Week 4: Fundamentals of Semiconductor Lasers

Question 1

What differences in operating regimes do semiconductor lasers have when compared to other semiconductor devices?

Answer 1

Semiconductor lasers operate with really high levels of carriers so they’re referred to as degenerate. They also operate at non-equilibrium.

Question 2

What are the similarities and differences between a semiconductor laser and other types of lasers?

Answer 2

Semiconductor lasers and other lasers are similar in that they can be highly monochromatic, coherent, and directional.

Semiconductor lasers are unique in that they transition between bands, unlike other lasers which have discrete atomic energy levels. Semiconductor lasers have high gain coefficients and short cavities. They are efficient to modulate and can be modulated very quickly.

Question 3

In general terms, how is the laser action produced?

Answer 3

The laser action is produced by using an electrical pump to drive a current so that holes and electrons are directly injected into the active region.

Question 4

What are the three competing processes in any laser?

Answer 4

Spontaneous emission: an electron from the conduction band transitions to the valence band, emitting a photon.

Stimulated emission: this occurs when a preexisting photon stimulates an electron transition from the conduction band to the valence band, causing another photon with the same energy, direction, phase, and wavelength to be emitted along with the original photon.

Absorption (stimulated absorption): a preexisting photon stimulates an electron to transition from the valence band to the conduction band, and the photon is lost (absorbed) in this transition.

Question 5

What four things determine the radiative transition rates?

Answer 5

1. Occupancy of electronic levels in the conduction and valance bands
2. Density of incident photons
3. Density of possible optical transitions
4. Transition probability of an electron-hole pair defined by probability coefficients A_21, B_12, B_21

Question 6

What do the integrals for the electron density and hole density as a function of temperature need to be evaluated numerically for semiconductor lasers?

Answer 6

When we approximate the integrals for hole and electron density previously, we were doing so under the non-generate assumption where there are a low number of carriers and the operating mode is in equilibrium. This assumption allowed us to approximate the Fermi-Dirac statistics with classical statistics.

However, in the case of semiconductor lasers, the carrier density is very high so the semiconductor laser is considered to be degenerate. It also operates in non-equilibrium levels. This means we can’t use this statistical approximation anymore and we have to evaluate the integrals numerically.

Question 7

How does the integral for the density of electrons as a function of temperature change for semiconductor lasers?

Answer 7

There’s a new term corresponding to the quasi fermi levels for electrons semiconductor lasers. We now have:

n(T) = integral[E_c, inf] { dE g_c(E) f_c(E) } where
f_c(E) = 1 / (1 + exp((E - F_c)/k_b*T)

F_c: quasi fermi level for electrons

Question 8

How does the integral for the density of holes as a function of temperature change for semiconductor lasers?

Answer 8

There’s a new term corresponding to the quasi fermi levels holes in semiconductor lasers. We now have:

p(T) = integral[-inf, E_v] { dE g_v(E) (1 - f_v(E) } where
f_v(E) = 1 / (1 + exp((E - F_v)/k_b*T)

F_v: quasi fermi level for holes

Question 9

What do we know about the relationship between F_c and F_v when current is flowing?

Answer 9

When current is flowing, we know that F_c does not equal F_v!

Question 10

Where does relative electron density peak on a graph of energy levels versus relative density? What about relative hole density?

Answer 10

The relative electron density peaks close to the bottom of the conduction band. The relative hole density peaks near the top of the valence band.

Question 11

On a graph of energy level versus the relative density, how does the curvature of the valence band compare to the curvature of the conduction band? What does this mean for the density of state in the conduction band and valence band?

Answer 11

The curvature of the conduction band is much greater than the curvature of the valence band.

This means that the effective mass of the holes is several orders of magnitude larger than the effective mass of the electrons. This large difference in effective mass means that the density of states in valence band for most semiconductors will be much bigger than the destiny of states in the conduction band.

Question 12

What happens to the wave functions from nearby states as a result of the highly doped semiconductors used in semiconductor lasers?

