Question 9.14: FREE CARRIER ABSORPTION COEFFICIENT AND CONDUCTIVITY From Ch...

Consider the conduction losses suffered by a propagating EM wave which experiences an imaginary permittivity given by Equation 9.54. We need the real part of \sigma _{ac}. We can write the ac conductivity as

\varepsilon_{r}^{\prime \prime}=\frac{\operatorname{Re}\left(\sigma_{ ac }\right)}{\varepsilon_{o} \omega}           [9.54]

\sigma_{ ac }=\frac{\sigma_{o}}{1+(\omega \tau)^{2}}-j \frac{\sigma_{o} \tau \omega}{1+(\omega \tau)^{2}}

and then use the real part in Equation 9.54 to find

\varepsilon_{r}^{\prime \prime}=\frac{\sigma_{o}}{\varepsilon_{o} \omega\left[1+(\omega \tau)^{2}\right]}

For ω > 1 ∕ τ the above equation becomes,

\varepsilon_{r}^{\prime \prime}=\frac{\sigma_{o}}{\varepsilon_{o} \omega(\omega \tau)^{2}}            [9.63]            Imaginary relative permittivity and dc conductivity

The relationship between the imaginary part \varepsilon_{r}^{\prime \prime} of the relative permittivity and the extinction coefficient K is given by Equation 9.60

2nK =  \varepsilon_{r}^{\prime \prime}

where n is the refractive index (the real part of N). Since the absorption coefficient from Example 9.13 is

\alpha=2 k^{\prime \prime}=2 k_{o} K=2\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\varepsilon_{r}^{\prime \prime}}{2 n}\right)

we have,

\alpha=\left(\frac{\omega}{c}\right) \frac{\varepsilon_{r}^{\prime \prime}}{n}        [9.64]              Absorption and imaginary relative permittivity

Substituting for \varepsilon_{r}^{\prime \prime} from Equation 9.63 gives,

\alpha=\frac{\sigma_{o}}{c n \varepsilon_{o}(\omega \tau)^{2}}               [9.65]        Absorption and conductivity

This is the well-known (highly simplified) classical free carrier absorption equation. For the n-type Ge sample, the frequency ω is

\omega=\frac{2 \pi c}{\lambda}=\left[\frac{2 \pi\left(3 \times 10^{8}  m  s ^{-1}\right)}{\left(10 \times 10^{-6}  m \right)}\right]=1.88 \times 10^{14}  rad  s ^{-1}

Equation 9.63 gives,

\varepsilon_{r}^{\prime \prime}=\frac{\sigma_{o}}{\varepsilon_{o} \omega(\omega \tau)^{2}}=\frac{\left(1 \times 10^{-3} \Omega  m \right)^{-1}}{\left(8.85 \times 10^{-12}  F  m ^{-1}\right)\left(1.88 \times 10^{14}  rad  s ^{-1}\right)^{3}\left(2.5 \times 10^{-13}  s \right)^{2}}

i.e., \quad  \varepsilon_{r}^{\prime \prime}=2.72 \times 10^{-4}

The absorption coefficient α due to free carriers is given by

i.e., \quad  \alpha=42.6  m ^{-1}
Ge is used as an optical window in various infrared application in the wavelength range approximately 2–16 μm. It is clear that the conductivity of Ge should be as low as possible to reduce free carrier absorption. The mean scattering time τ used above for electrons is inferred from the electron drift mobility equation \mu_{e}=e \tau / m_{e}^{*}  \text { by  taking }  \mu_{e}=3900  cm ^{2}  V ^{-1}  s ^{-1} and m_{e}^{*} \approx 0.12  m_{e} (Table 5.1 in Chapter 5). Notice that ω > 1 ∕ τ.