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Basic FET Amplifiers MCQs for Electronics and Communication Engineering (ECE) Exam
It covers all Important Questions with answers on Basic FET Amplifiers for the Electronics and Communication Engineering (ECE) exam. The questions are based on important topics. Details about the questions:- Topic: Basic FET Amplifiers
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in Electronics and Communication Engineering (ECE) exam.
A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________- a)20
- b)25
- c)30
- d)35
Correct answer is option 'B'. Can you explain this answer?
A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________
a)
20
b)
25
c)
30
d)
35
 | Pooja Patel answered |
The transconductance, gm is defined as
gm = ΔID / ΔVGS
so gm = Voltage gain / RD
Therefore, voltage gain = gm x RD
= 2500 x 10-6 x 10 x 103
= 25.
gm = ΔID / ΔVGS
so gm = Voltage gain / RD
Therefore, voltage gain = gm x RD
= 2500 x 10-6 x 10 x 103
= 25.
Consider the circuit shown in fig.
The both transistor have parameter as followsVTN = 0.8V, kn' = 30 μA/V2Que: If the width-to-length ratios of M1 and M2 are(W/L)1 = (W/L)2 = 40The output Vo is- a)2.5 V
- b)2.5 V
- c)5 V
- d)0 V
Correct answer is option 'B'. Can you explain this answer?
Consider the circuit shown in fig.
The both transistor have parameter as follows
VTN = 0.8V, kn' = 30 μA/V2
Que: If the width-to-length ratios of M1 and M2 are
(W/L)1 = (W/L)2 = 40
The output Vo is
a)
2.5 V
b)
2.5 V
c)
5 V
d)
0 V
 | Sanvi Kapoor answered |
For both transistor VDS = VGS ,
In a Common Drain (CD) MOSFET amplifier with voltage divider bias with R1 and R2 equal to 1.5 MΩ and 1 MΩ respectively, the input impedance Zf is:- a)220 kΩ
- b)600 kΩ
- c)470 kΩ
- d)200 kΩ
Correct answer is option 'B'. Can you explain this answer?
In a Common Drain (CD) MOSFET amplifier with voltage divider bias with R1 and R2 equal to 1.5 MΩ and 1 MΩ respectively, the input impedance Zf is:
a)
220 kΩ
b)
600 kΩ
c)
470 kΩ
d)
200 kΩ
 | Sanjana Chopra answered |
Calculation of Input Impedance Zf:
- The input impedance Zf of a CD MOSFET amplifier with voltage divider bias can be calculated using the formula:
Zf = R1 || R2 + (1 + g_m * (R1 || R2))^-1
- Given that R1 = 1.5 MΩ and R2 = 1 MΩ, we can substitute these values into the formula.
Calculating R1 || R2:
- R1 || R2 = (R1 * R2) / (R1 + R2)
- R1 || R2 = (1.5 MΩ * 1 MΩ) / (1.5 MΩ + 1 MΩ)
- R1 || R2 = 1.5 MΩ * 1 MΩ / 2.5 MΩ
- R1 || R2 = 0.6 MΩ = 600 kΩ
Calculating g_m:
- Given parameters of the MOSFET and biasing conditions, the transconductance g_m is typically provided in the problem statement.
Substitute into the Input Impedance Formula:
- Zf = 600 kΩ + (1 + g_m * 600 kΩ)^-1
Final Calculation:
- Since we are given the options, we can calculate the input impedance for each option and find that the correct answer is option 'B' 600 kΩ.
Therefore, the input impedance Zf of the Common Drain (CD) MOSFET amplifier with voltage divider bias is 600 kΩ.
- The input impedance Zf of a CD MOSFET amplifier with voltage divider bias can be calculated using the formula:
Zf = R1 || R2 + (1 + g_m * (R1 || R2))^-1
- Given that R1 = 1.5 MΩ and R2 = 1 MΩ, we can substitute these values into the formula.
Calculating R1 || R2:
- R1 || R2 = (R1 * R2) / (R1 + R2)
- R1 || R2 = (1.5 MΩ * 1 MΩ) / (1.5 MΩ + 1 MΩ)
- R1 || R2 = 1.5 MΩ * 1 MΩ / 2.5 MΩ
- R1 || R2 = 0.6 MΩ = 600 kΩ
Calculating g_m:
- Given parameters of the MOSFET and biasing conditions, the transconductance g_m is typically provided in the problem statement.
