Power Amplifiers and 555 Timer and Voltage Regulators MCQ Quiz - Objective Question with Answer for Power Amplifiers and 555 Timer and Voltage Regulators - Download Free PDF

Last updated on May 8, 2025

Latest Power Amplifiers and 555 Timer and Voltage Regulators MCQ Objective Questions

Power Amplifiers and 555 Timer and Voltage Regulators Question 1:

In a transistor amplifier, the composition of the current in any branch during operation is:

  1. the sum of AC and DC
  2. DC only
  3. AC only
  4. the multiplication of AC and DC

Answer (Detailed Solution Below)

Option 1 : the sum of AC and DC

Power Amplifiers and 555 Timer and Voltage Regulators Question 1 Detailed Solution

Concept

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In a transistor amplifier, each branch (base, collector, emitter) carries a current composed of:

Total Current DC bias current AC signal current

  • The DC component is required to bias the transistor and set its operating point (Q-point).
  • The AC component represents the input signal that gets amplified.


This combined current ensures linear amplification without distortion, provided the transistor remains in its active region.

So, during normal amplifier operation, the current is a sum of AC and DC components, not one or the other alone.

Power Amplifiers and 555 Timer and Voltage Regulators Question 2:

PLL operates in which domain?

  1. Time
  2. Frequency
  3. Phase
  4. All of these

Answer (Detailed Solution Below)

Option 4 : All of these

Power Amplifiers and 555 Timer and Voltage Regulators Question 2 Detailed Solution

Explanation:

Phase-Locked Loop (PLL)

Definition: A Phase-Locked Loop (PLL) is an electronic circuit that synchronizes an output signal's phase and frequency with a reference signal. PLLs are widely used in communication systems, signal processing, and control systems to generate stable and precise frequency signals.

  • Time Domain: In the time domain, a PLL ensures that the timing of the output signal matches the timing of the reference signal. By adjusting the VCO, the PLL can correct any timing errors, ensuring that the output signal remains synchronized with the reference signal.
  • Frequency Domain: In the frequency domain, a PLL locks the frequency of the VCO to the frequency of the reference signal. This is crucial for applications requiring stable and precise frequency control, such as in communication systems where accurate carrier frequencies are necessary.
  • Phase Domain: In the phase domain, a PLL maintains a constant phase relationship between the output signal and the reference signal. This phase alignment is essential for coherent signal processing and minimizing phase noise, which can degrade system performance.

Therefore, a PLL operates in all three domains, making Option 4 the correct choice

Power Amplifiers and 555 Timer and Voltage Regulators Question 3:

In FM demodulation using PLL, the signal is retrieved from:

  1. VCO
  2. Filter
  3. Phase detector
  4. Amplifier

Answer (Detailed Solution Below)

Option 3 : Phase detector

Power Amplifiers and 555 Timer and Voltage Regulators Question 3 Detailed Solution

Explanation:

In FM Demodulation Using PLL

Definition: Frequency Modulation (FM) demodulation using a Phase-Locked Loop (PLL) is a method where the frequency of the received signal is tracked and converted back to the original information signal. The PLL is an electronic circuit that consists of a phase detector, a low-pass filter, and a voltage-controlled oscillator (VCO).

Working Principle: In FM demodulation using PLL, the incoming FM signal is first fed into the phase detector of the PLL. The phase detector compares the phase of the incoming signal with the phase of the signal generated by the VCO. The difference in phase generates a voltage signal that is proportional to the frequency deviation of the FM signal. This voltage signal is then passed through a low-pass filter to remove high-frequency components and noise. The filtered signal is used to adjust the VCO, which in turn changes its frequency to match the frequency of the incoming FM signal. The output of the VCO is the demodulated signal, which is the original information signal.

Components of PLL:

  • Phase Detector: Compares the phase of the incoming FM signal with the VCO signal and generates a voltage proportional to the phase difference.
  • Low-Pass Filter: Filters the output of the phase detector to remove high-frequency noise and provides a smooth control voltage to the VCO.
  • Voltage-Controlled Oscillator (VCO): Generates a signal whose frequency is controlled by the input voltage. The VCO adjusts its frequency to match the frequency of the incoming FM signal.

