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?

The correct answer is B: 1000. In cascaded amplifiers, overall voltage gain = A₁ × A₂ = 50 × 20 = 1000. The bandwidth is determined by the stage with lowest bandwidth (100 kHz).

The decimal equivalent of binary number 10110.101 is:

The correct answer is A: 22.625. 10110.101 = 1×16 + 0×8 + 1×4 + 1×2 + 0×1 + 1×0.5 + 0×0.25 + 1×0.125 = 16 + 4 + 2 + 0.5 + 0.125 = 22.625

A clamping diode in logic circuits is used to:

The correct answer is A: Limit voltage swings to prevent logic levels from exceeding safe limits. Clamping diodes protect circuits by limiting the maximum voltage that can be applied to a node. They conduct when voltage exceeds V_DD + V_D_forward or goes below V_SS - V_D_forward, preventing overstress to gate oxide and junctions.

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)?

The correct answer is B: 150 mW. 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.

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?

The correct answer is 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.

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Electronics (ECE)
Electronic Devices

Analog/digital electronics, communication

30 Q 4 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 30

Topics in Electronics (ECE)

All Electronic Devices 100 Digital Electronics 100 Analog Circuits 100 Signals & Systems 100

Q.11 Hard Electronic Devices

The transconductance (g_m) of a MOSFET in saturation is given by:

A g_m = μ_n C_ox (W/L)(V_GS - V_T)

B g_m = (1/2)μ_n C_ox (W/L)(V_GS - V_T)²

C g_m = μ_n C_ox (W/L)(V_DS - V_GS)

D g_m = √(2μ_n C_ox (W/L)I_D)

Correct Answer: D. g_m = √(2μ_n C_ox (W/L)I_D)

EXPLANATION

In saturation, I_D = (1/2)μ_n C_ox (W/L)(V_GS - V_T)², so g_m = ∂I_D/∂V_GS = μ_n C_ox (W/L)(V_GS - V_T) = √(2μ_n C_ox (W/L)I_D).

Test

Q.12 Hard Electronic Devices

The avalanche multiplication factor M in reverse-biased junction is expressed as:

A M = 1 / (1 - (VR/BV)n)

B M = (VR/BV)n

C M = VR - BV

D M = BV / VR

Correct Answer: A. M = 1 / (1 - (VR/BV)n)

EXPLANATION

The avalanche factor M = 1/(1-(VR/BV)ⁿ) shows multiplication increases exponentially as reverse voltage approaches breakdown voltage BV, where n≈3-6.

Test

Q.13 Hard Electronic Devices

In a MOSFET differential amplifier, the common-mode gain is reduced by using:

A Larger channel length (L)

B Larger width-to-length ratio (W/L)

C Current mirror load and larger source degeneration

D Smaller output impedance

Correct Answer: C. Current mirror load and larger source degeneration

EXPLANATION

Current mirror loads increase output impedance, while source degeneration resistors provide common-mode feedback, both reducing common-mode gain.

Test

Q.14 Hard Electronic Devices

The noise figure of a BJT amplifier at low frequencies is primarily determined by:

A 1/f noise in base region

B Shot noise in base current

C Thermal noise in base resistance

D Flicker noise only

Correct Answer: B. Shot noise in base current

EXPLANATION

At low frequencies, shot noise associated with base current (IB) dominates, following Poisson statistics: i²n = 2q×IB×Δf.

Test

Q.15 Hard Electronic Devices

In practical MOSFET applications for RF circuits, which parasitic effect limits frequency response most?

A Substrate capacitance only

B Gate-drain capacitance (Miller effect)

C Source resistance variations

D Bulk resistance

Correct Answer: B. Gate-drain capacitance (Miller effect)

EXPLANATION

The gate-drain capacitance (Cgd) creates Miller effect, which effectively multiplies the capacitive reactance in feedback, severely limiting high-frequency gain.

Test

Q.16 Hard Electronic Devices

A JFET in pinch-off condition means:

A Gate-source junction is reverse biased and channel is closed

B Drain current equals zero regardless of drain voltage

C Gate voltage equals source voltage

D Drain current is maximum and independent of VDS

Correct Answer: A. Gate-source junction is reverse biased and channel is closed

EXPLANATION

At pinch-off, VGS reaches VP (pinch-off voltage), channel closes, and further VDS increase has minimal effect on ID (saturation region).

Test

Q.17 Hard Electronic Devices

When a BJT is used as a switch in saturated condition, the overdrive factor is defined as:

A Ratio of actual base current to minimum base current needed

B Ratio of collector current to base current

C Ratio of VCE(sat) to VBE(sat)

D Ratio of collector voltage to emitter voltage

Correct Answer: A. Ratio of actual base current to minimum base current needed

EXPLANATION

Overdrive factor β = IB(actual) / IB(min) ensures transistor stays saturated even with parameter variations.

Test

Q.18 Hard Electronic Devices

The avalanche breakdown voltage (BV) in a reverse-biased junction approximately follows:

A BV ∝ ND (doping concentration)

B BV ∝ √ND

C BV ∝ ND^(-3/4)

D BV is independent of doping concentration

Correct Answer: C. BV ∝ ND^(-3/4)

EXPLANATION

Breakdown voltage BV ∝ ND^(-3/4) approximately; lower doping increases depletion width, requiring higher field and higher voltage for breakdown.

Test

Q.19 Hard Electronic Devices

The Miller effect in a common-emitter BJT amplifier causes:

A Decrease in input impedance

B Effective multiplication of base-collector capacitance by (1+Av) at input

C Increase in voltage gain

D Phase shift of 180°

Correct Answer: B. Effective multiplication of base-collector capacitance by (1+Av) at input

EXPLANATION

Miller effect: CBE capacitance reflected to input = CμC·(1+|Av|); dominates high-frequency response and limits bandwidth in CE configuration.

Test

Q.20 Hard Electronic Devices

In a Gunn diode, negative resistance occurs due to:

A Impact ionization breakdown

B Differential mobility in different valley structures of semiconductor

C Tunneling phenomenon

D Avalanche multiplication

Correct Answer: B. Differential mobility in different valley structures of semiconductor

EXPLANATION

Gunn effect: electrons transfer from high-mobility Γ valley to low-mobility L valley at threshold field, causing dI/dV < 0 (negative resistance).

Test

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Electronics (ECE) Electronic Devices

Electronics (ECE)
Electronic Devices