Charge density and potential in a semiconductor


by Gogsey
Tags: charge, density, potential, semiconductor
Gogsey
Gogsey is offline
#1
Jan21-09, 08:20 PM
P: 160
Couple of electrcity questions here.

4. (a) Show that in a steady state, any isotropic material that obeys Ohmís Law contains
no regions with net charge.
(b) In certain types of anisotropic materials, the conductivity  is not a scalar but
instead a tensor that can be represented as a symmetric 3x3 matrix; any such
tensor can be rotated into a frame of reference in which it is diagonal, i.e. only the
diagonal components are non-zero. Assuming this frame of reference, calculate
the divergence of J in the material in terms of the components of  and the
components of the electric field. Is such a material also guaranteed to be free of
regions of net charge? If so, why? If not, give a counterexample.

5. In a silicon junction diode, the region of the planar junction between n-type and p-type
semiconductors can be approximately represented as two adjoining slabs of charge, one
negative and one positive. Away from the junction, outside these charge layers, the
potential is constant, with a value of Vn in the n-type material and Vp in the p-type
material. Given that the difference between Vp and Vn is 0.3 V, and that the thickness
of each of the two slabs of charge is 0.01 cm, find the charge density in each of the
two slabs, and make a graph of the potential V as a function of position through the
junction. What is the strength of the electric field at the midplane?


For the top one, we know that the material is uniform in propreties throughout the material. We aso know that in the steady state, the charges density is constant. Either rate in + rate out, or the is nom flow of electrons and there for the material is electrically neutral.

So charge density is constant, then, then its divergence is zero. Now we know thw conservation of charge is "del" J = -dp/dt. Do we just have to proove that both sides are zero, since they are both constant? Therefore charge density never changes? Also, mathematically, do we just show that they are both zero, and thats good enoough?

For the second question, we can use Gauss' law to get an expression of the field, but we don't know the charge. We can integrate potential twice to get the charge, but we on;y have a number not a function. Can we integrate the number twice with respect to t, just as we would do with gravity to get 1/2 gt^2, or in this case 0.3/2t^2?
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gabbagabbahey
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#2
Jan22-09, 02:52 AM
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P: 5,004
For 4(a), write Ohm's law in the form [itex]\vec{J}=\sigma\vec{E}[/itex]. What do you get when you take the divergence of each side of the equation?

For 4(b), [itex]J_i=\delta_{ij}(\sigma_{ij}E_j)[/itex] and [itex]\vec{\nabla}\cdot\vec{J}=\partial_iJ_i[/itex]

For 5, Use gauss' law to get an expression for the fields in n and p type materials in terms of the two unknown charge densities [itex]\sigma_{-}[/itex] and [itex]\sigma_{+}[/itex] assume that the charge densities are constant and integrate the fields over the thickness of each slab to find an expression for the potential difference. Then set that equal to 0.3V and assume that the overall charge on the junction is zero, so that [itex]\sigma_{+}=-\sigma_{-}[/itex] and solve for the charge densities.


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