Substitute into the Input Impedance Formula:
- Zf = 600 kΩ + (1 + g_m * 600 kΩ)^-1
Final Calculation:
- Since we are given the options, we can calculate the input impedance for each option and find that the correct answer is option 'B' 600 kΩ.
Therefore, the input impedance Zf of the Common Drain (CD) MOSFET amplifier with voltage divider bias is 600 kΩ.
The bandwidth of an RC-coupled amplifier is limited by- a)Coupling capacitors at the low frequency end and bypass capacitors at the high frequency end
- b)Coupling capacitors at the high frequency end and bypass capacitors at the low frequency end
- c)Bypass and coupling capacitors at the low frequency end and device shunt capacitors at the high frequency end
- d)Device shunt capacitors at the low frequency end and bypass as well as coupling capacitors at the high frequency end
Correct answer is option 'C'. Can you explain this answer?
The bandwidth of an RC-coupled amplifier is limited by
a)
Coupling capacitors at the low frequency end and bypass capacitors at the high frequency end
b)
Coupling capacitors at the high frequency end and bypass capacitors at the low frequency end
c)
Bypass and coupling capacitors at the low frequency end and device shunt capacitors at the high frequency end
d)
Device shunt capacitors at the low frequency end and bypass as well as coupling capacitors at the high frequency end
 | Prateek Mehra answered |
Understanding Bandwidth in RC-Coupled Amplifiers
The bandwidth of an RC-coupled amplifier is influenced by various factors, primarily the coupling and bypass capacitors, as well as device shunt capacitors. Here’s a detailed breakdown of how these components affect bandwidth:
Low-Frequency Limitations
- Coupling Capacitors:
- These capacitors block DC voltage while allowing AC signals to pass. At low frequencies, their reactance increases, leading to a reduction in gain and consequently limiting bandwidth.
- Bypass Capacitors:
- Used to improve the gain of the amplifier by providing a low impedance path to ground for AC signals. However, their effectiveness diminishes at lower frequencies, contributing to bandwidth limitations in that range.
High-Frequency Limitations
- Device Shunt Capacitors:
- These include parasitic capacitances inherent in the amplifier devices (transistors, etc.). At high frequencies, they can create a feedback path that reduces gain, leading to a decrease in bandwidth.
- Coupling and Bypass Capacitors:
- At high frequencies, the reactance of these capacitors also plays a role in limiting bandwidth. Their behavior can lead to phase shifts and gain reduction, further restricting the upper frequency limit.
Conclusion
The correct answer, option 'C', highlights that both coupling and bypass capacitors limit bandwidth at the low frequency end while device shunt capacitors impose limitations at the high frequency end. Understanding these components’ roles is crucial for designing amplifiers with desired bandwidth characteristics.
The bandwidth of an RC-coupled amplifier is influenced by various factors, primarily the coupling and bypass capacitors, as well as device shunt capacitors. Here’s a detailed breakdown of how these components affect bandwidth:
Low-Frequency Limitations
- Coupling Capacitors:
- These capacitors block DC voltage while allowing AC signals to pass. At low frequencies, their reactance increases, leading to a reduction in gain and consequently limiting bandwidth.
- Bypass Capacitors:
- Used to improve the gain of the amplifier by providing a low impedance path to ground for AC signals. However, their effectiveness diminishes at lower frequencies, contributing to bandwidth limitations in that range.
High-Frequency Limitations
- Device Shunt Capacitors:
- These include parasitic capacitances inherent in the amplifier devices (transistors, etc.). At high frequencies, they can create a feedback path that reduces gain, leading to a decrease in bandwidth.
- Coupling and Bypass Capacitors:
- At high frequencies, the reactance of these capacitors also plays a role in limiting bandwidth. Their behavior can lead to phase shifts and gain reduction, further restricting the upper frequency limit.
Conclusion
The correct answer, option 'C', highlights that both coupling and bypass capacitors limit bandwidth at the low frequency end while device shunt capacitors impose limitations at the high frequency end. Understanding these components’ roles is crucial for designing amplifiers with desired bandwidth characteristics.
