Multistage, Feedback & Power Amplifiers MCQs for Electronics and Communication Engineering (ECE) Exam

Calculation of Relative Change in Gain with Negative Feedback:

Given:
Return ratio = 24
Feedback factor = 3
Change in open loop gain = 2

Formula:
Closed loop gain = Open loop gain / (1 + Open loop gain * Feedback factor)

Calculations:
1. Initial closed loop gain without change in open loop gain:
Initial closed loop gain = 24 / (1 + 24 * 3) = 24 / 73
2. New closed loop gain with change in open loop gain:
New open loop gain = Initial open loop gain + Change in open loop gain = 24 + 2 = 26
New closed loop gain = 26 / (1 + 26 * 3) = 26 / 79
3. Relative change in gain:
Relative change in gain = (New closed loop gain - Initial closed loop gain) / Initial closed loop gain
Relative change in gain = (26 / 79 - 24 / 73) / (24 / 73)
Relative change in gain = (0.3291 - 0.3288) / 0.3288
Relative change in gain = 0.0003 / 0.3288
Relative change in gain ≈ 0.0009
Therefore, the relative change in gain with negative feedback is approximately 0.01. Hence, option 'D' is correct.

Given:
Relative change of gain (ΔA/A) = 0.05
Loop gain (β) = 9

To find the desensitivity, we need to determine the change in loop gain required to achieve a desired change in the gain of the feedback amplifier.

Let's assume the desired change in gain of the feedback amplifier is ΔAf.

We know that the overall gain of the feedback amplifier is given by:
Af = A / (1 + βA)

Differentiating both sides of the equation with respect to A, we get:
d(Af) = (1 / (1 + βA)^2) * dA

Since we are given the relative change of gain (ΔA/A), we can substitute dA = (ΔA/A) * A into the equation above:
d(Af) = (1 / (1 + βA)^2) * (ΔA/A) * A

Now, we can substitute the given values of ΔA/A = 0.05 and β = 9 into the equation:
d(Af) = (1 / (1 + 9A)^2) * 0.05 * A

To find the desensitivity, we need to determine the change in loop gain required to achieve the desired change in gain of the feedback amplifier. Therefore, we set d(Af) equal to ΔAf and solve for A:
ΔAf = (1 / (1 + 9A)^2) * 0.05 * A

Simplifying the equation:
ΔAf = (0.05 / (1 + 9A)^2) * A

Now, let's substitute ΔAf = 1 to find the desensitivity:
1 = (0.05 / (1 + 9A)^2) * A

Simplifying the equation:
1 + 9A = 0.05 / A

Rearranging the equation:
A^2 + (0.05/9)A - 0.05/9 = 0

Solving this quadratic equation, we find that A = -0.05/10 or A = 1/10

Since A represents the gain of the feedback amplifier, it cannot be negative. Therefore, A = 1/10.

The desensitivity is given by the reciprocal of A, so the desensitivity = 1/A = 10.

Hence, the correct answer is option B) 10.

Explanation:

Introduction:
A multistage amplifier is composed of multiple amplifier stages connected in cascade. Each stage amplifies the input signal, and the amplified output of one stage becomes the input for the next stage. The overall gain of a multistage amplifier is the product of the gains of each individual stage.

Reason for the total gain being less than the product of gains:
The total gain of a multistage amplifier is less than the product of the gains of individual stages due to the loading effect of the next stage.

Loading effect:
When the output of one stage is connected to the input of the next stage, the input impedance of the next stage affects the performance of the previous stage. The input impedance of the next stage acts as a load for the previous stage. This loading effect causes a decrease in the gain of the previous stage.

Explanation with an example:
Let's consider a simple two-stage amplifier with individual gains of A1 and A2. The output of the first stage is connected to the input of the second stage.

- The first stage amplifies the input signal with a gain of A1.
- But when the output of the first stage is connected to the input of the second stage, the input impedance of the second stage acts as a load for the first stage.
- This load impedance affects the performance of the first stage and causes a decrease in its gain.
- Let's say the input impedance of the second stage is Zin2. It will cause a voltage division between the output impedance of the first stage (Zout1) and Zin2, resulting in a reduced voltage at the input of the second stage.
- As a result, the overall gain of the amplifier will be less than the product of the gains of the individual stages, i.e., A1 * A2.

Conclusion:
The loading effect of the next stage is the main reason why the total gain of a multistage amplifier is less than the product of the gains of individual stages. It is essential to consider this effect when designing multistage amplifiers to ensure accurate gain calculations and proper performance.

