Oscillators and Feedback Amplifier MCQ Quiz - Objective Question with Answer for Oscillators and Feedback Amplifier - Download Free PDF

Last updated on Jun 26, 2025

Latest Oscillators and Feedback Amplifier MCQ Objective Questions

Oscillators and Feedback Amplifier Question 1:

In a three stage cascade amplifier, each stage has a gain of 10 dB and noise figure of 10 dB. The overall noise figure is

  1. 10.99
  2. 10
  3. 1.09
  4. 10.9

Answer (Detailed Solution Below)

Option 1 : 10.99

Oscillators and Feedback Amplifier Question 1 Detailed Solution

Explanation:

Three-Stage Cascade Amplifier and Overall Noise Figure Calculation

Problem Statement: In a three-stage cascade amplifier, each stage has a gain of 10 dB and a noise figure of 10 dB. We are tasked with calculating the overall noise figure of this cascade system.

Solution:

To calculate the overall noise figure (Foverall) of a cascade amplifier, we use Friis's formula for noise figure in cascaded systems:

Friis's Formula:

Foverall = F1 + (F2 - 1)/G1 + (F3 - 1)/G1 × G2 + ...

Where:

  • Fn = Noise figure of the nth stage (in linear scale).
  • Gn = Gain of the nth stage (in linear scale).

Step 1: Convert Gains and Noise Figures from dB to Linear Scale

The given gain (G) and noise figure (F) for each stage are 10 dB:

  • G = 10 dB → G (linear) = 10G(dB)/10 = 1010/10 = 10.
  • F = 10 dB → F (linear) = 10F(dB)/10 = 1010/10 = 10.

Step 2: Apply Friis's Formula

Since there are three stages, Friis's formula for the overall noise figure becomes:

Foverall = F1 + (F2 - 1)/G1 + (F3 - 1)/G1 × G2

Substitute the values for F1, F2, F3, G1, and G2:

  • F1 = 10 (linear).
  • F2 = 10 (linear).
  • F3 = 10 (linear).
  • G1 = G2 = 10 (linear).

Thus:

Foverall = 10 + (10 - 1)/10 + (10 - 1)/10 × 10

Foverall = 10 + 0.9 + 0.09

Foverall = 10.99

Final Answer: The overall noise figure of the three-stage cascade amplifier is 10.99 (linear scale).

Oscillators and Feedback Amplifier Question 2:

In transistor oscillators, FET and BJT are used. Instability is achieved by:

  1. Negative feedback
  2. Positive feedback
  3. Using a tank circuit
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Positive feedback

Oscillators and Feedback Amplifier Question 2 Detailed Solution

Explanation:

  • An amplifier can be converted into an oscillator by doing some change in the amplifier circuit as:
  • Connect the output of the amplifier to the input by a positive feedback circuit.
  • The phase-shifted the output by 180° and feed this phase shift output to the input via a feedback circuit.
  • An arrangement of the RC tuned circuit is connected as a load to the amplifier.
  • Oscillator circuit block diagram.

 

F1 Shubham 9.10.20 Pallavi D14

Oscillators and Feedback Amplifier Question 3:

An R-C coupled amplifier has a mid-frequency gain of 400. It has a upper and lower 3 db frequencies of 15 kHz and 100 Hz respectively. If a negative feedback with β = 0.01 is incorporated in the amplifier circuit determine the gain with feedback Avf and new bandwidth BWnew.

  1. Avf = 100 and BWnew = 55 kHz
  2. Avf = 100 and BWnew = 74.98 kHz
  3. Avf = 80 and BWnew = 55 kHz
  4. Avf = 80 and BWnew = 74.98 kHz

Answer (Detailed Solution Below)

Option 4 : Avf = 80 and BWnew = 74.98 kHz

Oscillators and Feedback Amplifier Question 3 Detailed Solution

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Explanation:

R-C Coupled Amplifier with Feedback

Definition: An R-C coupled amplifier is a type of amplifier where resistors and capacitors are used for coupling the output of one amplifier stage to the input of the next stage. Negative feedback is often introduced into amplifier circuits to improve stability, linearity, and bandwidth, while reducing distortion and the effects of parameter variations.

In this problem, an R-C coupled amplifier has a mid-frequency gain of 400 with upper and lower 3 dB frequencies of 15 kHz and 100 Hz, respectively. Negative feedback with a feedback factor β = 0.01 is incorporated into the circuit, and we are tasked with calculating the gain with feedback (Avf) and the new bandwidth (BWnew).

