The bandwidth of a common emitter amplifier can be increased by
Answer: D
Emitter degeneration without bypass capacitor increases input impedance and reduces gain but significantly increases bandwidth due to negative feedback.
Q.2Medium
In a differential amplifier, the common mode rejection ratio (CMRR) is defined as
Answer: A
CMRR = Ad/Ac, where Ad is differential mode gain and Ac is common mode gain. High CMRR (>80dB typical) is desired to reject common mode noise.
Q.3Medium
The input impedance of a voltage follower (common collector) amplifier is
Answer: B
CC amplifier has very high input impedance (Zin ≈ β·re) making it suitable as a buffer stage between high impedance sources and low impedance loads.
Q.4Medium
At what frequency does the gain of a RC coupled amplifier reduce to 70.7% of its mid-band gain?
Answer: B
The -3dB cutoff frequency is where the magnitude of gain reduces to 1/√2 (≈ 70.7%) of maximum gain. This defines the bandwidth of the amplifier.
Q.5Medium
In an RC coupled amplifier, the lower cutoff frequency is primarily determined by
Answer: A
Lower cutoff frequency fL ≈ 1/(2π·CC·Rin) where CC is coupling capacitor and Rin is input impedance. Larger CC gives lower fL.
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Q.6Medium
For a feedback amplifier with feedback fraction β, the loop gain is defined as
Answer: A
Loop gain = Aβ, where A is open-loop gain and β is feedback fraction. For stability, loop gain should be <1 for phase margin requirements.
Q.7Medium
In a Class A amplifier, the maximum theoretical efficiency is
Answer: A
Maximum theoretical efficiency of Class A = π/4 ≈ 78.5% when maximum swing is used, but practically 25-50% is achieved due to quiescent current and losses.
Q.8Medium
In a Colpitts oscillator, the frequency of oscillation is determined by
Answer: C
Colpitts oscillator frequency f = 1/(2π√(L·Ceq)) where Ceq = (C1·C2)/(C1+C2). Voltage divider is formed by C1 and C2 for feedback.
Q.9Medium
The gain-bandwidth product (GBP) of an op-amp is approximately constant. For a 741 op-amp with GBP ≈ 1 MHz, the open-loop gain at 100 kHz is approximately
Answer: A
GBP = Aol × f. For 741: 1 MHz = Aol × 100 kHz, therefore Aol = 1 MHz / 100 kHz = 10 V/V. GBP is approximately constant over frequency.
Q.10Medium
In a saturated BJT transistor used as a switch, the voltage drop VCE(sat) is typically
Answer: B
In saturation, VCE(sat) ≈ 0.1-0.3 V for silicon BJT (not zero due to residual resistance). VBE(on) ≈ 0.7V in saturation. This is used in switching applications.
Q.11Medium
A BJT amplifier exhibits thermal runaway due to increased leakage current. Which design modification can best prevent this phenomenon?
Answer: B
Voltage divider bias with low impedance provides stiff bias independent of transistor parameters, preventing thermal runaway by maintaining stable base voltage despite temperature changes.
Q.12Medium
Which configuration provides the highest input impedance and lowest output impedance among op-amp circuits?
Answer: C
Voltage follower has Zin ≈ infinity (determined by op-amp input impedance) and Zout ≈ 0, making it ideal for impedance transformation.
Q.13Medium
For the oscillator to sustain oscillations, the loop gain must satisfy which condition (assuming phase shift is 360° or 0°)?
Answer: B
Barkhausen criterion requires loop gain ≥ 1 (or |A×β| ≥ 1) for sustained oscillations; less than 1 causes damping, more than 1 causes amplitude growth until saturation.
Q.14Medium
A Class B amplifier has two transistors operating in push-pull configuration. What is the maximum theoretical efficiency?
Answer: B
Class B has maximum efficiency of π/4 ≈ 78.5%. Each transistor conducts 180° of the signal cycle, eliminating idle current waste compared to Class A.
Q.15Medium
In an RC phase shift oscillator, how many RC sections are typically used to achieve 180° phase shift?
Answer: C
Three RC sections are used, each providing 60° phase shift. Total = 3 × 60° = 180°, which combined with 180° from inverting amp gives 360° (0°) for sustained oscillation.
Q.16Medium
The Miller effect in a Common Emitter amplifier results in:
Answer: C
Miller effect causes effective input capacitance to increase by factor (1 + |Av|), reducing input impedance and bandwidth. This limits high-frequency response.
Q.17Medium
In a Wein bridge oscillator, the frequency of oscillation is given by f = 1/(2πRC). At what condition does oscillation start?
Answer: B
Wein bridge oscillator requires loop gain ≥ 3 to overcome the attenuation of the bridge network (which has 31 attenuation at oscillation frequency).
Q.18Medium
In a transimpedance amplifier (TIA), which parameter determines the transimpedance gain?
Answer: B
Transimpedance gain = Vout/Iin = -Rf for inverting configuration. The feedback resistor directly sets the input-to-output conversion.
Q.19Medium
Which oscillator topology has the poorest frequency stability due to dependency on passive component tolerances?
Answer: B
RC phase shift oscillator depends entirely on RC values which have poor tolerance and temperature coefficients, resulting in frequency drift.
Q.20Medium
For maximum power transfer from a source to load through an amplifier, what should be the relationship between source and load impedance?
Answer: B
Maximum power transfer theorem states that maximum power is delivered when load impedance equals complex conjugate of source impedance, especially in AC circuits.
