This is the case where an electrical field of arbitrary origin attracts
the majority carriers . | ||||||||||

Starting with the Poisson equation for doped semiconductors and all dopants ionized, we have seen that we can approximate the situation by | ||||||||||

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In contrast to the case of quasi-neutrality, we now have (and the sign is important) | ||||||||||

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This allows the approximation | ||||||||||

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and the Poisson equation reduces to | ||||||||||

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Using the Debye length
,
or L_{D}= {(ee_{0} kT )/(e^{2}N_{D})}^{1/2} ,N_{D} = ee_{0}kT/e^{2}L
^{2} _{D}the Poisson equation for accumulation
can be rewritten as | ||||||||||

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While this looks like a simple differential equation, it is not all that easy to solve it. |

What we would need first, are defined boundary conditions
so we can tackle the differential equation. There are no obvious candidates, so we have to think a little harder now. |
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Accumulation means that we have some surface charge
r that we put on the surface of the semiconductor (with our fictive
thin insulating layer in between). | ||||||||||||||||||

We thus need to refomulate the differential equation so that surface charge
can be included. the (not overly obvious) way to do this is to introduce the electrical field
strength
as a new variable besides E( x)D.E_{C} |
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For that we use the relation | ||||||||||||||||||

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We also made use of the equaltity dD with E_{ C}/d x
= e · E(x ) field strength. E(x)
= | ||||||||||||||||||

Inserting and separating the variables (and omitting the "(" for
clarity) gives x) | ||||||||||||||||||

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Tricky, but worth it. Now we can integrate both sides. The integrations run from far inside the bulk ,
i.e. from , to some value of E
= 0, and that means from EdD to some corresponding value E_{C}
= 0dD.E_{C} | ||||||||||||||||||

Omitting the integration (which is trivial), we obtain | ||||||||||||||||||

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i.e. an equation relating the amount of band bending at some position to the electrical
field strenght at this point, which isx | ||||||||||||||||||

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For n-type semiconductors, which we are considering, D
is negative and large (i.e. E_{C}D
) - and we may neglect the E_{C} >> kT– 1, obtaining | ||||||||||||||||||

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While this is fine, we still don't have the solution we want. We must now remember that there
is a simple relation tying surface charge to volume charge. | ||||||||||||||||||

This is Gauss
law, stating that the flux of the electrical field through a surface S is the integral
over the components of perpendicular to the surface. E | ||||||||||||||||||

The charge is usually expressed in terms of charge density r(. Gauss law then states:
x,y,z) | ||||||||||||||||||

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With = normal vector of the surface nS, da = surface increment, dV
= volume increment. For more details use the link. |
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For our case it means that we could replace the total charge
r contained in a slice between (where there is no charge and the field strength is x = ¥
) and E = E_{
bulk} = 0, by a xsurface (or better areal) charge
s at _{ area}(x) given by x | ||||||||||||||||||

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The total amount of band-bending induced by a real external surface charge
s is simply _{ex}D which we call E_{ C}(x
= 0)D: E_{C}^{0} | ||||||||||||||||||

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So we have all we need. The +/- sign came from the two solutions of the square root; we have to
pick the correct one depending on the situation (holes or electrons considered). | ||||||||||||||||||

Band-Bending and Surface Charge

© H. Föll (Semiconductors - Script)