Calculate the thickness of a semiconductor that will absorb 90 percent of the incident photon energy.
Consider silicon and assume that in the first case the incident wavelength is \lambda=1.0 \mu \mathrm{m} and in the second case, the incident wavelength is \lambda=0.5 \mu \mathrm{m}.
From Figure 14.4, the absorption coefficient is \alpha \approx 10^{2} \mathrm{~cm}^{-1} for \lambda=1.0 \mu \mathrm{m}. If 90 percent of the incident flux is to be absorbed in a distance d, then the flux emerging at x=d will be 10 percent of the incident flux. We can write
\frac{I_{v}(d)}{I_{v 0}}=0.1=e^{-\alpha d}
Solving for the distance d, we have
d=\frac{1}{\alpha} \ln \left(\frac{1}{0.1}\right)=\frac{1}{10^{2}} \ln (10)=0.0230 \mathrm{~cm}
In the second case, the absorption coefficient is \alpha \approx 10^{4} \mathrm{~cm}^{-1} for \lambda=0.5 \mu \mathrm{m}. The distance d, then, in which 90 percent of the incident flux is absorbed, is
d=\frac{1}{10^{4}} \ln \left(\frac{1}{0.1}\right)=2.30 \times 10^{-4} \mathrm{~cm}=2.30 \mu \mathrm{m}
Comment
As the incident photon energy increases, the absorption coefficient increases rapidly, so that the photon energy can be totally absorbed in a very narrow region at the surface of the semiconductor.
