Question

The donor and acceptor impurities in an abrupt junction silicon diode are $1\times {10}^{6}c{m}^{-3}$ and $5\times {10}^{18}c{m}^{-3}$, respectively. Assume that the intrinsic carrier concentration in silicon $n_i=1.5\times {10}^{10}c{m}^{-3}$ at $300K,\frac{kT}{q}=26mV$ and the permittivity of silicon ${ϵ}_{Si}=1.04\times {10}^{-12}F/cm$. The built-in potential and the depletion width of the diode under thermal equilibrium condition, respectively, are

• A. $0.7V$ and $1\times {10}^{-4}cm$
• B. $0.86V$ and $3.3\times {10}^{-5}cm$
• C. $0.86V$ and $1\times {10}^{-4}cm$
• D. $0.7V$ and $3.3\times {10}^{-5}cm$

Answer 1

The correct option is B $0.86V$ and $3.3\times {10}^{-5}cm$

Ans : (d)

Built –in potential is given by
$V_{bi}=V_T\ln \left[\frac{N_A{N}_D}{n_i^2}\right]$
$V_{bi}=0.026\ln \left[\frac{{10}^{16}\times 5\times {10}^{18}}{2.25\times {10}^{20}}\right]$
$V_{bi}=0.86V$
and depletion width $W$ is given by,
$W=\sqrt{\frac{2ϵ}{q}[\frac{1}{N_A}+\frac{1}{N_D}]V_{bi}}$
$W=\sqrt{\frac{2\times 1.04\times {10}^{-12}}{1.6\times {10}^{-19}}[\frac{1}{5\times {10}^{18}}+\frac{1}{{10}^{16}}]\times 0.86}$
$W=3.3\times {10}^{-5}cm$