91Ó°ÊÓ

The built-in voltage (\(V_{bi}\)) for a silicon pn junction can be calculated using the formula:\[V_{bi} = \frac{kT}{q} \ln\left(\frac{N_a N_d}{n_i^2}\right)\]where:- \(k\) is Boltzmann's constant \(= 1.38 \times 10^{-23} \text{ J/K}\),- \(T\) is the temperature in Kelvin (\(300\, K\)),- \(q\) is the charge of an electron \(= 1.6 \times 10^{-19} \text{ C}\),- \(n_i\) is the intrinsic carrier concentration for silicon \(\approx 1.5 \times 10^{10} \text{ cm}^{-3}\),- \(N_a\) is the acceptor concentration \(= 5 \times 10^{16} \text{ cm}^{-3}\),- \(N_d\) is the donor concentration \(= 10^{15} \text{ cm}^{-3}\).Plug in the values:\[V_{bi} = \frac{(1.38 \times 10^{-23} \times 300)}{1.6 \times 10^{-19}} \ln\left(\frac{5 \times 10^{16} \times 10^{15}}{(1.5 \times 10^{10})^2}\right)\]\[V_{bi} \approx 0.0259 \ln\left(\frac{5 \times 10^{31}}{2.25 \times 10^{20}}\right)\]\[= 0.0259 \times 5.17 \approx 0.134 \text{ V}\]

The depletion width \(W\) when there is no reverse bias voltage \((V_R = 0)\) can be calculated using:\[W = \sqrt{\frac{2\varepsilon_s(V_{bi} + V_R)}{q}\left(\frac{1}{N_a} + \frac{1}{N_d}\right)}\]where:- \(\varepsilon_s\) is the permittivity of silicon \(= 11.7 \varepsilon_0\), and \(\varepsilon_0 = 8.85 \times 10^{-14} \text{ F/cm}\), and \(V_R = 0\).Substitute \(V_{bi} = 0.134 \text{ V}\) from the previous step:\[W = \sqrt{\frac{2 \times 11.7 \times 8.85 \times 10^{-14} \times 0.134}{1.6 \times 10^{-19}}\left(\frac{1}{5 \times 10^{16}} + \frac{1}{10^{15}}\right)}\]\[\approx \sqrt{\frac{2.7674 \times 10^{-12}}{1.6 \times 10^{-19}}\left(\frac{1}{5 \times 10^{16}} + \frac{1}{10^{15}}\right)}\]After calculating, \(W \approx 4.12 \times 10^{-5} \text{ cm}\).

Repeat the calculation using \(V_R = 5 \, \text{V}\) in the formula for \(W\):\[W = \sqrt{\frac{2\varepsilon_s(V_{bi} + V_R)}{q}\left(\frac{1}{N_a} + \frac{1}{N_d}\right)}\]\[W = \sqrt{\frac{2 \times 11.7 \times 8.85 \times 10^{-14} \times (0.134 + 5)}{1.6 \times 10^{-19}}\left(\frac{1}{5 \times 10^{16}} + \frac{1}{10^{15}}\right)}\]\[\approx \sqrt{\frac{2.7674 \times 10^{-12} \times 5.134}{1.6 \times 10^{-19}}\left(\frac{1}{5 \times 10^{16}} + \frac{1}{10^{15}}\right)}\]Calculate to get \(W \approx 8.04 \times 10^{-5} \text{ cm}\).

The maximum electric field \(|E_{max}|\) at \(V_R = 0\) can be found using:\[|E_{max}| = \frac{2V_{bi}}{W}\]Using \(V_{bi} = 0.134 \, \text{V}\) and \(W = 4.12 \times 10^{-5} \text{ cm}\) from Step 2:\[|E_{max}| = \frac{2 \times 0.134}{4.12 \times 10^{-5}} \times 10^4 \, \text{V/cm}\]Calculate this to obtain \(|E_{max}| \approx 6.50 \times 10^{3} \, \text{V/cm}\).

Using the previously found formula for \(|E_{max}|\), and substitute \(W = 8.04 \times 10^{-5} \text{ cm}\) from Step 3:\[|E_{max}| = \frac{2(V_{bi} + V_R)}{W}\]\[|E_{max}| = \frac{2(0.134 + 5)}{8.04 \times 10^{-5}} \times 10^4 \, \text{V/cm}\]After calculating, \(|E_{max}| \approx 1.278 \times 10^{4} \, \text{V/cm}\).