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 Jun 12, 2025
Latest Power Amplifiers and 555 Timer and Voltage Regulators MCQ Objective Questions
Power Amplifiers and 555 Timer and Voltage Regulators Question 1:
Which of following class of amplifiers provides highest efficiency ?
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 1 Detailed Solution
Explanation:
Class F Amplifiers
Definition: Class F amplifiers are a type of tuned radio frequency (RF) amplifiers that achieve maximum efficiency by shaping the voltage and current waveforms using harmonic tuning. These amplifiers utilize harmonic resonators to manipulate the waveform, ensuring the voltage and current waveforms are non-overlapping, thus minimizing power dissipation.
Working Principle: The Class F amplifier works by employing harmonic tuning circuits that allow specific harmonics of the input signal to resonate while suppressing others. This tuning helps shape the output waveform to approximate a square wave for voltage and a half-sine waveform for current. By ensuring that the voltage and current waveforms are out of phase (non-overlapping), the power dissipation in the transistor is reduced, leading to higher efficiency.
Efficiency: Class F amplifiers are among the most efficient types of amplifiers, with theoretical efficiencies reaching up to 90% or more. This high efficiency is achieved by reducing the overlap between voltage and current waveforms during operation, minimizing energy loss as heat.
Correct Option Analysis:
The correct option is:
Option 4: Class F
Class F amplifiers provide the highest efficiency among all amplifier classes due to their unique waveform shaping technique, which minimizes the overlap between voltage and current waveforms, significantly reducing power dissipation
Power Amplifiers and 555 Timer and Voltage Regulators Question 2:
In a transistor amplifier, the composition of the current in any branch during operation is:
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 2 Detailed Solution
Concept
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 3:
PLL operates in which domain?
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 3 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 4:
In FM demodulation using PLL, the signal is retrieved from:
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 4 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 5:
Center frequency of VCO depends on:
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 5 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
Top Power Amplifiers and 555 Timer and Voltage Regulators MCQ Objective Questions
The efficiency of direct coupled class B Power amplifier is
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 6 Detailed Solution
Download Solution PDF- 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
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 7 Detailed Solution
Download Solution PDF- 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.
Important Points
An astable multivibrator is a free-running oscillator.
Time period of square pulse (T0) = 0.69 (RA + 2RB)C
Clock frequency = 1/T0
= \(\frac{1.45}{(R_A + 2R_B)C}\)
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 ________.
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 8 Detailed Solution
Download Solution PDF
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
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 9 Detailed Solution
Download Solution PDFThe 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
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 10 Detailed Solution
Download Solution PDFThere are three classifications of Push-Pull amplifier:
Class A: Collector 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°)
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
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 11 Detailed Solution
Download Solution PDFPeak 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.
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 12 Detailed Solution
Download Solution PDF- 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.
- 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.
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 13 Detailed Solution
Download Solution PDFConcept:
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 ______.
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 14 Detailed Solution
Download Solution PDF- 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:
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 _______.
Answer (Detailed Solution Below)
Power Amplifiers and 555 Timer and Voltage Regulators Question 15 Detailed Solution
Download Solution PDFThe collector current of a class C amplifier is a half-sine wave.
- 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 |