Semiconductor Electronics: Materials, Devices and Simple Circuits — NEET Physics MCQs with answers

Semiconductor Electronics: Materials, Devices and Simple Circuits is a Class 12 Physics chapter in the NEET (UG) syllabus. NEET720 has 197 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Extrinsic semiconductors and doping · Energy bands and classification of solids · p-n junction and diode characteristics · Carrier concentration and conductivity · Special-purpose diodes · Intrinsic and extrinsic semiconductors · p-n junction formation · Biasing and I-V characteristics · Diode circuits · Zener diode · Rectifiers · Logic gates · Optoelectronic junction devices

8 free Semiconductor Electronics: Materials, Devices and Simple Circuits practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Energy bands and classification of solids

   Light of wavelength 1500 nm falls on thin slabs of pure germanium (band gap 0.7 eV) and pure silicon (band gap 1.1 eV) at room temperature. Take hc = 1240 eV nm. Which statement about the creation of electron-hole pairs by this light is correct?

   • A.Both slabs generate electron-hole pairs, because any photon absorbed by a semiconductor frees an electron.
   • B.Only silicon generates pairs, because its larger band gap lets it absorb more energy.
   • C.Only germanium generates pairs; silicon is essentially transparent to this light.
   • D.Neither slab generates pairs, because 0.83 eV is far smaller than the ionisation energy of an isolated atom.
   Show answer and explanation

   Answer: C. Only germanium generates pairs; silicon is essentially transparent to this light.

   Photon energy = 1240/1500 = 0.83 eV. This exceeds 0.7 eV (Ge) but is less than 1.1 eV (Si), so only Ge absorbs it by band-to-band excitation.

   The photon energy is E = hc/λ = 1240 eV nm / 1500 nm ≈ 0.83 eV. An electron-hole pair is created only when a valence-band electron receives at least the band-gap energy. For germanium, 0.83 eV > 0.7 eV, so pairs are generated. For silicon, 0.83 eV < 1.1 eV, so the photon cannot lift an electron across the gap and silicon is transparent to this wavelength (apart from negligible impurity effects). The longest wavelength silicon can absorb in this way is 1240/1.1 ≈ 1130 nm.

   Common mistake: Thinking that a wider band gap makes a material a better absorber of all light.

   Key point: A photon creates an electron-hole pair only if hc/λ ≥ E_g; larger gaps need shorter wavelengths.

2. Question 2 · medium · Carrier concentration and conductivity

   Pure silicon at 300 K has n_i = 1.5 × 10^16 m^-3. It is doped uniformly with donors of concentration 3 × 10^21 m^-3, all of which are ionised. In the doped sample, the ratio of electron concentration to hole concentration, n_e/n_h, is closest to:

   • A.4 × 10^10
   • B.2 × 10^5
   • C.7.5 × 10^10
   • D.1
   Show answer and explanation

   Answer: A. 4 × 10^10

   n_e ≈ N_D = 3 × 10^21 m^-3 and n_h = n_i²/n_e = 7.5 × 10^10 m^-3, so n_e/n_h = 4 × 10^10.

   Since N_D ≫ n_i, n_e ≈ 3 × 10^21 m^-3. The mass-action law n_e n_h = n_i² gives n_h = (1.5 × 10^16)²/(3 × 10^21) = 2.25 × 10^32/3 × 10^21 = 7.5 × 10^10 m^-3. The ratio n_e/n_h = 3 × 10^21/7.5 × 10^10 = 4 × 10^10. Equivalently, n_e/n_h = (n_e/n_i)² = (2 × 10^5)² = 4 × 10^10: doping raises the majority carriers and suppresses the minority carriers by the same factor.

   Common mistake: Reporting N_D/n_i as the majority-to-minority ratio.

   Key point: In extrinsic material the majority-to-minority ratio equals the square of (majority/n_i).

3. Question 3 · medium · Intrinsic and extrinsic semiconductors

   In n-type silicon the donor level lies about 0.045 eV below the bottom of the conduction band, while the band gap is about 1.1 eV; at 300 K, k_BT ≈ 0.026 eV. Which statement best explains why, at room temperature, the electron concentration is practically equal to the donor concentration even though very few electrons cross the band gap?

