1A Junction Field-Effect Transistor (JFET) has three terminals. What are they?
Construction and Characteristics of JFETs
Easy
A.Drain, Source, Gate
B.Anode, Cathode, Gate
C.Positive, Negative, Ground
D.Emitter, Base, Collector
Correct Answer: Drain, Source, Gate
Explanation:
The three terminals of a JFET are the Drain (D), Source (S), and Gate (G). These terminals control the flow of current through the device's channel.
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2In an n-channel JFET, the width of the channel is controlled by applying a:
Construction and Characteristics of JFETs
Easy
A.Positive voltage to the source
B.Negative voltage to the gate with respect to the source
C.Negative voltage to the drain
D.Positive voltage to the gate with respect to the source
Correct Answer: Negative voltage to the gate with respect to the source
Explanation:
Applying a negative voltage to the gate of an n-channel JFET reverse-biases the gate-source p-n junction. This increases the depletion region, which narrows the conductive channel and reduces current flow.
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3What is the name for the gate-to-source voltage () at which the drain current () in a JFET is reduced to practically zero?
Construction and Characteristics of JFETs
Easy
A.Pinch-off Voltage ()
B.Cutoff Voltage ()
C.Threshold Voltage ()
D.Breakdown Voltage ()
Correct Answer: Cutoff Voltage ()
Explanation:
The gate-to-source cutoff voltage, , is the specific negative value of (for an n-channel JFET) that completely 'pinches off' the channel and stops the flow of drain current.
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4A key structural feature of a Depletion-Type MOSFET (D-MOSFET) is the presence of a:
Depletion-Type MOSFET
Easy
A.Break in the channel between the drain and source
B.Second gate terminal
C.Physical channel between the drain and source
D.Direct connection between the gate and channel
Correct Answer: Physical channel between the drain and source
Explanation:
Unlike an E-MOSFET, a D-MOSFET is fabricated with a physical conductive channel already existing between the drain and source. This allows current to flow even with zero gate voltage.
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5A Depletion-Type MOSFET (D-MOSFET) is known as a 'normally on' device because it conducts current when:
Depletion-Type MOSFET
Easy
A. is very large and positive
B. is very large and negative
C.The drain voltage is zero
D. V
Correct Answer: V
Explanation:
D-MOSFETs are considered 'normally on' because their pre-existing physical channel allows drain current to flow as soon as a drain-source voltage is applied, even with the gate-source voltage at zero.
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6In the schematic symbol for an n-channel Depletion-Type MOSFET, the line representing the channel is drawn as:
Depletion-Type MOSFET
Easy
A.A dashed or broken line
B.A solid, unbroken line
C.An arrow pointing inwards
D.A wavy line
Correct Answer: A solid, unbroken line
Explanation:
The solid line in the D-MOSFET symbol signifies the presence of a continuous, pre-existing physical channel between the drain and source terminals.
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7What is the primary structural difference in an Enhancement-Type MOSFET (E-MOSFET) compared to a D-MOSFET?
Enhancement-Type MOSFET
Easy
A.It has a thicker gate oxide layer
B.It is made only from p-type material
C.It has no pre-existing physical channel
D.It has four terminals instead of three
Correct Answer: It has no pre-existing physical channel
Explanation:
The key difference is that an E-MOSFET is built without a physical channel. A channel must be induced (enhanced) by applying a sufficient gate voltage before current can flow from drain to source.
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8For an n-channel Enhancement-Type MOSFET (E-MOSFET) to turn on and conduct, the gate-to-source voltage () must be:
Enhancement-Type MOSFET
Easy
A.Greater than the threshold voltage ()
B.Equal to zero
C.Negative
D.Less than the threshold voltage ()
Correct Answer: Greater than the threshold voltage ()
Explanation:
An E-MOSFET is a 'normally off' device. A conductive channel is only formed when the gate-to-source voltage exceeds a minimum value called the threshold voltage (), allowing current to flow.
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9In the schematic symbol for an n-channel Enhancement-Type MOSFET, the line representing the channel is drawn as:
Enhancement-Type MOSFET
Easy
A.A double line
B.An arrow pointing outwards
C.A solid, unbroken line
D.A dashed or broken line
Correct Answer: A dashed or broken line
Explanation:
The dashed or broken line in the E-MOSFET symbol signifies that there is no pre-existing physical channel; one must be induced by the gate voltage for the device to conduct.
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10In a JFET fixed-bias configuration, how is the gate-to-source voltage () established?
Fixed-Bias Configuration
Easy
A.It is always fixed at 0V
B.By the voltage drop across the drain resistor ()
C.By the voltage drop across the source resistor ()
D.By a separate DC voltage source () connected to the gate
Correct Answer: By a separate DC voltage source () connected to the gate
Explanation:
The 'fixed-bias' name comes from the fact that it uses a dedicated DC voltage source, , to set a fixed value for , which is independent of the drain circuit.
