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Analysis of unbiased PN junction and physics

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see attachment please

please, please, please I need help to solve this Q

I solved parts a,b, and c still need help in other parts

the reference book I am using ( the physics of solar cell (nelson) ) Ch6

Here is some website can help
http://ecee.colorado.edu/~bart/book/book/chapter4/ch4_3.htm
http://my.ece.msstate.edu/faculty/donohoe/ece3313notes6.pdf

I tray solve it. I found this web side helpful
I will let you if I found any things may help

Thanks

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Solution Summary

Analysis of physics for unbiased PN junction and derivation of electric field and voltage profile across depletion region

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Problem 2:
N_A represents the volume concentration of acceptor impurity atoms. In this case the volume concentration of Boron impurity atoms (Boron being the acceptor element having one less electron, than Si in its outer shell, to form complete covalent bonds with Si lattice atoms, thus leaving a hole for each such bond). In our example N_A=3×〖10〗^16 〖cm〗^(-3)

N_D represents the volume concentration of donor impurity atoms. In this case the volume concentration of Phosphorous impurity atoms (Phosphorous being the donor element having one extra electron, than Si in its outer shell, which on bonding with Si lattice atoms, leaves a free electron for each such bond). In our example N_D=5×〖10〗^15 〖cm〗^(-3)

A diagram showing the P and N type Si semi-conductor is shown below

The charge density ρ (assuming no free carriers) is simply the sum of electrons, , holes, p , ionized acceptor ions, N_A^- , and ionized donor ions, N_D^+ , multiplied by the electronic charge e, at any point.

ρ=〖e(N〗_A^-+N_D^++n+p)

Since in the depletion region n=p=0

And since at x=x_1/2 we are in the N type region so no acceptors, hence N_A^-=0

Therefore
ρ(x=x_1/2)=〖e.(N〗_A^-+N_D^++n+p)=0+eN_D^++0+0
ρ(x=x_1/2)=eN_D^+

Assuming 100% ionization so that all donor sites are ionized then N_D^+=N_D. The source of the charge density at x_1⁄2 is thus ionised donor sites

Charge density at x=x_1/2 ,
ρ(x_1/2)=eN_D

This question asks us to determine the built in potential (we assume this means at ambient room temperature T=290K

We first derive the thermal equilibrium voltage V_T at this temperature from

V_T=kT/e
Where Boltzmann's constant k=1.38×〖10〗^(-23) J.K^(-1)

Electronic charge e=1.602×〖10〗^(-19) C

At T=290K
V_T=kT/e=(1.38×〖10〗^(-23)×290)/(1.602×〖10〗^(-19) )=0.02498V

To determine the built in potential V_bi we use

V_bi=V_T.Ln((N_D.N_A)/〖n_i〗^2 ) Source-1

Where the impurity donor concentration N_D=5×〖10〗^15 〖cm〗^(-3), impurity acceptor concentration N_A=3×〖10〗^16 〖cm〗^(-3) and intrinsic carrier concentration n_i=〖10〗^10 〖cm〗^(-3)

Putting in numbers
V_bi=0.02498.Ln((5×〖10〗^15×3×〖10〗^16)/(〖10〗^10 )^2 )

V_bi=0.70V (to 2 sig fig.)

An Excel spreadsheet has been used to derive the plot ...

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