Field Effect Transistors Questions and Answers

Q: Match each parameter with the device it belongs to. List I A. Pinch-off voltage B. Threshold voltage C. Current gain List II P. Junction field effect transistor Q. Enhancement MOSFET R. Bipolar transistor Select the correct match.

The pairing is A-P, B-Q, C-R. Each parameter presupposes a structure: a channel that can be pinched off, an inversion layer that appears above a threshold, or a base current that is amplified. Recognising which structure a parameter assumes is a quick way to place an unfamiliar specification in the right family.

Q: Two devices are compared at one milliamp. ParameterBipolarField effect Transconductance40 mS2 mSInput impedance5 k10 MGain with 5k load20010 Which application favours the field effect device?

Reading the table, the field effect device loses on gain by twenty to one but wins on input impedance by two thousand, so it belongs wherever the source must not be loaded. The other three applications all favour the bipolar column. A comparison table becomes a selection guide once the application names which row is binding.

Q: The gate voltage at which a junction field effect transistor's channel closes is the ______ voltage.

This is the pinch-off voltage. Threshold is the enhancement MOSFET's equivalent with the opposite sense, breakdown is a destructive limit, and saturation describes the region beyond pinch-off rather than a voltage. Four terms for turning points in a characteristic, and attaching each to its own device and mechanism is what keeps them from being interchanged.

Q: Which of the following is not a field effect transistor parameter? Current gain Pinch-off voltage Transconductance On-resistance

Current gain has no meaning for a device whose input current is zero - there is nothing to take a ratio to. The other three are standard field effect specifications describing the transfer characteristic, its slope and the conducting channel. A ratio whose denominator is zero is a reliable sign that a parameter has been carried across from the wrong family.

Q: Its transconductance at that point is

Differentiating the square law at the operating point:\[ g_{m} = 2k(V_{GS}-V_{T}) = 2(0.5)(3) = 3 \ \mathrm{mS} \]giving 3 millisiemens. Because the transconductance is the derivative of a square, it is linear in the overdrive voltage while the current is quadratic in it - so doubling the overdrive doubles the transconductance but quadruples the current. Keeping track of which quantity is which power avoids applying the wrong factor.

Q: Driving a \( 3.3 \ \mathrm{k\Omega} \) drain load, that stage's gain is about

Multiplying transconductance by load:\[ A_{v} = -0.003 \times 3300 \approx -10 \]giving about -10. A bipolar stage at the same 4.5 milliamps would give a transconductance of 180 millisiemens and a gain of nearly 600, which puts the family difference in perspective. Working the same load through both device types is the quickest way to feel how large that difference is.

Q: A junction field effect transistor stage draws its full saturation current whatever the input. Which fault fits?

With no source resistor there is no self bias at all, the gate sits at the source potential, and the device runs at its saturation current. An open gate resistor would leave the gate floating and the behaviour erratic rather than fixed. A low drain resistor would not affect the current in saturation, and a forward-biased gate would draw current and behave quite differently. Asking which component establishes the bias polarity localises this fault at once. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

91.

Match each parameter with the device it belongs to.

List I

A. Pinch-off voltage

B. Threshold voltage

C. Current gain

List II

P. Junction field effect transistor

Q. Enhancement MOSFET

R. Bipolar transistor

Select the correct match.

A.A-R, B-Q, C-P
B.A-P, B-R, C-Q
C.A-P, B-Q, C-R
D.A-Q, B-P, C-R

Answer: C

The pairing is A-P, B-Q, C-R. Each parameter presupposes a structure: a channel that can be pinched off, an inversion layer that appears above a threshold, or a base current that is amplified. Recognising which structure a parameter assumes is a quick way to place an unfamiliar specification in the right family.

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92.

Consider the following pairs of property and the family it favours.

Very high input impedance - Field effect

High transconductance per milliamp - Bipolar

Easy paralleling in power stages - Field effect

Freedom from static damage - Field effect

How many of the pairs given above are correctly matched?

A.Only one
B.Only two
C.All four
D.Only three

Answer: D

The first three are correctly paired and the fourth is not, so three match. Static damage is a field effect weakness rather than a strength, and follows directly from the insulated gate that gives the family its high input impedance. That the same structural feature produces the greatest advantage and the greatest fragility is worth holding together.

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93.

Two devices are compared at one milliamp.

Parameter	Bipolar	Field effect
Transconductance	40 mS	2 mS
Input impedance	5 k	10 M
Gain with 5k load	200	10

Which application favours the field effect device?

