# Optical Gain in Semiconductors

This is a continuation from the previous tutorial - band-to-band optical transitions in semiconductors.

By following a line of reasoning similar to that used in the optical absorption and amplification tutorial while associating $$R_\text{a}(\nu)$$ with $$N_1W_{12}(\nu)$$ and $$R_\text{e}(\nu)$$ with $$N_2W_{21}(\nu)$$, we can write down the absorption and gain coefficients contributed by direct band-to-band transitions in a semiconductor as

$\tag{13-30}\alpha(\nu)=\frac{h\nu}{I(\nu)}[R_\text{a}(\nu)-R_\text{e}(\nu)]=\frac{c^2}{8\pi{n}^2\nu^2\tau_\text{sp}}[f_\text{v}(E_1)-f_\text{c}(E_2)]\rho(\nu)$

and

$\tag{13-31}g(\nu)=\frac{h\nu}{I(\nu)}[R_\text{e}(\nu)-R_\text{a}(\nu)]=\frac{c^2}{8\pi{n}^2\nu^2\tau_\text{sp}}[f_\text{c}(E_2)-f_\text{v}(E_1)]\rho(\nu)$

respectively.

By definition, $$g(\nu)=-\alpha(\nu)$$.

The relations in (13-30) and (13-31) are valid for carriers in either an equilibrium state or a quasi-equilibrium state because their validity follows from that of the relations in (13-26) and (13-27) [refer to the band-to-band optical transitions in semiconductors tutorial].

If the carriers in the conduction and valence bands are in thermal equilibrium, both $$f_\text{c}$$ and $$f_\text{v}$$ are characterized by the same Fermi level $$E_\text{F}$$. For an intrinsic semiconductor with no impurity doping, this Fermi level is located very close to the middle of the bandgap.

For band-to-band absorption to occur, the photon energy must be $$h\nu=E_2-E_1\ge{E}_\text{g}$$, implying that $$E_2-E_\text{F}\ge{E}_\text{g}/2$$ and $$E_\text{F}-E_1\ge{E}_\text{g}/2$$.

At $$T=300\text{ K}$$, we have $$k_\text{B}T=25.9\text{ meV}$$, which is at least one order of magnitude smaller than the bandgaps of most semiconductors, except those that have very small bandgaps such as some alloys containing HgSe or HgTe.

For an intrinsic semiconductor in thermal equilibrium, $$f(E_1)\approx1$$ and $$f(E_2)\approx0$$ if the bandgap of the semiconductor is significantly larger than $$k_\text{B}T$$. Then, according to (13-30), its intrinsic absorption spectrum is given by

$\tag{13-32}\alpha_0(\nu)=\frac{c^2}{8\pi{n}^2\nu^2\tau_\text{sp}}\rho(\nu)$

Consequently, the gain and absorption spectra of a semiconductor in an equilibrium or a quasi-equilibrium state at a given temperature $$T$$ can be expressed in terms of its intrinsic absorption spectrum at the same temperature as

$\tag{13-33}g(\nu)=-\alpha(\nu)=\alpha_0(\nu)[f_\text{c}(E_2)-f_\text{v}(E_1)]$

This relation is valid for gain and absorption contributed by direct band-to-band transitions under the condition that $$h\nu\ge{E}_\text{g}\gg{k}_\text{B}T$$.

Though it is obtained for an intrinsic semiconductor, it can be generalized to an extrinsic semiconductor doped with impurities by taking $$\alpha_0(\nu)$$ to be the absorption coefficient measured when the semiconductor is in thermal equilibrium with its environment at a given temperature $$T$$.

Because both $$f_\text{c}(E_2)$$ and $$f_\text{v}(E_1)$$ have a minimum value of 0 and a maximum value of 1, this relation implies that $$-\alpha_0(\nu)\le{g}(\nu)\le\alpha_0(\nu)$$ and $$-\alpha_0(\nu)\le\alpha(\nu)\le\alpha_0(\nu)$$ at any frequency in any condition, which can be clearly seen in Figure 13-7(a) discussed below.

