Field Effect Transistors Questions and Answers

Q: At \( V_{GS} = -2 \ \mathrm{V} \) that device's transconductance falls to

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.

Q: Why does a field effect transistor make a good current source?

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.

Q: Which field effect configuration corresponds to the bipolar common emitter?

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.

Q: What is a common-source stage's voltage gain?

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.

Q: A stage has \( g_{m} = 3 \ \mathrm{mS} \) and \( R_{D} = 4.7 \ \mathrm{k\Omega} \). Its gain is about

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.

Q: How does a source resistor left unbypassed affect the gain?

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.

Q: What is a common-source stage's input impedance?

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.

21.

Why is a field effect transistor's transconductance lower than a bipolar device's at the same current?

A.Its transfer law is square rather than exponential
B.Its gate draws no current and so provides no drive
C.Its channel resistance is higher than a bipolar base's
D.Its input capacitance limits the current it can control

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.

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22.

A device has \( I_{DSS} = 8 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). Its transconductance at zero gate voltage is

A.1 mS
B.4 mS
C.2 mS
D.8 mS

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.

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23.

At \( V_{GS} = -2 \ \mathrm{V} \) that device's transconductance falls to

A.1 mS
B.8 mS
C.2 mS
D.4 mS

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.

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24.

Why does a field effect transistor make a good current source?

A.Its gate draws no current from the controlling circuit
B.Its transconductance is constant over the whole range
C.Its channel resistance is very low in the ohmic region
D.Its drain current is nearly independent of drain voltage

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.

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25.

Which field effect configuration corresponds to the bipolar common emitter?

A.Common source
B.Common gate
C.Common drain
D.None of the three

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.

Discuss in Forum

26.

What is a common-source stage's voltage gain?

A.Minus the supply voltage divided by the drain voltage
B.Minus the transconductance times the drain load
C.Minus the drain load divided by the source resistor
D.Minus the drain current times the load resistance

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.

Discuss in Forum

27.

A stage has \( g_{m} = 3 \ \mathrm{mS} \) and \( R_{D} = 4.7 \ \mathrm{k\Omega} \). Its gain is about

A.-140
B.-4.7
C.-14
D.-1.6

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.

Discuss in Forum

28.

How does a source resistor left unbypassed affect the gain?

A.It leaves the gain unchanged, being outside the signal path
B.It raises the gain by increasing the transconductance
C.It converts the stage into a current source
D.It reduces the gain in the usual feedback way

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.

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29.

Adding \( R_{S} = 470 \ \Omega \) unbypassed to that stage gives a gain of about

A.-5.8
B.-14
C.-10
D.-1

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.

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30.

What is a common-source stage's input impedance?

A.The channel resistance in the ohmic region of operation
B.Essentially the gate bias resistor
C.The transconductance times the drain load resistance
D.The source resistor multiplied by the 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.

Discuss in Forum

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