Question

When a silicon diode having a doping concentration of $N_A=9\times {10}^{16}c{m}^{-3}$ on p-side and $N_D=1\times {10}^{16}c{m}^{-3}$ on n-side is reverse biased, the total depletion width is found to be $3\mu m$. Given that the permittivity of silion is $1.04\times {10}^{-12}F/cm$, the depletion width on the p-side and the maximum electric field in the depletion region, respectively are

• A. $0.3\mu m$ and $0.42\times {10}^{5}V/cm$
• B. $2.7\mu m$ and $2.3\times {10}^{5}V/cm$
• C. $0.3\mu m$ and $4.15\times {10}^{5}V/cm$
• D. $2.1\mu m$ and $0.42\times {10}^{5}V/cm$

Answer 1

The correct option is C $0.3\mu m$ and $4.15\times {10}^{5}V/cm$

Ans : (b)

We know that
$\frac{N_A}{N_D}=\frac{x_n}{x_p}$
So, $\frac{N_A}{N_D}+1=\frac{x_n}{x_p}+1$
$\frac{N_A+N_D}{N_D}=\frac{x_n+x_p}{x_p}$
$\frac{{10}^{17}}{{10}^{16}}=\frac{3\mu m}{x_p}$
$\therefore x_p=0.3\mu m$
and
$E_{max}=\frac{q{N}_A{x}_p}{{ϵ}_s}=\frac{q{N}_D{x}_n}{{ϵ}_s}$
$\therefore E_{max}=\frac{1.6\times {10}^{-19}\times 9\times {10}^{16}\times 0.3\times {10}^{-4}}{1.04\times {10}^{-12}}$
$E_{max}=4.15\times {10}^{5}V/cm$