Question 9.17: FUNDAMENTAL ABSORPTION A GaAs infrared LED emits at about 86...
FUNDAMENTAL ABSORPTION A GaAs infrared LED emits at about 860 nm. A Si photodetector is to be used to detect this radiation. What should be the thickness of the Si crystal that absorbs most of this radiation?
According to Figure 9.23, at \lambda \approx 0.8 \mu m , \text { Si has } \alpha \approx 6 \times 10^{4} m ^{-1} , so the absorption depth
\delta=\frac{1}{\alpha}=\frac{1}{6 \times 10^{4} m ^{-1}}=1.7 \times 10^{-5} m \quad \text { or } \quad 17 \mu m
If the crystal thickness is δ, then 63 percent of the radiation will be absorbed. If the thickness is 2δ, then the fraction of absorbed radiation, from Equation 9.71, will be
I(x)=I_{o} \exp (-\alpha x) [9.71]
Fraction of absorbed radiation = 1-\exp [-\alpha(2 \delta)]=0.86 \quad \text { or } \quad 86 \%