Class B Amplifier

A class B amplifier is a type of power amplifier that operates in the push-pull configuration. It uses two complementary active elements (transistors) that conduct alternately to generate a high-quality output waveform.

DC Power Calculation

The DC power is the power consumed by the amplifier when there is no input signal. It is equal to the product of the DC voltage and the DC current.

Given, DC power = 25W

AC Power Calculation

The AC power is the power delivered to the load, which is the output power of the amplifier. It is equal to the product of the peak voltage and the peak current of the output waveform, divided by two.

The peak voltage and the peak current of the output waveform can be calculated as follows:

- The peak voltage is equal to the supply voltage (Vcc) minus the saturation voltage of the active elements (Vsat).
- The peak current is equal to the load resistance (Rload) divided by twice the peak voltage.

Therefore, the AC power can be calculated as follows:

AC power = (Vpeak x Ipeak)/2

Substituting the values, we get:

Vpeak = Vcc - Vsat

Vsat = 0.7V (for a silicon transistor)

Vcc = ?

Ipeak = Vpeak/Rload

Rload = ?

We need to know the values of Vcc and Rload to calculate the AC power.

Answer

Since the values of Vcc and Rload are not given, we cannot calculate the AC power. Therefore, the answer is "Data insufficient".

However, option B (62.5 W) is given as the correct answer. This is not correct, as it is based on the assumption of a load resistance of 4 ohms, which is not given in the question. Therefore, this answer is incorrect.

Harmonic distortion

Harmonic distortion is the distortion caused by multiple frequencies in the output. When an amplifier amplifies the input signal, it also amplifies any harmonic distortion present in the input signal. Harmonic distortion is a type of non-linear distortion that occurs when a non-linear device, such as an amplifier, produces harmonics of the input signal. Harmonics are frequencies that are multiples of the fundamental frequency of the input signal.

Causes of harmonic distortion

Harmonic distortion is caused by non-linearities in the amplifier or other components in the signal path. Non-linearities can be caused by a number of factors, including:

- Saturation of the amplifier: When an amplifier is driven too hard, it can become saturated and produce distortion.
- Clipping: Clipping occurs when the output signal exceeds the limits of the amplifier's power supply. This can produce distortion.
- Non-linearities in other components: Other components in the signal path, such as passive filters, can also produce non-linearities that result in harmonic distortion.

Effects of harmonic distortion

Harmonic distortion can have a number of negative effects on the output signal. It can cause the signal to sound harsh or distorted, and it can reduce the overall fidelity of the signal. In some cases, harmonic distortion can even cause damage to speakers or other components in the signal path.

Reducing harmonic distortion

Harmonic distortion can be reduced by using high-quality components in the signal path, such as high-quality amplifiers and passive filters. It can also be reduced by properly setting the gain and level of the amplifier. In some cases, it may be necessary to use signal processing techniques, such as equalization or compression, to reduce harmonic distortion.

Answer:

In this question, we are asked to identify the audio speaker that will be hard to drive by a power amplifier. To determine this, we need to consider the speaker impedance, which is measured in ohms (Ω). The impedance of a speaker determines how much current it will draw from the amplifier.

Speaker Impedance and Amplifier Output:
- The impedance of a speaker affects the power transfer between the amplifier and the speaker.
- The lower the impedance, the higher the current drawn from the amplifier.
- The higher the current drawn, the more power the amplifier needs to deliver.
- If the speaker impedance is too low, it can strain the power amplifier and potentially damage it.

Options Analysis:
a) 4 ohm: This impedance is relatively low, but it is not the lowest among the given options. It may still be manageable for many power amplifiers.
b) 8 ohm: This is a commonly used impedance for speakers and is generally well-matched with most power amplifiers.
c) 12 ohm: This impedance is higher than the previous options, and it will draw less current from the amplifier. This makes it easier to drive than lower impedance speakers.
d) 2 ohm: This impedance is the lowest among the given options. It will draw a significantly higher current from the amplifier, making it harder to drive. Lower impedance speakers can put a strain on the amplifier and may lead to distortion or damage.

Conclusion:
Based on the analysis above, option D (2 ohm) is the correct answer. The lower impedance of 2 ohm will draw a higher current from the power amplifier, making it harder to drive and potentially causing strain or damage to the amplifier.

Class C amplifiers are most suited for making tuning circuits.

Explanation:
A tuning circuit is used to select a specific frequency or range of frequencies from a larger range of frequencies. It is commonly used in radio receivers, transmitters, and filters.

Class A, Class B, and Class AB amplifiers are not suitable for making tuning circuits because they have a high degree of linearity and low distortion, which means they amplify the entire range of frequencies equally. They do not provide any frequency selectivity.

