All questions of Amplifiers for UPSC CSE Exam

Calculation of Voltage Gain in FET Circuit:
1. Given:
- Transconductance (gm) = 2500 µS = 2.5 mS
- Drain resistance (RD) = 10 kΩ = 10000 Ω
2. Formula for Voltage Gain (Av):
Av = -gm * RD
3. Calculation:
Av = -2.5 mS * 10 kΩ
Av = -25
4. Result:
Therefore, the voltage gain of the FET circuit is 25.

Explanation:
The voltage gain of a FET circuit is determined by the product of the transconductance (gm) and the drain resistance (RD). In this case, with a transconductance of 2500 µS (2.5 mS) and a drain resistance of 10 kΩ, the voltage gain is calculated to be -25. The negative sign indicates that the output signal is inverted with respect to the input signal. Hence, the correct answer is option 'B' with a voltage gain of 25.

To find the input resistance of the network, we can use the formula:

Input resistance (Rin) = (Voltage gain / Current gain) * (Feedback factor / Amplifier gain)

Given:
Amplifier gain (Av) = 100
Feedback factor (β) = 5

The voltage gain (Av) can be calculated as the ratio of output voltage (Vo) to input voltage (Vi):

Av = Vo / Vi

Since it is a voltage series feedback network, the output voltage (Vo) and input voltage (Vi) are related by the formula:

Vo = Vi / (1 + β * Av)

Substituting the given values:

Av = 100
β = 5

Vo = Vi / (1 + 5 * 100)
Vo = Vi / 501

Now, we can calculate the output voltage (Vo) using the formula:

Vo = Av * Vi

Vo = 100 * 4V
Vo = 400V

Substituting the calculated values into the equation for Vo:

400V = Vi / 501

Solving for Vi:

Vi = 400V * 501
Vi = 200,400V

Now, we can substitute the calculated values into the formula for input resistance (Rin):

Rin = (Av / Ai) * (β / Av)

Rin = (100 / (2mA / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = 1,000 * 0.05
Rin = 50

Therefore, the input resistance of the network is 50 ohms.

Ri/Ro correct

Introduction:
The common drain amplifier, also known as a source follower, is a type of field-effect transistor (FET) amplifier. It is commonly used to provide a high input impedance and low output impedance. The voltage gain of the common drain amplifier can be determined by analyzing its small-signal equivalent circuit.

Explanation:
The voltage gain of a common drain amplifier is determined by the ratio of the output voltage to the input voltage. In the common drain configuration, the input is applied to the gate terminal, and the output is taken from the drain terminal. The source terminal is connected to a fixed reference voltage.

Operating Principle:
To understand the voltage gain of the common drain amplifier, it is important to understand its operating principle. When a small AC signal is applied to the gate terminal, it modulates the channel current flowing through the transistor. This modulated current flows through the load resistor connected to the drain terminal, producing an output voltage across it.

Factors Affecting Voltage Gain:
The voltage gain of a common drain amplifier is affected by several factors, including the transconductance (gm) of the transistor, the load resistor (RL), and the drain resistor (RD). The transconductance represents the ability of the transistor to convert a small change in the input voltage into a corresponding change in the output current.

Analysis of Small-Signal Equivalent Circuit:
To determine the voltage gain, we can analyze the small-signal equivalent circuit of the common drain amplifier. The small-signal equivalent circuit includes the transistor, the load resistor, and the drain resistor. By applying nodal analysis to this circuit, we can derive an expression for the voltage gain.

Voltage Gain Calculation:
The voltage gain (Av) of the common drain amplifier can be calculated using the following equation:

Av = -gm * RL

Where:
- Av is the voltage gain
- gm is the transconductance of the transistor
- RL is the load resistor

Conclusion:
In conclusion, the voltage gain of a common drain amplifier is always slightly less than 1. This is because the transconductance of the transistor is typically less than 1, and the voltage gain is directly proportional to the transconductance. Therefore, the correct answer is option B) 1.

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.

To understand why the gain of the amplifier without feedback must have been 600, we need to analyze the relationship between gain and negative feedback.

1. Gain reduction due to negative feedback:
When negative feedback is applied to an amplifier, it reduces the gain of the amplifier. In this case, the gain of the amplifier was reduced from 200 to 100.