Solution:

The gain with feedback (Avf) and the new bandwidth (BWnew) can be calculated using the following formulae:

  • Gain with feedback:

    Avf = A / (1 + βA)

    Where:

    • A = Mid-frequency gain of the amplifier without feedback
    • β = Feedback factor
  • New Bandwidth:

    BWnew = BW × (1 + βA)

    Where:

    • BW = Bandwidth of the amplifier without feedback

Step 1: Calculate the gain with feedback (Avf)

Given:

  • Mid-frequency gain without feedback, A = 400
  • Feedback factor, β = 0.01

Using the formula for Avf:

Avf = A / (1 + βA)

Substitute the given values:

Avf = 400 / (1 + 0.01 × 400)

Avf = 400 / (1 + 4)

Avf = 400 / 5

Avf = 80

Step 2: Calculate the new bandwidth (BWnew)

Given:

  • Lower 3 dB frequency, fL = 100 Hz
  • Upper 3 dB frequency, fH = 15 kHz
  • Bandwidth without feedback, BW = fH − fL

Calculate BW:

BW = fH − fL

BW = 15,000 − 100

BW = 14,900 Hz

Using the formula for BWnew:

BWnew = BW × (1 + βA)

Substitute the values:

BWnew = 14,900 × (1 + 0.01 × 400)

BWnew = 14,900 × (1 + 4)

BWnew = 14,900 × 5

BWnew = 74,500 Hz or 74.98 kHz

Final Results:

  • Gain with feedback, Avf = 80
  • New bandwidth, BWnew = 74.98 kHz

Correct Option: Option 4: Avf = 80 and BWnew = 74.98 kHz

Important Information

To further understand the analysis, let’s evaluate why the other options are incorrect:

Option 1: Avf = 100 and BWnew = 55 kHz

This option is incorrect because the gain with feedback (Avf) is calculated as 80, not 100. Additionally, the new bandwidth (BWnew) is 74.98 kHz, not 55 kHz.

Option 2: Avf = 100 and BWnew = 74.98 kHz

While the new bandwidth (BWnew) is correctly stated as 74.98 kHz, the gain with feedback (Avf) is incorrect. The actual Avf is 80, not 100.

Option 3: Avf = 80 and BWnew = 55 kHz

While the gain with feedback (Avf) is correctly stated as 80, the new bandwidth (BWnew) is incorrect. The actual BWnew is 74.98 kHz, not 55 kHz.

Conclusion:

The correct option is Option 4, as it accurately represents both the gain with feedback (Avf = 80) and the new bandwidth (BWnew = 74.98 kHz). The calculations demonstrate the effect of negative feedback in reducing the amplifier gain while simultaneously increasing its bandwidth, which are desirable outcomes in many amplifier applications.

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Oscillators and Feedback Amplifier Question 4:

When a crystal vibrates, L represents:

  1. electrical equivalent of crystal mass
  2. electrical equivalent of heat
  3. electrical equivalent of mechanical friction
  4. electrical equivalent of elasticity

Answer (Detailed Solution Below)

Option 1 : electrical equivalent of crystal mass

Oscillators and Feedback Amplifier Question 4 Detailed Solution

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Explanation:

Electrical Equivalent of Crystal Mass (L):

Definition: When a crystal vibrates, its mechanical properties such as mass, elasticity, and friction have corresponding electrical equivalents. The parameter L specifically represents the electrical equivalent of crystal mass. This concept arises from the analogy between mechanical systems and electrical systems, where mechanical quantities like mass, stiffness, and damping are mapped to electrical quantities such as inductance, capacitance, and resistance.

Understanding the Concept:

Crystals are widely used in oscillatory circuits due to their ability to maintain stable frequencies. The vibrations of a crystal can be modeled as an equivalent electrical circuit, known as the "electrical equivalent circuit." This circuit includes components such as inductance (L), capacitance (C), and resistance (R), which correspond to the mechanical properties of the vibrating crystal:

  • Inductance (L): Represents the electrical equivalent of the crystal's mass. In mechanical systems, mass resists changes in motion, analogous to inductance in electrical systems, which resists changes in current.
  • Capacitance (C): Represents the electrical equivalent of the crystal's elasticity or compliance. Elasticity determines how easily a crystal can deform and return to its original shape, similar to capacitance storing energy in an electric field.
  • Resistance (R): Represents the electrical equivalent of mechanical friction or damping. Friction dissipates energy in mechanical systems, analogous to resistance dissipating energy in electrical systems.