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11What is a primary disadvantage of the fixed-bias configuration for FETs?
Fixed-Bias Configuration
Easy
A.High complexity
B.Poor Q-point stability
C.Requires an AC power source
D.Low input impedance
Correct Answer: Poor Q-point stability
Explanation:
While simple to design, the fixed-bias circuit's operating point (Q-point) is very sensitive to variations in the FET's parameters (like and ), making it unstable.
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12In a JFET self-bias configuration, the negative gate-to-source voltage () is developed by the voltage drop across which component?
Self-Bias Configuration
Easy
A.The drain resistor ()
B.The main supply ()
C.The gate resistor ()
D.The source resistor ()
Correct Answer: The source resistor ()
Explanation:
In self-bias, the drain current () flows through , creating a positive voltage at the source (). Since the gate is at 0V, , resulting in a negative that biases the FET.
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13Why is the gate resistor () in a self-bias circuit usually a very large value (e.g., 1 MΩ)?
Self-Bias Configuration
Easy
A.To set the drain current precisely
B.To provide a large voltage drop at the gate
C.To limit the gate current to a safe level
D.To keep the gate at approximately 0 V DC and provide a high input impedance
Correct Answer: To keep the gate at approximately 0 V DC and provide a high input impedance
Explanation:
connects the gate to ground. Since negligible DC current flows into the gate, there is no voltage drop across , holding the gate at 0V. Its large value ensures a high input impedance for the amplifier.
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14What is the primary function of resistors and in a FET voltage-divider biasing circuit?
Voltage-Divider Biasing
Easy
A.To provide AC coupling to the input signal
B.To decrease the input impedance of the circuit
C.To directly limit the drain current
D.To set a specific, positive DC voltage at the gate
Correct Answer: To set a specific, positive DC voltage at the gate
Explanation:
The resistors and form a voltage divider that sets a fixed, positive DC voltage at the gate terminal (). This makes the overall biasing less dependent on the FET's characteristics.
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15Which biasing configuration for a JFET generally offers the best Q-point stability against variations in device parameters?
Voltage-Divider Biasing
Easy
A.Voltage-divider biasing
B.Fixed-bias
C.Self-bias
D.Gate-bias
Correct Answer: Voltage-divider biasing
Explanation:
Voltage-divider biasing provides the most stable operating point because the gate voltage is set firmly by the divider, making the circuit's Q-point largely independent of the FET's transfer characteristics.
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16On a JFET datasheet, what does the parameter represent?
Understanding the datasheets of FET
Easy
A.The drain-to-source leakage current when the FET is off
B.The maximum safe drain current the device can handle
C.The current flowing into the gate terminal
D.The drain current when the gate is shorted to the source ( V)
Correct Answer: The drain current when the gate is shorted to the source ( V)
Explanation:
stands for Drain-to-Source Saturation Current. It is a key parameter that represents the maximum drain current a JFET will pass, which occurs when is zero.
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17What kind of information is found in the 'Absolute Maximum Ratings' section of a FET datasheet?
Understanding the datasheets of FET
Easy
A.Graphs of the transfer characteristics
B.Recommended resistor values for biasing
C.Typical operating conditions for best performance
D.Voltage, current, and power limits that must not be exceeded
Correct Answer: Voltage, current, and power limits that must not be exceeded
Explanation:
The 'Absolute Maximum Ratings' section specifies the stress limits (e.g., maximum , , power dissipation) beyond which the device may be permanently damaged. These are not recommended operating values.
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18According to their datasheets, the 2N5457, 2N5458, and 2N5459 devices are all examples of which type of transistor?
Understanding the datasheet of FETs 2N5457, 2N5458, 2N5459
Easy
A.Bipolar Junction Transistors (BJTs)
B.N-Channel JFETs
C.P-Channel JFETs
D.Enhancement-Type MOSFETs
Correct Answer: N-Channel JFETs
Explanation:
The 2N5457/58/59 series are well-known general-purpose N-Channel Junction Field-Effect Transistors, primarily used for amplification and switching applications.
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19A major trend in modern microprocessors to create smaller and more power-efficient transistors is the use of which three-dimensional structure?
recent trends in electronics
Easy
A.UJT
B.BJT
C.JFET
D.FinFET
Correct Answer: FinFET
Explanation:
FinFETs are multi-gate, non-planar transistors where the gate wraps around the channel (the 'fin'). This structure provides better control, reducing leakage current and enabling the creation of smaller, faster, and more energy-efficient integrated circuits.
