Charge density and potential in a semiconductor

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SUMMARY

The discussion focuses on charge density and potential in semiconductors, specifically addressing isotropic and anisotropic materials. In isotropic materials obeying Ohm's Law, the charge density remains constant, leading to zero divergence of the current density (J). For anisotropic materials, the conductivity is represented as a symmetric 3x3 tensor, and the divergence of J can be expressed in terms of the electric field components. Additionally, the charge density in a silicon junction diode is calculated, revealing a potential difference of 0.3 V between n-type and p-type materials, with the charge densities derived using Gauss' law.

PREREQUISITES
  • Understanding of Ohm's Law and its mathematical representation
  • Familiarity with Gauss' Law in electrostatics
  • Knowledge of charge density and electric field relationships
  • Basic concepts of semiconductor physics, particularly n-type and p-type materials
NEXT STEPS
  • Study the derivation of the divergence of current density in anisotropic materials
  • Learn about the mathematical treatment of charge density in semiconductor junctions
  • Explore the implications of tensor conductivity in advanced materials
  • Investigate the graphical representation of electric potential in semiconductor junctions
USEFUL FOR

Electrical engineers, semiconductor physicists, and students studying materials science who are interested in the behavior of charge density and electric fields in semiconductor devices.

Gogsey
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Couple of electricity questions here.

4. (a) Show that in a steady state, any isotropic material that obeys Ohm’s 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 that's 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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For 4(a), write Ohm's law in the form \vec{J}=\sigma\vec{E}. What do you get when you take the divergence of each side of the equation? :wink:

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

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 \sigma_{-} and \sigma_{+} 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 \sigma_{+}=-\sigma_{-} and solve for the charge densities.
 

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