The parameter L is critical in the equivalent circuit because it reflects the inertial properties of the crystal, which influence its vibrational frequency. A higher mass corresponds to higher inductance, which in turn affects the resonant frequency of the crystal.

Applications:

Understanding the electrical equivalent of crystal mass is essential in designing and analyzing crystal oscillators, which are used in:

  • Timing devices (e.g., clocks and watches).
  • Communication systems (e.g., radio transmitters and receivers).
  • Microprocessors and computers (e.g., providing clock signals).
  • Frequency control and stabilization in various electronic systems.

Correct Option Analysis:

The correct option is:

Option 1: Electrical equivalent of crystal mass.

The parameter L corresponds to the electrical equivalent of the crystal's mass. This is consistent with the modeling of mechanical systems as electrical circuits, where inductance represents the inertial properties of the vibrating crystal. Recognizing this equivalence is crucial for the design and optimization of devices that rely on crystal oscillators.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Electrical equivalent of heat.

This option is incorrect because heat is not directly related to the parameter L in the context of vibrating crystals. Heat dissipation in a vibrating crystal is associated with energy loss due to friction or resistance, which is modeled by the electrical equivalent resistance (R) in the equivalent circuit. However, this does not pertain to the inductance (L), which represents mass.

Option 3: Electrical equivalent of mechanical friction.

This option is also incorrect. Mechanical friction is analogous to electrical resistance (R) in the equivalent circuit. Resistance represents the energy dissipated due to frictional forces in the mechanical domain, and it is separate from inductance (L), which specifically represents the crystal's mass.

Option 4: Electrical equivalent of elasticity.

This option is incorrect because elasticity is represented by capacitance (C) in the equivalent circuit. Capacitance corresponds to the ability of the system to store energy and return to its original state, much like how elasticity allows a crystal to deform and recover. Inductance (L), on the other hand, is tied to mass, not elasticity.

Conclusion:

The parameter L in the context of vibrating crystals is the electrical equivalent of the crystal's mass. This understanding is pivotal in designing circuits that use crystals for stable frequency generation and control. By analyzing the mechanical properties of the crystal and their electrical equivalents, engineers can create precise and reliable oscillatory systems for various applications.

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Oscillators and Feedback Amplifier Question 5:

Which of the following statements about oscillator circuits is FALSE?

  1. A circuit that generates a sine wave without any input is called a linear oscillator.
  2. A circuit that generates a non-sinusoidal wave without any input is called a linear oscillator.
  3. The frequency of an oscillator depends on the RC or LC network.
  4. Multivibrators are used for generating non-sinusoidal waveforms.

Answer (Detailed Solution Below)

Option 2 : A circuit that generates a non-sinusoidal wave without any input is called a linear oscillator.

Oscillators and Feedback Amplifier Question 5 Detailed Solution

The correct answer is: 2) A circuit that generates a non-sinusoidal wave without any input is called a linear oscillator.

Explanation:

  1. Linear Oscillators (Option 1 - TRUE)

    • Generate sinusoidal waveforms (e.g., sine waves).

    • Examples: LC oscillators (Hartley, Colpitts), RC oscillators (Wien Bridge, Phase Shift).

  2. Non-Linear Oscillators (Option 2 - FALSE)

    • Generate non-sinusoidal waveforms (e.g., square, triangle, sawtooth waves).

    • Not called linear oscillators—they are called relaxation oscillators or multivibrators.

  3. Frequency Determination (Option 3 - TRUE)

    • Oscillator frequency depends on RC (resistor-capacitor) or LC (inductor-capacitor) networks.

  4. Multivibrators (Option 4 - TRUE)

    • Used to generate square waves, pulses, or other non-sinusoidal signals.

    • Examples: Astable, Monostable, and Bistable multivibrators.

Top Oscillators and Feedback Amplifier MCQ Objective Questions

In which type of power amplifier does the output current flow for the entire cycle of input signal?