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20Which wide-bandgap semiconductor material is becoming increasingly popular for high-power and high-frequency applications, often replacing silicon?
recent trends in electronics
Easy
A.Gallium Nitride (GaN)
B.Polyester
C.Germanium (Ge)
D.Copper (Cu)
Correct Answer: Gallium Nitride (GaN)
Explanation:
Gallium Nitride (GaN) is a modern semiconductor material that can operate at much higher voltages, temperatures, and frequencies than traditional silicon. This makes it ideal for new trends in power supplies, electric vehicles, and 5G communications.
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21An n-channel JFET has parameters mA and V. Using Shockley's equation, determine the drain current when the gate-source voltage is -2 V, assuming the device is in the saturation region.
Construction and Characteristics of JFETs
Medium
A.3 mA
B.6 mA
C.12 mA
D.9 mA
Correct Answer: 3 mA
Explanation:
The drain current in the saturation region is given by Shockley's equation: . Plugging in the values: mA.
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22For a p-channel JFET, the pinch-off voltage is +5 V. At what drain-source voltage () does the device enter the saturation region if the gate-source voltage () is +2 V?
Construction and Characteristics of JFETs
Medium
A.-3 V
B.3 V
C.-7 V
D.7 V
Correct Answer: -3 V
Explanation:
For a JFET, saturation begins when . The pinch-off condition occurs at . For a p-channel JFET, is negative. So, V. The device enters saturation when becomes more negative than -3 V.
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23An n-channel D-MOSFET has mA and V. If it is operated with a positive gate-source voltage of V, what is the resulting drain current ?
Depletion-Type MOSFET
Medium
A.12.5 mA
B.8.0 mA
C.16.0 mA
D.4.5 mA
Correct Answer: 12.5 mA
Explanation:
A D-MOSFET can operate in enhancement mode with . The current is calculated using the same transconductance equation: mA.
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24An n-channel E-MOSFET has a threshold voltage V and a conduction parameter . To achieve a drain current of mA in the saturation region, what gate-source voltage () must be applied?
Enhancement-Type MOSFET
Medium
A.4 V
B.3 V
C.5 V
D.2 V
Correct Answer: 4 V
Explanation:
The drain current for an E-MOSFET in saturation is . We need to solve for : . This simplifies to . Taking the square root, (since must be ). Therefore, V.
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25How does the construction of a D-MOSFET differ from an E-MOSFET, allowing it to operate for both positive and negative gate voltages (for n-channel)?
Depletion-Type MOSFET
Medium
A.The D-MOSFET uses a different gate oxide material.
B.The D-MOSFET substrate is made of a different semiconductor material.
C.The D-MOSFET has no physical channel initially.
D.The D-MOSFET is fabricated with a physical channel between the drain and source.
Correct Answer: The D-MOSFET is fabricated with a physical channel between the drain and source.
Explanation:
The key difference is that a D-MOSFET is built with a physical n-type channel already diffused between the drain and source. This allows current to flow even with . A negative depletes this channel, while a positive enhances it. An E-MOSFET has no initial channel and requires a positive to create one.
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26A JFET fixed-bias circuit is supplied with V and V. The JFET parameters are mA and V. If the drain resistor is 1 kΩ, what is the drain-source voltage ?
Fixed-Bias Configuration
Medium
A.8 V
B.10 V
C.2 V
D.6 V
Correct Answer: 10 V
Explanation:
The correct option follows directly from the given concept and definitions.
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27A JFET with mA and V is used in a self-bias circuit with a source resistor kΩ. Which equation correctly represents the bias line for this circuit on the vs transfer characteristic plane?
Self-Bias Configuration
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
In a self-bias configuration, the gate is at 0V (since ) and the source is at . The gate-source voltage is . Rearranging this equation to express as a function of gives the bias line equation: . With kΩ, this becomes .
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28In a JFET voltage-divider bias circuit, V, MΩ, kΩ, kΩ. If the drain current is measured to be 2.5 mA, what is the operating gate-source voltage ?
Voltage-Divider Biasing
Medium
A.+5.18 V
B.+2.18 V
C.-3.00 V
D.-0.82 V
Correct Answer: -0.82 V
Explanation:
The correct option follows directly from the given concept and definitions.
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29A FET datasheet specifies a maximum power dissipation of mW at °C and a derating factor of 3.2 mW/°C above 25°C. What is the maximum power the device can safely dissipate at an ambient temperature of 75°C?
Understanding the datasheets of FET
Medium
A.560 mW
B.400 mW
C.240 mW
D.160 mW
Correct Answer: 240 mW
Explanation:
The temperature increase above the reference is . The reduction in power dissipation is the derating factor multiplied by this temperature change: mW. The maximum power at 75°C is the initial max power minus the reduction: mW.
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30In a JFET self-bias configuration, what is the effect of increasing the source resistance on the Q-point drain current ()?
