JEE Previous Year Questions

How JEE Tests This Chapter

Semiconductors deliver 1-2 questions to nearly every JEE Main paper, from tight templates:

1. Carrier arithmetic: mass-action law, doped conductivities $\sigma = e(n_e\mu_e + n_h\mu_h)$, ppm-to-density conversions.
2. Diode circuit logic: ideal-diode on/off analysis of networks; real-diode cut-in subtraction; dynamic resistance from graphs.
3. Rectifiers: ripple frequencies, peak outputs from rms, filter time constants.
4. Zener regulators: $I_s$, $I_Z$, $I_L$ bookkeeping; regulation ranges.
5. Optoelectronics: LED wavelength ↔ band gap; photodiode logic; solar-cell characteristics.
6. Logic gates: identify from truth tables/waveforms; simplify cascades; universal-gate constructions.

All questions below are JEE-style previous-year questions with worked solutions. Years are attached only where attribution is certain; otherwise tags read [JEE Mains] / [JEE Advanced].

Constants: hc ≈ 1240 eV nm; e = $1.6 \times 10^{-19}$ C; cut-ins 0.2 V (Ge), 0.7 V (Si).

JEE PYQ Worked Set A: Carriers & Conductivity

PYQ 1. A Ge sample is doped with $10^{22}$ m⁻³ donors ($n_i = 2.4 \times 10^{19}$ m⁻³). The hole concentration is: [JEE Mains]

Solution:

1. $n_e \approx 10^{22}$ m⁻³.
2. $n_h = \dfrac{n_i^2}{n_e} = \dfrac{5.76 \times 10^{38}}{10^{22}} = 5.76 \times 10^{16}$ m⁻³.

PYQ 2. Pure Si at 300 K has $n_i = 1.5 \times 10^{16}$ m⁻³. Doping raises $n_h$ to $4.5 \times 10^{22}$ m⁻³. The electron concentration becomes: [JEE Mains]

Solution:

1. $n_e = \dfrac{n_i^2}{n_h} = \dfrac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^{9}$ m⁻³ — p-type with electrons crushed.

PYQ 3. The conductivity of intrinsic Ge ($n_i = 2.4 \times 10^{19}$ m⁻³, $\mu_e$ = 0.39, $\mu_h$ = 0.19 m²/Vs): [JEE Mains]

Solution:

1. $\sigma = en_i(\mu_e + \mu_h) = 1.6 \times 10^{-19} \times 2.4 \times 10^{19} \times 0.58$.
2. $\sigma \approx 2.2$ S/m (resistivity ≈ 0.45 ohm m).

PYQ 4. A potential difference across an intrinsic sample drives electron and hole drift. If $\mu_e = 3\mu_h$, the fraction of current carried by electrons is: [JEE Mains]

Solution:

1. With $n_e = n_h$: currents split as mobilities.
2. Fraction = $\dfrac{\mu_e}{\mu_e + \mu_h} = \dfrac{3}{4}$ = 75%.

JEE PYQ Worked Set B: Diode Circuits

PYQ 5. Two ideal diodes and resistors: D1 (forward) in series with 100 ohm; D2 (reverse) in series with 200 ohm; both branches parallel across 6 V. Current from the battery: [JEE Mains]

Solution:

1. D2 branch: open. D1 branch: $I = \dfrac{6}{100} = 60$ mA.
2. Total: 60 mA.

PYQ 6. A Si diode (0.7 V) in series with 1 kohm and a 3 V cell. Current and diode power: [JEE Mains]

Solution:

1. $I = \dfrac{3 - 0.7}{1000} = 2.3$ mA.
2. Diode power: $P = 0.7 \times 2.3 \times 10^{-3} \approx 1.6$ mW.

PYQ 7. In the circuit, an ideal diode's branch contains 30 ohm; a parallel resistor 60 ohm; supply 6 V through a series 10 ohm. Find the current through the 30 ohm when the diode conducts. [JEE Advanced pattern]

Solution:

1. Conducting diode → 30 ∥ 60 = 20 ohm; total = 30 ohm.
2. Battery current: 6/30 = 0.2 A; voltage across the pair: 0.2 × 20 = 4 V.
3. Through 30 ohm: $\dfrac{4}{30} \approx 0.133$ A.
4. Takeaway: replace conducting diodes with wires, blocked ones with breaks — then it's pure network analysis.

PYQ 8. A diode's current jumps from 5 mA to 15 mA as V goes 0.70 → 0.74 V. Its dynamic resistance: [JEE Mains]

Solution:

1. $r_d = \dfrac{0.04}{10 \times 10^{-3}} = 4$ ohm.

JEE PYQ Worked Set C: Rectifiers, Zener & Optoelectronics

PYQ 9. A full-wave rectifier runs from 50 Hz mains. The least ripple frequency a filter must suppress: [JEE Mains]

Solution:

1. Full-wave ripple = 2 × 50 = 100 Hz.

PYQ 10. A Zener ($V_Z$ = 8 V) with $R_s$ = 2 kohm on a 20 V supply feeds a 4 kohm load. $I_Z$: [JEE Mains]

Solution:

1. $I_s = \dfrac{20 - 8}{2000} = 6$ mA; $I_L = \dfrac{8}{4000} = 2$ mA.
2. $I_Z = 4$ mA.

PYQ 11. The supply in PYQ 10 sags to 12 V. Does regulation survive? [JEE Advanced pattern]

Solution:

1. $I_s = \dfrac{12 - 8}{2000} = 2$ mA = exactly $I_L$.
2. $I_Z = 0$ — the Zener is at the edge; any further sag loses regulation.
3. Takeaway: regulation demands $I_s > I_L$; the boundary case is the exam's favourite.

PYQ 12. An LED made of a semiconductor with $E_g$ = 1.9 eV emits at approximately: [JEE Mains]

Solution:

1. $\lambda = \dfrac{1240}{1.9} \approx 653$ nm — red.

PYQ 13. A photodiode's photocurrent doubles when illumination doubles. This is because: [JEE Mains]

Solution:

1. Each above-gap photon yields one electron-hole pair; pair rate ∝ photon rate ∝ intensity.
2. Photocurrent ∝ intensity — linear detection, the photodiode's defining property.

JEE PYQ Worked Set D: Logic Gates

PYQ 14. The Boolean expression for the circuit 'A and B into NAND; output into NOT' is: [JEE Mains]

Solution:

1. NAND: $\overline{A\cdot B}$; NOT inverts: $Y = \overline{\overline{A\cdot B}} = A \cdot B$.
2. The pair reduces to a plain AND gate.

PYQ 15. Inputs A and B feed a NOR gate; its output and A feed a second NOR. Simplify. [JEE Advanced pattern]

Solution:

1. First: $X = \overline{A + B}$.
2. Second: $Y = \overline{A + X} = \overline{A + \overline{A + B}}$.
3. Case-check: A = 1 → Y = 0. A = 0 → $X = \bar{B}$ → $Y = \overline{\bar{B}} = B$.
4. So $Y = \bar{A}B$ — a 'B AND NOT A' function.
5. Takeaway: when algebra thickens, brute-force the four input pairs — two-variable circuits yield instantly to a truth table.

PYQ 16. For the waveforms A: 1 in (0-4 s), 0 after; B: 1 in (2-6 s): give the OR and NAND outputs in each interval. [JEE Mains]

Solution:

1. Intervals: (0-2): A=1,B=0; (2-4): 1,1; (4-6): 0,1; (>6): 0,0.
2. OR: 1, 1, 1, 0.
3. NAND: 1, 0, 1, 1.
4. Takeaway: tabulate intervals first; gates then read off row by row.