(a) The resistivities of the P-region and N-region of a germanium diode are 6 Ω-cm and 4Ω-cm, respectively. Calculate the contact potential Vo and potential energy barrier Eo. (b) If the doping densities of both P and N-regions are doubled, determine Vo and Eo. Given that q = 1.602 × 10^–19 C,

Question

(a) The resistivities of the P-region and N-region of a germanium diode are 6 Ω-cm and 4Ω-cm, respectively. Calculate the contact potential [latex]V_{o}[/latex] and potential energy barrier [latex]E_{o}[/latex]. (b) If the doping densities of both P and N-regions are doubled, determine [latex]V_{o} and E_{o}[/latex]. Given that [latex]q = 1.602 × 10^{–19} C, n_{i} = 2.5 × 10^{13}/cm^{3}, \mu_{p} = 1800 cm^{2}/V-s, \mu_{n} = 3800 cm^{2} /V-s and V_{T} = 0.026 V at 300 K.[/latex]

Accepted Answer

(a) Resistivity, [latex] ho =\frac{1}{\sigma }=\frac{1}{N_{A} q\mu _{p} }6\Omega -cm[/latex]

Therefore, [latex]N_{A}=\frac{1}{6q\mu _{p}}=\frac{1}{6 imes 1.602 imes 10^{-19} imes 1800 }=0.579 imes 10^{15}/cm^{3} [/latex]

Similarly, [latex]N_{D}=\frac{1}{4q\mu _{n}}=\frac{1}{4 imes 1.602 imes 10^{-19} imes 3800 }=0.579 imes 10^{15}/cm^{3}[/latex]

Therefore, [latex]V_{0}=V_{T} \ln\frac{N_{D}N_{A}}{n^{2}_{i} } =0.026\ln \frac{0.579 imes 0.411 imes 10^{30} }{(2.5 imes 10^{13} )^{2} } =0.1545 V[/latex]

Hence [latex]E_{0} = 0.1545 eV[/latex]

(b) [latex]V_{0}=0.026\frac{2 imes 0.579 imes 10^{15} imes 2 imes 0.411 imes 10^{15} }{(2.5 imes 10^{13} )^{2} }=0.1906V[/latex]

Therefore, [latex]E_{0} = 0.1906 eV[/latex]

Question 4.11: (a) The resistivities of the P-region and N-region of a germ...

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