A.Amplifying from a high-impedance sensor
B.Providing the greatest voltage gain from one stage of amplification
C.Driving a low-impedance load without an additional buffer stage
D.Achieving the widest bandwidth at a given voltage gain

Answer: A

Reading the table, the field effect device loses on gain by twenty to one but wins on input impedance by two thousand, so it belongs wherever the source must not be loaded. The other three applications all favour the bipolar column. A comparison table becomes a selection guide once the application names which row is binding.

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94.

The gate voltage at which a junction field effect transistor's channel closes is the ______ voltage.

A.saturation
B.pinch-off
C.threshold
D.breakdown

Answer: B

This is the pinch-off voltage. Threshold is the enhancement MOSFET's equivalent with the opposite sense, breakdown is a destructive limit, and saturation describes the region beyond pinch-off rather than a voltage. Four terms for turning points in a characteristic, and attaching each to its own device and mechanism is what keeps them from being interchanged.

Discuss in Forum

95.

Which of the following is not a field effect transistor parameter?

Current gain

Pinch-off voltage

Transconductance

On-resistance

A.Transconductance
B.On-resistance
C.Current gain
D.Pinch-off voltage

Answer: C

Current gain has no meaning for a device whose input current is zero - there is nothing to take a ratio to. The other three are standard field effect specifications describing the transfer characteristic, its slope and the conducting channel. A ratio whose denominator is zero is a reliable sign that a parameter has been carried across from the wrong family.

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96.

A MOSFET has \( V_{T} = 2 \ \mathrm{V} \) and \( k = 0.5 \ \mathrm{mA/V^{2}} \). At \( V_{GS} = 5 \ \mathrm{V} \) its drain current is

A.1.5 mA
B.12.5 mA
C.9 mA
D.4.5 mA

Answer: D

Applying the enhancement-mode square law to the overdrive voltage:

\[ \begin{aligned} I_{D} &= k(V_{GS}-V_{T})^{2} \\ &= 0.5(3)^{2} = 4.5 \ \mathrm{mA} \end{aligned} \]

giving 4.5 mA. Using the full five volts rather than the three volts of overdrive gives 12.5 mA, which is the standard error. Subtracting the threshold before squaring is the one step that matters, and it is easy to overlook when the numbers are convenient.

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97.

Its transconductance at that point is

A.3 mS
B.4.5 mS
C.1.5 mS
D.5 mS

Answer: A

Differentiating the square law at the operating point:

\[ g_{m} = 2k(V_{GS}-V_{T}) = 2(0.5)(3) = 3 \ \mathrm{mS} \]

giving 3 millisiemens. Because the transconductance is the derivative of a square, it is linear in the overdrive voltage while the current is quadratic in it - so doubling the overdrive doubles the transconductance but quadruples the current. Keeping track of which quantity is which power avoids applying the wrong factor.

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98.

Driving a \( 3.3 \ \mathrm{k\Omega} \) drain load, that stage's gain is about

A.-15
B.-10
C.-3
D.-33

Answer: B

Multiplying transconductance by load:

\[ A_{v} = -0.003 \times 3300 \approx -10 \]

giving about -10. A bipolar stage at the same 4.5 milliamps would give a transconductance of 180 millisiemens and a gain of nearly 600, which puts the family difference in perspective. Working the same load through both device types is the quickest way to feel how large that difference is.

Discuss in Forum

99.

A junction device is self biased with \( R_{S} = 1 \ \mathrm{k\Omega} \), \( I_{DSS} = 10 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). The drain current settles near

A.10 mA
B.0.6 mA
C.2.2 mA
D.5 mA

Answer: C

The bias line and the transfer curve must agree, so the current satisfies both:

\[ \begin{aligned} I_{D} \\ &= 10\left(1 - \tfrac{-I_{D} \times 1000}{-4}\right)^{2} \end{aligned} \]

whose solution is about 2.2 mA, giving a gate-source voltage near -2.2 volts. Solving a device curve against a circuit line is exactly the graphical method's algebraic equivalent, and iterating two or three times converges quickly.

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100.

A junction field effect transistor stage draws its full saturation current whatever the input. Which fault fits?

A.The gate bias resistor has gone open circuit
B.The drain load resistor has fallen in value
C.The gate has been forward biased by a leaky capacitor
D.The source resistor has gone open circuit

Answer: D

With no source resistor there is no self bias at all, the gate sits at the source potential, and the device runs at its saturation current. An open gate resistor would leave the gate floating and the behaviour erratic rather than fixed. A low drain resistor would not affect the current in saturation, and a forward-biased gate would draw current and behave quite differently. Asking which component establishes the bias polarity localises this fault at once. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Discuss in Forum

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