Question 5.4: An abrupt Si p-n junction (A = 10^-4 cm²) has the following ...
An abrupt Si p-n junction (A = 10^{-4} cm^2) has the following properties at 300 K:
p side n side
\space \\N_a = 10^{17} cm^{-3}\quad \quad\quad\quad \quad N_d = 10^{15} \\ \space \\ \tau_n = 0.1 \mu s \quad \quad \quad\quad\quad\quad \quad \tau_p = 10 \mu s \\ \space \\ \mu_p = 200 cm^2 /V – s \quad\quad\quad \quad \mu_n = 1300 \\ \space \\ \mu_n = 700 \quad\quad\quad \quad\quad \quad \quad \quad \mu_p = 450
The junction is forward biased by 0.5 V. What is the forward current?
What is the current at a reverse bias of −0.5 V?
I=qA\Big(\frac{D_p}{L_p}p_n+\frac{D_n}{L_n}n_p\Big)(e^{qV/kT}-1)=I_0(e^{qV/kT}-1) \\ \space \\ p_n=\frac{n_i^2}{n_n}=\frac{(1.5\times 10^{10})^2}{10^{15}}=2.25\times 10^5 \space cm^{-3}\\ \space \\ n_p=\frac{n_i^2}{p_p}=\frac{(1.5\times 10^{10})^2}{10^{17}}=2.25\times 10^3 \space cm^{-3}
For minority carriers,
D_p=\frac{kT}{q}\mu_p=0.0259\times 450=11.66 \space cm^2/s \space on \space the \space n \space side \\ \space \\ D_n=\frac{kT}{q}\mu_n=0.0259\times 700=18.13 \space cm^2/s \space on \space the \space p \space side \\ \space \\ L_p=\sqrt{D_p\tau_p}=\sqrt{11.66\times 10\times 10^{-6}}=1.08\times 10^{-2} \space cm \\ \space \\ L_n=\sqrt{D_n\tau_n}=\sqrt{18.13\times 0.1\times 10^{-6}}=1.35\times 10^{-3} \space cm \\ \space \\ I_0=qA\Big(\frac{D_p}{L_p}p_n+\frac{D_n}{L_n}n_p\Big)=1.6\times 10^{-19}\times 0.0001\Big(\frac{11.66}{0.0108}2.25\times 10^5+\frac{18.13}{0.00135}2.25 \times 10^3\Big)\\ \space \\ =4.370\times 10^{-15}A\\ \space \\ I=I_0(e^{0.5/0.0259}-1)\approx 1.058\times 10^{-6}A \space in \space forward \space bias. \\ \space \\ I=-I_0=-4.37\times 10^{-15}A \space in \space reverse\space bias.