Amplifiers & Feedback
Change Setup
📘 Comprehensive Syllabus & Examination Guide
Amplifiers & Feedback
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Mapped Subjects & Topic Question Distribution
Total Question Pool
100%
20 MCQs
Combined Active Syllabus
Amplifiers & Feedback
20 MCQs
Topic Pool
📊 Question Pool Structure
20 MCQs across fundamental, intermediate, and advanced concept tiers.
⚡ Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
⚖️ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Amplifiers & Feedback, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
Ready to test your knowledge?
Launch interactive 1-by-1 practice with instant feedback, bookmarking, and step-by-step rationales.
Solved Blueprint Examples
Showing 10 solved representative questions
📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Amplifiers & Feedback
Easy • Electronics Engineering
A Class-A power amplifier is biased such that collector current flows for:
A
The entire 360 degrees of the input AC signal cycle
B
Exactly 180 degrees of the input cycle
C
Between 180 and 360 degrees
D
Less than 180 degrees
✓ Correct Answer: A - The entire 360 degrees of the input AC signal cycle
📖 Step-by-Step Solution & Conceptual Rationale:
In Class-A operation, the Q-point is centered in the active region, ensuring the transistor conducts for the full 360° conduction angle. It offers maximum linearity and minimum harmonic distortion, but has low maximum theoretical efficiency (25% for direct-coupled resistive load, 50% for transformer-coupled load).
Sample Question 2
Amplifiers & Feedback
Easy • Electronics Engineering
In a Class-B push-pull amplifier, each active transistor conducts for:
A
Exactly 180 degrees (one half-cycle) of the input waveform
B
360 degrees
C
90 degrees
D
270 degrees
✓ Correct Answer: A - Exactly 180 degrees (one half-cycle) of the input waveform
📖 Step-by-Step Solution & Conceptual Rationale:
Class-B stages bias transistors at the cutoff boundary (Q-point at ICQ = 0). One transistor conducts for the positive 180° half-cycle and the complementary transistor conducts for the negative 180° half-cycle, combining to deliver a full cycle with theoretical maximum efficiency of pi / 4 ≈ 78.5%.
Sample Question 3
Amplifiers & Feedback
Easy • Electronics Engineering
What is the primary cause of 'Crossover Distortion' in a standard complementary-symmetry Class-B push-pull amplifier?
A
Thermal runaway of the power transistors
B
The deadband region around zero volts where neither transistor conducts until input signal exceeds the forward base-emitter threshold voltage (VBE ≈ 0.7 V)
C
Collector saturation voltage VCE_sat
D
Miller effect capacitance
✓ Correct Answer: B - The deadband region around zero volts where neither transistor conducts until input signal exceeds the forward base-emitter threshold voltage (VBE ≈ 0.7 V)
📖 Step-by-Step Solution & Conceptual Rationale:
A BJT does not conduct until base-emitter forward voltage exceeds ~0.6-0.7 V. Near the zero-crossing of the input sine wave, both transistors remain in cutoff simultaneously, creating a flat, non-conducting zero-volt notch (crossover distortion) in the output waveform.
Sample Question 4
Amplifiers & Feedback
Easy • Electronics Engineering
How is crossover distortion eliminated in modern audio power output stages?
A
By operating in Class-AB mode, where diodes or VBE multipliers provide a small forward trickle bias to keep both transistors slightly conducting at zero signal
B
By using series inductors
C
By cooling the transistors to absolute zero
D
By running in Class-C mode
✓ Correct Answer: A - By operating in Class-AB mode, where diodes or VBE multipliers provide a small forward trickle bias to keep both transistors slightly conducting at zero signal
📖 Step-by-Step Solution & Conceptual Rationale:
Class-AB introduces a small quiescent pre-bias (~1.4 V across both bases via two series diodes or a VBE multiplier) so each transistor conducts for slightly more than 180° (typically 185°-200°), smoothing the transition through zero and eliminating crossover distortion.
Sample Question 5
Amplifiers & Feedback
Medium • Electronics Engineering
Class-C amplifiers have a conduction angle of less than 180 degrees. They achieve high efficiencies (> 85-90%) and are predominantly used in:
A
High-fidelity audio power amplifiers
B
Tuned Radio Frequency (RF) power transmitters and power oscillators using an LC tank load to reconstruct sinusoidal waves
C
Precision instrumentation DC amplifiers
D
Operational amplifier input stages
✓ Correct Answer: B - Tuned Radio Frequency (RF) power transmitters and power oscillators using an LC tank load to reconstruct sinusoidal waves
📖 Step-by-Step Solution & Conceptual Rationale:
Class-C conducts in short pulses (< 180°), producing severe harmonic distortion. However, when loaded with a high-Q parallel LC tank tuned to the fundamental carrier frequency, the tank flywheel effect filters out harmonics, generating pure high-power sinusoidal RF output at very high efficiency.
