. Gate current is zero either way, and the remaining objections do not arise. That two members of one family need opposite gate polarities is the first thing to establish before choosing a bias circuit for either.

Answer: A

A divider from the supply holds the gate above the source by more than the threshold voltage, often combined with a source resistor for stability. Self bias gives the wrong polarity and a negative supply would keep the device off. Because the required polarity differs between the modes, the bias circuit rather than the device is what has to be redrawn when substituting one for the other. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: B

A large resistor from drain to gate makes the gate voltage equal the drain voltage, which guarantees saturation and gives a stable operating point in one component. It is used with enhancement MOSFETs and is the field effect analogue of collector feedback bias. Since no gate current flows, the resistor causes no drop at all and can be as large as convenient. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

The source rises above the grounded gate by the resistive drop:

\[ \begin{aligned} V_{GS} &= -I_{D}R_{S} \\ &= -4\mathrm{m} \times 500 \\ &= -2 \ \mathrm{V} \end{aligned} \]

giving -2 volts. The sign is the whole point: it is negative because the gate is below the source, which is the reverse bias a depletion device requires. A positive answer would mean the gate junction was conducting.

Answer: D

A rise in drain current raises the source voltage, which makes the gate more negative of the source and reduces the current - negative feedback in the bias loop from a single resistor. The drain voltage and transconductance do change but neither closes the loop. It is the same mechanism as an emitter resistor in a bipolar stage, and it works for the same reason.

Answer: A

Both the saturation current and the pinch-off voltage vary two or three to one across a part number, so the operating point must be tolerant of a whole area of the transfer characteristic rather than a point. Thresholds are stable in service, transconductance does not follow the supply, and gate leakage is negligible either way. The spread is if anything worse than beta's, which is why graphical design against a maximum and minimum curve is conventional here. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: B

Drawing the self-bias line across both extreme transfer curves shows the range of operating points the design will actually produce, and the source resistor is chosen so that range is acceptable. Averaging the curves hides exactly the information wanted. The output load line answers a different question. Designing against a range rather than a nominal value is what makes a circuit manufacturable, and here the graphical method makes the range visible. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

A current source in the source lead fixes the drain current outright, leaving the gate voltage to settle wherever the device requires - which removes the parameter spread from the answer entirely. Self bias and drain feedback both reduce the spread without eliminating it, and a divider with no source resistor is the least stable of all. Fixing the quantity you care about directly, and letting the device choose the rest, is the most robust approach available.

Answer: D

The channel is a genuine resistance, conducting either way with no offset voltage, and the drain and source are interchangeable - which is exactly what passing an analogue signal requires, and precisely what a bipolar transistor with its junction drop cannot offer. The threshold does drift with temperature, and the gate draws no current at all. A switch that introduces no voltage of its own is the property that matters here. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.