Current through p-type silicon

[hogrampage]

Homework Statement

A 1-cm cube of p-type silicon (ρ = 0.1Ω-cm) acquires a linear electron distribution in the x-direction, such that n = 10¹⁴/cm³ at one side and n = 10⁵/cm³ at the opposite side.

Wires are attached to the sides of the cube via ohmic contacts, and a 0.1mV voltage source is applied. Find the values of the electron, hole, and total currents that flow in the external circuit.

Homework Equations

ρ = R$\frac{A}{L}$ = $\frac{E}{J}$
E = $\frac{V}{L}$
J_e = qnμ_eE + qD_e$\frac{dn}{dx}$

J_h = qpμ_hE + qD_h$\frac{dp}{dx}$
i = $\frac{V}{R}$
n₀p₀ = n²_i

The Attempt at a Solution

I started by finding R:

R = ρ$\frac{L}{A}$ = (0.1Ω-cm)$\frac{(1cm)}{(1cm^{2})}$ = 0.1Ω

Then, the total current is:

I = $\frac{V}{R}$ = $\frac{(0.1mV)}{(0.1Ω)}$ = 1mA

p₀ = $\frac{n^{2}_{i}}{n_{0}}$ = $\frac{(1.5x10^{10})^{2}}{10^{14}}$ = 225x10⁴/cm³

p₁ = $\frac{n^{2}_{i}}{n_{1}}$ = $\frac{(1.5x10^{10})^{2}}{10^{5}}$ = 225x10¹³/cm³

Now, $\frac{dp}{dx}$ = 225x10¹³/cm⁴ and $\frac{dn}{dx}$ = -1x10¹⁴/cm⁴, since dx = 1cm.

Which values would I use for n and p in the following equations, assuming the above steps are correct?

J_e = qnμ_eE + qD_e$\frac{dn}{dx}$

J_h = qpμ_hE + qD_h$\frac{dp}{dx}$

 

[KataKoniK]

The values for n and p in the above equations would be n = 1014/cm3 and p = 105/cm3, as given in the problem statement. These values represent the electron and hole concentrations at the two opposite sides of the cube.