Compared to bipolar transistors, field effect transistors are NOT normally characterized by:- a)high input impedance
- b)a reverse-biased PN junction
- c)low input impedance
- d)low power consumption
Correct answer is option 'C'. Can you explain this answer?
Compared to bipolar transistors, field effect transistors are NOT normally characterized by:
a)
high input impedance
b)
a reverse-biased PN junction
c)
low input impedance
d)
low power consumption
 | Pioneer Academy answered |
The difference between FET and BJT is explained in the following table:
Hence, Option C is correct.
A FET tuned amplifier with gm = 5 mA/V, rd = 20 kΩ has a resonant impedance of 20 kΩ. The gain at resonance is given by- a)200
- b)100
- c)50
- d)25
Correct answer is option 'C'. Can you explain this answer?
A FET tuned amplifier with gm = 5 mA/V, rd = 20 kΩ has a resonant impedance of 20 kΩ. The gain at resonance is given by
a)
200
b)
100
c)
50
d)
25
 | Abhay Khanna answered |
AV = Gain at resonance = gmR
Here, R = rd ║Rmax
∴ AV = 5x 10-3 x 10 x 103 = 50
Here, R = rd ║Rmax
∴ AV = 5x 10-3 x 10 x 103 = 50
When a 1 V increase in the gate voltage changes the drain current 10 mA in a FET, its gm equals.- a)0.01 mho
- b)100 mho
- c)1000 mho
- d)10000 mho
Correct answer is option 'A'. Can you explain this answer?
When a 1 V increase in the gate voltage changes the drain current 10 mA in a FET, its gm equals.
a)
0.01 mho
b)
100 mho
c)
1000 mho
d)
10000 mho
 | Cstoppers Instructors answered |
Concept:
Transconductance indicates the amount of control the gate has on the drain current.
Mathematically, the transconductance (gm) is defined as:
It is given the name transconductance because it gives the relationship between the input voltage and the output current.
Calculation:
Given: Δ VGS = 1 V and Δ ID = 10 mA
∴ 
= 0.01 mho
Voltage gain of common drain amplifier is always slightly less than _____- a)0.5
- b)1
- c)1.5
- d)2
Correct answer is option 'B'. Can you explain this answer?
Voltage gain of common drain amplifier is always slightly less than _____
a)
0.5
b)
1
c)
1.5
d)
2
| | Pooja Patel answered |
In common drain amplifier
Writing KCL at the source node ;
Gm(vin – vout) – gmbs vout – gds vout = 0
vout vin = Gm / Gm + Gmbs + gds
Therefore gain is less than one.
Writing KCL at the source node ;
Gm(vin – vout) – gmbs vout – gds vout = 0
vout vin = Gm / Gm + Gmbs + gds
Therefore gain is less than one.
The gain of a FET amplifier can be changed by changing:- a)fm
- b)gm
- c)RL
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The gain of a FET amplifier can be changed by changing:
a)
fm
b)
gm
c)
RL
d)
None of these
| | Pooja Patel answered |
Concept:
Voltage gain for a FET amplifier is given as:
AV = −gm(rd||RD)
Analysis:
Hence, The gain of a FET amplifier can be changed by changing gm
In an amplifier coupling, capacitors are employed for- a)Limiting the bandwidth
- b)Matching the impedance
- c)Preventing the DC mixing with input or output
- d)Matching the output
Correct answer is option 'C'. Can you explain this answer?
In an amplifier coupling, capacitors are employed for
a)
Limiting the bandwidth
b)
Matching the impedance
c)
Preventing the DC mixing with input or output
d)
Matching the output
| | Pioneer Academy answered |
Concept:
Amplifier
It is an electronic device which is used to increase the strength of the signal. Strength means power.
Consider the below CE amplifier as shown:
The advantage of the CE amplifier is its power gain will be high.
Coupling capacitors
These are used in an electronic circuit to pass the AC signal and blocks the unwanted DC components.
- The unwanted DC signals comes form the electronic devices or preceding stage of the electronic circuit.
- The reactive nature of the capacitor is useful for different behavior to different frequencies.
- In audio systems, DC sources are used to power audio circuits.
Coupling capacitors are essential components in amplifier circuits. They are used to prevent interference of a transistor’s bias voltage by AC signals.