Power Amplifiers and 555 Timer and Voltage Regulators Question 4:

Center frequency of VCO depends on:

  1. Input current
  2. Capacitor only
  3. Supply voltage
  4. External R and C

Answer (Detailed Solution Below)

Option 4 : External R and C

Power Amplifiers and 555 Timer and Voltage Regulators Question 4 Detailed Solution

Explanation:

Voltage-Controlled Oscillator (VCO)

Definition: A Voltage-Controlled Oscillator (VCO) is an electronic oscillator whose oscillation frequency is controlled by a voltage input. The frequency of the output signal varies with the applied input voltage, making VCOs crucial components in various electronic devices such as phase-locked loops (PLLs), frequency synthesizers, and communication systems.

Correct Option Analysis:

The correct option is:

Option 4: External R and C

This option correctly identifies the components that the center frequency of a VCO depends on. In most VCO designs, external resistors (R) and capacitors (C) are used to set the oscillation frequency. The value of these components determines the resonant frequency of the circuit, which is modulated by the input control voltage to produce the desired oscillation frequency.

The resonant frequency f of an LC circuit (inductor-capacitor circuit) or an RC circuit (resistor-capacitor circuit) is given by:

For an LC Circuit:

f = 1 / (2π√(LC))

Where:

  • L is the inductance
  • C is the capacitance

For an RC Circuit:

f = 1 / (2πRC)

Where:

  • R is the resistance
  • C is the capacitance

In both cases, the external R and C components play a crucial role in determining the center frequency of the VCO. By adjusting these components, designers can set the desired frequency range for the oscillator

Power Amplifiers and 555 Timer and Voltage Regulators Question 5:

The output of a phase detector is a:

  1. Sine wave
  2. DC voltage
  3. Pulse train
  4. Square wave

Answer (Detailed Solution Below)

Option 2 : DC voltage

Power Amplifiers and 555 Timer and Voltage Regulators Question 5 Detailed Solution

Explanation:

Phase Detector Output

Definition: A phase detector is an essential component in various electronic systems, particularly in phase-locked loops (PLLs), which are used to synchronize an output signal's phase with a reference signal. The phase detector compares the phase of two input signals and generates an output that represents the phase difference between these signals.

Working Principle: The primary function of a phase detector is to generate a signal that is proportional to the phase difference between two input signals. This output signal can be used to adjust the frequency of a voltage-controlled oscillator (VCO) to maintain phase synchronization in a PLL system. Phase detectors can be implemented using different circuit designs, including mixers, digital logic gates, or specialized integrated circuits.

Correct Option Analysis:

The correct option is:

Option 2: DC voltage

This option is correct because the output of a phase detector in a phase-locked loop is typically a DC voltage. The phase detector produces a voltage that is proportional to the phase difference between the input signals. This DC voltage is then used to control the frequency of the VCO, ensuring that the output signal remains in phase with the reference signal. The output voltage is zero when the input signals are in perfect phase alignment, and it increases or decreases as the phase difference changes.

Additional Information:

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

Option 1: Sine wave

This option is incorrect. A sine wave is a continuous wave that oscillates smoothly, representing a periodic oscillation. While phase detectors work with sine waves as input signals, their output is not typically a sine wave. The output of a phase detector is intended to represent the phase difference, which is more effectively done with a DC voltage or a pulse train.

Option 3: Pulse train

This option is partially correct but not typically the final output used in PLL systems. A pulse train can be produced by certain types of phase detectors, especially digital ones, where the width or frequency of the pulses represents the phase difference. However, in most practical applications, this pulse train is further processed to produce a DC voltage that can be used to control the VCO. Hence, the pulse train is more of an intermediate step rather than the final output.

Option 4: Square wave

This option is incorrect. A square wave is a type of periodic waveform that alternates between two levels at a constant frequency. While a square wave can be one of the input signals to a phase detector, the output of the phase detector is not a square wave. The phase detector's output needs to indicate the phase difference, which a square wave does not effectively do.

Conclusion:

Understanding the function and output of a phase detector is crucial for its application in phase-locked loops and other synchronization systems. The correct output of a phase detector is typically a DC voltage, which reflects the phase difference between the input signals. This DC voltage is then used to control the VCO, maintaining phase alignment. While pulse trains can be intermediate outputs in some phase detector designs, the final output used in PLL systems is a DC voltage. Options like sine waves and square waves are not appropriate as they do not represent the phase difference in a usable form for controlling the VCO.