Answer 12

Because semiconductor lasers are highly doped and operate in the degenerate regime, the wave functions from nearby states overlap which creates quasi-extended states. The impurity levels broaden and get closer to the band edge, meaning that there are band tails.

Question 13

What are band tail states?

Answer 13

A narrowing of the band gap modifies the density of states with the net result being that it increases the number of available states at energies lower than the band states. The spectrum shifts and broadens and because of the larger density of states, a higher injection is required for quasi-fermi level position.

Question 14

What determines the density of photons in equilibrium?

Answer 14

The density of photons in equilibrium is determined by Planck’s law which gives us the equilibrium distribution of frequency per radiation interval.

u(v)dv = ((8pihv^3n^3)/c^3) * (dv / (exp(hv/k_b*T) - 1) in units of [W/m^2]

Question 15

Explain the density of allowed optical transitions and how that impacts the radiative transition rate.

Answer 15

The density of allowed optical transitions is the number of optical transitions per volume per energy. This is limited by the k selection rules. Each level in the conduction band can only recombine with one level in the valence band because you have to have the same k and the same spin.

Question 16

How do you express the density of optical transitions in terms of the density of states in the valence and conduction band?

Answer 16

You can express the density of optical transitions in terms of the density of states in the valence and conduction bands by:

rho_opt = (1/2) * [ (1/g_c + 1/g_v ]^-1

Question 17

What do each of the Einstein coefficients that impact the radiative transition rate correspond to?

Answer 17

B_12: absorption coefficient with units [V/time-energy]

B_21: stimulated emission coefficient with units of [V/time-energy]

A_21: spontaneous emission coefficient with units [1/time]

Question 18

How would you calculate the rate of absorption?

Answer 18

The rate of absorption is:

r_12 = B_12 * f_v * (1-f_c) * rho_opt(E) * rho_photon(E)

f_v: probability of finding an electron in the valence band

1-f_c: probability of finding a hole in the conduction band

rho_opt(E): probability of optical transition as a function of energy

rho_photon(E): density of photons

Question 19

How would you calculate the rate of stimulated emission?

Answer 19

The rate of stimulated emission is:

r_21 = B_21 * f_c * (1-f_v) * rho_opt(E) * rho_photon(E)

f_c: probability of finding an electron in the conduction band

1-f_v: probability of finding a hole in the valence band

rho_opt(E): probability of optical transition

rho_photon(E): density of photons

Question 20

How would you calculate the rate of spontaneous emission?

Answer 20

The rate of stimulated emission is:

r_21_sp = A_21 * f_c * (1-f_v) * rho_opt(E) * rho_photon(E)

f_c: probability of finding an electron in the conduction band

1-f_v: probability of finding a hole in the valence band

rho_opt(E): probability of optical transition

Question 21

What do we know about the radiative transition rates and quasi-fermi levels when in thermal equilibrium?

Answer 21

When in thermal equilibrium, we know that the quasi-fermi levels, F_c and F_v, are equal. We also know that the upward rate of transition has to be equal to the downward rate of transition. This means that the rate of absorption is equal to the rate of spontaneous emission plus the rate of simulated emission.

F_c = F_v

and

r_21 + r_21_sp = r_12

Question 22

Explain how the Einstein coefficients are related. If you have one of these coefficients, how would you solve for the others?

Answer 22

The Einstein coefficients are related through the relationship that can be derived from the quasi fermi levels and transition rates in thermal equilibrium.

We start by combining the transition rate equation with the quasi fermi equations at thermal equilibrium. This gives us:

rho(E) = A_21 / ((B_12 * exp[(E_c - E_v)/ k_b*T]) - B)

E_c: energy that’s in the conduction band
E_v: energy that’s in the valence band

We know that the energy density per unit frequency at equilibrium is Planck’s law. So we set the expression for rho(E) equal to Plank’s law and solve to get:

B_12 = B_21
A_21 = Z(E) * B_21

where Z(E) = ((8 * pi * n^3) / (h^3 * c^3)) * E^2

Question 23

How would you calculate the net stimulated emission rate?