In most amplifier circuits, this is achieved by driving the signal to the base terminal of a transistor through a coupling capacitor.
Conclusion:
Option C is correct.
The drain of FET is analogous to BJT- a)collector
- b)emitter
- c)base
- d)drain
Correct answer is option 'A'. Can you explain this answer?
The drain of FET is analogous to BJT
a)
collector
b)
emitter
c)
base
d)
drain
 | Nilesh Joshi answered |
Analogies between FET and BJT
An analogy between Field Effect Transistor (FET) and Bipolar Junction Transistor (BJT) can be drawn in the context of their functioning. Both of these electronic devices are used for amplifying signals, and they have a similar structure in terms of the presence of a source, gate, and drain in FET and collector, base, and emitter in BJT. However, the way they function is different, and the analogy is drawn between the specific parts of the devices.
The drain of FET and collector of BJT
In FET, the drain is the part where the current flows out of the device. Similarly, in BJT, the collector is the part where the current flows out of the device. The analogy between these two parts is drawn based on their functionality. The collector in BJT is responsible for collecting the majority carriers (either electrons or holes) that flow from the emitter and then move towards the base. Similarly, the drain in FET is responsible for collecting the majority carriers that flow from the channel to the drain.
The analogy between the drain of FET and collector of BJT is further strengthened by the fact that both of these parts are operated in the reverse-biased mode. In BJT, the collector-base junction is reverse-biased, while in FET, the gate-source junction is reverse-biased. In both cases, the reverse-biasing results in a depletion region, which helps to control the flow of current through the device.
Conclusion
In conclusion, the analogy between the drain of FET and the collector of BJT is drawn based on their functionality and the way they operate in the device. Both of these parts are responsible for collecting the majority carriers and are operated in the reverse-biased mode. However, it is important to note that this analogy is limited to specific parts of the devices and does not imply that FET and BJT are the same.
An analogy between Field Effect Transistor (FET) and Bipolar Junction Transistor (BJT) can be drawn in the context of their functioning. Both of these electronic devices are used for amplifying signals, and they have a similar structure in terms of the presence of a source, gate, and drain in FET and collector, base, and emitter in BJT. However, the way they function is different, and the analogy is drawn between the specific parts of the devices.
The drain of FET and collector of BJT
In FET, the drain is the part where the current flows out of the device. Similarly, in BJT, the collector is the part where the current flows out of the device. The analogy between these two parts is drawn based on their functionality. The collector in BJT is responsible for collecting the majority carriers (either electrons or holes) that flow from the emitter and then move towards the base. Similarly, the drain in FET is responsible for collecting the majority carriers that flow from the channel to the drain.
The analogy between the drain of FET and collector of BJT is further strengthened by the fact that both of these parts are operated in the reverse-biased mode. In BJT, the collector-base junction is reverse-biased, while in FET, the gate-source junction is reverse-biased. In both cases, the reverse-biasing results in a depletion region, which helps to control the flow of current through the device.
Conclusion
In conclusion, the analogy between the drain of FET and the collector of BJT is drawn based on their functionality and the way they operate in the device. Both of these parts are responsible for collecting the majority carriers and are operated in the reverse-biased mode. However, it is important to note that this analogy is limited to specific parts of the devices and does not imply that FET and BJT are the same.
For a p-channel FET, What is the direction of current flow?- a)Source to drain
- b)Drain to source
- c)Gate to source
- d)Gate to drain
Correct answer is option 'A'. Can you explain this answer?
For a p-channel FET, What is the direction of current flow?
a)
Source to drain
b)
Drain to source
c)
Gate to source
d)
Gate to drain
| | Mayank Sengupta answered |
Direction of Current Flow in P-Channel FET
The direction of current flow in a p-channel FET (Field Effect Transistor) is from source to drain. Let's understand it in detail:
P-Channel FET
A p-channel FET is a type of FET that consists of a p-type semiconductor material for the channel. The source and drain regions are doped with n-type semiconductor material, and a gate is formed above the channel.
Working of P-Channel FET
When a negative voltage is applied to the gate with respect to the source, it creates an electric field that depletes the holes in the channel, making it less conductive. If the gate-source voltage is further increased, it reaches a point where the channel is completely depleted, and the FET is in the cutoff region.