Self-Bias Configuration
Medium
A. becomes equal to .
B. decreases.
C. remains unchanged.
D. increases.
Correct Answer: decreases.
Explanation:
The Q-point is the intersection of the transfer curve and the bias line . Increasing makes the slope of the bias line () less steep (closer to the horizontal axis). This causes the intersection point on the transfer curve to shift to a lower value of and a more negative value of .
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31An E-MOSFET is used as a switch controlled by a microcontroller. The MOSFET has V and is used to control a 12V lamp. The microcontroller output is 0 V for 'OFF' and 5 V for 'ON'. How does the MOSFET behave when the microcontroller output is 5 V?
Enhancement-Type MOSFET
Medium
A.It remains in cutoff.
B.It turns on and operates in the saturation region.
C.It turns on and operates in the ohmic (triode) region.
D.It remains off because V.
Correct Answer: It turns on and operates in the ohmic (triode) region.
Explanation:
When the microcontroller outputs 5 V, V. Since this is greater than V, the MOSFET turns on. When used as a switch, the goal is to have a very low voltage drop across it (). Since a large is applied and will be very small, the condition for ohmic operation () is met. This low ensures minimal power loss in the switch.
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32A designer needs a JFET for an amplifier and wants to bias it near the middle of its typical current range for good signal swing. The required drain current is about 5 mA. Which of the 2N5457, 2N5458, or 2N5459 series would be the most suitable choice based on their typical ranges?
Understanding the datasheet of FETs 2N5457, 2N5458, 2N5459
Medium
A.2N5457 ( range: 1-5 mA)
B.Any of them will work equally well.
C.2N5458 ( range: 2-9 mA)
D.2N5459 ( range: 4-16 mA)
Correct Answer: 2N5458 ( range: 2-9 mA)
Explanation:
The goal is to select a device where the desired operating current (5 mA) is roughly in the middle of its specified range. For the 2N5457, 5 mA is at the maximum limit. For the 2N5459, 5 mA is near the minimum limit. For the 2N5458, whose range is 2-9 mA, 5 mA is almost exactly in the center, making it the most suitable choice for predictable biasing and performance.
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33Why is voltage-divider biasing often preferred for FET amplifiers even though it requires more components than self-bias?
Voltage-Divider Biasing
Medium
A.It provides higher voltage gain.
B.It consumes less power.
C.It provides a higher input impedance.
D.It provides a Q-point that is less sensitive to variations in FET parameters.
Correct Answer: It provides a Q-point that is less sensitive to variations in FET parameters.
Explanation:
The main advantage of voltage-divider bias is stability. The gate voltage is held relatively constant by the divider resistors. This makes the operating point (, ) much less dependent on the FET's parameters (, ), which can vary significantly between devices. This is crucial for mass production and predictable circuit performance.
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34A self-biased JFET circuit has a Q-point of mA and V. What must be the value of the source resistor ?
Self-Bias Configuration
Medium
A.8 kΩ
B.1 kΩ
C.4 kΩ
D.2 kΩ
Correct Answer: 2 kΩ
Explanation:
In a self-bias circuit, the relationship between the Q-point parameters is given by the bias line equation . We can rearrange this to solve for : . Substituting the given values: Ω or 2 kΩ.
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35What is a major disadvantage of the fixed-bias configuration for JFETs that makes it less common in practical applications?
Fixed-Bias Configuration
Medium
A.It requires a negative power supply for n-channel JFETs.
B.It cannot be used for amplifier circuits.
C.It has very low input impedance.
D.Its Q-point is highly unstable and dependent on device parameter variations.
Correct Answer: Its Q-point is highly unstable and dependent on device parameter variations.
Explanation:
In fixed-bias, is set to a constant value. Since the JFET transfer characteristic ( vs ) varies significantly from one device to another (due to manufacturing spread in and ), a fixed will result in a widely varying drain current . This makes the Q-point unstable and unpredictable.
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36The transconductance () of a JFET is a measure of its effectiveness as an amplifier. How does change as the gate-source voltage () approaches the pinch-off voltage ()?
Construction and Characteristics of JFETs
Medium
A. becomes negative.
B. remains constant.
C. increases and reaches its maximum value, .
D. decreases and approaches zero.
Correct Answer: decreases and approaches zero.
Explanation:
Transconductance is defined by the relationship . As gets closer to , the term approaches zero. Consequently, decreases and becomes zero at pinch-off. The maximum transconductance, , occurs at .
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37A datasheet for a 2N5457 JFET lists the 'Forward Transfer Admittance', , with a typical value of 3000 µS. What does this parameter represent?
Understanding the datasheet of FETs 2N5457, 2N5458, 2N5459
Medium
A.The DC resistance of the channel.
B.The AC transconductance, .
C.The gate leakage current.
D.The input capacitance of the gate.
