Answer to Question #251536 in Optics for Ancient green

Question #251536
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
2021-10-17T16:54:15-0400

We know that the band bending is the difference of the work function of the p-type and n-type material. so,

"qV_{bi}=q\\phi_{p}-q\\phi_{n}"

We know that total band bending is the intrinsic level is the build in potential,

"\\Rightarrow E_{ip}-E_{in}=qV_{bi}"

"\\Rightarrow (E_{ip}-E_F)-(E_{in}-E_F)=qV_{bi}"

In the case of equilibrium,

"E_{ip}-E_F=q\\phi_p"

and "E_F-E_{in}=q\\phi_n"

Hence, we can write it as,

"\\Rightarrow q\\phi_p+q\\phi_n=qV_{bi}"

From the Fermi Dirac statistics,

"p_{Po}=n_i e^{\\frac{E_i-E_F}{RT}}"

"n_{no}=n_i e^{\\frac{E_i-E_F}{RT}}"

In the case of full ionization,

"P_{po}=N_A" and "n_{no}=N_D"

So, we can write,

"qF_{p}+qF_{n}=KT [ \\ln(\\frac{N_A}{n_i})+\\ln(\\frac{N_D}{n_i})]"

Hence, we can write,


"qV_{bi}=KT\\ln{\\frac{N_AN_D}{n_i^2}}"

Now, calculating the depletion region width

"V_{bi}=\\int_{-x_p}^{x_n}E.dx"


"E=\\frac{qN_A}{\\epsilon} x_p"

Hence, "V_{bi}=(\\frac{x_p+x_n}{2})\\frac{qN_A}{\\epsilon}x_p"

But, "N_A x_p = N_D x_n"

Hence, "\\frac{2\\epsilon}{qN_A}=\\frac{N_A+N_D}{N_A}x_p^2"

So, "x_p=\\sqrt{\\frac{2\\epsilon}{q}\\frac{N_D}{N_A}\\frac{1}{N_A+N_D}V_{bi}}"


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