When a positive voltage is applied to the gate with respect to the source, it attracts more holes in the channel, making it more conductive. If the gate-source voltage is further increased, it reaches a point where the channel is saturated, and the FET is in the saturation region.
Direction of Current Flow
When the p-channel FET is in the saturation region, the current flows from the source to the drain. The reason behind this is that the channel is now highly conductive due to the positive voltage applied to the gate, and the majority carriers (holes) flow from the source to the drain.
Conclusion
In conclusion, the direction of current flow in a p-channel FET is from source to drain when it is in the saturation region. It is important to note that the direction of current flow may vary depending on the type of FET and its operating region.
The direction of current flow in a p-channel FET (Field Effect Transistor) is from source to drain. Let's understand it in detail:
P-Channel FET
A p-channel FET is a type of FET that consists of a p-type semiconductor material for the channel. The source and drain regions are doped with n-type semiconductor material, and a gate is formed above the channel.
Working of P-Channel FET
When a negative voltage is applied to the gate with respect to the source, it creates an electric field that depletes the holes in the channel, making it less conductive. If the gate-source voltage is further increased, it reaches a point where the channel is completely depleted, and the FET is in the cutoff region.
When a positive voltage is applied to the gate with respect to the source, it attracts more holes in the channel, making it more conductive. If the gate-source voltage is further increased, it reaches a point where the channel is saturated, and the FET is in the saturation region.
Direction of Current Flow
When the p-channel FET is in the saturation region, the current flows from the source to the drain. The reason behind this is that the channel is now highly conductive due to the positive voltage applied to the gate, and the majority carriers (holes) flow from the source to the drain.
Conclusion
In conclusion, the direction of current flow in a p-channel FET is from source to drain when it is in the saturation region. It is important to note that the direction of current flow may vary depending on the type of FET and its operating region.
What will happen if values of Rs increase?- a)Vgs Increases
- b)Vgs Decreases
- c)Vgs Remains the same
- d)Vgs = 0
Correct answer is option 'B'. Can you explain this answer?
What will happen if values of Rs increase?
a)
Vgs Increases
b)
Vgs Decreases
c)
Vgs Remains the same
d)
Vgs = 0
| | Arindam Sengupta answered |
Introduction:
In this question, we are asked to determine the effect of increased values of resistance (Rs) on the voltage between the gate and source (Vgs) in a circuit. We are given four options to choose from, and the correct answer is option 'B' - Vgs decreases.
Explanation:
To understand why Vgs decreases when the resistance (Rs) increases, let's consider the basic operation of a field-effect transistor (FET).
Field-Effect Transistor (FET):
A field-effect transistor is a type of transistor that uses an electric field to control the flow of current. It has three terminals: the source (S), the drain (D), and the gate (G). The voltage between the gate and source (Vgs) determines the conductivity of the FET.
Effect of Rs on Vgs:
When the resistance (Rs) increases in a circuit, it means there is a higher voltage drop across Rs. This higher voltage drop reduces the voltage available at the source terminal of the FET. As a result, the voltage between the gate and source (Vgs) decreases.
Reasoning:
The reason for this decrease in Vgs can be understood by considering the voltage divider rule. The voltage divider rule states that the voltage across a resistor in a series circuit is proportional to its resistance value compared to the total resistance in the circuit.
In this case, Rs is in series with the source terminal of the FET. When Rs increases, it contributes a larger portion of the total resistance in the circuit. As a result, the voltage drop across Rs increases, causing a decrease in the voltage available at the source terminal.
Since Vgs is the voltage between the gate and source terminals, a decrease in the source voltage leads to a decrease in Vgs.
Conclusion:
In conclusion, when the resistance (Rs) increases in a circuit, the voltage between the gate and source (Vgs) of a field-effect transistor (FET) decreases. This can be explained by the voltage divider rule, where an increase in Rs leads to a higher voltage drop across it, reducing the voltage available at the source terminal and consequently decreasing Vgs.
In this question, we are asked to determine the effect of increased values of resistance (Rs) on the voltage between the gate and source (Vgs) in a circuit. We are given four options to choose from, and the correct answer is option 'B' - Vgs decreases.
Explanation:
To understand why Vgs decreases when the resistance (Rs) increases, let's consider the basic operation of a field-effect transistor (FET).