Correct Answer: The AC transconductance, .
Explanation:
Forward Transfer Admittance, denoted as , is the common-source AC forward transfer admittance. In the low-frequency region where capacitive effects are negligible, this parameter is equivalent to the AC transconductance, . It represents the ratio of the change in drain current to the change in gate-source voltage (). Its unit, Siemens (S) or mhos, is the unit of conductance.
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38Consider a voltage-divider biased JFET circuit. If the JFET is replaced with another one of the same part number but with a higher value, what is the likely effect on the Q-point drain current ()?
Voltage-Divider Biasing
Medium
A. will remain almost the same.
B. will increase significantly.
C. will decrease significantly.
D. will drop to zero.
Correct Answer: will remain almost the same.
Explanation:
This demonstrates the stability of voltage-divider bias. The gate voltage is fixed by the resistors. The source voltage is . The relation is . If a higher FET causes to increase, increases, making more negative. This more negative counteracts the initial increase in , providing negative feedback and keeping relatively stable.
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39On an n-channel JFET datasheet, the parameter is listed as -25 V. What is the physical meaning of this rating?
Understanding the datasheets of FET
Medium
A.It is the threshold voltage for conduction.
B.It is the maximum gate voltage for normal operation.
C.It is the pinch-off voltage.
D.It is the gate-source breakdown voltage with drain shorted to source.
Correct Answer: It is the gate-source breakdown voltage with drain shorted to source.
Explanation:
stands for Breakdown Voltage from Gate to Source, with the drain shorted to the source. It specifies the maximum reverse-bias voltage that can be applied to the gate-source p-n junction before avalanche breakdown occurs, which would permanently damage the device. The negative sign indicates the polarity for reverse bias on an n-channel JFET.
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40FinFETs are a recent advancement in MOSFET technology used in modern microprocessors. What is the primary structural feature of a FinFET that allows for better electrostatic control of the channel compared to a traditional planar MOSFET?
recent trends in electronics
Medium
A.The source and drain are placed vertically instead of horizontally.
B.The channel is a three-dimensional 'fin' and the gate wraps around it on multiple sides.
C.A thicker gate oxide layer to prevent leakage.
D.The use of a Gallium Nitride (GaN) substrate.
Correct Answer: The channel is a three-dimensional 'fin' and the gate wraps around it on multiple sides.
Explanation:
In a FinFET, the traditional flat (planar) channel is replaced by a three-dimensional silicon fin that rises vertically from the substrate. The gate electrode is wrapped around this fin on three sides (left, top, and right). This multi-gate structure gives the gate much stronger control over the channel, which significantly reduces short-channel effects and leakage currents, allowing for smaller and more efficient transistors.
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41In an N-channel JFET, as the operating temperature increases, the drain current () for a fixed tends to decrease. This is primarily because:
Construction and Characteristics of JFETs
Hard
A.The built-in potential of the p-n junction increases, which narrows the channel for a given .
B.The gate-to-channel junction leakage current () increases, effectively increasing the reverse bias.
C.The magnitude of the pinch-off voltage () decreases, and the mobility of charge carriers () decreases.
D.The channel resistance increases due to increased carrier scattering, while the pinch-off voltage remains constant.
Correct Answer: The magnitude of the pinch-off voltage () decreases, and the mobility of charge carriers () decreases.
Explanation:
Two main effects occur with increasing temperature in a JFET. First, the carrier mobility () decreases due to increased lattice scattering, which reduces conductivity and thus . Second, the magnitude of the pinch-off voltage () decreases. Both effects contribute to reducing the drain current for a given . The decrease in mobility is typically the more dominant effect.
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42The standard Shockley equation for a JFET, , is an ideal model. In reality, for a JFET operating in the saturation region, if is increased significantly, the drain current slightly increases. How is this effect, known as channel length modulation, typically modeled?
Construction and Characteristics of JFETs
Hard
A.By replacing with a voltage-dependent term .
B.By multiplying the entire Shockley equation by a factor of , where and is the Early voltage.
C.By multiplying the Shockley equation by a term , where is the channel length modulation parameter.
D.By adding a term proportional to to the pinch-off voltage, .
Correct Answer: By multiplying the entire Shockley equation by a factor of , where and is the Early voltage.
Explanation:
Similar to the Early effect in BJTs, channel length modulation in FETs accounts for the finite output resistance in the saturation region. It is modeled by multiplying the ideal current equation by a factor , where is the channel length modulation parameter and is the inverse of the Early Voltage, . This factor models the slight increase in with due to the effective shortening of the channel length.
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43A D-MOSFET is biased with V. Its datasheet specifies mA and V. Assuming the device operates in the saturation region, what is the approximate drain current ?
