 Solution

ID: past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2021/section-b/6/d/solution

Let the neutral edges be

- x_{p}

and

x_{n}

and take the exterior electric field to vanish. Integrating

d E_{x} / d x = ρ / ε_{s} = e αx / ε_{s}

between the edges gives

x_{n}^{2} = x_{p}^{2}

, so

x_{n} = x_{p} = w /2

. This is also the condition of equal and opposite integrated depletion charges.

A second integration with

E_{x} (w /2) = 0

yields

E_{x} (x) = \frac{e α}{2 ε _{s}} (x^{2} - \frac{w ^{2}}{4}), ∣ x ∣ \leq \frac{w}{2} .

It is negative inside the layer, pointing from n to p, and zero at both edges. Its largest magnitude occurs at the metallurgical junction:

x = 0, ∣ E_{x} ∣_{m a x} = \frac{e α w ^{2}}{8 ε _{s}} .

For reference, choosing

V (- w /2) = 0

V (x) = \frac{e α}{2 ε _{s}} (\frac{w ^{2} x}{4} - \frac{x ^{3}}{3} + \frac{w ^{3}}{12}) .

 New to topics? Read the docs here!