Semiconductor ElectronicsNEET Physics · Class 12 · NCERT Chapter 14

Solution

Band gap: conductor (0) < semiconductor (~1 eV) < insulator (>3 eV).

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Pentavalent (5 valence). 4 form covalent bonds; 1 is free.

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p-type: holes are majority.

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Si: 0.7 V; Ge: 0.3 V.

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Half-wave: same as input. 50 Hz.

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Full-wave: 2 × input = 100 Hz.

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NAND alone can implement NOT, AND, OR, etc.

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LED: forward bias; recombination emits photons.

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Photodiode: reverse bias; light increases reverse current.

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Operates in reverse breakdown to clamp output voltage.

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Mass-action law: $n_e{n}_h=n_i^2$ at equilibrium.

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NAND: NOT(1 AND 0) = NOT 0 = 1.

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Forward bias: barrier reduces, depletion narrows.

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Full-wave: $2V_m/\pi \approx 0.636V_m$.

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At 0 K, no thermal excitation across gap. Behaves like an insulator.

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OR: Y = 1 when A or B (or both) = 1.

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Trivalent (3 valence): one bond is incomplete → hole → p-type.

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In reverse bias, only minority carriers cross the depletion region.

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Bridge: 4 diodes, no centre tap needed.

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Solar cell: light incident, no external bias. Generates EMF.

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NOR: NOT(0 OR 0) = NOT 0 = 1.

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Intrinsic: thermal generation creates equal e and h: $n_e=n_h=n_i$.

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V_out = V_Z (constant) for V_in > V_Z.

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The barrier opposes further diffusion of majority carriers.

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AND: 1 AND 1 = 1.

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Centre-tapped uses 2 diodes; bridge uses 4.

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I_0 nearly constant in reverse bias (until breakdown).

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Donor electrons shift Fermi level towards CB.

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E_photon < E_g: not enough energy to promote an electron.

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NAND = NOT AND = $\overline{A\cdot B}$.

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PIV (half-wave) = V_m. (For bridge: V_m. For centre-tap: 2 V_m.)

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Depletion = depleted of free carriers; only fixed ions remain.