Question #251533
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
1
Expert's answer
2021-10-17T16:54:22-0400

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

qVbi=qϕpqϕnqV_{bi}=q\phi_{p}-q\phi_{n}

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

EipEin=qVbi\Rightarrow E_{ip}-E_{in}=qV_{bi}

(EipEF)(EinEF)=qVbi\Rightarrow (E_{ip}-E_F)-(E_{in}-E_F)=qV_{bi}

In the case of equilibrium,

EipEF=qϕpE_{ip}-E_F=q\phi_p

and EFEin=qϕnE_F-E_{in}=q\phi_n

Hence, we can write it as,

qϕp+qϕn=qVbi\Rightarrow q\phi_p+q\phi_n=qV_{bi}

From the Fermi Dirac statistics,

pPo=nieEiEFRTp_{Po}=n_i e^{\frac{E_i-E_F}{RT}}

nno=nieEiEFRTn_{no}=n_i e^{\frac{E_i-E_F}{RT}}

In the case of full ionization,

Ppo=NAP_{po}=N_A and nno=NDn_{no}=N_D

So, we can write,

qFp+qFn=KT[ln(NAni)+ln(NDni)]qF_{p}+qF_{n}=KT [ \ln(\frac{N_A}{n_i})+\ln(\frac{N_D}{n_i})]

Hence, we can write,


qVbi=KTlnNANDni2qV_{bi}=KT\ln{\frac{N_AN_D}{n_i^2}}

Now, calculating the depletion region width

Vbi=xpxnE.dxV_{bi}=\int_{-x_p}^{x_n}E.dx


E=qNAϵxpE=\frac{qN_A}{\epsilon} x_p

Hence, Vbi=(xp+xn2)qNAϵxpV_{bi}=(\frac{x_p+x_n}{2})\frac{qN_A}{\epsilon}x_p

But, NAxp=NDxnN_A x_p = N_D x_n

Hence, 2ϵqNA=NA+NDNAxp2\frac{2\epsilon}{qN_A}=\frac{N_A+N_D}{N_A}x_p^2

So, xp=2ϵqNDNA1NA+NDVbix_p=\sqrt{\frac{2\epsilon}{q}\frac{N_D}{N_A}\frac{1}{N_A+N_D}V_{bi}}

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