0 like 0 dislike
22 views

When a silicon diode having a doping concentration of $N_{A}= 9\times 10^{16}cm^{-3}$ on  $p$-side and $N_{D}= 1 \times 10^{16}cm^{-3}$ on $n$-side is reverse biased, the total depletion width is found to be $3 \mu$. Given that the permittivity of silicon is $1.04 \times 10^{-12}$ F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are

1. $2.7 \mu m$ and $2.3 \times 10^{5} V/cm$
2. $0.3 \mu m$ and $4.15 \times 10^{5} V/cm$
3. $0.3 \mu m$ and $0.42 \times 10^{5} V/cm$
4. $2.1 \mu m$ and $0.42 \times 10^{5} V/cm$

edited | 22 views