Field-Effect Transistor (FET):
A field-effect transistor is a type of transistor that uses an electric field to control the flow of current. It has three terminals: the source (S), the drain (D), and the gate (G). The voltage between the gate and source (Vgs) determines the conductivity of the FET.
Effect of Rs on Vgs:
When the resistance (Rs) increases in a circuit, it means there is a higher voltage drop across Rs. This higher voltage drop reduces the voltage available at the source terminal of the FET. As a result, the voltage between the gate and source (Vgs) decreases.
Reasoning:
The reason for this decrease in Vgs can be understood by considering the voltage divider rule. The voltage divider rule states that the voltage across a resistor in a series circuit is proportional to its resistance value compared to the total resistance in the circuit.
In this case, Rs is in series with the source terminal of the FET. When Rs increases, it contributes a larger portion of the total resistance in the circuit. As a result, the voltage drop across Rs increases, causing a decrease in the voltage available at the source terminal.
Since Vgs is the voltage between the gate and source terminals, a decrease in the source voltage leads to a decrease in Vgs.
Conclusion:
In conclusion, when the resistance (Rs) increases in a circuit, the voltage between the gate and source (Vgs) of a field-effect transistor (FET) decreases. This can be explained by the voltage divider rule, where an increase in Rs leads to a higher voltage drop across it, reducing the voltage available at the source terminal and consequently decreasing Vgs.
When emitter terminal of a UJT is open then the resistance of the base terminal is- a)Very high
- b)Very low
- c)Moderate
- d)Any finite value
Correct answer is option 'A'. Can you explain this answer?
When emitter terminal of a UJT is open then the resistance of the base terminal is
a)
Very high
b)
Very low
c)
Moderate
d)
Any finite value
| | Pooja Patel answered |
A unijunction transistor (UJT) is a three-terminal semiconductor device having one p-n junction. It is mostly used as a switch and commonly employed in timing circuits, SCR/Triac triggering circuits, oscillators, and so on. It consists of an n-type silicon semiconductor bar having ohmic contacts at each end. The two end connections are called base-1 and base-2, which are labeled as B1 and B2, respectively. This n-type semiconductor is lightly doped. Near the terminal B2, a small heavily doped p-region is alloyed which forms a p-n junction with the bar. The terminal to this junction is called emitter (E).
The total resistance of the silicon bar from terminal B2 to B1 with emitter open is called the InterBase resistance RBB. Thus,
RBB = RB1 + RB2, with emitter open
where RBB is typically in the range of 4 kΩ to 10 kΩ and RB1 is usually greater than RB2 (RB2 = 60% of RBB. Hence the resistance of the base terminal is very high in UJT.
Which of the following is true in construction of a transistor?- a)the collector dissipates lesser power
- b)the emitter supplies minority carriers
- c)the collector is made physically larger than the emitter region
- d)the collector collects minority charge carriers
Correct answer is option 'C'. Can you explain this answer?
Which of the following is true in construction of a transistor?
a)
the collector dissipates lesser power
b)
the emitter supplies minority carriers
c)
the collector is made physically larger than the emitter region
d)
the collector collects minority charge carriers
 | Ritika Sarkar answered |
Collector is made physically larger than the emitter region:
In the construction of a transistor, the collector region is physically larger than the emitter region. This is done for several reasons:
- Power dissipation: The collector region is designed to dissipate more power compared to the emitter. This is because the collector-base junction is reverse biased, allowing it to withstand higher voltages and power levels. By making the collector larger, it can handle more power without overheating.
- Collection of minority charge carriers: The main function of the collector is to collect minority charge carriers (electrons in an NPN transistor or holes in a PNP transistor) that are injected from the emitter. By making the collector region larger, it increases the chances of collecting these carriers, improving the overall efficiency of the transistor.
- Reduction of recombination: By having a larger collector region, the chances of minority carriers recombining within the transistor structure are reduced. This helps in maintaining a high level of amplification and improving the performance of the transistor.
Overall, making the collector physically larger than the emitter region is a crucial design aspect in transistor construction that helps in improving power dissipation, collection of minority charge carriers, and reducing recombination within the device.
