Practice question

**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 semiconductor device configuration,