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Suppose a pure Si crystal has $5 \times 10^{28} \text{atoms} / m^3$. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes, given that $n_i = 1.5 \times 10^{16} \;/m^3$

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Given that $n_i = 1.5 \times 10^{16} \;/m^3$, the thermally generated electrons are negligibly small compared to those produced by doping. Therefore, $n_e \approx N_D$
Since $n_e\;n_h=n_i^2$, the number of holes $n_h = \large \frac {2.25 \times 10^{32}}{5 \times 10^{22}}$$ \approx 9 \times 10^9 \;m^{-3}$
answered Jul 13, 2014 by balaji.thirumalai
edited Jul 13, 2014 by balaji.thirumalai
 

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