Field Effect Transistors Questions and Answers

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.

Answer: C

A source follower's input impedance is set by a bias resistor and can be megohms, against an emitter follower's beta times the load. Its voltage gain is actually further from unity, because the transconductance is smaller, and its output impedance is higher for the same reason. Choosing between the two means deciding whether input impedance or output impedance matters more, and they pull in opposite directions. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

The follower's gain uses the same degeneration factor:

\[ \begin{aligned} A_{v} &= \frac{0.004 \times 1000}{1 + 0.004 \times 1000} \\ &= 0.8 \end{aligned} \]

giving 0.8, noticeably below an emitter follower's 0.99 because the transconductance is ten times smaller. A gain above one is impossible, which rules out 1.25 immediately. Unity is a limit these followers approach far less closely than their bipolar counterparts.

Answer: A

Looking into the source gives about one over the transconductance, in parallel with the source resistor - 250 ohms at four millisiemens, against tens of ohms for an emitter follower. There is no source-impedance division by beta here, because there is no beta. The absence of current gain removes both the loading on the source and the ability to lower the output impedance, which is the same fact seen from two sides. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

With the gate common, the gate-drain capacitance no longer bridges input and output across an inverting gain, so there is no Miller multiplication - precisely the common-base argument. The capacitance itself is a device property unchanged by configuration, and the transconductance does not rise with frequency. Bandwidth is about which capacitance sees which gain, in either device family.