Question 5.1: An abrupt Si p-n junction has Na = 10^18 cm^-3 on one side a...
An abrupt Si p-n junction has N_{d} = 10^{18} cm^{-3} on one side and N_{d} =5× 10^{15} cm^{-3} on the other.
(a) Calculate the Fermi level positions at 300 K in the p and n regions.
(b) Draw an equilibrium band diagram for the junction and determine the contact potential V_{0} from the diagram.
(c) Compare the results of part (b) with V_{0} as calculated from Eq. (5–8).
V_{0}=\frac{kT}{q} \ln \frac{N_{a}} {n^{2}_{i}/N_{d} } =\frac{kT}{q} \ln \frac{N_{a} N_{d}}{n^{2}_{i}} (5–8)
(a) E_{ip} -E_{F} =kT\ln \frac{P{p} }{n_{i} } =0.0259\ln \frac{10^{18}}{(1.5 \times 10^{10})} =0.467 eV
E_{F} -E_{in} =kT\ln \frac{n_{n} }{n_{i} } =0.0259\ln \frac{5\times 10^{15}}{(1.5 \times 10^{10})} =0.329 eV(b) qV_{0}=0.467 + 0.329 = 0.796 eV
(c) qV_{0}=kT\ln \frac{N_{a}N_{d} }{n^{2}_{i} } =0.0259\ln \frac{5 \times 10^{33}}{2.25\times 10^{20}} = 0.796eV
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