A silicon diode with area (A=1.0 $imes10^{-4}$,\text{cm}^2) has donor concentration (N_D=1.0 $imes10^{16}$,\text{cm}^{-3}) and acceptor concentration (N_A=1.0 $imes10^{14}$,\text{cm}^{-3}). Its built‑in potential is (0.65,\text{V}). Under a reverse bias of (5,\text{V}), calculate the junction capacitance (C). Use (${\epsilon }_{si}$=11.7,varepsilon_0).

Step-by-step solution

1. 1. Compute (${\epsilon }_{si}$=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}approx1.01 $imes10^{-14}$).
3. 3. Evaluate (\frac{$2{\epsilon }_{si}$}{q}\times\frac{N_D+N_A}{N_DN_A}approx1.31 $imes10^{-7}$).
4. 4. Add the reverse bias to the built‑in potential: ($V_{bi}$+V_R=0.65+5=5.65,\text{V}).
5. 5. Determine the depletion width: (W=sqrt{1.31 $imes10^{-7}$\times5.65}approx8.59 $imes10^{-4}$,\text{cm}).
6. 6. Finally, compute the capacitance: (C=\frac{${\epsilon }_{si}A$}{W}=\frac{1.035 $imes10^{-12}$\times1.0 $imes10^{-4}$}{8.59 $imes10^{-4}$}approx1.21 $imes10^{-13}$,\text{F}).

Concept tested: Junction capacitance under reverse bias

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