Depletion-Type MOSFET
Hard
A.10 mA
B.12.7 mA
C.7.65 mA
D.The device is in cutoff, so .
Correct Answer: 12.7 mA
Explanation:
The D-MOSFET can operate in both depletion mode () and enhancement mode (). The same transconductance equation applies: . Here, is equivalent to . Plugging in the values:
mA.
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44Consider a depletion-type MOSFET (D-MOSFET). Which statement most accurately compares its input impedance () when operating in depletion mode () versus enhancement mode ()?
Depletion-Type MOSFET
Hard
A. is high in depletion mode but significantly lower in enhancement mode due to the formation of a conducting channel.
B. is extremely high in both modes and virtually identical, as the gate is insulated by .
C. is high in depletion mode but drops to a low value in enhancement mode due to forward biasing the substrate junction.
D. is extremely high in depletion mode, but can decrease in enhancement mode if becomes large enough to cause gate oxide breakdown.
Correct Answer: is extremely high in both modes and virtually identical, as the gate is insulated by .
Explanation:
The key feature of any MOSFET is the insulated gate, typically made of silicon dioxide (). This insulating layer prevents significant DC current from flowing into the gate, regardless of whether is positive or negative (within its operating limits). Therefore, the input impedance is extremely high (typically to ) in both depletion and enhancement modes.
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45In an n-channel enhancement-type MOSFET, the threshold voltage () is specified as 2 V when the source and substrate (body) are both grounded. If the source terminal is connected to a potential of +3 V while the substrate remains grounded, how will the effective threshold voltage change?
Enhancement-Type MOSFET
Hard
A.It will remain exactly 2 V because is a fixed device parameter.
B.It will decrease significantly due to the increased forward bias on the source-substrate junction.
C.It will become negative, allowing the device to conduct with .
D.It will increase to a value greater than 2 V due to the body effect.
Correct Answer: It will increase to a value greater than 2 V due to the body effect.
Explanation:
This phenomenon is known as the body effect. The threshold voltage depends on the source-to-body voltage, . When is positive (source potential is higher than body), the depletion region under the channel widens, requiring a larger gate voltage to form the inversion layer. This increases the effective threshold voltage. The new threshold voltage is given by , where is the threshold voltage for . Since , will increase.
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46You need to design a simple switch using an E-MOSFET that is "normally closed" (i.e., conducting when the control signal is 0 V). Which of the following configurations would achieve this?
Enhancement-Type MOSFET
Hard
A.A P-channel E-MOSFET with its gate connected to ground via a pull-down resistor and the control signal being a positive voltage pulse.
B.An N-channel E-MOSFET cannot be used as a normally closed switch as it requires to turn on.
C.A P-channel E-MOSFET with its source connected to and its gate tied to ground. The control signal must be a high voltage to turn it off.
D.An N-channel E-MOSFET with its gate tied to a positive supply voltage via a large pull-up resistor and the control signal applied to the gate through a capacitor.
Correct Answer: A P-channel E-MOSFET with its source connected to and its gate tied to ground. The control signal must be a high voltage to turn it off.
Explanation:
A P-channel E-MOSFET turns on when its gate-source voltage () is more negative than its threshold voltage (, which is negative). By connecting the source to and the gate to ground, . As long as is more negative than , the device will be ON by default. Applying a high control voltage (close to ) to the gate makes , turning the device OFF.
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47A JFET fixed-bias circuit is designed with V. The JFET has parameters mA and V, resulting in a Q-point of ( V, mA). If this JFET is replaced by another one from the same family with mA and V, what will be the new drain current, ?
Fixed-Bias Configuration
Hard
A.2.0 mA
B.2.5 mA
C.3.6 mA
D.4.0 mA
Correct Answer: 3.6 mA
Explanation:
In a fixed-bias configuration, the gate-source voltage is determined solely by the biasing voltage source, so V. For the new JFET, we use its parameters with the fixed :
mA. This significant change highlights the poor Q-point stability of the fixed-bias configuration.
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48For an E-MOSFET in a fixed-bias configuration with V, k, and M, the gate is connected directly to . The MOSFET has V and mA/V. Determine the operating region of the MOSFET.
Fixed-Bias Configuration
Hard
A.Ohmic (triode) region
B.Breakdown region
C.Saturation region
D.Cutoff region
Correct Answer: Ohmic (triode) region
Explanation:
In this configuration, V. The condition for saturation is , which means must be V. The drain voltage is . Let's calculate the saturation current: mA. If this current were to flow, would be V, which is impossible. The circuit cannot supply this much current and maintain the saturation condition. Therefore, the MOSFET is driven hard into the ohmic region, where is very small and the current is limited by the external resistor .
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49A JFET self-bias circuit is designed using a resistor . Consider two JFETs from the same family. JFET1: mA, V. JFET2: mA, V. For a large value of , what is the primary behavior of the Q-point (, )?