   • A.Donor electrons already occupy the conduction band at 0 K, so no energy is needed to free them.
   • B.The donor level lies above the bottom of the conduction band, so donor electrons simply fall into it.
   • C.Doping with donors lowers the band gap of silicon to about 0.045 eV.
   • D.The donor ionisation energy is comparable to k_BT, so almost every donor is ionised, while the 1.1 eV gap is over 40 times k_BT and thermal excitation across it remains rare.
   Show answer and explanation

   Answer: D. The donor ionisation energy is comparable to k_BT, so almost every donor is ionised, while the 1.1 eV gap is over 40 times k_BT and thermal excitation across it remains rare.

   Freeing a donor electron costs only about 0.045 eV (≈ 1.7 k_BT) but crossing the gap costs 1.1 eV (≈ 42 k_BT), so donors ionise almost completely while intrinsic excitation stays small.

   The probability of thermal excitation falls extremely steeply with the ratio of required energy to k_BT. For a donor, 0.045/0.026 ≈ 1.7, so at 300 K practically all donors give up their electron to the conduction band. For band-to-band excitation, 1.1/0.026 ≈ 42, so only a tiny fraction of valence electrons cross; this is why n_i is small. Therefore n_e ≈ N_D in doped material, and the gap itself is unchanged by light doping.

   Common mistake: Believing doping reduces the band gap.

   Key point: Donor levels lie a few hundredths of an eV below the conduction band, so they are almost fully ionised at room temperature.

4. Question 4 · medium · Carrier concentration and conductivity

   Intrinsic germanium at 300 K has n_i = 2.4 × 10^19 m^-3, electron mobility 0.39 m^2 V^-1 s^-1 and hole mobility 0.19 m^2 V^-1 s^-1. Taking e = 1.6 × 10^-19 C, its electrical conductivity is closest to:

   • A.1.50 S m^-1
   • B.0.77 S m^-1
   • C.2.23 S m^-1
   • D.4.45 S m^-1
   Show answer and explanation

   Answer: C. 2.23 S m^-1

   σ = e n_i (μ_e + μ_h) = (1.6 × 10^-19)(2.4 × 10^19)(0.58) ≈ 2.23 S m^-1.

   For intrinsic material n_e = n_h = n_i, so σ = e(n_e μ_e + n_h μ_h) = e n_i (μ_e + μ_h). Here e n_i = 1.6 × 10^-19 × 2.4 × 10^19 = 3.84 C m^-3, and μ_e + μ_h = 0.39 + 0.19 = 0.58 m^2 V^-1 s^-1, giving σ = 3.84 × 0.58 ≈ 2.23 S m^-1 (resistivity ≈ 0.45 Ω m). Electrons and holes drift in opposite directions but, having opposite charges, their currents add.

   Common mistake: Ignoring the hole contribution.

   Key point: σ = e(n_e μ_e + n_h μ_h); both carriers contribute additively.

5. Question 5 · medium · Carrier concentration and conductivity

   A p-type silicon sample has hole concentration 5 × 10^22 m^-3 and hole mobility 0.05 m^2 V^-1 s^-1 (e = 1.6 × 10^-19 C). The intrinsic concentration of silicon is of order 10^16 m^-3. The resistivity of the sample is closest to:

   • A.400 Ω m
   • B.2.5 × 10^-3 Ω m
   • C.1.25 × 10^-3 Ω m
   • D.It cannot be found without the exact value of n_i, because the minority electrons also contribute.
   Show answer and explanation

   Answer: B. 2.5 × 10^-3 Ω m

   σ ≈ e n_h μ_h = (1.6 × 10^-19)(5 × 10^22)(0.05) = 400 S m^-1, so ρ = 1/σ = 2.5 × 10^-3 Ω m.

   In strongly p-type material the minority electron concentration is n_i²/n_h ~ 10^32/5 × 10^22 ~ 10^9 m^-3, about 10^13 times smaller than n_h, so it can be neglected. Then σ ≈ e n_h μ_h = 1.6 × 10^-19 × 5 × 10^22 × 0.05 = 8000 × 0.05 = 400 S m^-1. The resistivity is ρ = 1/400 = 2.5 × 10^-3 Ω m.