Population Inversion

The concept of population inversion in a semiconductor cannot be simply defined as that the conduction band is more populated than the valence band because for a positive optical gain coefficient it is neither necessary nor possible to have more electrons in the conduction band than in the valence band.

The population of electrons in an energy band is subject to the requirement of Fermi distribution and the availability of energy states in the band structure. Therefore, a practical definition of population inversion in a semiconductor is when the electron and hole concentrations in the semiconductor lead to a positive optical gain coefficient.

Because $$\alpha_0(\nu)\ge0$$ for any frequency $$\nu$$, the sign of $$g(\nu)$$ given in (13-33) is determined by that of the quantity $$f_\text{c}(E_2)-f_\text{v}(E_1)$$. It can be shown that

$\tag{13-34}f_\text{c}(E_2)-f_\text{v}(E_1)=f_\text{c}(E_2)[1-f_\text{v}(E_1)]\left[1-\text{e}^{(h\nu-\Delta{E}_\text{F})/k_\text{B}T}\right]$

where $$\Delta{E}_\text{F}=E_\text{Fc}-E_\text{Fv}$$.

Because $$1\ge{f}_\text{c}(E_2)(1-f_\text{v}(E_1))\ge0$$, the sign of the quantity $$f_\text{c}(E_2)-f_\text{v}(E_1)$$ is solely determined by the sign of the quantity $$h\nu-\Delta{E}_\text{F}$$.

Therefore, the condition for a positive gain coefficient that $$g(\nu)\gt0$$ at any given optical frequency $$\nu$$ is that the separation between the quasi-Fermi levels be larger than the photon energy at the frequency $$\nu$$:

$\tag{13-35}\Delta{E}_\text{F}=E_\text{Fc}-E_\text{Fv}\gt{h}\nu\gt{E}_\text{g}$

This condition dictates the distributions of electrons and holes for a positive optical gain coefficient. It can thus be considered as the condition for population inversion in a semiconductor.

As mentioned in the electron and hole concentrations tutorial, the quasi-Fermi levels completely quantify the electron and hole concentrations in a semiconductor that is maintained in a quasi-equilibrium state by electrical or optical pumping.

Furthermore, (12-43) and (12-46) [refer to the electron and hole concentrations tutorial] indicate that the product $$np$$ strongly depends on the value of $$\Delta{E}_\text{F}$$. Therefore, the condition given in (13-35) determines the electron and hole concentrations needed for a positive gain coefficient.

When a semiconductor is pumped to have a positive gain coefficient, the pumped electron and hole concentrations are normally a few orders of magnitude higher than the intrinsic electron and hole concentrations so that $$n\gg{n}_0$$ and $$p\gg{p}_0$$.

In this situation, we consider the electron and hole concentrations in a semiconductor gain medium to be practically the same as the excess carrier density $$N$$ defined in (12-55) [refer to the carrier recombination tutorial] so that

$\tag{13-36}n\approx{p}\approx{N}$

The minimum carrier density required for a gain is known as the transparency carrier density, $$N_\text{tr}$$. According to (13-35), it is determined by the condition that $$\Delta{E}_\text{F}=E_\text{g}$$, which implies that $$E_\text{Fc}-E_\text{c}=E_\text{Fv}-E_\text{v}$$.

By using (12-41), (12-42) [refer to the electron and hole concentrations tutorial], and (13-36), we find that the transparency carrier density is given by

$\tag{13-37}N_\text{tr}=N_\text{c}(T)F_{1/2}(\xi_\text{tr})=N_\text{v}(T)F_{1/2}(-\xi_\text{tr})$

where $$\xi_\text{tr}=(E_\text{Fc}-E_\text{c})/k_\text{B}T=(E_\text{Fv}-E_\text{v})/k_\text{B}T$$.

With known values of $$N_\text{c}$$ and $$N_\text{v}$$ at a given temperature, the value of $$N_\text{tr}$$ can be found by solving this relation to find the parameter $$\xi_\text{tr}$$.

In terms of the carrier density, the condition for population inversion in a semiconductor can be regarded as $$N\gt{N}_\text{tr}$$.

