Question

Consider a silicon p-n junction with a uniform acceptor doping concentration of ${10}^{17}c{m}^{-3}$ on the p-side and a uniform donor doping concentration of ${10}^{16}c{m}^{-3}$ on the n-side. No external voltage is applied to the diode.
Given: $kT/q=26mV,n_i=1.5\times {10}^{10}c{m}^{-3},{ϵ}_{Si}=12{\epsilon }_0,{\epsilon }_0=8.85\times {10}^{-14}F/cm$, and $q=1.6\times {10}^{-19}C$.
The charge per unit junction area ($nCc{m}^{-2}$ in the depletion region on the p-side is

• A. -4.83

Answer 1

The correct option is A -4.83
$V_0=V_T\ln \frac{N_A{N}_D}{n_i^2}$
$=26\times {10}^{-3}\ln \frac{{10}^{16}\times {10}^{17}}{(1.5\times {10}^{10}{)}^2}$
$V_0=0.757V$
$W=\sqrt{\frac{2\epsilon }{q}(\frac{1}{N_A}+\frac{1}{N_D})V_0}$
$=\sqrt{\frac{2\times 8.854\times {10}^{-16}\times 12}{1.6\times {10}^{-19}}[\frac{1}{{10}^{16}}+\frac{1}{{10}^{17}}]0.757}$
$W=3.3255\mu cm$
$W_P=\frac{W{N}_D}{N_A+N_D}=\frac{3.3255\times {10}^{-6}\times {10}^{16}}{{10}^{16}+{10}^{17}}$
=$0.3023\mu cm$
Charge per unit junction area in the depletion layer on p-side is
$=-q{N}_A{W}_p$
$=-1.6\times {10}^{-19}\times {10}^{17}\times 0.3023\times {10}^{-6}$
$=-4.8368nC/c{m}^2$