Question

The number of silicon atoms per $m ^{3}$ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per $m ^{3}$ of arsenic and $5 \times 10^{20}$ per $m ^{3}$ atoms of indium. Given that $n_{i}=1.5 \times 10^{16} m ^{-3}$. Number of electrons and holes (in per metre cube of sample) are respectively,

• A. $4.95 \times 10^{22}, 4.54 \times 10^{9}$
• B. $4.54 \times 10^{9}, 4.54 \times 10^{9}$
• C. $4.54 \times 10^{9}, 4.95 \times 10^{22}$
• D. $4.95 \times 10^{22}, 4.95 \times 10^{22}$

Answer 1

Answer: (A)

$N=5 \times 10^{28}$ atoms $/ m ^{3},$
$ A=$ acceptor, indium,
$D=$ donor, arsenic
$N_{A} =0.05 \times 10^{22} $ atoms $ / m ^{3} $
$\Rightarrow N_{D} =5 \times 10^{22} $ atoms $ / m ^{3} $
$n_{i} =1.5 \times 10^{16} m ^{-2}$
$\Rightarrow N_{D}-N_{A}=n_{e}-n_{h}$ and $n_{e} n_{h}=n_{i}^{2}$
$\therefore N_{D}-N_{A}=n_{e}-\frac{n_{i}^{2}}{n_{e}}$
$\Rightarrow n_{e}^{2}-\left(N_{D}-N_{A}\right) n_{e}-n_{i}^{2}=0$
$\Rightarrow n_{e}=\frac{\left(N_{D}-N_{A}\right)+\sqrt{\left(N_{D}-N_{A}\right)^{2}+4 n_{i}^{2}}}{2}$
On substituting values, we get,
$n_{e}=4.95 \times 10^{22} / m ^{3}$
$\therefore n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.95 \times 10^{22}}$
$=4.54 \times 10^{9} \,m ^{-3}$
Observe that doping level $\frac{10^{22}}{10^{28}}=10^{-6}\, ppm$ nearly