Field Effect Transistors Questions and Answers

Answer: A

An exponential characteristic is far steeper than a square-law one, so a bipolar device gives about forty millisiemens per milliamp while a field effect device gives a few. The absence of gate current is the reason for the high input impedance, not the low transconductance. Channel resistance and capacitance affect other things. One difference in the underlying law produces the family's central trade: enormous input impedance bought with much less gain per milliamp.

Answer: B

The maximum transconductance follows from the square law's slope:

\[ \begin{aligned} g_{m0} &= \frac{2 I_{DSS}}{|V_{P}|} \\ &= \frac{16}{4} \\ &= 4 \ \mathrm{mS} \end{aligned} \]

giving 4 millisiemens, against the forty a bipolar device would give at eight milliamps. The factor of two in the numerator comes from differentiating the square, and omitting it is the usual slip here.

Answer: C

Transconductance falls linearly towards pinch-off:

\[ \begin{aligned} g_{m} &= g_{m0}\left(1 - \tfrac{-2}{-4}\right) \\ &= 4 \times 0.5 \\ &= 2 \ \mathrm{mS} \end{aligned} \]

so 2 millisiemens. Halving is right here because the transconductance is the derivative of a square and so is linear in gate voltage, even though the current itself falls to a quarter. Knowing which quantity is linear and which is squared avoids applying the wrong factor.

Answer: D

Beyond pinch-off the current is set by the gate and barely moves with drain voltage, which is exactly what a current source requires - a single junction field effect transistor with its gate tied to its source gives a two-terminal source of the saturation current. The absence of gate current is convenient but separate, the transconductance varies, and the ohmic region is the wrong one. A flat characteristic is a high output impedance, and that is all a current source is.

Answer: A

Common source is the analogue: it inverts, provides voltage gain, and is the default amplifier configuration. Common gate corresponds to common base and common drain to the emitter follower. The correspondence is exact enough that every result about impedance reflection and Miller effect carries across, which makes the second family much quicker to learn than the first. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: B

The gain is minus the transconductance times the drain load, exactly parallel to the bipolar case but with a much smaller transconductance, so the gains are correspondingly modest - tens rather than hundreds. The second option is the degenerated form. The parallel with the bipolar expression is worth noticing, since it means the same intuitions about loading and degeneration apply unchanged. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

Multiplying transconductance by load:

\[ A_{v} = -0.003 \times 4700 \approx -14 \]

giving about -14, an order of magnitude less than a bipolar stage with the same load resistor. That modest figure is the price of the enormous input impedance, and it is why field effect input stages are usually followed by bipolar gain rather than being asked to provide it all. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

An unbypassed source resistor divides the gain by one plus the transconductance times that resistance, exactly as an emitter resistor does in a bipolar stage. It is very much in the signal path, cannot raise the transconductance, and does not turn the stage into a current source. Because the transconductance is small, the same resistor produces far less degeneration here than it would in a bipolar stage, which is worth allowing for. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: A

The degeneration factor uses the transconductance and the source resistance:

\[ \begin{aligned} A_{v} &= -\frac{0.003 \times 4700}{1 + 0.003 \times 470} \\ &\approx -5.8 \end{aligned} \]

so about -5.8. The factor of one plus 1.41 is modest because the transconductance is small; the same resistor in a bipolar stage at the same current would reduce the gain far more. How much feedback a given resistor applies depends entirely on the device's transconductance.

Answer: B

Since the gate draws no current, the input impedance is whatever resistor sets the gate's direct potential - commonly a megohm or more. There is no beta multiplication to compute and no device parameter involved. That the designer chooses the input impedance directly, rather than inheriting it from the device, is the family's most useful practical property. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.