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#trivalent impurity

2 public questions tagged with this topic.

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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Doping, Charge Carriers and Conductivity

A pure Ge crystal has \( 4 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) and is doped with 2 ppm of trivalent impur

**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. 2 ppm = 2 × 10⁻⁶ . Number of acceptor atoms = 2 × 10⁻⁶ × 4 × 10²⁸ = 8 × 10²² m⁻³ , contributing holes for p-type conductivity. Substituting values gives 8 × 10²² m⁻³, which matches expected behaviour for this semiconductor device configuration, confirming doping, dep

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

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