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Class 12 Semiconductor Electronics worksheet

CBSE Physics. Intrinsic and doped semiconductors, carrier concentrations, p-n junctions, diodes, rectifiers. Every sheet has 20 questions in three sections and an answer key, and the numbers change every time.

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Sample questions

Real questions from this chapter, as the app sets them. Tap Answer to see the answer and working.

Section A · multiple choice

  1. 1.

    Silicon has an intrinsic carrier concentration of 1 × 10¹⁶ m⁻³. It is doped with 9 × 10²⁰ donor atoms per m³. The hole concentration is:

    1. 1.11 × 10¹¹ m⁻³
    2. 9 × 10²⁰ m⁻³
    3. 1 × 10¹⁶ m⁻³
    4. 1.11 × 10⁻⁵ m⁻³
    Answer

    (a) 1.11 × 10¹¹ m⁻³

    nₑnₕ = nᵢ²; nₑ ≈ N_D

  2. 2.

    Doping pure germanium with indium makes it:

    1. n-type, with electrons as majority carriers
    2. an insulator
    3. p-type, with holes as majority carriers
    4. p-type, with electrons as majority carriers
    Answer

    (c) p-type, with holes as majority carriers

    indium is trivalent

  3. 3.

    The input to a full-wave rectifier is 1000 Hz AC. The frequency of the output ripple is:

    1. 0 Hz
    2. 1000 Hz
    3. 2000 Hz
    4. 500 Hz
    Answer

    (c) 2000 Hz

    both halves of each cycle give a pulse

Section B · short answer

  1. 4.

    Pure silicon has nᵢ = 1 × 10¹⁶ m⁻³. It is doped with 6 × 10²³ pentavalent atoms per m³. Find the electron and hole concentrations, and name the type.

    Answer

    nₑ ≈ 6 × 10²³ m⁻³; nₕ = 1.67 × 10⁸ m⁻³; n-type

    nₕ = nᵢ²/nₑ

  2. 5.

    Silicon has 5 × 10²⁸ atoms per m³. It is doped with 1 atom of arsenic for every 10⁸ silicon atoms, and nᵢ = 1 × 10¹⁶ m⁻³. Find the number of donor atoms per m³ and the hole concentration.

    Answer

    5 × 10²⁰ m⁻³; 2 × 10¹¹ m⁻³

    N_D = 5 × 10²⁸ ÷ 10⁸; nₕ = nᵢ²/N_D

Section C · long answer

  1. 6.

    Silicon with nᵢ = 5 × 10¹⁵ m⁻³ is doped with 3 × 10²⁰ pentavalent atoms per m³. Find the majority and minority carrier concentrations, name the type, and say what happens to the band diagram.

    Answer

    electrons ≈ 3 × 10²⁰ m⁻³; holes = 8.33 × 10¹⁰ m⁻³; n-type; a donor level appears just below the conduction band

    minority = nᵢ²/majority

Fresh sheets for your child

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