Example 13-3

Find the transparency carrier density for GaAs at $$300\text{ K}$$. When GaAs is injected with this concentration of electron-hole pairs, where are its quasi-Fermi levels located?

To find $$N_\text{tr}$$, we have to find the value of $$\xi_\text{tr}$$ first by solving the second relation in (13-37).

From (12-24) [refer to the electron and hole concentrations tutorial], $$N_\text{c}\propto(m_\text{e}^*)^{3/2}$$ and $$N_\text{v}\propto(m_\text{h}^*)^{3/2}$$. Therefore, we find the following relation from (13-37):

$\tag{13-38}\frac{F_{1/2}(\xi_\text{tr})}{F_{1/2}(-\xi_\text{tr})}=\left(\frac{m_\text{h}^*}{m_\text{e}^*}\right)^{3/2}$

For GaAs, $$m_\text{e}^*=0.067m_0$$ and $$m_\text{h}^*=0.52m_0$$ [refer to Table 12-2 in the current density in semiconductors tutorial]. Then

$\frac{F_{1/2}(\xi_\text{tr})}{F_{1/2}(-\xi_\text{tr})}=\left(\frac{0.52}{0.067}\right)^{3/2}=21.62$

The solution of this relation is found from the value of $$F_{1/2}(\xi)$$ as a function of $$\xi$$ plotted in Figure 12-3 [refer to the electron and hole concentrations in semiconductors tutorial] to be $$\xi_\text{tr}=1.99$$ for $$F_{1/2}(\xi_\text{tr})=F_{1/2}(1.99)\approx2.81$$ and $$F_{1/2}(-\xi_\text{tr})=F_{1/2}(-1.99)\approx0.13$$.

From Example 12-2 [refer to the electron and hole concentrations in semiconductors tutorial], we have $$N_\text{c}=4.35\times10^{23}\text{ m}^{-3}$$ and $$N_\text{v}=9.41\times10^{24}\text{ m}^{-3}$$ for GaAs at $$300\text{ K}$$.

We then find from (13-37) that the transparency carrier density for GaAs at $$300\text{ K}$$ is

$N_\text{tr}=2.81\times4.35\times10^{23}\text{ m}^{-3}=0.13\times9.41\times10^{24}\text{ m}^{-3}=1.22\times10^{24}\text{ m}^{-3}$

Because $$\xi_\text{tr}=1.99$$ at this injected carrier concentration, the quasi-Fermi levels are located at

$E_\text{Fc}=E_\text{c}+1.99k_\text{B}T=E_\text{c}+51.6\text{ meV}$

$E_\text{Fv}=E_\text{v}+1.99k_\text{B}T=E_\text{v}+51.6\text{ meV}$

Though $$E_\text{Fc}-E_\text{Fv}=E_\text{g}$$ for a semiconductor at transparency with $$N=N_\text{tr}$$, we find that $$E_\text{Fc}$$ and $$E_\text{Fv}$$ do not respectively lie at the conduction-band and valence-band edges due to the fact that $$m_\text{e}^*\ne{m}_\text{h}^*$$.

Because $$m_\text{h}^*\gt{m}_\text{e}^*$$ for GaAs, $$E_\text{Fc}$$ lies above the conduction-band edge and $$E_\text{Fv}$$ lies above the valence-band edge while they are subject to the condition $$E_\text{Fc}-E_\text{Fv}=E_\text{g}$$, as shown in Figure 13-6 below.

Carrier Dependence of Gain

Both the spectrum and the magnitude of the optical gain coefficient in a semiconductor are a function of the excess carrier density $$N$$.

To find the carrier dependence of the gain spectrum, one starts with a given value of $$N\approx{n}\approx{p}$$ to find the corresponding values of the quasi-Fermi levels $$E_\text{Fc}$$ and $$E_\text{Fv}$$ from (12-41) and (12-42) [refer to the electron and hole concentrations in semiconductors tutorial] through the following relaton:

$\tag{13-39}N=N_\text{c}(T)F_{1/2}(\xi_\text{c})=N_\text{v}(T)F_{1/2}(\xi_\text{v})$

where $$\xi_\text{c}=(E_\text{Fc}-E_\text{c})/k_\text{B}T$$ and $$\xi_\text{v}=(E_\text{v}-E_\text{Fv})/k_\text{B}T$$.