2. Gain increase with feedback:
Now, we want to raise the gain with the same feedback to 150. This means that we need to increase the gain of the amplifier without feedback by a certain factor.

3. Relationship between gain with feedback and gain without feedback:
The gain with feedback (A_f) can be calculated using the following formula:
A_f = A / (1 + Aβ)
where A is the gain without feedback and β is the feedback factor.

4. Solving for the gain without feedback:
Given that the gain with feedback (A_f) is 150 and the gain with feedback (A) was previously 100, we can rearrange the formula to solve for the gain without feedback (A):
150 = A / (1 + Aβ)
150(1 + Aβ) = A
150 + 150Aβ = A
150Aβ - A = -150
A(150β - 1) = -150
A = -150 / (150β - 1)

5. Calculating the gain without feedback:
Now, we need to find the value of β that corresponds to the given gain reduction. The gain was reduced from 200 to 100, so the gain reduction is 200 - 100 = 100.

Using the formula for gain reduction due to negative feedback:
Gain reduction = A / (1 + Aβ) - A
100 = 200 / (1 + 200β) - 200
100(1 + 200β) = 200 - 200(1 + 200β)
100 + 20000β = 200 - 200 - 40000β
20000β + 40000β = 100
60000β = 100
β = 100 / 60000
β = 1/600

Substituting the value of β into the equation for A:
A = -150 / (150(1/600) - 1)
A = -150 / (1/4 - 1)
A = -150 / (-3/4)
A = 150 * 4/3
A = 200

6. Conclusion:
The gain without feedback (A) must have been 200 in order to achieve a gain of 150 with the given feedback. Therefore, the correct answer is option D) 600.

To solve this problem, we can use the formula for the reduction in distortion due to negative feedback:

Rd = 1 - (1 + β) * Hd

Where Rd is the reduction in distortion, β is the feedback factor, and Hd is the distortion without feedback.

Given that the reduction in distortion is desired to be from 8% to 1%, we can calculate the initial distortion without feedback (Hd) as:

Hd = 8% = 0.08

And the desired distortion with feedback (Hd') as:

Hd' = 1% = 0.01

We are also given that the feedback factor (β) is 10%.

Using the formula, we can calculate the reduction in distortion (Rd) as:

Rd = 1 - (1 + 0.1) * 0.08
= 1 - 1.08 * 0.08
= 1 - 0.0864
= 0.9136

Next, we can calculate the gain of the amplifier with reduced distortion (A2) using the formula:

A2 = A1 / (1 + β * A1 * Rd)

Where A1 is the gain of the amplifier with original distortion.

Since we want to find the relationship between A1 and A2, we can rearrange the formula to:

A2 = A1 / (1 + β * A1 * Rd)
A2 * (1 + β * A1 * Rd) = A1
A2 + β * A1 * Rd * A2 = A1
A2 - A1 = - β * A1 * Rd * A2
(A2 - A1) / (A1 * A2) = - β * Rd
(A2 - A1) / (A1 * A2 * Rd) = - β

Therefore, A1 * A2 = -(A2 - A1) / (β * Rd)

Substituting the values we have calculated:

A1 * A2 = -(A2 - A1) / (0.1 * 0.9136)
= -(A2 - A1) / 0.09136

An amplifier has an open loop voltage gain of "A." The open loop voltage gain, also known as the gain without any feedback applied, is a measure of how much the amplifier amplifies the input voltage. It is typically expressed as a ratio or in decibels.

The open loop voltage gain can be calculated by dividing the output voltage by the input voltage:

Open Loop Voltage Gain (A) = Output Voltage / Input Voltage

For example, if an amplifier has an output voltage of 10V and an input voltage of 1V, the open loop voltage gain would be 10.

It is important to note that the open loop voltage gain is not a fixed value and can vary depending on the frequency of the input signal. Amplifiers often have a frequency response curve that shows how the gain changes with frequency.

Additionally, the open loop voltage gain is usually much higher than the closed loop voltage gain when feedback is applied. Feedback is commonly used to stabilize the amplifier's performance and reduce distortion.