   Common mistake: Forgetting to invert conductivity to get resistivity.

   Key point: In extrinsic material σ ≈ e × (majority concentration) × (majority mobility).

6. Question 6 · medium · Intrinsic and extrinsic semiconductors

   A bar of intrinsic silicon is connected across a battery. A student argues: "Electrons drift towards the positive terminal and holes drift towards the negative terminal. Since they move in opposite directions, their currents partly cancel, so the total current is |I_e − I_h|." What is the correct assessment?

   • A.The argument is wrong: holes behave as positive charges, so both drifts give conventional current in the same direction and the total current is I_e + I_h.
   • B.The argument is right, and in intrinsic material the net current is zero because n_e = n_h.
   • C.The argument is wrong because holes do not move at all; only conduction electrons carry current in a semiconductor.
   • D.The argument is wrong because both electrons and holes drift towards the positive terminal.
   Show answer and explanation

   Answer: A. The argument is wrong: holes behave as positive charges, so both drifts give conventional current in the same direction and the total current is I_e + I_h.

   Opposite charges moving in opposite directions produce current in the same direction, so I = I_e + I_h.

   Conventional current is the flow of positive charge. Electrons (negative) moving towards the positive terminal constitute a current directed from the positive to the negative terminal through the bar. Holes (effectively positive) moving towards the negative terminal give current in that same direction. Hence the currents add: I = I_e + I_h, which is why σ = e(n_e μ_e + n_h μ_h) contains a sum.

   Common mistake: Subtracting the hole current because holes move the other way.

   Key point: Electron and hole currents always add in a semiconductor.

7. Question 7 · medium · Intrinsic and extrinsic semiconductors

   Consider the following statements about extrinsic silicon. (i) A few dopant atoms per million silicon atoms can raise the conductivity by several orders of magnitude. (ii) An n-type crystal carries a net negative charge because it contains excess free electrons. (iii) In p-type silicon every ionised acceptor is an immobile negative ion fixed in the lattice. (iv) In p-type silicon the hole concentration exceeds n_i while the electron concentration falls below n_i. Which statements are correct?

   • A.(i) and (ii) only
   • B.(ii) and (iv) only
   • C.(i), (ii) and (iii) only
   • D.(i), (iii) and (iv) only
   Show answer and explanation

   Answer: D. (i), (iii) and (iv) only

   Doped crystals are electrically neutral: every free electron in n-type material is balanced by a fixed positive donor ion. The other three statements are correct.

   (i) is correct: doping at the ppm level multiplies the carrier concentration by around 10^6. (ii) is false: each pentavalent atom that donates an electron becomes a positive ion, so charge balance is maintained. (iii) is correct: an acceptor that has taken an electron is a negative ion bound in the lattice and cannot drift. (iv) is correct: with n_h > n_i, the mass-action law forces n_e = n_i²/n_h < n_i.

   Common mistake: Thinking n-type material is negatively charged.

   Key point: Doped semiconductors are neutral overall; dopant ions are immobile.

8. Question 8 · easy · Intrinsic and extrinsic semiconductors

   Which pair of impurities, each added alone in small amounts to pure germanium, would both produce p-type material?

   • A.Arsenic and antimony
   • B.Indium and gallium
   • C.Indium and antimony
   • D.Phosphorus and boron
   Show answer and explanation

   Answer: B. Indium and gallium

   p-type material needs trivalent (group 13) acceptors such as B, Al, Ga or In; both indium and gallium qualify.

   Germanium is tetravalent. A trivalent atom forms only three complete covalent bonds, leaving a vacancy (hole) that can accept an electron from a neighbouring bond; such acceptors make the material p-type. Indium and gallium are both trivalent. Arsenic, antimony and phosphorus are pentavalent and act as donors (n-type).

   Common mistake: Mixing up which valence gives which type.

   Key point: Group 13 dopants → p-type; group 15 dopants → n-type.

Questions about Semiconductor Electronics: Materials, Devices and Simple Circuits for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (197 active practice questions in this chapter today).