#doping

Semiconductors, Types and Energy Bands

The conductivity of an extrinsic semiconductor increases due to:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. Doping introduces impurities (pentavalent or trivalent) that provide additional charge carriers (electrons or holes), significantly enhancing conductivity compared to intrinsic semiconductors. Substituting values gives Addition of impurities, which matches expecte

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Doping, Charge Carriers and Conductivity

In a p-type semiconductor, the acceptor impurities create:

**Doping** is adding impurity to pure semiconductor to increase carriers, pentavalent (P, As, Sb) donates extra electron, 1 ppm doping in Ge with 4×10²⁸ atoms/m³ gives donor density N_d = 4×10²⁸×10⁻⁶ =4×10²² m⁻³ for 1 ppm, acceptor atoms p-type trivalent B, Al, Ga. Overall charge neutrality maintained because donor ion core positive but electron negative, net neutral. Trivalent acceptor impurities (e.g., B, Al) in a p-type semiconductor lack one electron per atom, creating holes that act as majority carriers by accepting electrons from the lattice. Substituting values gives Holes, which matche

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Doping, Charge Carriers and Conductivity

A Si crystal with \( 5 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) is doped with 0.5 ppm of pentavalent impurity.

**Number of carriers** from doping: Ge crystal 4×10²⁸ atoms/m³ doped 2 ppm trivalent gives acceptor atoms 8×10²² m⁻³, holes ≈ that, for 1.5 ppm 6×10²² m⁻³, for 0.5 ppm pentavalent Si 5×10²⁸ atoms/m³ gives 2.5×10²² donors/m³. Acceptor atom effectively negative when accepts electron, donor positive when donates, but crystal neutral. 0.5 ppm = 0.5 × 10⁻⁶ . Number of donor atoms = 0.5 × 10⁻⁶ × 5 × 10²⁸ = 2.5 × 10²² m⁻³ , each contributing one electron. Substituting values gives 2.5 × 10²² m⁻³, which matches expected behaviour for this semiconductor device configuration, confirming doping, depletio

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Doping, Charge Carriers and Conductivity

In an n-type semiconductor, the minority carriers are:

**Doping** is adding impurity to pure semiconductor to increase carriers, pentavalent (P, As, Sb) donates extra electron, 1 ppm doping in Ge with 4×10²⁸ atoms/m³ gives donor density N_d = 4×10²⁸×10⁻⁶ =4×10²² m⁻³ for 1 ppm, acceptor atoms p-type trivalent B, Al, Ga. Overall charge neutrality maintained because donor ion core positive but electron negative, net neutral. In an n-type semiconductor, electrons are majority carriers due to pentavalent doping, while holes, generated intrinsically or reduced by recombination, are minority carriers. Substituting values gives Holes, which matches expect

Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity

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Doping, Charge Carriers and Conductivity

The majority carriers in an n-type semiconductor are:

**Number of carriers** from doping: Ge crystal 4×10²⁸ atoms/m³ doped 2 ppm trivalent gives acceptor atoms 8×10²² m⁻³, holes ≈ that, for 1.5 ppm 6×10²² m⁻³, for 0.5 ppm pentavalent Si 5×10²⁸ atoms/m³ gives 2.5×10²² donors/m³. Acceptor atom effectively negative when accepts electron, donor positive when donates, but crystal neutral. In an n-type semiconductor, pentavalent doping increases the number of conduction electrons, making them the majority carriers, while holes are minority carriers ( n_e gg n_h ). Substituting values gives Electrons, which matches expected behaviour for this semiconduc

Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity

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Semiconductors, Types and Energy Bands

In an n-type semiconductor, the charge of the donor ion core is:

**Energy bands in semiconductors** consist of valence band filled at 0 K and conduction band empty, gap E_g small ~1 eV (Si 1.1 eV, Ge 0.7 eV), insulators large gap >3 eV (C diamond 5.4 eV), conductors overlapping. Intrinsic semiconductor at 0 K behaves as insulator because no thermal excitation, at T>0 K electrons jump to conduction band leaving holes, conductivity increases with temperature. A pentavalent donor (e.g., P) donates one electron, leaving a positively charged ion core (+q) after ionization, balanced by the free electron’s negative charge. Substituting values gives Positive, which

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Semiconductors, Types and Energy Bands

In an extrinsic semiconductor, the overall charge neutrality is maintained because:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. The charge of additional carriers (electrons in n-type or holes in p-type) is balanced by the equal and opposite charge of ionized dopant cores (positive in n-type, negative in p-type). Substituting values gives Charge of carriers equals that of ionized cores, whi

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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PHYSICS

A pure Si crystal is doped with 1 ppm of pentavalent impurity. If the Si atom density is 5 × 10²⁸ m^{-3, the number

Given: A pure Si crystal is doped with 1 ppm of pentavalent impurity. If the Si atom density is 5 × 10²⁸ m^{-3, the number of donor atoms per cubic meter is: These values define the system as per NCERT data. Formula: 1 ppm = 1 part per million = 10⁻⁶. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Number of donor atoms = 10⁻⁶ × 5 × 10²⁸= 5 × 10²² m^{-3 . These contribute electrons, assuming full ionization at room temperature. Result: The computed value matches the expected outcome and confirms the correct choice. Units a

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons

Given: A Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons per cubic meter is: These values define the system as per NCERT data. Formula: 0.5 ppm = 0.5 × 10⁻⁶. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Number of donor atoms = 0.5 × 10⁻⁶ × 5 × 10²⁸= 2.5 × 10²² m^{-3, each contributing one electron. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

Which impurity type is used to create an n-type semiconductor?

An n-type semiconductor is formed by doping a tetravalent semiconductor (like Si or Ge) with a pentavalent impurity (e.g., As, Sb, P), which donates an extra electron for conduction.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A p-type semiconductor is doped with:

A p-type semiconductor is created by doping a tetravalent semiconductor (e.g., Si) with a trivalent impurity (e.g., B, Al, In), which accepts electrons and creates holes as majority carriers.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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