Answer to Question #251533 in Optics for Ancient green

Question

Derive an expression for xp in terms of Vbi. Hence, find a numerical value for |à ƒ ƒ °   ¸Ãƒ ƒ ƒ °  ‘ š| and the width of the depletion region

Answer 1

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