  1. Class C
  2. Class AB
  3. Class B
  4. Class A

Answer (Detailed Solution Below)

Option 4 : Class A

Oscillators and Feedback Amplifier Question 6 Detailed Solution

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  • The transistor amplifier in which collector current flows for the entire cycle of input AC signal is called class A amplifier.
  • The transistor amplifier in which collector current flows for the half-cycle of an AC signal is called a class B amplifier.
  • The transistor amplifier in which collector current flows for less than half the cycle of an AC signal is called a class C amplifier

 

Power Amplifier

Conduction Angle

Maximum Efficiency

Figure of Merit

Class A

360°

50%

2

Class B

180°

78.5%

0.4

Class AB

180° - 360°

50 – 78.5%

0.4 – 2

Class C

< 180°

≥ 90°

< 0.25

For the non-inverting amplifiers as shown, find the closed-loop voltage gain.

quesOptionImage2302

  1. 100
  2. 10
  3. 101
  4. 11

Answer (Detailed Solution Below)

Option 4 : 11

Oscillators and Feedback Amplifier Question 7 Detailed Solution

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Concept of Virtual Ground:

  • The differential input voltage Vid between the noninverting and inverting input terminals is essentially zero.
  • This is because even if the output voltage is few volts, due to a large open-loop gain of the op-amp, the difference voltage Vid at the input terminals is almost zero.

F1 Harish Batula 13.5.21 Pallavi D3

Where Vid is differential voltage, Vin1 is noninverting voltage, Vin2 is inverting voltage.

If the output voltage is 10 V and A i.e., the open-loop gain is 104 then, 

out = A Vid

Vid = V out / A

= 10 / 104

= 1 mV.

Hence Vid is very small, for analyzing the circuit assumed to be zero.

Vid = Vin1 - Vin2

(Vin1 - Vin2) = V out / A

= V out / ∞ = 0

Calculation:

Circuit diagram:

F1 Harish Batula 13.5.21 Pallavi D4

Two terminals of Op-Amp i.e.Inverting Terminal and Non-Inverting Terminal are at equipotential.

Apply KCL at node 1Vpp,

1Vpp010k+1VppV0100k=0

10Vpp+1VppV0100K=0

11VppV0=0

Closed-loop gain is given by the ratio of output to the input.

so the Closed-loop voltage gain is given by,

V0Vpp=11

As per Barkhausen criterion for oscillation, the magnitude of the loop gain BA must be:

  1. Less than 1
  2. Equal to 1
  3. Greater than 1
  4. Equal to 1.5

Answer (Detailed Solution Below)

Option 2 : Equal to 1

Oscillators and Feedback Amplifier Question 8 Detailed Solution

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Barkhausen stability criteria

The Barkhausen stability criteria is a mathematical requirement used in electronics to predict whether a linear electronic circuit may oscillate.

It is commonly utilized in the design of electronic oscillators as well as general negative feedback circuits such as op-amps to keep them from oscillating.

Barkhausen's criteria is a necessary but not sufficient condition for oscillation:

Barkhausen Conditions For Oscillation:

F1 Engineering Mrunal 13.03.2023 D6

It states that if 'A' is the gain of the amplifying element in the circuit and β(s) is the feedback path transfer function, so βA is the loop gain around the circuit's feedback loop, the circuit will maintain steady-state oscillations only at frequencies for which:

  1. The loop gain is equal to one in absolute magnitude, which means that |βA| = 1
  2. The phase shift through the loop is either zero or an integer multiple of 2π or 3600.

Important Points

Parameters

Positive Feedback

Negative Feedback

The relation between input and output

In phase

Out of phase

Overall Gain

Greater than the gain of the system where feedback is not present.

Smaller than the gain of the system where feedback is absent.

Effective input

The sum of applied input and fed-back signal

The difference between applied input and the feedback signal

Stability

Less

Comparatively More

Phase shift

0° or 360°

180°

Sensitivity

Low

High

Use

In oscillators

In amplifiers

An RC phase shift oscillator uses capacitors of (16π) pf. Find the value of resistance ‘R’ to produce frequency of 1000 kHz.

  1. 50 kΩ
  2. 500 kΩ
  3. 25 kΩ
  4. 250 kΩ

Answer (Detailed Solution Below)

Option 2 : 500 kΩ

Oscillators and Feedback Amplifier Question 9 Detailed Solution

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RC phase shift oscillator:

F1 Savita Engineering 2-7-22 D1

  • It consists of three pairs of RC combinations, each providing a 60° phase shift, thus a total of 180° phase shift.
  • RC oscillators are used to generate low or audio-frequency signals. Hence they are also known as audio-frequency oscillators.

The frequency of oscillation is given by:

f=12πRC6 Hz 

Calculation:

Given, f = 1000 kHz

C=16π pF

106=12πR×16π×1012×6

R=12×1012×106

R = 500 kΩ 

Among the following, identify the incorrect statement.