Top Power Amplifiers and 555 Timer and Voltage Regulators MCQ Objective Questions

The efficiency of direct coupled class B Power amplifier is

  1. 5%
  2. 50%
  3. 25%
  4. 78.5%

Answer (Detailed Solution Below)

Option 4 : 78.5%

Power Amplifiers and 555 Timer and Voltage Regulators Question 6 Detailed Solution

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  • In Power amplifiers, the improved output ac power level is the result of a transfer of energy from the applied dc supplies.
  • It is the applied dc power that permits an ac power output to be greater than the input ac power. In other words, there is an "exchange" of dc power to the ac domain that results in higher output ac power.
  • Conversion efficiency is defined as:

           \(\eta = \frac{P_{o(ac)}}{P_i(dc)}\)

          Po(ac) is the ac power to the load

          Pi(dc) is the dc power supplied.

For a class B power amplifier maximum theoretical maximum efficiency is 78.5 %

Power Amplifier

Conduction Angle

Maximum Efficiency

Class A

360 °

50 %

Class B

180 °

78.5 %

Class AB

180 ° - 360 °

50-78.5 %

Class C

< 180°

≥ 90%

Astable multivibrator is

  1. a square wave clock
  2. equivalent to a flip-flop
  3. a one shot multivibrator
  4. Monostable multivibrator

Answer (Detailed Solution Below)

Option 1 : a square wave clock

Power Amplifiers and 555 Timer and Voltage Regulators Question 7 Detailed Solution

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  • An astable multivibrator is a square wave generator. 
  • Astable multivibrators generally have an even 50% duty cycle.
  • It means 50% of the cycle time the output is “HIGH” and the remaining 50% of the cycle time the output is “OFF”.
  • A free-running multivibrator that has no stable states but switches continuously between two states this action produces a train of square wave pulses at a fixed frequency or constant frequency.

09.10.1018.004

Important Points

An astable multivibrator is a free-running oscillator. 

3

Time period of square pulse (T0) = 0.69 (RA + 2RB)C

Clock frequency = 1/T

\(\frac{1.45}{(R_A + 2R_B)C}\)

Monostable multivibrator:

  • The Monostable Multivibrator circuit has only ONE stable state making it a “one-shot” pulse generator. 
  • It is used as a pulse stretcher.
  • Time period of monostable multivibrator (T) =  RC ln3

 

Bistable multivibrator:

  • It operates in a similar fashion to flip-flops producing one of two stable outputs that are the complement of each other.
  • It is equivalent to a flip flop.
  • It is used as a frequency divider.

The conduction angle of class-AB amplifiers lies between ________.

  1. 180° and 360°
  2. 90° and 360°
  3. 90° and 180°
  4. 270° and 360°

Answer (Detailed Solution Below)

Option 1 : 180° and 360°

Power Amplifiers and 555 Timer and Voltage Regulators Question 8 Detailed Solution

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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

The transistor amplifier with 85% efficiency is likely to be

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

Answer (Detailed Solution Below)

Option 4 : Class C

Power Amplifiers and 555 Timer and Voltage Regulators Question 9 Detailed Solution

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The maximum efficiency of a class A amplifier is 50%

The maximum efficiency of a class B amplifier is 78.5%

The maximum efficiency of a class C amplifier is 90%.

Hence, The transistor amplifier with 85% efficiency is likely to be Class C.

Note:

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

In class B amplifier, the output current flows for

  1. less than half input cycle
  2. more than half input cycle
  3. half input cycle
  4. entire input cycle

Answer (Detailed Solution Below)

Option 3 : half input cycle

Power Amplifiers and 555 Timer and Voltage Regulators Question 10 Detailed Solution

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​There are three classifications of Push-Pull amplifier:

Class ACollector current flows at all times during the full cycle of signal (i.e. 360°)

Class B: Collector current flows only during the positive half cycle of the input signal (i.e. 180°)

Class C: Collector current flows for less than half cycle of the input signal (typical value 80° - 120°)

 

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Features of Push-Pull:

  • Class AB type Push – Pull amplifiers suffer from the cross – over distortion.
  • Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
  • Due to the Class B operation, their collector efficiency is quite high (> 50 %)
  • It is capable of generating high gains.
  • There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.

