Design and sketch a photoconductor using a $5-\mu \mathrm{m}$ -thick film of $\mathrm{CdS}$, assuming that $\tau_{n}=\tau_{p}=10^{-6} \mathrm{~s}$ and $N_{d}=10^{14} \mathrm{~cm}^{-3}$. The dark resistance (with $g_{\text {op }}=0$ ) should be $10 \mathrm{M} \Omega$, and the device must fit in a square $0.5 \mathrm{~cm}$ on a side; therefore, some sort of folded or zigzag pattern is in order. With an excitation of $g_{\text {op }}=10^{21} \mathrm{EHP} / \mathrm{cm}^{3}-\mathrm{s}$, what is the resistance change?