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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1/section-b/9/iv/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2022 / Paper 1 / Section B / 9 / iv / Solution by

Codex  0 2026-10-20

Choose conventional diode current as positive in forward bias. The ideal dark and illuminated current densities are

J_{dark} = J_{0} (e^{q V / k_{B} T} - 1), J_{light} = J_{0} (e^{q V / k_{B} T} - 1) - J_{ph} .

(1)

The dark reverse-bias asymptote is $- J_{0}$ before breakdown. Illumination shifts the curve down by

J_{ph}

: the short-circuit point is

(0, - J_{ph})

, and the open-circuit point is

(V_{oc}, 0)

. The power-generating region has positive voltage and negative diode-convention current. A plot using positive delivered current would flip the vertical sign; its physical intercepts are the same.

Figure 1.
Ideal dark and illuminated silicon-cell current–voltage characteristics
. Original calculation using the supplied photon flux and saturation current at an explicitly assumed 298 K. A separate reverse-bias panel resolves the tiny dark saturation current; there is no breakdown term in this model. **Details**.

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