Understanding Low-Pass and High-Pass Networks
Low-pass (LP) and high-pass (HP) networks are essential components in electrical engineering, particularly in signal processing. They are distinguished by their frequency response characteristics.
Low-Pass (LP) Networks
- Functionality: LP networks are designed to allow low frequencies to pass through while attenuating higher frequencies.
- DC Response: They pass direct current (DC) and low-frequency signals effectively.
- Frequency Attenuation: As the frequency increases beyond a certain cutoff point, the LP network reduces the signal amplitude significantly.
High-Pass (HP) Networks
- Functionality: HP networks do the opposite; they allow high frequencies to pass while attenuating low frequencies.
- DC Response: HP networks block DC (zero frequency) and low-frequency signals.
- Frequency Attenuation: Signals with frequencies below a specific cutoff frequency are significantly reduced in amplitude.
Explanation of Option B
- Correct Statement: Option B accurately states that LP networks pass DC and low frequencies while attenuating high frequencies, and the opposite holds true for HP networks.
- Comparison: This characterization aligns with the basic definitions of LP and HP filters in electrical engineering.
Conclusion
Understanding these characteristics is crucial for designing circuits that meet specific frequency requirements, ensuring optimal signal processing in various applications.

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.

Understanding Voltage Amplifier and Transconductance Amplifier
The problem compares the open circuit voltage of a voltage amplifier (AV) with the short circuit transconductance (Gm) of a transconductance amplifier. To understand why the correct answer is R0, we need to delve into the definitions and relationships between these parameters.

Definitions
- **Open Circuit Voltage (AV)**: This is the output voltage of the voltage amplifier when the output is not loaded, meaning the load resistance is infinite.
- **Short Circuit Transconductance (Gm)**: This represents the output current per unit of input voltage change, measured when the output is shorted (i.e., output resistance is zero).

Relationship Between Parameters
1. **Internal Resistance (Ri)**:
- Both amplifiers have the same internal resistance, which affects how much of the input signal is transferred to the output.
2. **Output Resistance (R0)**:
- This is crucial in determining how the amplifier behaves under load. For the voltage amplifier, it can be seen as providing a certain voltage output when loaded.

Ratio Derivation
- The open circuit voltage for a voltage amplifier can be expressed as:
**AV = Vout / Vin**.
- The short circuit transconductance for a transconductance amplifier can be expressed as:
**Gm = Iout / Vin**.
- Since both amplifiers have the same output resistance (R0), the relationship between the output current and voltage can be simplified.

Conclusion
Thus, the ratio of open circuit voltage (AV) to short circuit transconductance (Gm) simplifies to the output resistance R0, leading to the final answer: **R0**. This indicates that under equivalent conditions, the output performance of both amplifiers is fundamentally tied to their output resistance.

Solution:

Given data:

Return Ratio = 24

Feedback Factor = 3

Change in open loop gain = 2

Relative change in gain with negative feedback = ?

Formula used:

Relative change in gain with negative feedback = ΔA / (1 + Aβ)

Where,

ΔA = Change in open-loop gain

A = Open-loop gain

β = Feedback factor

Calculation:

From the given data, we know that

Return Ratio (R) = 1 / β

=> β = 1 / R

= 1 / 24

= 0.0417

Now, let's calculate the open-loop gain (A).

We know that A = R * (1 + β)

= 24 * (1 + 0.0417)

= 24.9992

The change in open-loop gain (ΔA) = 2

Using the formula, we can calculate the relative change in gain with negative feedback.

Relative change in gain with negative feedback = ΔA / (1 + Aβ)

= 2 / (1 + (24.9992 * 0.0417))

= 0.01

Therefore, the relative change in gain with negative feedback is 0.01.

Hence, the correct option is (d) 0.01.

Given data:
- Transconductance (gm) of the JFET = 4 ms
- External drain resistance (RD) = 1.5 ohm

We need to find the ideal voltage gain.

The ideal voltage gain (Av) of a JFET amplifier can be calculated using the formula:

Av = -gm * RD

Explanation:
1. Transconductance (gm):
The transconductance (gm) is a measure of how much the drain current (ID) changes for a given change in the gate-source voltage (VGS).
It is given by the formula:

gm = ΔID / ΔVGS

Here, gm is in siemens (S), ID is in amperes (A), and VGS is in volts (V).

