Question

An abrupt silicon $p-n$ junction has dopant concentration of $N_A=2\times {10}^{16}c{m}^{-3}$ and $N_D=2\times {10}^{15}c{m}^{-3}$ at $T-{300}^{\circ }K$. The maximum electric field in the space charge region when the reverse bias voltage $V_R=9$ will be_____$\times {10}^{4}V/cm.({\epsilon }_{si}=11.7\times 8.85\times {10}^{-14}F/cm,V_T=25.9mV,n_i=1.5\times {10}^{10}/c{m}^3)$

Answer 1

$\begin{array}{cc}\text{Now}{V}_{bi}& =V_{T}ln\left(\frac{N_A{N}_D}{n_i^2}\right)=\left(0.0259\right)ln\left(\frac{2\times {10}^{16}2\times {10}^{15}}{(1.5\times {10}^{10}{)}^2}\right)=0.671V\\ W& ={\left[\frac{2R(V_{ln}+V_R)}{q}[\frac{1}{N_A}+\frac{1}{N_D}]\right]}^{\frac{1}{2}}\\ & ={\left[\frac{2\left(11.7\right)\times (8.85d\times {10}^{-14}(0.671)}{1.6\times {10}^{-19}}[\frac{1}{2\times {10}^6}+\frac{1}{2\times {10}^{15}}]\right]}^{\frac{1}{2}}\\ W& =0.691\times {10}^{-4}cm\\ \text{now},E_{max}& =\frac{2\left(V_{bi}\right)}{W}=\frac{2\times 0.671}{0.691\times {10}^{-4}}=1.94\times {10}^{4}V/cm\end{array}$