Question
A semiconductor has equal electron and hole concentration of 6 × 108/m3. On doping with certain impurity, electron concentration increases to 9 × 1012/m3.
  1. Identify the new semiconductor obtained after doping.
  2. Calculate the new hole concentration.

Answer

  1. The doped semiconductor is n-type.

  2. $\text{n}_\text{e}\text{n}_\text{h}=\text{n}^2_\text{i}$

$\Rightarrow\text{n}_\text{h}=\frac{\text{n}^2_\text{i}}{\text{n}_\text{e}}$

$=\frac{(6\times10^8)^2}{9\times10^{12}}=4\times10^4\text{per m}^3$

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