Electronics and Communication questions reward anyone who is comfortable moving between the time domain and the frequency domain. This set covers network theory, analog and digital circuits, signals and systems, control systems, communication systems, and electromagnetics. Numerical solutions keep the units visible at every step, since a dropped factor is the most common reason a correct method still produces a wrong option.
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.
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.
