50 questions on this topic of Electronic Devices (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (A) V₀ - V
Under forward bias, the applied voltage opposes the built-in potential barrier V₀. Consequently, the net height of the potential barrier decreases from V₀ to (V₀ - V).
Answer: (A) Both Statement I and Statement II are correct
The reverse saturation current is determined by the rate of thermal generation of minority carriers within diffusion length of the depletion region. Because the internal electric field sweeps these carriers across the junction regardless of how large the reverse bias is (prior to breakdown), the current saturates and remains virtually constant.
Answer: (A) Cⱼ decreases as Vᵣ increases, roughly as (V₀ + Vᵣ)⁻¹/²
The depletion layer capacitance is analogous to a parallel plate capacitor, Cⱼ = εA / d, where d is the depletion layer width. As reverse voltage Vᵣ increases, the depletion layer widens according to d ∝ √(V₀ + Vᵣ), which causes the junction capacitance Cⱼ to decrease inversely with the square root of the total reverse barrier voltage.
Answer: (A) Extremely high, of the order of megaohms (10⁶ Ω)
In the reverse-bias region prior to breakdown, the change in reverse current ΔI with change in reverse voltage ΔV is practically negligible (nearly zero), meaning ΔV/ΔI is extremely large, typically on the order of megaohms (10⁶ Ω).
Answer: (A) 5.0 Ω
Dynamic resistance is defined as r = ΔV / ΔI. Here, ΔV = 0.77 V - 0.72 V = 0.05 V, and ΔI = 15.0 mA - 5.0 mA = 10.0 mA = 10 × 10⁻³ A. Therefore, r = 0.05 / (10 × 10⁻³) = 5.0 Ω.
Answer: (B) 7.0 × 10¹⁶ m/s²
The force on the electron is F = eE. The acceleration is a = eE / mₑ = (1.6 × 10⁻¹⁹ C × 4.0 × 10⁵ V/m) / (9.1 × 10⁻³¹ kg) ≈ 6.4 × 10⁻¹⁴ / 9.1 × 10⁻³¹ ≈ 7.03 × 10¹⁶ m/s².
Answer: (B) 2.0 mA
Dynamic resistance is r_d = ΔV/ΔI. Therefore, ΔI = ΔV/r_d = (0.75 - 0.70) V / 25 Ω = 0.05 V / 25 Ω = 2.0 × 10⁻³ A = 2.0 mA.
Answer: (B) 2.5 Ω
Dynamic resistance r_d = ΔV / ΔI. Here, ΔV = 0.65 - 0.60 = 0.05 V, and ΔI = (30 - 10) mA = 20 × 10⁻³ A. Thus, r_d = 0.05 / (20 × 10⁻³) = 2.5 Ω.
Answer: (B) Holes diffuse from the p-side to the n-side leaving behind unneutralized immobile negative acceptor ions
Due to the concentration gradient, holes diffuse from the p-region into the n-region, leaving behind uncompensated immobile negative acceptor ions near the junction on the p-side. Similarly, electrons diffuse from the n-side to the p-side, leaving behind immobile positive donor ions on the n-side.
Answer: (B) electrons cross the junction from the n-side to the p-side and become minority carriers near the junction boundary
Under forward bias, the lowered barrier enables electrons from the n-side (where they are majority) to cross the depletion layer into the p-side. On the p-side, these arriving electrons act as injected minority carriers near the edge of the depletion region before recombining.
Answer: (D) I_diff flows from p-side to n-side; I_drift flows from n-side to p-side
Diffusion current is due to majority carriers (holes diffusing from p to n, electrons from n to p), both producing conventional current from the p-side to the n-side. Drift current is caused by the built-in electric field (directed from n to p), which sweeps minority electrons from p to n and minority holes from n to p, resulting in a conventional drift current directed from the n-side to the p-side. At equilibrium, I_diff = I_drift, so net current is zero.
Answer: (C) 4.3 mA
Because the diodes are in parallel with opposing polarities, one diode will be forward-biased while the other is reverse-biased. The reverse-biased diode acts as an open circuit, while the forward-biased diode maintains a drop of 0.7 V. The voltage across the 1.0 kΩ resistor is V = 5.0 V - 0.7 V = 4.3 V, giving a total current of I = 4.3 V / 1000 Ω = 4.3 mA.