  1. An oscillator is a circuit that convers DC to AC.
  2. All oscillators generate sine wave.
  3. An oscillator is an amplifier that supplies its own input signal.
  4. In-phase feedback is called positive feedback.

Answer (Detailed Solution Below)

Option 2 : All oscillators generate sine wave.

Oscillators and Feedback Amplifier Question 10 Detailed Solution

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Phase shift oscillator:

  • The phase shift oscillator is a linear electronic circuit that produces a sine wave output.
  • It consists of an inverting amplifier element such as a transistor or op-amp with its output feedback to its input through a phase shift network consisting of resistors and capacitors in a ladder network.
  • The feedback network shifts the phase of the amplifier output by 1800 at the oscillation frequency to give positive feedback. 

 

Wein bridge oscillator :

  • The Wein bridge oscillator uses two RC networks connected together to produce a sinusoidal oscillator.
  • The Wein bridge oscillator uses a feedback circuit consisting of a series RC circuit connected with a parallel RC of the same component values producing a phase delay or phase advance depending upon the circuit frequency
  • At the resonant frequency, the phase shift is 00.

 

Clapp oscillator:

  • The Clapp oscillator is an LC oscillator that uses a particular combination of an inductor and three capacitors to set the oscillator frequency.
  • The Clapp is often drawn as a Colpitts oscillator that has an additional capacitor placed in series with the inductor.
  • This comes under linear or harmonic oscillators, which produces a sine wave output.
  • Clapp oscillator is also one kind of phase shift oscillator containing L, C elements, and a transistor or op-amp along with feedback, so it provides phase shift.

 

​Relaxation oscillator:

  • A relaxation oscillator is a nonlinear electronic oscillator circuit that produces nonsinusoidal repetitive output signals such as a triangle wave or square wave.
  • Relaxation oscillators are generally used to produce low-frequency signals.
  • These oscillators will not provide any phase shift in their output
  • Examples of Relaxation oscillators are Astable multivibrator, flyback or sweep oscillator, etc.

In an amplifier with the negative feedback, bandwidth is _________ and voltage gain is _________.

  1. Bandwidth is decreased by a factor (1+Aβ) and voltage gain decreases
  2. Bandwidth is decreased by factor β and voltage gain remains same
  3. Bandwidth is increased by the factor (1+Aβ) and voltage gain is reduced
  4. Bandwidth remains the same and voltage gain increases

Answer (Detailed Solution Below)

Option 3 : Bandwidth is increased by the factor (1+Aβ) and voltage gain is reduced

Oscillators and Feedback Amplifier Question 11 Detailed Solution

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The feedback amplification factor is given by Af=A1+Aβ

where A is open-loop gain and βA is loop gain.

As feedback increases the gain decreases thereby bandwidth increases.

The negative feedback in amplifiers causes

1. reduced the voltage gain and increases the stability in gain

2. increases the bandwidth by the factor (1+Aβ) to maintain constant gain-bandwidth product

3. Reduces the distortion and noise in the amplifier

4. but the signal to noise ratio is not affected

Which oscillator is characterized by a split capacitor in its tank circuit?

  1. RC phase shift oscillator
  2. Colpitts oscillator 
  3. Wien bridge oscillator 
  4. None of the above 

Answer (Detailed Solution Below)

Option 2 : Colpitts oscillator 

Oscillators and Feedback Amplifier Question 12 Detailed Solution

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Colpitts oscillator:

  • The Colpitts oscillator consists of one inductor and one split capacitor in the tank circuit.
  • A capacitor with a center tap is used in the feedback system of the Colpitts oscillator
  • It is used for the generation of sinusoidal output signals with very high frequencies

 

Electrician 25 12Q Oscillators Hindi - Final images Q4

26 June 1

RC phase shift oscillator:

The circuit diagram of the RC phase shift oscillator is shown below:

F1 S.B Deepak 28.12.2019 D18

The frequency produced by the above phase shift oscillator is given by:

f=12πRC6

Wein bridge oscillator:

The circuit diagram of the Wein bridge oscillator is shown below:

F2 U.B Madhu 28.12.19 D 2

The frequency of oscillation is given by:

ωo=1RC

fo=12πRC

Type of Oscillator   

Approximate frequency range
Crystal oscillator Fixed frequency
Tickler feedback oscillator Nearly fixed frequency
Wien bridge oscillator 1 Hz to 1 MHz
Phase-shift oscillator 1 Hz to 10 MHz
Hartley's oscillator  10 kHz to 100 MHz
Colpitt's oscillator  10 kHz to 100 MHz

Match the oscillator circuits in column A with their respective characteristics in column B. 

