Practice question

**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, depletion and rectifier principles.