What is the primary advantage of a Darlington pair configuration in BJT amplifiers?
Answer: A
Darlington pairs consist of two transistors in cascade, providing extremely high current gain (β² product) and very high input impedance due to the two base-emitter junctions.
Q.62Medium
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.63Easy
A BJT is operated in the saturation region. Which of the following statements is true?
Answer: B
In saturation, both junctions are forward biased: base-emitter junction is forward biased for current conduction, and base-collector junction is also forward biased, allowing maximum collector current.
Q.64Medium
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.65Easy
In a two-stage cascaded amplifier, the first stage has gain A₁ = 50 and bandwidth BW₁ = 100 kHz, while the second stage has A₂ = 20 and BW₂ = 200 kHz. What is the overall voltage gain?
Answer: B
In cascaded amplifiers, overall voltage gain = A₁ × A₂ = 50 × 20 = 1000. The bandwidth is determined by the stage with lowest bandwidth (100 kHz).
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Q.66Easy
A voltage-controlled voltage source (VCVS) has transconductance gm = 50 mS. If the input voltage is 10 mV, what is the output current?
Answer: A
Output current Id = gm × Vgs = 50 mS × 10 mV = 50 × 10⁻³ × 10 × 10⁻³ = 0.5 × 10⁻³ A = 0.5 mA
Q.67Medium
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.68Medium
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.69Medium
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.70Medium
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.71Medium
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.72Medium
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.73Medium
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.74Hard
For a photodiode amplifier using trans-impedance configuration, increasing the feedback resistance Rf by 10 times will affect the noise figure and gain as:
Answer: B
Increasing Rf increases gain proportionally (V_out = I_in × Rf). However, thermal noise of Rf (4kTRf/Δf) increases, degrading noise figure.
Q.75Medium
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.76Hard
A current mirror circuit using matched BJTs has output impedance of approximately:
Answer: B
BJT current mirror output impedance = 1/gm (Early effect) × (1+λ), where λ is Early effect parameter. This is typically in the range of MΩ, providing high impedance.
Q.77Hard
In an instrumentation amplifier using three op-amps, what is the primary function of the first stage (two non-inverting amplifier configuration)?
Answer: A
The input stage of a 3-op-amp instrumentation amplifier provides very high input impedance (prevents loading) and adjustable gain through a single external resistor, which is then amplified by a differential stage.
Q.78Hard
A common-source FET amplifier with active load has voltage gain Av = -gm/gm,load. To maximize gain, which statement is true?
Answer: B
Gain is inversely proportional to load transconductance. To maximize |Av|, we need to minimize gm,load, which is achieved by reducing W/L ratio of the load transistor.
Q.79Medium
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.80Medium
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.