Note that $$\xi_\text{c}=-\xi_\text{v}=\xi_\text{tr}$$ in the case when $$N=N_\text{tr}$$, as seen in (13-37).

In general, however, $$\xi_\text{c}\ne-\xi_\text{v}$$ when $$N\ne{N}_\text{tr}$$. Therefore, the values of $$\xi_\text{c}$$ and $$\xi_\text{v}$$ have to be found separately from (13-39) in order to find $$E_\text{Fc}$$ and $$E_\text{Fv}$$.

According to (13-35), we find that $$E_\text{Fc}-E_\text{Fv}\gt{E}_\text{g}$$ when $$N\gt{N}_\text{tr}$$, but $$E_\text{Fc}-E_\text{Fv}\lt{E}_\text{g}$$ when $$N\lt{N}_\text{tr}$$.

Then, $$g(\nu)$$ as a function of the optical frequency $$\nu$$ can be found using (13-31) or (13-33), and $$\alpha(\nu)$$ can be similarly calculated.

Figure 13-7(a) shows the gain and absorption spectra of GaAs as a function of photon energy at various levels of excess carrier density $$N$$.

It can be seen that an optical gain associated with direct band-to-band transitions occurs in a range of photon energies larger than the bandgap only when $$N\gt{N}_\text{tr}$$.

Both the spectral range and the peak value of the optical gain coefficient increase with the carrier density.

A semiconductor gain medium has a very broad gain bandwidth. In typical operating conditions of semiconductor lasers and amplifiers, the gain bandwidth is in the range of $$2k_\text{B}T$$ to $$4k_\text{B}T$$ in photon energy.

At room temperature, this gain bandwidth is 50-100 meV in photon energy, corresponding to a frequency range of 12-24 THz.

In most cases, the peak gain coefficient increases almost linearly with carrier density, as shown in Figure 13-7(b). Therefore, we can express the peak value of the optical gain coefficient approximately as

$\tag{13-40}g_\text{max}(N)=\sigma(N-N_\text{tr})$

where the coefficient $$\sigma$$ is an equivalent gain cross section similar to the transition cross section described in the optical transitions for laser amplifiers tutorial.

The value of this gain coefficient depends on the composition of the semiconductor material and the operating temperature. It is typically in the range of $$1$$ to $$5\times10^{-20}\text{ m}^2$$ for the common semiconductor laser materials of GaAs and InGaAsP at room temperature.

The value of $$N_\text{tr}$$ is on the order of $$10^{24}\text{ m}^{-3}$$ for such semiconductors.

Note that, as can be seen in Figure 13-7(a), the peak of the gain spectrum does not occur at a fixed frequency but moves to a higher frequency as the carrier density is increased. Therefore, $$g_\text{max}$$ as expressed in (13-40) represents the gain coefficient at different optical frequencies as the value of $$N$$ varies.

The linear relation in (13-40) is normally a very good approximation, but it does not hold strictly. In particular, as the carrier density increases to a certain level, the peak gain coefficient tends to increase less than linearly with carrier density.

Example 13-4

A GaAs sample at $$300\text{ K}$$ is injected with excess electron-hole pairs of a concentration $$N=2.83\times10^{24}\text{ m}^{-3}$$. Take $$\tau_\text{sp}=500\text{ ps}$$ for GaAs at $$300\text{ K}$$.

(a) Find the sample's quasi-Fermi levels.

(b) Find its optical gain coefficient at 850 nm wavelength. The refractive index of GaAs at 850 nm is $$n=3.65$$.

(c) Use this gain coefficient to calculate an equivalent gain cross section. At this injection level, the gain peak occurs very close to, though not exactly at, 850 nm.

(a)

By using $$N_\text{c}=4.35\times10^{23}\text{ m}^{-3}$$ and $$N_\text{v}=9.41\times10^{24}\text{ m}^{-3}$$ found in Example 12-2 [refer to the electron and hole concentrations in semiconductors tutorial] for GaAs at $$300\text{ K}$$, we find that (13-39) leads to

$F_{1/2}(\xi_\text{c})=\frac{N}{N_\text{c}}=6.51\qquad\text{and}\qquad{F}_\text{1/2}(\xi_\text{v})=\frac{N}{N_\text{v}}=0.30$

for $$N=2.83\times10^{24}\text{ m}^{-3}$$.

From the value of $$F_{1/2}(\xi)$$ as a function of $$\xi$$ plotted in Figure 12-3 [refer to the electron and hole concentrations in semiconductors tutorial], we find that $$\xi_\text{c}=4$$ and $$\xi_\text{v}=-1.1$$.

Therefore, $$E_\text{Fc}-E_\text{c}=\xi_\text{c}k_\text{B}T=4\times25.9\text{ meV}=103.6\text{ meV}$$ and $$E_\text{Fv}-E_\text{v}=-\xi_\text{v}k_\text{B}T=1.1\times25.9\text{ meV}=28.5\text{ meV}$$.

As shown in Figure 13-8 below, both quasi-Fermi levels lie above their respective band edges with $$E_\text{Fc}$$ located at 103.6 meV above $$E_\text{c}$$ and $$E_\text{Fv}$$ located at 28.5 meV above $$E_\text{v}$$.

We also find that $$E_\text{Fc}-E_\text{Fv}\gt{E}_\text{g}$$ because $$N\gt{N}_\text{tr}$$.

(b)

According to  the results obtained in Example 13-2 (b) [refer to the band-to-band optical transitions for semiconductors tutorial], the optical transition at $$\lambda=850\text{ nm}$$ takes place between $$E_2=E_\text{c}+31\text{ meV}$$ and $$E_1=E_\text{v}-4\text{ meV}$$.

Using the values of $$E_\text{Fc}$$ and $$E_\text{Fv}$$ found above, we find that $$E_2-E_\text{Fc}=-72.6\text{ meV}$$ and $$E_1-E_\text{Fv}=-32.5\text{ meV}$$. These relations are shown in Figure 13-8.

Using (12-35) and (12-36) [refer to the electron and hole concentrations for semiconductors tutorial], We then find that

$f_\text{c}(E_2)=\frac{1}{\text{e}^{(E_2-E_\text{Fc})/k_\text{B}T}+1}=\frac{1}{\text{e}^{-72.6/25.9}+1}=0.9428$

$f_\text{v}(E_1)=\frac{1}{\text{e}^{(E_1-E_\text{Fv})/k_\text{B}T}+1}=\frac{1}{\text{e}^{-32.5/25.9}+1}=0.7781$

We have found from Example 13-2 [refer to the band-to-band optical transitions in semiconductors tutorial] that $$\rho(\nu)=7.63\times10^{10}\text{ m}^{-3}\text{ Hz}^{-1}$$ for GaAs at 850 nm.

By using (13-31), the optical gain coefficient at 850 nm can then be found:

\begin{align}g(\nu)&=\frac{c^2}{8\pi{n}^2\nu^2\tau_\text{sp}}[f_\text{c}(E_2)-f_\text{v}(E_1)]\rho(\nu)\\&=\frac{\lambda^2}{8\pi{n}^2\tau_\text{sp}}[f_\text{c}(E_2)-f_\text{v}(E-1)]\rho(\nu)\\&=\frac{(850\times10^{-9})^2}{8\pi\times3.65^2\times500\times10^{-12}}\times(0.9428-0.7781)\times7.63\times10^{10}\text{ m}^{-1}\\&=5.42\times10^4\text{ m}^{-1}\end{align}

(c)

By taking the value of $$N_\text{tr}=1.22\times10^{24}\text{ m}^{-3}$$ found in Example 13-3 and using (13-40), we find that

$\sigma=\frac{g}{N-N_\text{tr}}=\frac{5.42\times10^4}{2.83\times10^{24}-1.22\times10^{24}}\text{ m}^2=3.37\times10^{-20}\text{ m}^2$

The next tutorial covers the topic of spontaneous emission in semiconductors