Past exam of the quantum condensed matter course of the University of Cambridge / 2021 / Section B / 6 / e

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- Solution e

Solution

The electric potential difference is the negative integral of the field:

V_{b} - V_{ext} = - \int_{- w /2}^{w /2} E_{x} (x) d x = \frac{e α}{2 ε _{s}} \int_{- w /2}^{w /2} (\frac{w ^{2}}{4} - x^{2}) d x = \frac{e α w ^{3}}{12 ε _{s}} .

Consequently

w (V_{ext}) = [\frac{12 ϵ _{0} ϵ ( V _{b} - V _{ext} )}{e α}]^{1/3}, w (0) = (\frac{12 ϵ _{0} ϵ V _{b}}{e α})^{1/3} .

The linearly graded junction depletion width has this cube-root dependence because of the linearly growing space charge: the charge on one side scales as

w^{2}

w^{2}

, and its potential integral as

w^{3}

. The formula assumes

V_{b} - V_{ext} > 0

and that the depletion approximation remains applicable.

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