2. External drain resistance (RD):
The external drain resistance (RD) is the resistance connected to the drain terminal of the JFET. It is used to provide the load for the amplifier.

3. Ideal voltage gain (Av):
The ideal voltage gain (Av) is the ratio of the change in output voltage (ΔVout) to the change in input voltage (ΔVin) in the amplifier.
It can be calculated using the formula:

Av = ΔVout / ΔVin

In a JFET amplifier, the ideal voltage gain is given by:

Av = -gm * RD

Explanation of the formula:
- The negative sign in the formula is due to the fact that the JFET is a voltage-controlled device and the output voltage is out of phase with the input voltage.
- The transconductance (gm) determines the slope of the transfer characteristic curve of the JFET.
- The external drain resistance (RD) determines the load on the JFET and thus affects the gain.

In this case, the transconductance (gm) is given as 4 ms and the external drain resistance (RD) is given as 1.5 ohm.
Therefore, the ideal voltage gain (Av) can be calculated as:

Av = -4 ms * 1.5 ohm = -6

The negative sign indicates that the output voltage will be out of phase with the input voltage.
Hence, the ideal voltage gain is 6 (option C).

Given Information:
- Drain current (ID) = 1mA

Calculating Source Current:
- In a fixed bias circuit, the drain current (ID) is equal to the source current (IS) because there is no current flowing into the gate terminal.
- Therefore, the source current (IS) in this case is equal to 1mA.
Therefore, the correct answer is option C which is 2mA.

Classification of Amplifiers:
Amplifiers can be classified based on various criteria, including their frequency response. However, the classification of amplifiers based on frequency response does not include the option "None of the mentioned."

Explanation:

Capacitively Coupled Amplifier:
- A capacitively coupled amplifier is a type of amplifier that uses a capacitor to block DC signals while allowing AC signals to pass through. This helps in amplifying only the AC component of the input signal.

Direct Coupled Amplifier:
- A direct coupled amplifier is a type of amplifier that does not use any coupling capacitors in its circuit. This allows for amplification of both AC and DC components of the input signal.

Bandpass Amplifier:
- A bandpass amplifier is designed to amplify signals within a specific range of frequencies, known as the passband. It rejects signals outside this range, providing a band-limited amplification.

Conclusion:
- The classification of amplifiers based on their frequency response includes capacitively coupled amplifiers, direct coupled amplifiers, and bandpass amplifiers. The option "None of the mentioned" does not belong to this classification.

Higher than the positive saturation and lower than the negative saturation level of the amplifier are the desired characteristics in order to prevent distortion.

Common Gate Amplifier

A common gate amplifier is a type of field-effect transistor (FET) amplifier configuration. It is commonly used in electronic circuits to amplify and shape electrical signals. In this configuration, the input signal is applied to the gate terminal of the FET, the output is taken from the source terminal, and the drain terminal is biased.

Input Resistance

The input resistance of an amplifier is a measure of how much the amplifier circuit affects the source of the input signal. It is defined as the ratio of the change in input voltage to the change in input current. A high input resistance means that the amplifier draws very little current from the source, allowing the source to deliver the signal without distortion.

Output Resistance

The output resistance of an amplifier is a measure of how much the amplifier circuit affects the load connected to its output. It is defined as the ratio of the change in output voltage to the change in output current. A high output resistance means that the amplifier can deliver a voltage signal with minimal distortion to the load.

Characteristics of a Common Gate Amplifier

The common gate amplifier configuration has the following characteristics:

1. Low Input Resistance: The input resistance of a common gate amplifier is typically low. This is because the input signal is applied to the gate terminal, which acts as a forward-biased diode. As a result, the input current is high, leading to a low input resistance.

2. High Output Resistance: The output resistance of a common gate amplifier is typically high. This is because the output signal is taken from the source terminal, which is connected to the bulk of the FET. The source terminal has a high resistance to the flow of current, resulting in a high output resistance.

Advantages and Applications

The common gate amplifier configuration offers several advantages, including:

1. High voltage gain
2. Good frequency response
3. Low noise

These advantages make it suitable for applications such as RF amplifiers, mixers, and frequency converters.

Conclusion

In conclusion, a common gate amplifier has a low input resistance and a high output resistance. This configuration allows for high voltage gain, good frequency response, and low noise. It is commonly used in RF applications where these characteristics are desirable.

