Q. 1
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Class-A power amplifier is biased such that collector current flows for:
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Step-by-Step Explanation & Concept 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).
Q. 2
Electronics Engineering
Difficulty: Easy
(1 Mark)
In a Class-B push-pull amplifier, each active transistor conducts for:
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Step-by-Step Explanation & Concept 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%.
Q. 3
Electronics Engineering
Difficulty: Easy
(1 Mark)
What is the primary cause of 'Crossover Distortion' in a standard complementary-symmetry Class-B push-pull amplifier?
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Step-by-Step Explanation & Concept 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.
Q. 4
Electronics Engineering
Difficulty: Easy
(1 Mark)
How is crossover distortion eliminated in modern audio power output stages?
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Step-by-Step Explanation & Concept 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.
Q. 5
Electronics Engineering
Difficulty: Medium
(1 Mark)
Class-C amplifiers have a conduction angle of less than 180 degrees. They achieve high efficiencies (> 85-90%) and are predominantly used in:
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Step-by-Step Explanation & Concept 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.
Q. 6
Electronics Engineering
Difficulty: Easy
(1 Mark)
Applying negative feedback to an amplifier produces which of the following overall performance improvements?
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Step-by-Step Explanation & Concept 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).
Q. 7
Electronics Engineering
Difficulty: Medium
(1 Mark)
In a Voltage-Series negative feedback amplifier (Series-Shunt topology):
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Step-by-Step Explanation & Concept 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.
Q. 8
Electronics Engineering
Difficulty: Medium
(1 Mark)
In a Current-Shunt negative feedback amplifier (Shunt-Series topology):
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Step-by-Step Explanation & Concept 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.
Q. 9
Electronics Engineering
Difficulty: Easy
(1 Mark)
Barkhausen's Criterion states that for a linear feedback circuit to maintain sustained sinusoidal oscillations without external input, the loop gain must satisfy:
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Step-by-Step Explanation & Concept 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.
Q. 10
Electronics Engineering
Difficulty: Medium
(1 Mark)
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?
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Step-by-Step Explanation & Concept 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.
Q. 11
Electronics Engineering
Difficulty: Medium
(1 Mark)
What is the minimum amplifier voltage gain (|Av|) required for an RC phase-shift oscillator using an ideal 3-section ladder network to sustain oscillations?
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Step-by-Step Explanation & Concept Rationale
At the frequency where phase shift across the 3-section RC network is exactly 180°, the transfer attenuation factor beta = Vout / Vin = -1 / 29. For loop gain |A * beta| >= 1, the amplifier must provide an inverting gain of |Av| >= 29.
Q. 12
Electronics Engineering
Difficulty: Easy
(1 Mark)
In a Wien Bridge oscillator, the lead-lag feedback network has a transfer ratio beta = 1/3 and a phase shift of 0 degrees at resonant frequency f0 = 1 / (2*pi*R*C). The non-inverting amplifier must provide a closed-loop gain of:
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Step-by-Step Explanation & Concept Rationale
At resonance, the reactive components in the series-parallel RC Wien bridge balance, giving zero phase shift and attenuation beta = 1/3. For Barkhausen criterion |A * beta| >= 1, the non-inverting op-amp stage must have a gain Av = 1 + Rf/R1 >= 3.
Q. 13
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Hartley oscillator uses which type of resonant tank circuit in its feedback loop?
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Step-by-Step Explanation & Concept Rationale
A Hartley oscillator is distinguished by an inductive voltage divider (tapped inductor L1 and L2 with mutual inductance M) tuned by capacitor C: Leq = L1 + L2 + 2M, f0 = 1 / (2*pi*sqrt(Leq * C)).
Q. 14
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Colpitts oscillator uses which configuration in its resonant frequency-determining tank?
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Step-by-Step Explanation & Concept Rationale
A Colpitts oscillator uses a capacitive voltage divider (C1 and C2 in series across inductor L). Equivalent capacitance Ceq = (C1 * C2) / (C1 + C2), and oscillation frequency f0 = 1 / (2*pi*sqrt(L * Ceq)). It offers superior high-frequency stability over Hartley.
Q. 15
Electronics Engineering
Difficulty: Hard
(1 Mark)
A Clapp oscillator is a superior refinement of the Colpitts oscillator achieved by:
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Step-by-Step Explanation & Concept Rationale
In the Clapp oscillator, capacitor C3 is chosen much smaller than C1 and C2 (C3 << C1, C2). The net capacitance Ceq ≈ C3, making the resonant frequency almost completely immune to transistor internal junction capacitances (Cpi, Cmu), yielding exceptional frequency stability.
Q. 16
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Quartz Crystal oscillator exhibits extraordinary frequency stability (parts per million drift) primarily because:
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Step-by-Step Explanation & Concept Rationale
Based on the Piezoelectric Effect, mechanical vibration and electrical signals couple directly in quartz. Its equivalent RLC circuit has very large inductance L, very tiny series capacitance Cs, and tiny resistance R, resulting in astronomical Q-factors (10^4 to 10^6) that lock frequency precisely against temperature and voltage variations.
Q. 17
Electronics Engineering
Difficulty: Medium
(1 Mark)
The electrical equivalent circuit of a piezoelectric quartz crystal consists of:
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Step-by-Step Explanation & Concept Rationale
The motional arm comprises R (mechanical losses), L (vibrating mass), and Cs (elastic compliance). The parallel capacitance Cp is the electrostatic capacitance formed by the quartz dielectric between its metal plated electrodes. It possesses both series resonant (fs) and parallel anti-resonant (fp) frequencies.
Q. 18
Electronics Engineering
Difficulty: Medium
(1 Mark)
In a multistage amplifier, cascading N identical amplifier stages each having lower cutoff frequency fL and upper cutoff frequency fH causes the overall bandwidth to:
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Step-by-Step Explanation & Concept Rationale
For N identical cascaded stages: fL_overall = fL / sqrt(2^(1/N) - 1) (increases), and fH_overall = fH * sqrt(2^(1/N) - 1) (decreases). Because the band edges close in from both sides, overall bandwidth shrinks as more stages are cascaded.
Q. 19
Electronics Engineering
Difficulty: Easy
(1 Mark)
Which coupling method in multistage amplifiers allows amplification of direct current (DC) and extremely slow-varying signals?
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Step-by-Step Explanation & Concept Rationale
Coupling capacitors block DC signals completely, and transformers cannot transfer unvarying DC magnetic flux. Direct coupling connects the output of one stage directly (or via Zener diode) to the input of the next without reactive elements, passing signals from 0 Hz (DC) upwards.
Q. 20
Electronics Engineering
Difficulty: Medium
(1 Mark)
The major design disadvantage of Direct-Coupled BJT amplifiers is:
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Step-by-Step Explanation & Concept Rationale
Because there are no DC blocking capacitors, any drift in VBE or ICBO with temperature in the first stage is directly amplified by all subsequent stages, causing severe DC drift of the output baseline. Differential pairs with matched thermal tracking are used to cancel this drift.
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