Germanium (Ge) is doped with Aluminum (Al). I | vedclass

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(A) In a $p$-type semiconductor,the concentration of holes $(n_h)$ is approximately equal to the concentration of acceptor atoms $(N_A)$.Given: $n_h \

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$10^{17} 	ext{ m}^{-3}$

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(A) In a $p$-type semiconductor,the concentration of holes $(n_h)$ is approximately equal to the concentration of acceptor atoms $(N_A)$.Given: $n_h \approx N_A = 10^{21} 	ext{ m}^{-3}$.The intrinsic carrier concentration is $n_i = 10^{19} 	ext{ m}^{-3}$.Using the law of mass action for semiconductors: $n_e \cdot n_h = n_i^2$.Therefore,the concentration of electrons $(n_e)$ is:$n_e = \frac{n_i^2}{n_h} = \frac{(10^{19})^2}{10^{21}} = \frac{10^{38}}{10^{21}} = 10^{17} 	ext{ m}^{-3}$.

Germanium $(Ge)$ is doped with Aluminum $(Al)$. If the concentration of acceptor atoms is $N_A \cong 10^{21} \text{ atoms/m}^3$ and the intrinsic carrier concentration is $n_i = 10^{19} \text{ m}^{-3}$,then the concentration of electrons is:

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