A silicon p–n junction has a donor concentration (N_D=5.0 $imes10^{15}$,\text{cm}^{-3}) on the n‑side and an acceptor concentration (N_A=5.0 $imes10^{13}$,\text{cm}^{-3}) on the p‑side. The built‑in potential is ($V_{bi}$=0.75,\text{V}). Using (${\epsilon }_{si}$=11.7,varepsilon_0) with (varepsilon_0=8.85 $imes10^{-14}$,\text{F/cm}) and (q=1.6 $imes10^{-19}$,\text{C}), calculate the total depletion width (W).

Step-by-step solution

1. 1. Compute the semiconductor permittivity: (${\epsilon }_{si}$=11.7\times8.85 $imes10^{-14}$=1.035 $imes10^{-12}$,\text{F/cm}).
2. 2. Calculate the factor (\frac{N_D+N_A}{N_DN_A}=\frac{5.0 $imes10^{15}$+5.0 $imes10^{13}$}{5.0 $imes10^{15}$\times5.0 $imes10^{13}$}approx2.02 $imes10^{-14}$).
3. 3. Compute the term (\frac{$2{\epsilon }_{si}$}{q}\times\frac{N_D+N_A}{N_DN_A}=2\times1.035 $imes10^{-12}$\times2.02 $imes10^{-14}$/1.6 $imes10^{-19}\approx 2$.61 $imes10^{-7}$).
4. 4. Multiply by ($V_{bi}$): (2.61 $imes10^{-7}$\times0.75=1.96 $imes10^{-7}$).
5. 5. The depletion width is (W=sqrt{1.96 $imes10^{-7}$}approx4.42 $imes10^{-4}$,\text{cm}).

Concept tested: Depletion width of a p‑n junction

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked