Field Effect Transistors Questions and Answers

Q: Why is a source follower preferred to an emitter follower in some designs?

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.

Q: A source follower has \( g_{m} = 4 \ \mathrm{mS} \) and \( R_{S} = 1 \ \mathrm{k\Omega} \). Its gain is about

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.

Q: What is a source follower's output impedance?

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.

Q: Why does a common-gate stage have a wide bandwidth?

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.

Q: How does self bias set a junction field effect transistor's operating point?

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.

Q: Why is self bias unsuitable for an enhancement-mode MOSFET?

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.

Q: How is an enhancement-mode MOSFET usually biased?

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.

Q: What is drain feedback bias?

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.

Q: Why is self bias self-correcting?

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.

31.

Why is a source follower preferred to an emitter follower in some designs?

A.Its output impedance is lower than an emitter follower's
B.It introduces less phase shift at high frequencies
C.Its input impedance is far higher
D.Its voltage gain is closer to unity than a follower's

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.

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32.

A source follower has \( g_{m} = 4 \ \mathrm{mS} \) and \( R_{S} = 1 \ \mathrm{k\Omega} \). Its gain is about

A.0.99
B.0.5
C.1.25
D.0.8

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.

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33.

What is a source follower's output impedance?

A.The reciprocal of the transconductance, roughly
B.The source resistor divided by the transconductance
C.The gate bias resistor divided by the transconductance
D.The drain load in parallel with the source resistor

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.

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34.

Why does a common-gate stage have a wide bandwidth?

A.Its transconductance rises with frequency in that arrangement
B.No inverting gain bridges its input and output
C.Its gate capacitance is smaller than in other configurations
D.Its input impedance is low enough to swamp all capacitance

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.

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35.

How does self bias set a junction field effect transistor's operating point?

A.A resistor from drain to gate feeds back the drain voltage
B.A negative supply is applied directly to the gate terminal
C.A source resistor makes the gate negative of the source
D.A divider holds the gate at a fixed positive voltage

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.

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36.

Why is self bias unsuitable for an enhancement-mode MOSFET?

A.It cannot supply enough gate current to the device
B.It makes the operating point depend on the threshold voltage
C.It requires a source resistor larger than is practical
D.It produces the wrong polarity of gate bias

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.

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37.

How is an enhancement-mode MOSFET usually biased?

A.A divider making the gate positive of the source
B.A source resistor alone, as in self bias
C.A negative supply connected to the gate terminal
D.A resistor from source to gate with no divider

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.

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38.

What is drain feedback bias?

A.A source resistor bypassed for signals but not for direct current
B.A resistor from drain to gate holding the device in saturation
C.A resistor from drain to source setting the operating current
D.A divider on the gate fed from the drain rather than the supply

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.

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39.

A self-biased device carries \( 4 \ \mathrm{mA} \) through \( R_{S} = 500 \ \Omega \). Its gate-source voltage is

A.-4 V
B.-0.5 V
C.-2 V
D.+2 V

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.

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40.

Why is self bias self-correcting?

A.More current lowers the drain voltage and reduces the gain
B.More current increases the transconductance and the feedback
C.More current reduces the channel resistance and the drop
D.More current raises the source and reduces the bias

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.

Discuss in Forum

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