Column A

Column B

A.

Hartley oscillator 

I.

Two stage RC coupled amplifier  

B.

Crystal oscillator 

II.

LC tuned circuit 

C.

Wien bridge oscillator 

III.

Greater stability 

  1. A - I, B - III, C - II
  2. A - II, B - I, C - III
  3. A - III, B - I, C - II
  4. A - II, B - III, C - I

Answer (Detailed Solution Below)

Option 4 : A - II, B - III, C - I

Oscillators and Feedback Amplifier Question 13 Detailed Solution

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The correct option is 4

Concept:

A. Hartley oscillator - II. LC tuned circuit

Explanation: The Hartley oscillator is an electronic oscillator circuit in which the oscillation frequency is determined by an LC (inductor-capacitor) tank circuit. The frequency can be adjusted based on the values of the inductors and capacitors used.

B. Crystal oscillator - III. Greater stability

Explanation: A Crystal oscillator uses a quartz crystal for frequency control and offers excellent frequency stability due to the quartz crystal's high Q-factor. This makes a crystal oscillator more stable compared to the other oscillator circuits.

C. Wien bridge oscillator - I. Two-stage RC coupled amplifier

Explanation: The Wien Bridge Oscillator employs a feedback circuit with an RC (resistor-capacitor) network to produce sinusoidal oscillations. Its design can involve a two-stage RC coupled amplifier and it's often used for generating audio frequencies.

The negative feedback in an amplifier leads to which one of the following?

  1. Increase in current gain  
  2. Increase in voltage gain
  3. Decrease in voltage gain
  4. Decrease in bandwidth

Answer (Detailed Solution Below)

Option 3 : Decrease in voltage gain

Oscillators and Feedback Amplifier Question 14 Detailed Solution

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Concept:

Negative feedback circuit:

60741b71695ce5174c2923ee 16445025689711

 

The feedback amplification factor is given by:

 AAo1+Aoβ

Where,

Ao is open-loop gain

Aoβ is the loop gain.

Explanation:

The negative feedback in amplifiers causes:

  • Reduced the gain and increases the stability in G.
  • Increases the bandwidth to maintain constant gain-bandwidth product
  • Reduces the distortion and noise in the amplifier
  • The signal-to-noise ratio is not affected.
  • The voltage gain (Av) of an amplifier is defined as the ratio of output voltage to the input voltage.

Av = Vo/Vin

Here, Vo is the output voltage of an amplifier and Vin is the input voltage of an amplifier.

  • In a negative feedback amplifier, closed-loop voltage gain is given 

            Av = Vo / Vin = 1/(1+Aoβ)

Here, β = feedback factor,

Ao = open-loop gain of the amplifier.

  • This expression clearly shows that closed-loop voltage gain has reduced by introducing negative feedback
  • We know that a product of gain and bandwidth is inversely proportional so here bandwidth of amplifier will increase ;

            (gain × bandwidth = 0.35)

  • A negative feedback amplifier decreases the current gain.

The crystal oscillator is a constant frequency oscillator due to:

  1. Rigidity
  2. Vibrations
  3. Low Q
  4. High Q

Answer (Detailed Solution Below)

Option 4 : High Q

Oscillators and Feedback Amplifier Question 15 Detailed Solution

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  • One of the most important features of the crystal oscillator is its frequency stability as it has the ability to provide a constant frequency output under varying load conditions.
  • The stability of the crystal oscillator is closely related to its quality factor or Q.
  • High-Q crystal oscillator will oscillate at constant frequency because it produces oscillation only when it is nearer to its resonance frequency.
  • A typical Q for a crystal oscillator ranges from 104 to 106.

Important Points:

  • The crystal of crystal oscillator is usually made of the quartz material and provides a high degree of frequency stability and accuracy.
  • It uses a piezoelectric crystal and when an ac voltage is applied across a crystal it starts vibrating at the frequency of supply voltage this effect is known as piezoelectric effect and the crystal which exhibits this effect is known as piezoelectric crystals.
  • Conversely, when these crystals are placed under mechanical strain to vibrate, they produce an ac voltage
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