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#hole concentration

2 public questions tagged with this topic.

Semiconductors, Types and Energy Bands

In an intrinsic semiconductor, the number of free electrons (\( n_e \)) is equal 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. In an intrinsic semiconductor, thermal excitation generates equal numbers of free electrons ( n_e ) and holes ( n_h ), so n_e = n_h = n_i , where n_i is the intrinsic carrier concentration. Substituting values gives Number of holes ( n_h ), which matches expected

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

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Doping, Charge Carriers and Conductivity

A Ge crystal with \( 4 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) is doped with 1.5 ppm of trivalent impurity. T

**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. 1.5 ppm = 1.5 × 10⁻⁶ . Number of acceptor atoms = 1.5 × 10⁻⁶ × 4 × 10²⁸ = 6 × 10²² m⁻³ , each creating one hole, assuming full ionization. Substituting values gives 6 × 10²² m⁻³, which matches expected behaviour for this semicond

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

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