Answer to Question #210534 in Optics for anuj

Question #210534

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


1
Expert's answer
2021-06-28T04:57:02-0400

Given,

Temperature (T)= 300K

Doping concentration (NA)=1016cm3(N_A)=10^{16}cm^{-3}

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


Vbi=KBTeln(NnNp)V_{bi}=\frac{K_B T}{e}\ln (\frac{N_n}{N_p})


Vbi=KBTeln(NANDni2)\Rightarrow V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})


ni=1.5×1010cm3n_i= 1.5×10^{10}cm^{−3} ,in the quasi-neutral p-region

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

Now, substituting the values,

Vbi=KBTeln(NANDni2)V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})

Now, substituting the values,

Vbi=KBTeln(NANDni2)\Rightarrow V_{bi}=\frac{K_B T}{e}\ln (\frac{N_AN_D}{n_i^2})


0.7V=1.4×1023×3001.6×1019ln(1016ND(1.5×1010)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})


0.7=0.026lnND2.25×104\Rightarrow 0.7=0.026\ln{\frac{N_D}{2.25\times 10^{4}}}


26.92=lnND2.25×104\Rightarrow 26.92 =\ln{\frac{N_D}{2.25\times 10^{4}}}


e26.92=ND2.25×104\Rightarrow e^{26.92}=\frac{N_D}{2.25\times 10^4}


4.5×1011×2.25×104=ND\Rightarrow 4.5\times 10^{11}\times 2.25\times 10^4=N_D


ND=10.125×1015\Rightarrow N_D=10.125\times 10^{15}




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