Peak inverse voltage for a full wave rectifier is

  1. Em
  2. 2Em
  3. 3Em
  4. 4Em

Answer (Detailed Solution Below)

Option 2 : 2Em

Power Amplifiers and 555 Timer and Voltage Regulators Question 11 Detailed Solution

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Peak Inverse Voltage is the maximum voltage that appears across a reverse biased diode:

If Em is the maximum voltage of the sinusoidal input signal 

Then

For Half Wave rectifier \({\rm{PIV\;}} = {\rm{\;}}{{\rm{E}}_{\rm{m}}}\)

For Full Wave rectifier \({\rm{PIV\;}} = {\rm{\;}}2{{\rm{E}}_{\rm{m}}}\)

For Bridge Wave rectifier \({\rm{PIV\;}} = {\rm{\;}}{{\rm{E}}_{\rm{m}}}\)

In a class B power amplifier, output is ________ of the input full cycle.

  1. less than half  
  2. half
  3. more than half but less than full
  4. full 

Answer (Detailed Solution Below)

Option 2 : half

Power Amplifiers and 555 Timer and Voltage Regulators Question 12 Detailed Solution

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  • Class B amplifier is a type of power amplifier where the active device (transistor) conducts only for a one-half cycle of the input signal.
  • Class-B amplifiers use two or more transistors biased in such a way that each transistor only conducts during a one-half cycle of the input waveform
  • That means the conduction angle is 180° for a Class B amplifier.
  • Class B Amplifier also is known as a push-pull amplifier configuration.
  • The circuit of the Class B amplifier and input, output waveforms are shown in the figure below.

F1 U.B Madhu 8.11.19 D 8

  • Since the active device (transistor) is switched off for half the input cycle, the active device dissipates less power and hence the efficiency is improved.
  • The theoretical maximum efficiency of a Class B power amplifier is 78.5%.

In a class B amplifier, it is found that D.C. power is 25 W, find the A.C. power.

  1. 10 W
  2. 62.5 W
  3. 25 W
  4. 50 W

Answer (Detailed Solution Below)

Option 2 : 62.5 W

Power Amplifiers and 555 Timer and Voltage Regulators Question 13 Detailed Solution

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Concept:

The maximum efficiency of a class A amplifier is 50%

The maximum efficiency of a class B amplifier is 78.5%

The maximum efficiency of a class C amplifier is 90%.

Hence, The transistor amplifier with 85% efficiency is likely to be Class C.

Note:

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

 

Calculation: 

We know that figure of merit for class B power amplifier = .4

Figure of merit = dc power / ac power

⇒ .4 = 25/ AC Power

⇒ AC Power = 62.5 W

Push-Pull amplifier circuit is used as ______.

  1. Power Amplifier
  2. Audio Amplifier
  3. RF Amplifier
  4. Emitter follower

Answer (Detailed Solution Below)

Option 1 : Power Amplifier

Power Amplifiers and 555 Timer and Voltage Regulators Question 14 Detailed Solution

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  • Push-Pull is a power amplifier that is used to supply high power to the load.
  • It consists of two transistors in which one is NPN and another is PNP.
  • One transistor pushes the output on a positive half-cycle and the other pulls on a negative half cycle. This is why it is known as a push-pull amplifier.
  • The push-pull Amplifier circuit is as shown:

     F2 S.B Madhu 17.02.20 D1

Note:

There are three classifications of Push-Pull amplifier:

  • Class A amplifier
  • Class B amplifier
  • Class AB amplifier

The collector current of a class C amplifier is _______.

  1. square wave
  2. half sine wave
  3. sine wave
  4. pulse

Answer (Detailed Solution Below)

Option 2 : half sine wave

Power Amplifiers and 555 Timer and Voltage Regulators Question 15 Detailed Solution

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The collector current of a class C amplifier is a half-sine wave.

F1 Shraddha Jaideep 29.10.2021 D6

  • Class C power amplifier is a type of amplifier where the active element (transistor) conduct for less than one-half cycle of the input signal.
  • Less than one-half cycle means the conduction angle is less than 180° and its typical value is 80° to 120°.
  • The reduced conduction angle improves the efficiency to a great extends but causes a lot of distortion.
  • The theoretical maximum efficiency of a Class C amplifier is around 90%.
  • In a Class C Amplifier efficiency and distortion, both are maximum.

Important Points

Class

Class-A

Class-B

Class-C

Class-AB

Operating cycle

360°

180°

Less than 180°

More than 180° and less than 360°

Position of Q

Centre

On X-axis

Below X-axis

Above X-axis

Efficiency

25% or 50%

78.5%

High (almost 100 %)

50 % to 78.5 %

Distortion

Absent

Present more than class-A

Highest

Present less than class

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