Sample Question 6
Amplifiers & Feedback
Easy • Electronics Engineering
Applying negative feedback to an amplifier produces which of the following overall performance improvements?
A
Increases closed-loop gain and increases non-linear distortion
B
Stabilizes closed-loop gain against component parameter variations, widens bandwidth, reduces non-linear distortion, and attenuates internal noise
C
Causes instability and uncontrolled oscillation
D
Reduces input impedance to zero under all configurations
✓ Correct Answer: B - Stabilizes closed-loop gain against component parameter variations, widens bandwidth, reduces non-linear distortion, and attenuates internal noise
📖 Step-by-Step Solution & Conceptual Rationale:
Negative feedback trades gain for quality: Closed-loop gain Af = A / (1 + A*beta). It desensitizes gain (dAf/Af = (1/(1+A*beta)) * dA/A), extends upper cutoff frequency fHf = fH * (1 + A*beta), lowers lower cutoff fLf = fL / (1 + A*beta), and reduces non-linear harmonic distortion by (1 + A*beta).
Sample Question 7
Amplifiers & Feedback
Medium • Electronics Engineering
In a Voltage-Series negative feedback amplifier (Series-Shunt topology):
A
Input impedance increases by (1 + A*beta), and output impedance decreases by (1 + A*beta)
B
Input impedance decreases, and output impedance increases
C
Both input and output impedances increase
D
Both input and output impedances decrease
✓ Correct Answer: A - Input impedance increases by (1 + A*beta), and output impedance decreases by (1 + A*beta)
📖 Step-by-Step Solution & Conceptual Rationale:
Series mixing at the input opposes incoming voltage, raising input impedance: Rif = Ri * (1 + A*beta). Shunt sampling at the output senses output voltage, lowering output impedance: Rof = Ro / (1 + A*beta). This topology forms an ideal Voltage Amplifier.
Sample Question 8
Amplifiers & Feedback
Medium • Electronics Engineering
In a Current-Shunt negative feedback amplifier (Shunt-Series topology):
A
Input impedance decreases by (1 + A*beta), and output impedance increases by (1 + A*beta)
B
Input impedance increases, and output impedance decreases
C
Both input and output impedances decrease
D
Gain becomes infinite
✓ Correct Answer: A - Input impedance decreases by (1 + A*beta), and output impedance increases by (1 + A*beta)
📖 Step-by-Step Solution & Conceptual Rationale:
Shunt mixing at the input draws away incoming current, reducing input impedance: Rif = Ri / (1 + A*beta). Series sampling at the output senses load current, raising output impedance: Rof = Ro * (1 + A*beta). This represents an ideal Current Amplifier.
Sample Question 9
Amplifiers & Feedback
Easy • Electronics Engineering
Barkhausen's Criterion states that for a linear feedback circuit to maintain sustained sinusoidal oscillations without external input, the loop gain must satisfy:
A
Magnitude |A * beta| = 1, and total net phase shift around the closed loop must be 0 degrees (or an integer multiple of 360 degrees)
B
|A * beta| < 1 and phase shift = 90 degrees
C
|A * beta| = 0 and phase shift = 180 degrees
D
|A * beta| > 100 at all frequencies
✓ Correct Answer: A - Magnitude |A * beta| = 1, and total net phase shift around the closed loop must be 0 degrees (or an integer multiple of 360 degrees)
📖 Step-by-Step Solution & Conceptual Rationale:
Barkhausen's criteria for sustained oscillation: 1) The magnitude of the loop transmission |A(j*omega) * beta(j*omega)| must equal unity (1.0). 2) The total loop phase shift must be 0° or 360° (positive regenerative feedback). To initiate startup, loop gain is designed slightly > 1, stabilizing at 1 via amplitude limiting.
Sample Question 10
Amplifiers & Feedback
Medium • Electronics Engineering
An RC Phase-Shift oscillator using a single Common Emitter BJT requires three cascaded identical RC ladder sections in its feedback network. Each RC section provides what nominal phase shift at the oscillation frequency?
A
60 degrees (giving a total feedback shift of 180 degrees, which combines with the 180-degree CE inversion to yield 360 degrees)
B
90 degrees
C
45 degrees
D
120 degrees
✓ Correct Answer: A - 60 degrees (giving a total feedback shift of 180 degrees, which combines with the 180-degree CE inversion to yield 360 degrees)
📖 Step-by-Step Solution & Conceptual Rationale:
The CE transistor introduces 180° phase inversion. The passive 3-stage RC ladder network provides an additional 3 x 60° = 180° phase shift at frequency f0 = 1 / (2*pi*R*C * sqrt(6)). Total loop phase shift equals 180° + 180° = 360° (or 0°). The amplifier gain must be at least |Av| >= 29 to overcome the 1/29 attenuation of the RC network.
Practice All 20 Questions Interactively
Test your knowledge in real-time with continuous progress saving, instant scoring, and performance analytics.