Field Effect Transistors Questions and Answers

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.

Q: A MOSFET amplifier works when new but fails intermittently after handling. What should be suspected?

Intermittent failure following handling is the classic signature of a partly punctured gate oxide, which may conduct erratically rather than failing outright. Resistor drift and supply faults would not correlate with handling, and thresholds are stable in normal service. A fault whose onset coincides with a particular activity points at the mechanism that activity exposes, and for MOSFETs that mechanism is almost always static.

61.

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.

Discuss in Forum

62.

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

63.

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.

Discuss in Forum

64.

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.

Discuss in Forum

65.

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.

Discuss in Forum

66.

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

67.

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.

Discuss in Forum

68.

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

69.

A MOSFET amplifier works when new but fails intermittently after handling. What should be suspected?

A.Gate oxide damage from static discharge
B.A source resistor drifting in value with age
C.The drain supply falling below its nominal figure
D.The threshold voltage rising as the device ages

Answer: A

Intermittent failure following handling is the classic signature of a partly punctured gate oxide, which may conduct erratically rather than failing outright. Resistor drift and supply faults would not correlate with handling, and thresholds are stable in normal service. A fault whose onset coincides with a particular activity points at the mechanism that activity exposes, and for MOSFETs that mechanism is almost always static.

Discuss in Forum

70.

A source follower shows a gain of only 0.6 where 0.9 was expected. What is the likely cause?

A.The output is loaded by too high an impedance
B.The operating current is lower than intended
C.The gate bias resistor is smaller than specified
D.The drain is connected to too low a supply voltage

Answer: B

The follower's gain depends on the transconductance times the source resistance, and transconductance falls with drain current, so a low operating current directly reduces the gain. The gate resistor affects input impedance, the drain supply affects headroom, and a high load impedance would raise the gain rather than lower it. Tracing a shortfall back to the parameter it is proportional to is the fastest route to the cause.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Field Effect Transistors Questions and Answers || Field Effect Transistors MCQ Questions and Answers || Field Effect Transistors GK Questions and Answers || Field Effect Transistors GK MCQ Questions || Field Effect Transistors Multiple Choice Questions and Answers || Field Effect Transistors GK || GK on Field Effect Transistors || Analog Electronics Questions and Answers || Analog Electronics MCQ Questions and Answers || Analog Electronics GK Questions and Answers || GK on Analog Electronics

Field Effect Transistors Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications