= Solution
Choose conventional diode current as positive in forward bias. The ideal dark and illuminated current densities are
$$
J_{\rm dark}=J_0\left(e^{qV/k_BT}-1\right),\qquad
J_{\rm light}=J_0\left(e^{qV/k_BT}-1\right)-J_{\rm ph}.
$$
The \b[dark reverse-bias asymptote is $-J_0$] before breakdown. Illumination shifts the curve down by $J_{\rm ph}$: the short-circuit point is $(0,-J_{\rm ph})$, and the open-circuit point is $(V_{\rm oc},0)$. 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.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1-solar-iv.png]
{title=Ideal dark and illuminated silicon-cell current–voltage characteristics}
{description=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.}
Back to article page
