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Semiconductor Electron Density Analysis

1) A 1cm long piece of undoped silicon with a lifetime of 1ms is illuminated in the middle, generating 2x1019cm-2s-1 electron-hole pairs. 2) The excess electron density and current density throughout the material is calculated using the simple recombination model and diffusion equation, assuming mobility of 1000 cm2/V-s. 3) The solutions show the excess electron density and current density vary sinusoidally from the center of the silicon, reaching a maximum in the middle and reducing to zero at the contacts.

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51 views2 pages

Semiconductor Electron Density Analysis

1) A 1cm long piece of undoped silicon with a lifetime of 1ms is illuminated in the middle, generating 2x1019cm-2s-1 electron-hole pairs. 2) The excess electron density and current density throughout the material is calculated using the simple recombination model and diffusion equation, assuming mobility of 1000 cm2/V-s. 3) The solutions show the excess electron density and current density vary sinusoidally from the center of the silicon, reaching a maximum in the middle and reducing to zero at the contacts.

Uploaded by

fneojwgfje
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Example 2.

12 A 1cm long piece of undoped silicon with a lifetime of 1ms is


illuminated with light, generating Gopt = 2x1019cm-2s-1
electron-hole pairs in the middle of the silicon.

"ideal" Gopt "ideal"


ohmic ohmic
contact contact

x
0 L/2 L
This bar silicon has ideal Ohmic contacts on both sides.
Find the excess electron density throughout the material using
the simple recombination model and assuming that μn = μp =
1000 cm2/V-s.
Also find the resulting electron current density throughout the
material.
Solution Because of the symmetry one can treat each half separately
with half the number of electron-hole pairs generated on both
sides.
Then we solve the diffusion equation:
d 2n n − no
0 = Dn −
dx 2 τn
The general solution to the diffusion equation equals:
n − n o = A exp( x / L n ) + B exp(− x / L n )
where A and B need to be determined by applying the
boundary conditions.
Since an ideal contact implies that the material is in thermal
equilibrium at the contact,
n(x=0) = no
so that A = -B and, linking the current density due to the carrier
generation to the carrier gradient.

L dn G opt
J n (x = ) = qD n =q
2 dx x=L / 2 2
one finds that:
dn A L B L G opt
= exp( )− exp(− )=
dx x=L / 2 Ln 2 Ln Ln 2 Ln 2Dn
or
G opt Ln
A=
4Dn L
cosh( )
2 Ln
And the solution for the electron density becomes:
G opt Ln x
n − no = sinh( )
2Dn L Ln
cosh( )
2 Ln
The corresponding electron current density is then:
x
cosh( )
G opt Ln
J n ( x) =
2 L
cosh( )
2 Ln
Both the excess electron density and the electron current
density are plotted versus position on the graph below.
1E+17 2

Current density (A/cm2)


9E+16 1.8
8E+16 1.6
7E+16 1.4
n-no (cm-3)

6E+16 1.2
5E+16 1
4E+16 0.8
3E+16 0.6
2E+16 0.4
1E+16 0.2
0 0
0 0.2 0.4 0.6 0.8 1

Position (cm)

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