Answer to Question #210534 in Optics for anuj

Question

Consider a silicon p-n junction at T = 300 K with doping concentrations of NA = 10^16 cm-3 . Given that the built-in voltage Vbi = 0.7 V, determine the doping concentration of the n-side of the junction. 

b) Derive an expression for the width of the depletion region in terms of Vbi and the maximum electric field |𝐸𝑚| at x = 0

c) Derive an expression for xp in terms of Vbi. Hence, find a numerical value for |𝐸𝑚| and the width of the depletion region

Answer 1

Given,

Temperature (T)= 300K

Doping concentration $(N_A)=10^{16}cm^{-3}$

Built in voltage $(V_{bi})=0.7V$

$V_{bi}=\frac{K_B T}{e}\ln (\frac{N_n}{N_p})$

$\Rightarrow V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})$

$n_i= 1.5×10^{10}cm^{−3}$ ,in the quasi-neutral p-region

Let the doping of n-type Si be $(N_D)=?$

Now, substituting the values,

$V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})$

Now, substituting the values,

$\Rightarrow V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})$

$\Rightarrow 0.7V=\frac{1.4\times 10^{-23}\times 300}{1.6\times 10^{-19}}\ln (\frac{10^{16}N_D}{(1.5\times 10^{10})^2})$

$\Rightarrow 0.7=0.026\ln{\frac{N_D}{2.25\times 10^{4}}}$

$\Rightarrow 26.92 =\ln{\frac{N_D}{2.25\times 10^{4}}}$

$\Rightarrow e^{26.92}=\frac{N_D}{2.25\times 10^4}$

$\Rightarrow 4.5\times 10^{11}\times 2.25\times 10^4=N_D$

$\Rightarrow N_D=10.125\times 10^{15}$