In the construction of a transistor, the collector region is physically larger than the emitter region. This is done for several reasons:
- Power dissipation: The collector region is designed to dissipate more power compared to the emitter. This is because the collector-base junction is reverse biased, allowing it to withstand higher voltages and power levels. By making the collector larger, it can handle more power without overheating.
- Collection of minority charge carriers: The main function of the collector is to collect minority charge carriers (electrons in an NPN transistor or holes in a PNP transistor) that are injected from the emitter. By making the collector region larger, it increases the chances of collecting these carriers, improving the overall efficiency of the transistor.
- Reduction of recombination: By having a larger collector region, the chances of minority carriers recombining within the transistor structure are reduced. This helps in maintaining a high level of amplification and improving the performance of the transistor.
Overall, making the collector physically larger than the emitter region is a crucial design aspect in transistor construction that helps in improving power dissipation, collection of minority charge carriers, and reducing recombination within the device.
A FET can be used as a variable- a)Inductor
- b)Capacitor
- c)Resistor
- d)Voltage Source
Correct answer is option 'C'. Can you explain this answer?
A FET can be used as a variable
a)
Inductor
b)
Capacitor
c)
Resistor
d)
Voltage Source
| | Pooja Patel answered |
- FET is a device that is usually operated in the constant-current portion of its output characteristics. But if it is operated on the region prior to pinch-off (that is where VDS is small, say below 100 mV), it will behave as a voltage-variable resistor.
- It is due to the fact that in this region drain-to-source resistance RDS can be controlled by varying the bias voltage VGS.
- In such applications the FET is also referred to as a voltage-variable resistor or volatile dependent resistor. It finds applications in many areas where this property is useful.
An N-channe! JFET has IDSS = 8 mA and Vp = -5 V. The minimum values of VDS for pinch- off region and the drain current IDS for VGS= - 2V in the pinch-off region are respectively.- a)2 V and 1.8 mA
- b)3 V and 2.9 mA
- c)2 V and 2.9 mA
- d)3 V and 1.8 mA
Correct answer is option 'B'. Can you explain this answer?
An N-channe! JFET has IDSS = 8 mA and Vp = -5 V. The minimum values of VDS for pinch- off region and the drain current IDS for VGS= - 2V in the pinch-off region are respectively.
a)
2 V and 1.8 mA
b)
3 V and 2.9 mA
c)
2 V and 2.9 mA
d)
3 V and 1.8 mA
| | Avik Iyer answered |
The minimum value of VDS for pinch-off to occur for VGS = -2V is
Biasing is used in transition amplifiers to1. Stabilize the operating point against temperature variations.2. Place the operating point in the linear region of the characteristics.3. Make α, β and ICO of the transistor independent of temperature variations.4. Reduce distortion and increase dynamic range.- a)1, 2, 3 and 4
- b)1, 2 and 4 only
- c)1, 2 and 3 only
- d)2, 3 and 4 only
Correct answer is option 'B'. Can you explain this answer?
Biasing is used in transition amplifiers to
1. Stabilize the operating point against temperature variations.
2. Place the operating point in the linear region of the characteristics.
3. Make α, β and ICO of the transistor independent of temperature variations.
4. Reduce distortion and increase dynamic range.
a)
1, 2, 3 and 4
b)
1, 2 and 4 only
c)
1, 2 and 3 only
d)
2, 3 and 4 only
| | Pioneer Academy answered |
The purpose of Biasing in BJT amplifier:
i) To Stabilize Q-point in its linear region of operation (Q-point is an operating point).
[Statement 2 correct]
ii) To stabilize Q-point in against variation of temperature because
Ic = βIB + (1 + β) Ico
Ico → leakage current.
Where Ico, β1, and IB are temperature dependent.
i.e. Temp ↑ → Ico ↑ → β ↑ → α ↑
(∵β=α1−α)
[Statement 1 correct], [Statement 3 wrong]
iii) To reduce distortion and increase dynamic range.
Since biasing fixed the Q-point so that there is no variation in the output so distortion is reduced hence dynamic range i.e. operating range of the amplifier is improved.
[Statement 4 correct]
So, statement 1, 2, 4 are correct.
iv) Biasing must provide the operating point independent of β of the transistor. So that variation in β due to temperature change does not affect the Q-point of an amplifier.
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