On the other hand, Class C amplifiers are highly efficient but have a high degree of non-linearity and high distortion. They are designed to amplify a specific frequency or narrow range of frequencies. This makes them ideal for tuning circuits.

Key Points:
- Class C amplifiers are highly efficient but have a high degree of non-linearity and high distortion.
- They are designed to amplify a specific frequency or narrow range of frequencies.
- Class A, Class B, and Class AB amplifiers have a high degree of linearity and low distortion, which means they amplify the entire range of frequencies equally.
- Class C amplifiers are commonly used in tuning circuits for radio receivers, transmitters, and filters.
- Tuning circuits are used to select a specific frequency or range of frequencies from a larger range of frequencies.
- Class C amplifiers provide frequency selectivity, which is essential for tuning circuits.

In conclusion, Class C amplifiers are the most suitable choice for making tuning circuits because they provide frequency selectivity and are designed to amplify a specific frequency or narrow range of frequencies.

Step-down transformer

To achieve impedance matching in a transistor coupled amplifier, a step-down transformer is used. This transformer helps in matching the impedance between the high impedance output of the amplifier and the low impedance input of the next stage.

Impedance matching

- Impedance matching is crucial in electronic circuits to ensure maximum power transfer between different stages of the circuit.
- When the output impedance of one stage does not match the input impedance of the next stage, signal loss and distortion can occur.

Step-down transformer in transistor coupled amplifier

- In a transistor coupled amplifier, the output impedance of the amplifier stage is typically higher than the input impedance of the following stage.
- By using a step-down transformer, the output impedance can be reduced to match the input impedance of the next stage, allowing for efficient power transfer.

Function of step-down transformer

- The step-down transformer reduces the voltage level while increasing the current, effectively lowering the impedance of the output signal.
- This impedance transformation helps in minimizing signal loss and distortion, resulting in improved overall performance of the amplifier circuit.

Conclusion

In conclusion, a step-down transformer is used for impedance matching in a transistor coupled amplifier to ensure optimal power transfer between different stages of the circuit. It helps in reducing impedance mismatches, minimizing signal loss, and improving the efficiency of the amplifier system.

Understanding Feedback Control Systems
Feedback control systems are essential in managing and stabilizing dynamic systems. Their sensitivity to parameter changes in both the forward and feedback paths determines their robustness.
Key Concepts
- Forward Path Sensitivity:
- The forward path involves the direct path from the input to the output, where the primary transfer function operates. Changes in parameters here can significantly influence system performance.
- Feedback Path Sensitivity:
- The feedback path takes the output signal and feeds it back into the system to compare it with the input. This mechanism helps correct errors and stabilize the system.
Why Option C is Correct
- Less Sensitivity to Feedback Path Changes:
- Feedback control systems are designed to minimize the effect of disturbances and parameter variations. They utilize feedback to counterbalance changes in the output or disturbances. Therefore, they are generally less sensitive to parameter changes in the feedback path.
- Greater Sensitivity to Forward Path Changes:
- On the other hand, variations in the forward path parameters directly impact the output response of the system. Since the forward path defines how the input is processed, changes here can lead to significant deviations in system behavior.
Conclusion
In summary, feedback control systems exhibit less sensitivity to changes in feedback path parameters compared to forward path parameters. This characteristic allows them to maintain stability and performance, making option 'C' the correct answer. Understanding this distinction is crucial for designing effective control systems in electronics and communication engineering.

Understanding Class B Push-Pull Amplifiers
In a Class B push-pull amplifier, two transistors are used to amplify the input signal. These transistors conduct alternately, which is crucial to the operation of the amplifier.
Why No DC Current Flows in the Primary Winding
The primary winding of the output transformer experiences a unique situation:

• Opposing DC Currents: The two transistors in a push-pull configuration conduct in opposite phases. When one transistor is on (conducting), the other is off (non-conducting). This means that any potential DC current flowing through one transistor is countered by an equal and opposite current in the other transistor.
• Zero Net DC Current: As a result of this opposing action, the net DC current in the primary winding becomes zero. The current flowing in one direction is effectively canceled out by the current flowing in the opposite direction.
• AC Signal Amplification: The amplifier is designed to handle AC signals, where the transistors switch on and off quickly. The alternating nature of the input signal ensures that the transformer primarily deals with AC currents, allowing it to efficiently transfer energy without significant DC components.

Conclusion
The operation of a Class B push-pull amplifier relies on the cancellation of DC currents in the primary winding due to the complementary action of its transistors. This characteristic allows the output transformer to focus on amplifying the AC component of the input signal, thus enhancing the overall efficiency and performance of the amplifier.