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Consider a silicon p-n step junction (or abrupt junction) at room temperature. If the doping concentrations are Na = 10^17 cm^-3 and Nd = 10^14 cm^-3 , what is the depletion width (also known as space charge width)? Also find the maximum electric field (E0). Then derive the expression for the electric field E(x), and the potential field V(x). Do not use limits of integration. To find the constant of integration, use boundary conditions such as "E=E0 at x=0". Once you get expressions for E(x), you can use that for V(x). Use boundary conditions, "at x=-xp0, V(x)=0, and at x=xn0, V(x)=V0".

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opudodennis

Answer:

w=2.78 mm

Explanation:

Depletion region width is found by

[tex]w=x_p+x_N=\sqrt {\frac {2\epsilon_{si} v_{bi}}{q}(\frac {1}{N_D}+\frac {1}{N_A})}[/tex]

To obtain [tex]v_{bi}[/tex] which is built in voltage

[tex]v_{bi}=\frac {KT}{q}ln(\frac {N_D N_A}{n_i^{2}})[/tex]

Where [tex]q=1.6*10^{-19} and [tex]\epsilon_{si}=11\epsilon_o=11*(8.854*10^{-12})=9.74*10^{-11}[/tex]

Generally, for silicon, [tex]n_i=1.5*10^{10}/cm^{3}[/tex]

Therefore, [tex]v_{bi}=0.026 ln(\frac {10^{17} * 10^{14}}{(1.5*10^{10})^{2})}=0.637 Volts[/tex]

[tex]w=\sqrt {(\frac {1}{10^{17}}+\frac {1}{10^{14}})\frac {2*(9.74*10^{-11})*0.637}{1.6*10^{-19}*}[/tex]

w=2.78 mm

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