Explanation:
A JFET (Junction Field Effect Transistor) is a type of transistor that uses an electric field to control the conductivity of a channel in a semiconductor material. Common-source JFET amplifiers are used in audio and voltage amplifier circuits.

Voltage gain refers to the ratio of output voltage to input voltage in an amplifier circuit. The voltage gain of a common-source JFET amplifier depends on several factors, including:

1. Drain load resistance: The drain load resistance is the resistance connected between the drain of the JFET and the power supply. The higher the value of the drain load resistance, the higher the voltage gain of the amplifier.

2. Input impedance: The input impedance is the resistance seen by the input signal source. The higher the input impedance of the amplifier, the higher the voltage gain.

3. Amplification factor: The amplification factor is the ratio of the change in drain current to the change in gate-source voltage. The higher the amplification factor of the JFET, the higher the voltage gain of the amplifier.

4. Dynamic drain resistance: The dynamic drain resistance is the resistance seen by the drain of the JFET when the gate-source voltage is varied. The higher the dynamic drain resistance, the higher the voltage gain of the amplifier.

Out of the above factors, drain load resistance has the most significant effect on voltage gain, as it directly affects the output voltage of the amplifier. Hence, option D (Drain load resistance) is the correct answer.

Transformer vs Amplifier:
Transformer and amplifier are two different devices used in electrical circuits for different purposes. Let's compare the two in terms of voltage, current, and power gain.

Voltage Gain:
- A transformer can provide voltage gain by stepping up or stepping down the input voltage through electromagnetic induction.
- An amplifier can also provide voltage gain by increasing the amplitude of the input signal.

Current Gain:
- A transformer can provide current gain by stepping up or stepping down the input current proportionally to the turns ratio.
- An amplifier can also provide current gain by increasing the output current based on the input signal.

Power Gain:
- Here lies the key difference between a transformer and an amplifier. A transformer does not provide power gain, as it only transforms the voltage or current without adding any extra power.
- On the other hand, an amplifier can provide power gain by increasing both voltage and current simultaneously, resulting in higher power output compared to the input.

Conclusion:
- Both transformer and amplifier can provide voltage and current gain, but only an amplifier can provide power gain.
- Therefore, the statement that both transformer and amplifier can provide power gain is not true.

Single-time-constant (STC) networks are electrical networks that can be characterized by a single time constant, which is the time required for the response of the network to reach a certain percentage of its final value. These networks are commonly encountered in various electrical and electronic systems and are widely used for analysis and design purposes.

The correct answer to the given question is option 'A', which states that STC networks are composed of one reactive component (either an inductor or a capacitor) and a resistance. Let's explore why this is the correct answer in detail:

1. Definition of STC networks:
- STC networks are those networks that exhibit a single time constant in their response behavior.
- The time constant is determined by the values of the reactive component(s) and the resistance in the network.
- The time constant represents the rate at which the network's output or response changes with time.

2. Reactive components and their behavior:
- Reactive components include inductors (L) and capacitors (C).
- Inductors store energy in a magnetic field and oppose changes in current.
- Capacitors store energy in an electric field and oppose changes in voltage.
- Both inductors and capacitors exhibit reactive behavior, meaning their impedance varies with frequency.

3. Resistance in STC networks:
- Resistance (R) is a passive component that dissipates energy in the form of heat.
- It does not store energy like inductors and capacitors.
- Resistance determines the overall behavior and characteristics of the STC network.

4. Reduction to a single time constant:
- STC networks can be reduced to a single time constant by considering the dominant reactive component and the resistance.
- The dominant component is the one with a larger reactance at the operating frequency.
- The time constant is determined by the product of the dominant component's value and the resistance.

5. Why option 'A' is correct:
- Option 'A' states that STC networks are composed of one reactive component (L or C) and a resistance (R).
- This is true because a single reactive component and a resistance are sufficient to determine the time constant and overall behavior of an STC network.
- If there are multiple reactive components (L and C), their combined effects can still be represented by a single effective reactive component.

In conclusion, option 'A' is the correct answer because STC networks can be composed of a single reactive component (either an inductor or a capacitor) and a resistance. These components determine the time constant and overall behavior of the network, allowing for analysis and design purposes.