Self-Bias Configuration
Hard
A. stabilizes at a value largely independent of JFET parameters.
B. stabilizes near , causing to be very small and relatively stable.
C. stabilizes near 0 V, causing to approach .
D. becomes highly dependent on JFET parameters, making the bias unstable.
Correct Answer: stabilizes at a value largely independent of JFET parameters.
Explanation:
The self-bias circuit provides stability via negative feedback. The bias line is . A large makes this line more horizontal on the transfer characteristics graph. The intersection of this nearly horizontal line with different JFET transfer curves (due to parameter spread) will occur at roughly the same value. The circuit essentially forces the drain current to be stable, while the adjusts to accommodate the specific JFET.
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50You are designing a self-bias circuit for a JFET with mA and V. The design requires the Q-point to be set at . What is the required value of the source resistor ?
Self-Bias Configuration
Hard
A.293
B.500
C.1.0 k
D.354
Correct Answer: 293
Explanation:
The correct option follows directly from the given concept and definitions.
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51In a standard JFET self-bias amplifier, a large capacitor () is placed in parallel with the source resistor . What is the primary reason for including this bypass capacitor?
Self-Bias Configuration
Hard
A.To filter out noise coming from the power supply through the source terminal.
B.To decrease the input impedance of the amplifier for better matching with the signal source.
C.To provide an AC ground at the source, thus preventing AC signal degeneration and maximizing voltage gain.
D.To increase the DC drain current and shift the Q-point.
Correct Answer: To provide an AC ground at the source, thus preventing AC signal degeneration and maximizing voltage gain.
Explanation:
The resistor , essential for DC biasing, introduces negative feedback for AC signals (degeneration), which reduces the amplifier's voltage gain. A bypass capacitor is chosen to have a very low impedance at signal frequencies. It effectively shorts the source terminal to ground for AC signals, while not affecting the DC bias condition. This removes the degenerative feedback, maximizing the AC voltage gain.
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52A voltage-divider bias circuit is designed for an N-channel JFET. If the JFET is replaced with an N-channel E-MOSFET with a threshold voltage , what condition must be met for the E-MOSFET to be turned ON?
Voltage-Divider Biasing
Hard
A.The voltage at the gate, , must be less than the source voltage, .
B.The circuit will not work because E-MOSFETs require positive .
C.The drain resistor must be small enough to prevent the MOSFET from saturating.
D.The voltage divider must be designed such that .
Correct Answer: The voltage divider must be designed such that .
Explanation:
An N-channel JFET operates with . In contrast, an N-channel E-MOSFET requires to turn on. In a voltage-divider circuit, . Therefore, the condition to turn on the E-MOSFET is , which can be rearranged to . The gate voltage must be sufficiently positive to overcome both the source voltage () and the device's inherent threshold voltage.
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53In designing a voltage-divider bias circuit for a JFET, to make the Q-point almost independent of the JFET's parameters (), which design choice is most effective?
Voltage-Divider Biasing
Hard
A.Choose (gate voltage) to be much larger than the magnitude of .
B.Set the Q-point exactly at .
C.Choose to be very small to maximize the drain current.
D.Choose and to be very large to maximize input impedance.
Correct Answer: Choose (gate voltage) to be much larger than the magnitude of .
Explanation:
The drain current is given by . The parameter variation is captured in , which can range from 0 to . If the fixed gate voltage is designed to be much larger than the maximum magnitude of (i.e., ), the term becomes negligible in comparison. This makes , which means the drain current is determined almost entirely by the external, stable resistor values and supply voltage, making it independent of the JFET's characteristics.
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54An E-MOSFET voltage-divider bias circuit has V, k, k, k, k. The MOSFET has V and mA/V. Find the approximate drain current .
Voltage-Divider Biasing
Hard
A.5.00 mA
B.1.25 mA
C.2.68 mA
D.9.32 mA
Correct Answer: 2.68 mA
Explanation:
The correct option follows directly from the given concept and definitions.
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55A MOSFET datasheet lists the parameter "Total Gate Charge ()" as 50 nC (nanocoulombs) at V. If you are designing a high-frequency switching circuit that drives this MOSFET's gate with a driver capable of sourcing/sinking a constant 2 A, what is the theoretical minimum time it will take to turn the MOSFET on (i.e., charge the gate from 0 V to 10 V)?
Understanding the datasheets of FET
Hard
A.500 ps
B.100 ns
C.25 ns
D.This cannot be determined without knowing the gate capacitance.
Correct Answer: 25 ns
Explanation:
The Total Gate Charge () represents the total charge required to switch the device state. The fundamental relationship between charge (), current (), and time () is . The gate driver provides the current. Therefore, the minimum switching time can be estimated as .
ns. The parameter is a more practical measure for switching time calculation than simple capacitance because it accounts for the Miller effect.
