Question

The doping concentrations on the p-side and n-side of a silicon diode are $1\times {10}^{16}c{m}^{-3}$ and $1\times {10}^{17}c{m}^{-3}$ , respectively. A forward bias of $0.3V$ is applied tothe diode. At $T=300K$ , the intrincic carrier concentration of silicon $n_i=1.5\times {10}^{10}c{m}^{-3}$ and $\frac{kT}{q}=26mV$. The electron concentration at the edge of the deplection region on the p-side is

• A. $2.25\times {10}^{6}c{m}^{-3}$
• B. $2.3\times {10}^{9}c{m}^{-3}$
• C. $1\times {10}^{17}c{m}^{-3}$
• D. $1\times {10}^{16}c{m}^{-3}$

Answer 1

The correct option is B $2.3\times {10}^{9}c{m}^{-3}$

Ans : (a)

The electron concentration at the edge of the depletion region on the p-side is given by
$n_p=n_{p_0}exp\left[\frac{q{V}_f}{kT}\right]$
where, $n_{p_0}=\frac{n_i^2}{N_A}$
$n_{p_0}=\frac{(1.5\times {10}^{10}{)}^2}{{10}^{16}}=2.25\times {10}^{4}c{m}^{-3}$
We then have
$n_p=2.25\times {10}^{-4}exp\left[\frac{0.3}{0.026}\right]$
$n_p=2.25\times {10}^4\times 1.025\times {10}^5$
$n_p=2.306\times {10}^{9}c{m}^{-3}$