An op-amp has an open-loop gain of 100,000 and a feedback resistor ratio of Rf/Rin = 10. What is the closed-loop gain?
Answer: A
For a non-inverting amplifier, Acl ≈ 1 + (Rf/Rin) = 1 + 10 = 11. The feedback dominates over the open-loop gain.
Q.22Medium
In a Colpitts oscillator, the feedback network consists of:
Answer: A
Colpitts oscillator uses capacitive voltage divider with two capacitors in series across an inductor for frequency-determining feedback.
Q.23Medium
A Class B power amplifier has two transistors. What is the primary advantage over Class A?
Answer: A
Class B achieves ~78.5% efficiency by conducting only 180° of the cycle compared to Class A's ~25% efficiency, though it introduces crossover distortion.
Q.24Medium
In a push-pull amplifier configuration, what is the phase relationship between the two transistor inputs?
Answer: A
Push-pull amplifiers use complementary transistors driven 180° out of phase to handle positive and negative half-cycles separately.
Q.25Medium
An inverting amplifier has Rin = 10 kΩ and Rf = 100 kΩ. If the input signal is 100 mV, what is the output voltage?
Answer: A
For inverting amplifier, Vo = -(Rf/Rin) × Vin = -(100k/10k) × 0.1 = -10 × 0.1 = -1 V.
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Q.26Medium
Which parameter primarily determines the frequency stability of an RC oscillator?
Answer: A
RC oscillators derive their frequency from RC time constants. The frequency f = 1/(2πRC) approximately, making RC the dominant parameter.
Q.27Medium
A Class AB amplifier operates with a quiescent current (ICQ) that is:
Answer: A
Class AB uses a small quiescent current to bias transistors slightly into conduction, eliminating crossover distortion while maintaining better efficiency than Class A.
Q.28Medium
In a differential amplifier using matched transistors, the Common Mode Rejection Ratio (CMRR) is primarily limited by:
Answer: A
CMRR depends on the matching of transistor parameters and passive components. Mismatches in emitter resistances and transistor characteristics directly degrade CMRR.
Q.29Medium
For a Source Follower (Common Drain FET amplifier), what is the typical voltage gain range?
Answer: B
Source follower has voltage gain Av = gm·Rl/(1 + gm·Rl) ≈ 1 for practical cases, always less than 1. It provides high input impedance and low output impedance.
Q.30Medium
In a Wien Bridge oscillator, the frequency of oscillation is determined by:
Answer: B
Wien Bridge oscillator has frequency of oscillation: f = 1/(2πRC), where R and C are the series and parallel RC components in the bridge network.
Q.31Medium
What is the phase shift introduced by a first-order RC low-pass filter at the cutoff frequency (fc)?
Answer: B
At cutoff frequency, the impedances of R and C are equal, resulting in a phase shift of -45° for a single RC low-pass filter.
Q.32Medium
A trans-impedance amplifier (TIA) has feedback resistor Rf = 1 MΩ. If the input current is 1 μA, what is the output voltage?
Answer: A
Trans-impedance gain = Rf = 1 MΩ. Output voltage Vout = Iin × Rf = 1 μA × 1 MΩ = 1 V
Q.33Medium
In a Class A amplifier operating at quiescent point Q, if the load line intersects the characteristic curve at VCE = 6V and IC = 50 mA, what is the maximum output power swing (assuming linear operation)?
Answer: B
In Class A, maximum output power = 0.5 × VCE × IC = 0.5 × 6 × 50 = 150 mW. The factor 0.5 accounts for peak AC swing being limited to half the quiescent values.
Q.34Medium
Which of the following topologies provides the highest input impedance for an amplifier?
Answer: B
Source follower (Common Drain) provides extremely high input impedance (in gigaohms range) because the gate of the FET is the input, which has very high impedance characteristic of FET gates.
Q.35Medium
In a precision rectifier (ideal diode) circuit using op-amp, what is the key advantage over conventional diode rectifiers?
Answer: A
Precision rectifier uses op-amp feedback to compensate for the forward voltage drop (Vf ≈ 0.7V) of the diode, making it ideal for low-signal applications.
Q.36Medium
A cascode amplifier configuration is preferred over single-stage amplifier because it provides:
Answer: A
Cascode configuration cascades a common-emitter/source stage with a common-base/gate stage, providing high gain while significantly reducing Miller capacitance effect and improving bandwidth.
Q.37Medium
In a Push-Pull (Class B) amplifier, what is the primary disadvantage that Class AB configuration addresses?
Answer: A
Class B exhibits crossover distortion where both transistors are off near the zero-crossing point. Class AB adds small quiescent bias to keep transistors in active region, eliminating this distortion.
Q.38Medium
In a feedback amplifier, negative feedback is applied by feeding back output to which junction?
Answer: C
Negative feedback must be applied in opposition to the input signal. For inverting topology, feedback goes to inverting terminal; for non-inverting, to non-inverting terminal. The feedback signal opposes the original input.
Q.39Medium
In a BJT differential amplifier, if both inputs are grounded and the tail current source has finite output impedance, what is the effect on CMRR?
Answer: A
Finite output impedance of tail current source reduces the symmetry of the differential pair, degrading Common Mode Rejection Ratio. Higher impedance improves CMRR.
Q.40Medium
In a transimpedance amplifier used for photodiode detection, increasing feedback resistance Rf causes which effect?
Answer: B
In transimpedance configuration, gain = Rf. However, bandwidth is inversely proportional to Rf due to feedback capacitance (Cf ≈ Cp + Cin), creating a gain-bandwidth tradeoff.