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56A JFET datasheet specifies a maximum power dissipation () of 400 mW at an ambient temperature () of 25°C, and a thermal resistance from junction-to-ambient () of 312.5 °C/W. What is the maximum allowable drain current () if the device is operating at °C with V?
Understanding the datasheets of FET
Hard
A.16 mA
B.40 mA
C.32 mA
D.24 mA
Correct Answer: 24 mA
Explanation:
The correct option follows directly from the given concept and definitions.
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57According to a typical datasheet for the 2N5457 N-channel JFET, the Gate-Source Cutoff Voltage () can range from -0.5 V to -6.0 V. If you design a simple self-bias circuit with this JFET, this wide parameter range implies that:
Understanding the datasheet of FETs 2N5457, 2N5458, 2N5459
Hard
A.A fixed-bias configuration would be a more stable choice for this JFET.
B.The Q-point drain current () could vary significantly, necessitating a biasing scheme that provides good current stability.
C.The input impedance will vary significantly depending on the specific JFET used.
D.The circuit's voltage gain will be highly predictable and stable.
Correct Answer: The Q-point drain current () could vary significantly, necessitating a biasing scheme that provides good current stability.
Explanation:
The wide specified range for key parameters like and means that two different 2N5457 transistors can have vastly different transfer characteristics. This device-to-device variation will cause the operating point (Q-point) to shift significantly in simple biasing circuits. A robust design must use a biasing scheme like voltage-divider or source-follower bias to stabilize the drain current against these variations, ensuring predictable circuit performance.
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58The datasheets for the 2N5457, 2N5458, and 2N5459 JFETs show that they have progressively higher typical values for and . For instance, typical values might be (3mA, -3V), (6mA, -4V), and (9mA, -5V) respectively. Based on this trend, what can be inferred about their relative transconductance () at ?
Understanding the datasheet of FETs 2N5457, 2N5458, 2N5459
Hard
A.The 2N5459 will have the highest at .
B.The 2N5457 will have the highest at .
C.Transconductance is independent of and .
D.All three will have approximately the same at .
Correct Answer: The 2N5459 will have the highest at .
Explanation:
Maximum transconductance, , occurs at and is given by the formula . Using the provided typical values:
2N5457: mS
2N5458: mS
2N5459: mS
The calculation shows a clear trend: the 2N5459, with the highest current and pinch-off voltage, also possesses the highest transconductance, making it suitable for applications requiring higher gain.
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59In modern sub-22nm semiconductor nodes, planar MOSFETs have been largely replaced by FinFETs. What is the primary structural difference in a FinFET that provides superior electrostatic control over the channel compared to a planar MOSFET?
recent trends in electronics
Hard
A.The channel is made of a different material, like silicon-germanium (SiGe), to enhance mobility.
B.The channel is a thin, raised "fin" of silicon, and the gate wraps around it on three sides.
C.The use of a high-k dielectric material for the gate oxide.
D.The source and drain regions are elevated above the substrate to reduce resistance.
Correct Answer: The channel is a thin, raised "fin" of silicon, and the gate wraps around it on three sides.
Explanation:
The defining innovation of the FinFET is its three-dimensional structure. The channel is no longer a flat layer but a vertical fin of silicon. The gate electrode is wrapped around this fin on three sides (top and both sides). This multi-gate structure gives the gate much stronger control over the entire channel, significantly reducing short-channel effects like leakage current (off-state current) and allowing for better performance at smaller scales. While other options like high-k dielectrics are also used, the 3D fin is the key structural change.
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60As semiconductor technology scales below 5nm, FinFETs are being succeeded by Gate-All-Around (GAAFET) architectures, such as those using nanosheets or nanowires. What is the fundamental advantage of the GAAFET structure over the FinFET structure?
recent trends in electronics
Hard
A.GAAFETs eliminate the need for a gate oxide, improving reliability.
B.GAAFETs can be manufactured at a much lower cost than FinFETs.
C.The gate material in a GAAFET completely surrounds the channel, providing near-perfect electrostatic control and minimizing leakage.
D.GAAFETs operate at much higher voltages, making them suitable for power electronics.
Correct Answer: The gate material in a GAAFET completely surrounds the channel, providing near-perfect electrostatic control and minimizing leakage.
Explanation:
The Gate-All-Around (GAA) architecture is the next step in transistor evolution. In a GAAFET, the gate material fully encloses the channel (which may be a nanowire or a stack of nanosheets). This provides gate control from all four sides, as opposed to the three sides in a FinFET. This complete control is electrostatically ideal, allowing for maximum suppression of short-channel effects and enabling further scaling of transistors to smaller dimensions with better performance and lower power leakage.