The Miller effect in a common emitter amplifier causes
Answer: D
Miller effect refers to the multiplication of base-collector capacitance by (1+Av) at the input, increasing input capacitance, reducing input impedance, and decreasing bandwidth.
Q.2Hard
Negative feedback in an amplifier primarily results in
Answer: B
Negative feedback reduces gain by factor (1+Aβ) but provides benefits: improved linearity, reduced distortion, increased input impedance (series feedback), decreased output impedance (shunt feedback).
Q.3Hard
The Barkhausen criterion for oscillation states that
Answer: B
Barkhausen criterion: For sustained oscillations, |Aβ| = 1 (unity gain) AND total phase shift = 0° (or 360°). Both conditions must be satisfied simultaneously.
Q.4Hard
Which statement about thermal stability in BJT amplifiers is CORRECT?
Answer: C
Thermal stability improved by: (1) RE without bypass for DC stabilization, (2) lower supply voltage, (3) smaller β transistor, (4) heat sinking. VBE decreases with temperature (negative TC), increasing base current and ICO, causing thermal runaway.
Q.5Hard
In a Common Emitter amplifier, if the input impedance is 1 kΩ and the output impedance is 50 kΩ, what is the approximate voltage gain if β = 100 and gm = 0.04 S?
Answer: B
Voltage gain = gm × Rc (where Rc ≈ output impedance) = 0.04 × 50k = 2000 for a CE amplifier configuration.
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Q.6Hard
In a differential amplifier, Common Mode Rejection Ratio (CMRR) is measured as 80 dB. What is the numerical ratio of differential gain to common mode gain?
A feedback amplifier with closed-loop gain Acl = 10 has feedback fraction β = 0.1. What is the open-loop gain (A)?
Answer: B
From Acl = A/(1 + Aβ), we get 10 = A/(1 + A×0.1). Solving: 10(1 + 0.1A) = A → 10 = 0.9A → A ≈ 1000 (for high gain).
Q.8Hard
In a two-stage amplifier with individual gains A₁ = 50 and A₂ = 100, if negative feedback β = 0.01 is applied across both stages, what is the approximate closed-loop gain?
Answer: B
Overall open-loop gain = 50×100 = 5000. Closed-loop gain = A/(1+Aβ) = 5000/(1+5000×0.01) = 515000 ≈ 98 ≈ 100. With feedback applied, typical result is ~990.
Q.9Hard
In a Common Gate FET amplifier, how does the input impedance compare to Common Source?
Answer: A
CG configuration has low input impedance ≈ 1/gm because the input is presented at the source terminal, acting as a transimpedance amplifier.
Q.10Hard
In a feedback amplifier with loop gain L = Aβ, the closed-loop gain is approximately:
Answer: A
For negative feedback, Acl = A/(1+L) where L = Aβ is the loop gain. This formula shows how feedback reduces gain but improves stability.
Q.11Hard
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.12Hard
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.13Hard
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.14Hard
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.15Hard
What is the maximum theoretical efficiency of a Class AB amplifier?
Answer: C
Class AB combines Class A and Class B characteristics. Maximum efficiency approaches π/2(√2) ≈ 78.5% for Class B, but with Class A bias, typical maximum is around 88.5% under ideal conditions.
Q.16Hard
Which configuration minimizes the output impedance of an amplifier stage?
Answer: A
Series-shunt (voltage-series) feedback reduces output impedance by factor (1+Aβ). Other configurations either increase impedance or have minimal effect on output impedance.
Q.17Hard
In a log amplifier circuit, what is the primary source of error in practical implementations?
Answer: B
Log amplifier uses diode in feedback. Temperature changes cause Is (reverse saturation current) to vary exponentially, directly affecting the logarithmic transfer function. This is the dominant non-ideal effect.
Q.18Hard
For a bootstrapped voltage follower, what is the advantage over a standard buffer?
Answer: B
Bootstrapping increases input impedance by reducing effective base current drawn from source through capacitive feedback. This minimizes loading effects on high-impedance sources.
Q.19Hard
In a Wien bridge oscillator, what is the condition for sustained oscillation?
Answer: C
For sustained oscillation: (1) Barkhausen condition: |Aβ| = 1 (unity loop gain), (2) Phase condition: ∠Aβ = 0° or 360°. In Wien bridge, gain needed ≈ 3 to compensate losses.
