= Solution
Absorbed <photons> with <energy> at least the <band gap> generate <electron>–hole pairs. The built-in field of the <p-n junction> separates carriers generated in or reaching the <depletion region>; <minority carriers> generated nearby can diffuse into that collection region. Selective contacts carry <electrons> and holes into an external circuit before they recombine.
Define positive <electric current> into the positive-voltage terminal. In the ideal illuminated-diode model,
$$
I(V)=I_0\left[\exp\!\left(\frac{qV}{n k_BT}\right)-1\right]-I_{\rm ph}.
$$
The dark curve has $I_{\rm ph}=0$; illumination shifts it downward. At short circuit $I\simeq-I_{\rm ph}$; at open circuit $I=0$ and $V=V_{\rm oc}$. Between them the illuminated cell lies in \b[quadrant IV] and supplies <power> $P_{\rm out}=-VI>0$. The <maximum power point> maximizes the area of the rectangle with sides $V$ and $-I$:
$$
\boxed{\frac{d(-VI)}{dV}=0\quad\Longleftrightarrow\quad I+V\frac{dI}{dV}=0}.
$$
It is neither the open-circuit point nor the short-circuit point, both of which supply zero <power>.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2021/paper-1-solar-iv.png]
{title=Dark and illuminated diode curves and the photovoltaic maximum power point}
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