Calculate the generation rate of electron-hole pairs given an incident intensity of photons.
Consider gallium arsenide at T=300 \mathrm{~K}. Assume the photon intensity at a particular point is I_{\nu}(x)=0.05 \mathrm{~W} / \mathrm{cm}^{2} at a wavelength of \lambda=0.75 \mu \mathrm{m}. This intensity is typical of sunlight, for example.
The absorption coefficient for gallium arsenide at this wavelength is \alpha \approx 0.9 \times 10^{4} \mathrm{~cm}^{-1}. The photon energy, using Equation (14.1), is
\lambda=\frac{c}{\nu}=\frac{h c}{E}=\frac{1.24}{E} \mu \mathrm{m} (14.1)
E=h \nu=\frac{1.24}{0.75}=1.65 \mathrm{eV}
Then, from Equation (14.6) and including the conversion factor between joules and \mathrm{eV}, we have, for a unity efficiency factor,
\begin{aligned} g^{\prime}=\frac{\alpha I_{\nu}(x)}{h \nu}=\frac{\left(0.9 \times 10^{4}\right)(0.05)}{\left(1.6 \times 10^{-19}\right)(1.65)}=1.70 \times 10^{21} \mathrm{~cm}^{-3}-\mathrm{s}^{-1}\end{aligned}
If the incident photon intensity is a steady-state intensity, then, from Chapter 6 , the steadystate excess carrier concentration is \delta n=g^{\prime} \tau, where \tau is the excess minority carrier lifetime. If \tau=10^{-7} \mathrm{~s}, for example, then
\delta n=\left(1.70 \times 10^{21}\right)\left(10^{-7}\right)=1.70 \times 10^{14} \mathrm{~cm}^{-3}
Comment
This example gives an indication of the magnitude of the electron-hole generation rate and the magnitude of the excess carrier concentration. Obviously, as the photon intensity decreases with distance in the semiconductor, the generation rate also decreases.