How NEET Tests This Chapter

Semiconductors reliably supply 1-2 NEET questions from a compact pool:

  1. Classification & bands: gap orderings, majority/minority carriers, dopant identification.
  2. Junction & bias: depletion-width changes, barrier arithmetic, current scales (mA vs μ\muA).
  3. Rectifiers: output frequencies, circuit identification, filter roles.
  4. Extras (still in NEET syllabus): Zener regulation, LED/photodiode/solar-cell bias table, logic-gate truth tables and identification.
  5. Numericals: mass-action law, simple diode circuits with cut-in subtraction.

Everything below is NEET-style previous-year material with fully worked explanations. Years are attached only where attribution is certain; otherwise the tag is the generic [NEET].

Anchors: gaps C/Si/Ge = 5.4/1.1/0.7 eV; cut-ins Ge/Si = 0.2/0.7 V; ripple half/full = f/2f; bias table LED-forward, photodiode-reverse, solar-cell-none.

NEET PYQ Worked Set A: Materials & Carriers

PYQ 1. In a p-type semiconductor, the acceptor level lies: [NEET]

Solution:

  1. Slightly above the top of the valence band (EVE_V) — a small hop ionises the acceptor, releasing a hole.

PYQ 2. Si doped with phosphorus becomes: [NEET]

Solution:

  1. P is pentavalent → donor → n-type; electrons majority.

PYQ 3. The forbidden gap of an insulator, semiconductor and conductor compare as: [NEET]

Solution:

  1. Insulator > 3 eV; semiconductor < 3 eV; conductor ≈ 0 (overlap).

PYQ 4. A semiconductor at 0 K behaves as: [NEET]

Solution:

  1. An insulator — full valence band, empty conduction band, no carriers.

PYQ 5. For a doped sample, ne=5×1020n_e = 5 \times 10^{20} m⁻³ and ni=1.5×1016n_i = 1.5 \times 10^{16} m⁻³: find nhn_h and the type. [NEET]

Solution:

  1. nh=(1.5×1016)25×1020=4.5×1011n_h = \dfrac{(1.5 \times 10^{16})^2}{5 \times 10^{20}} = 4.5 \times 10^{11} m⁻³.
  2. nenhn_e \gg n_h: n-type.

NEET PYQ Worked Set B: Junction, Diode & Rectifier

PYQ 6. In an unbiased junction, the motion of holes p → n is due to: [NEET]

Solution:

  1. The concentration gradient (holes abundant on p, scarce on n) — diffusion, not the field (which opposes it).

PYQ 7. Under reverse bias, the depletion width and barrier height: [NEET]

Solution:

  1. Both increase — barrier V0+VV_0 + V, wider space-charge region.

PYQ 8. Which meter measures a diode's reverse current? [NEET]

Solution:

  1. Microammeter — reverse saturation is μ\muA scale (forward mA needs a milliammeter).

PYQ 9. A full-wave rectifier on 60 Hz mains gives ripple at: [NEET]

Solution:

  1. 2 × 60 = 120 Hz.

PYQ 10. A Si diode (0.7 V) with 300 ohm across 3.7 V forward: the current is: [NEET]

Solution:

  1. I=3.70.7300=10I = \dfrac{3.7 - 0.7}{300} = 10 mA.

PYQ 11. In the half-wave rectifier, during the diode's non-conducting half-cycle the output voltage is: [NEET]

Solution:

  1. Zero — the diode blocks; the entire (peak) secondary voltage appears across the DIODE in reverse.

NEET PYQ Worked Set C: Extras (Zener, Optoelectronics, Gates)

PYQ 12. Match the device to its bias: LED, photodiode, solar cell. [NEET]

Solution:

  1. LED: forward (radiative recombination of injected carriers).
  2. Photodiode: reverse (μ\muA photo-signal detectable atop μ\muA dark current).
  3. Solar cell: no bias (built-in field separates photo-carriers; generates power).

PYQ 13. A Zener (VZV_Z = 5 V, RsR_s = 1 kohm) on a 12 V supply with a 1 kohm load: find IZI_Z. [NEET]

Solution:

  1. Is=1251000=7I_s = \dfrac{12 - 5}{1000} = 7 mA; IL=51000=5I_L = \dfrac{5}{1000} = 5 mA.
  2. IZ=2I_Z = 2 mA.

PYQ 14. The gate whose output is 0 only when all inputs are 1: [NEET]

Solution:

  1. NANDY=ABY = \overline{A\cdot B}; the (1,1) → 0 signature.

PYQ 15. Inputs A = 0, B = 0 into a NOR gate give: [NEET]

Solution:

  1. Y=0+0=1Y = \overline{0 + 0} = 1 — NOR's lone 1, its signature row.

PYQ 16. An LED begins to emit only above ~1.8 V forward voltage while a Si diode conducts at 0.7 V. Why the higher turn-on? [NEET]

Solution:

  1. The LED's material has a larger band gap (1.8-3 eV for visible emission).
  2. The turn-on voltage tracks Eg/eE_g/e — more gap, more volts before injection floods the junction.
  3. Takeaway: turn-on voltage ≈ band gap in volts — red LEDs ~1.8 V, blue ~3 V.