Answer 23

r_st_net = r_21 - r_12 = B_21 * (f_c - f_v) * rho_c^-1 + rho_v^-1 *rho(E)

Remember that if we want net stimulated emission, this number must be positive. This means that f_c must be greater than f_v.

Question 24

How could you obtain E_v if you knew E_c and the photon energy?

Answer 24

We can use conservation of energy here so we know that:

E_v = E_c - hv

Question 25

What can you determine about the band gap energy from the quasi fermi levels and the photon energy when you want net stimulated emission?

Answer 25

We can use conservation of energy to obtain E_v = E_c - hv, and using the Boltzmann approximate with classical statistics, we can get a relationship to describe what we need for net stimulated emission.

exp[ E_c - hv - F_v / k_b*T ] < exp[ E_c - F_c / k_b * T ]

So we can deduce that for net stimulated emission:

F_c - F_v > hv
F_c - F_v= E_g

which tells us that for net stimulated emission, you need the separation of the quasi-fermi levels of the electrons and the holes to be greater than the band gap energy. Another way to think about this is that the difference between the quasi-fermi levels has to be greater than the energy of the photons that are exciting the net stimulated emission.

Question 26

How would you get the minority carrier concentration for net stimulated emission with holes as the minority carrier?

Answer 26

We can use an integral that needs to be numerically evaluated.

p = integral[-inf, E_v] { (1/2pi^2) (2m_v/h_bar^2) ((E_v-E)^1/2 / (1 + exp(F_v - E / k_b * T)) dE }

Question 27

How would you calculated the recombination rate and what does it tell you?

Answer 27

Recombination rate:

R = delta(P) / tau_p

delta(P): desired level of minority carriers needed to generate the stimulated emission

tau_p: minority carrier lifetime = 1 / B*n_n0

This tells you the rate at which you are losing minority carriers.

Question 28

How can we determine the number of minority carriers that are being generated by the light?

Answer 28

We start with defining the generation rate, G.

G = alpha * S

alpha: absorption coefficient
S: incident photon flux

We can set the recombination rate equal to the generation rate to get the number of carriers we need for net stimulated emission. So:

S = delta(P) / tau_p * alpha

Question 29

How do we obtain the intensity [W/m^2] from the generation rate and the recombination rate?

Answer 29

To get net stimulated emission, we can set the recombination rate R equal to the generation rate, which helps us solve for the incident photon flux, S.

Using this flux, we can obtain the intensity from the relationship:

I = S * hv where hv is the energy associated with a photon at a particular frequency.

Question 30

Gain in semiconductor lasers…

Answer 30

WTF is the video showing this… need more info!

Question 31

Practice Q1

Explain what quasi Fermi levels are and how they are different from Fermi level.

Answer 31

The Fermi level describes the energy level in equilibrium, where the probability of finding an electron or hole is 50%. The quasi Fermi levels have the same definition, but for the case of non-equilibrium.

Question 32

Practice Q2

Write down net stimulated emission condition for illumination with photons with energy equal to bandgap energy. You answer can be in terms F_c (quasi-Fermi level for conduction band, electrons), F_v (quasi-Fermi level for valence band, holes) and E_g (bandgap energy). Write an expression that equals zero.

Answer 32

F_c​−F_v​−E_g​=0

Question 33

Practice Q3

Write down an expression for the probability finding an electron in the conduction band [fv(E)f_v(E)fv​(E)] under non-equilibrium conditions. Your answer can be in terms of E (energy), F_v (quasi Fermi level for holes), k_b (Boltzmann constant) and T (temperature).

Note: Make sure you answer writes the exponential term as ‘exp’ and not ‘e’ or ‘E’.

Answer 33

1/(1 + exp((E-F_v )/(k_b * T)))