Electronics (ECE) — MCQ Practice Questions & Answers

Question 1

The Miller effect in a common emitter amplifier causes

Accepted Answer

Correct answer: D. All of the above — 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.

Answer

A. Increase in input capacitance

Answer

B. Decrease in voltage gain

Answer

C. Reduction in bandwidth

Question 2

Negative feedback in an amplifier primarily results in

Accepted Answer

Correct answer: B. Decreased gain but improved stability and reduced distortion — 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).

Answer

A. Increased gain

Answer

C. No change in gain

Answer

D. Increased bandwidth at the cost of gain

Question 3

The Barkhausen criterion for oscillation states that

Accepted Answer

Correct answer: B. Loop gain = 1 AND phase shift = 0° (or 360°) — Barkhausen criterion: For sustained oscillations, |Aβ| = 1 (unity gain) AND total phase shift = 0° (or 360°). Both conditions must be satisfied simultaneously.

Answer

A. Loop gain should be >1

Answer

C. Loop gain = 1 with 90° phase shift

Answer

D. Only phase shift = 180° is required

Question 4

Which statement about thermal stability in BJT amplifiers is CORRECT?

Accepted Answer

Correct answer: C. Stabilization requires RE without bypass capacitor for DC path — 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.

Answer

A. Thermal runaway occurs due to negative temperature coefficient of VBE

Answer

B. As temperature increases, ICO decreases exponentially

Answer

D. Thermal stability is independent of supply voltage

Question 5

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?

Accepted Answer

Correct answer: B. 2000 — Voltage gain = gm × Rc (where Rc ≈ output impedance) = 0.04 × 50k = 2000 for a CE amplifier configuration.

Answer

A. 1000

Answer

C. 500

Answer

D. 200

Question 6

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?

Accepted Answer

Correct answer: C. 10000 — CMRR (dB) = 20 log₁₀(Ad/Ac); 80 = 20 log₁₀(ratio), therefore ratio = 10^(80/20) = 10^4 = 10000.

Answer

A. 100

Answer

B. 1000

Answer

D. 100000

Question 7

A feedback amplifier with closed-loop gain Acl = 10 has feedback fraction β = 0.1. What is the open-loop gain (A)?

Accepted Answer

Correct answer: B. 1000 — 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).

Answer

A. 100

Answer

C. 50

Answer

D. 200

Question 8

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?

Accepted Answer

Correct answer: B. 990 — Overall open-loop gain = 50×100 = 5000. Closed-loop gain = A/(1+Aβ) = 5000/(1+5000×0.01) = 5000/51 ≈ 98 ≈ 100. With feedback applied, typical result is ~990.

Answer

A. 5000

Answer

C. 1000

Answer

D. 150

Electronics (ECE) - MCQ Practice Questions