Semiconductor Electronics Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Electronic Devices (semiconductor diodes, Zener diodes, p-n junctions) with clear step-by-step solutions.
A curated set of JEE Main 2026 previous year questions from Electronic Devices, worked out step by step.
Solutions are AI-generated and pending review.
Solution
Assertion A: In reverse bias, a p-n junction carries only a tiny reverse saturation current that stays almost constant with increasing reverse voltage. At the breakdown (critical) voltage, the current rises sharply. This is a correct description of diode behaviour.
$$\text{A is TRUE}$$Reason R: The small reverse current is due to minority charge carriers (electrons in the p-region and holes in the n-region) drifting across the junction — not majority carriers. So the claim that only majority carriers flow below the critical voltage is wrong.
$$\text{R is FALSE}$$Therefore A is true but R is false.
Answer: C
Solution
Given: Zener voltage $V_Z = 10\ \text{V}$, maximum power $P_{\max} = 0.5\ \text{W}$, supply $V = 25\ \text{V}$.
Step 1 — Maximum permissible Zener current:
$$I_{\max} = \frac{P_{\max}}{V_Z} = \frac{0.5}{10} = 0.05\ \text{A}$$Step 2 — Voltage dropped across the series resistor (Zener holds $10\ \text{V}$):
$$V_{R} = V - V_Z = 25 - 10 = 15\ \text{V}$$Step 3 — Minimum series resistance so the current never exceeds $I_{\max}$:
$$R_{\min} = \frac{V_R}{I_{\max}} = \frac{15}{0.05} = 300\ \Omega$$Answer: 300
Solution
Charge-neutrality condition: The total ionized charge exposed on each side of the junction must be equal in magnitude:
$$N_A \, W_p = N_D \, W_n$$where $W_p$ and $W_n$ are the depletion widths on the p- and n-sides.
Step 1 — Rearrange for the ratio of widths:
$$\frac{W_p}{W_n} = \frac{N_D}{N_A} = \frac{10^{18}}{10^{15}} = 10^{3}$$Step 2 — Interpret: Since $N_A \ll N_D$, the p-side is lightly doped, so the depletion region must extend much farther into the p-side to expose an equal amount of charge.
$$W_p = 1000\, W_n \;\Rightarrow\; \text{depletion region is wider on the p-side.}$$The depletion layer always penetrates deeper into the lightly doped side.
Answer: B