All questions of Amplifiers for Electrical Engineering (EE) Exam

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.

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.

The reverse transmission factor refers to the ratio of feedback by the output signal in a control system. It plays a crucial role in determining the stability and performance of the system. To understand this concept, let's break down the components of a control system and how they interact with each other.

1. Control system components:
a) Input signal: The desired or reference signal that the system aims to achieve.
b) Output signal: The measured or actual signal produced by the system.
c) Feedback signal: The portion of the output signal that is fed back to the system for comparison with the input signal.
d) Forward path: The path through which the input signal is processed to produce the output signal.
e) Feedback path: The path through which the feedback signal is combined with the input signal to generate an error signal.

2. Importance of feedback:
Feedback is essential in control systems as it allows for error correction and improves system performance. By comparing the output signal with the input signal, any discrepancies or errors can be detected, and appropriate adjustments can be made to minimize these errors.

3. Reverse transmission factor:
The reverse transmission factor, denoted as β, represents the ratio of feedback by the output signal. Mathematically, it can be expressed as:
β = Feedback signal / Output signal

4. Significance of the reverse transmission factor:
The reverse transmission factor influences the gain and stability of a control system. Here's why:

a) Gain: The gain of a system determines how much the output signal changes in response to a change in the input signal. The presence of feedback modifies the overall gain of the system. The reverse transmission factor affects the gain by determining the fraction of the output signal that is fed back into the system. A higher reverse transmission factor implies more feedback, leading to a reduction in the overall gain of the system.

b) Stability: Stability is a critical aspect of control systems. It refers to the ability of the system to maintain a steady output in the presence of disturbances or changes in the input signal. The reverse transmission factor plays a significant role in determining system stability. If the reverse transmission factor is too high, it can introduce instability into the system, causing oscillations or even system failure. Hence, it is crucial to carefully design the reverse transmission factor to ensure stability.

In summary, the reverse transmission factor represents the ratio of feedback by the output signal in a control system. It affects the gain and stability of the system and needs to be carefully considered during system design and analysis.

Ω is connected to its output. This means that the output voltage decreases by 20% when the load resistance is added.

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

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.

Given:
R1 = R2 = 1K (1 kiloohm)
VDD = 5V

To find:
Gate voltage (VG) for the voltage divider

Solution:

1. Voltage Divider Formula:
The voltage divider formula is given by:
VG = VDD * (R2 / (R1 + R2))

2. Substituting the given values:
VG = 5V * (1K / (1K + 1K))
= 5V * (1K / 2K)
= 5V * 0.5
= 2.5V

3. Answer:
The gate voltage for the voltage divider with R1 = R2 = 1K and VDD = 5V is 2.5V.

Therefore, the correct answer is option 'D' (2.5V).

Understanding Voltage Series Feedback Amplifier Gain
The gain of a voltage series feedback amplifier is primarily influenced by the ratio of two resistors in the feedback network. Here's a detailed explanation:
Open Loop Voltage Gain
- Open loop voltage gain refers to the gain of the amplifier without any feedback applied. While it sets the initial gain level, it does not directly determine the final gain when feedback is employed.
Feedback Voltage
- Feedback voltage is the voltage returned from the output to the input of the amplifier. Although crucial for stability and linearity, it does not dictate the gain but rather regulates the output based on the feedback network.
Ratio of Two Resistors
- The key factor that determines the gain in a voltage series feedback amplifier is the ratio of two resistors in the feedback circuit. This resistance ratio sets the amount of feedback applied to the amplifier, thus controlling the overall gain.
- For instance, if resistors R1 and R2 are used in a feedback configuration, the gain can be expressed as A = R2/(R1 + R2). This equation shows that the gain is directly influenced by how these resistors are chosen.
Gain of Feedback Circuit
- The gain of the feedback circuit is an important consideration, but it is not the primary determinant of the overall amplifier gain. Instead, it works in conjunction with the resistor ratio to establish the final output.
Conclusion
In summary, while several factors contribute to the performance of a voltage series feedback amplifier, it is the ratio of two resistors that fundamentally determines its gain. Understanding this relationship is essential for designing effective feedback systems in electrical engineering.

Internal Capacitors and Gain
Internal capacitors of a device do not cause the gain to fall off at low frequencies. In fact, internal capacitors are often used in electronic circuits to block DC voltage while allowing AC signals to pass through. They are designed to have a minimal impact on the gain of the circuit at low frequencies.

Coupling Capacitors and Gain
On the other hand, coupling capacitors are used to block DC components from one part of a circuit to another, allowing only the AC components to pass through. Coupling capacitors can cause the gain to fall off at high frequencies due to their impedance characteristics. As the frequency increases, the impedance of the coupling capacitor decreases, affecting the gain of the circuit at higher frequencies.

Conclusion
Therefore, the correct statement is that coupling capacitors cause the gain to fall off at high frequencies, not internal capacitors at low frequencies. It is crucial to understand the role of capacitors in a circuit to design and analyze electronic systems effectively.

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.

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.

Introduction:
In voltage shut feedback amplifiers, the output voltage is fed back to the input in such a way that it opposes any change in the input voltage. This feedback mechanism is used to stabilize and control the gain of the amplifier. There are various special cases of voltage shut feedback amplifiers, each with its own characteristics and applications.

Voltage Follower:
A voltage follower, also known as a unity gain amplifier or a buffer, is a special case of voltage shut feedback amplifier. In a voltage follower, the output voltage follows the input voltage, hence the name. The voltage gain of a voltage follower is unity, which means the output voltage is the same as the input voltage.

Current to Voltage Converter:
A current to voltage converter, also known as a transimpedance amplifier, is another special case of voltage shut feedback amplifier. It converts an input current into an output voltage. The output voltage is proportional to the input current, and the proportionality constant is determined by the feedback resistor.

Inverter:
An inverter, also known as an inverting amplifier, is a special case of voltage shut feedback amplifier where the output voltage is the inverse of the input voltage. The inverter has a negative voltage gain, which means it amplifies the input voltage with an opposite polarity. The voltage gain is determined by the feedback resistor and the input resistor.

Explanation:
The correct answer is option 'A', voltage follower. A voltage follower is not a special case of voltage shut feedback amplifier. It is a basic amplifier configuration where the output voltage is directly connected to the input. There is no feedback mechanism involved in a voltage follower, and the gain is unity. It is mainly used to isolate the input signal from the load and provide high input impedance.

Conclusion:
In summary, the special cases of voltage shut feedback amplifiers include voltage follower, current to voltage converter, and inverter. These configurations have different characteristics and applications. The voltage follower is not a special case of voltage shut feedback amplifier as it does not involve any feedback mechanism.

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.

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).