Week 2: Radiative Recombination in Semiconductors

Question 1

Electronic transitions have two main types. Describe both.

Answer 1

Both types of transitions involve an electron making a transition from an initial state to a final state.

Radiative transitions: in these types of transitions the energy difference from the electron state change is made up by the emission of a photon

Non-radiative: in these types of transitions the extra energy goes somewhere else and not to photon emission. Often this is heat loss.

Question 2

What is spontaneous emission?

Answer 2

Spontaneous emission is when a radiative transition from a higher to a lower energy level occurs without any type of perturbation.

Question 3

What is Auger recombination and is it a desired or non-desired effect?

Answer 3

Auger recombination is when an electron and a hole recombine, and any extra energy actually goes into knocking a secondary electron or hole deeper into the band. This is undesirable behavior.

Question 4

Describe the transitions that can be radiative in a semiconductor.

Answer 4

An electron and a hole recombine right at the band gap transition. An electron that’s higher in the band gap can transition down to the valence band. Or a hole that’s deeper in the valence band can transition.

Question 5

What kinds of impurity related transitions can you have? Are they usually radiative or usually non-radiative? Give an example of a non-radiative transition that is not impurity related.

Answer 5

Impurity related transitions are usually non-radiative. You can have an electron recombining with a hole and an impurity level, for example an acceptor recombination site. You can have a hole recombining with an electron in a donor recombination site (an electron within the band gap somewhere combines with a hole on the valence band). You can have a donor-acceptor pair recombine within the band gap. You can have recombination at a deep level. An example of a non-radiative transition that is not impurity related are intra-subband transitions. This is when an electron transitions within the conduction band or a hole transitions within the valence band.

Question 6

What rule applies to transitions across a band edge? How would you apply it to optical transitions?

Answer 6

The k selection rule which states that:

h_bar * k_initial = h_bar * k_final

which can be expanded out for optical transitions where a photon is released:

h_bar * k_c = h_bar * k_v + h_bar * k_photon

k_c: the momentum of the electron in the conduction band
k_v: the momentum of the electron in the valence band

Question 7

When and how can you estimate the values for the momentum of an electron and a hole? What about estimating the momentum for a photon from the visible light spectrum?

Answer 7

If the electron and hole are near the Brillouin zone boundary, we can approximate their momentum as ~= 2*pi / a where a is the lattice constant.

For visible light, k is approximately 2pi/lambda. Based on how small this number is, we can neglect the photon momentum and on the E vs k diagram, draw straight lines from electron to hole with the photon coming out side ways. We call this a vertical transition.

Question 8

When a transition occurs where the electron and the hole have different k vectors, what makes up the difference?

What is the new momentum conservation equation in this case?

Answer 8

A phonon is referred to as a lattice vibration and it makes up the difference when the hole and electron have different k vectors. This takes the transition from a first order process to a second-order process where you have to include the phonon. A phonon’s momentum is comparable to an electron.

h_bar * k_c = h_bar * k_v + h_bar * k_photon +/- h_bar * k_phonon

Transitions that require a phonon have much smaller transition rates than purely optical transitions.

Question 9

_________ band gap semiconductors are really good at light emission. Explain this type of semiconductor and give two examples.

Answer 9

Direct band gap semiconductors are really good at light transmission. In these types of semiconductors, the minimum of the conduction band lining up directly with the maximum of the valence band, making for vertical transitions of electrons and holes, while emitting a photon.

Two examples are gallium arsenide and indium phosphide.

Question 10

_____________ band gap semiconductors are poor light emitters. Explain this type of semiconductor and give an example.

Answer 10

Indirect band gap semiconductors do not allow for purely vertical transitions because the minimum of the conduction band does not line up with the maximum of the valence band. Silicon is an example.

Question 11

What determines the absorption for direct and indirect band gap semiconductors and how would you calculate the absorption coefficients?

Which one has weaker absorption and how do you know?

Answer 11

Absorption is determined by the threshold energy band gap. If you have a really tiny band gap, you’ll get lots of absorption at all energies while a large band gap means more energy needed to excite state changes. For indirect band gap semiconductors, absorption is usually small, unless the photon energy is larger than the direct gap value

Absorption for direct band gap:
alpha = A_1 (h * nu - E_g) ^ 1/2

Absorption for indirect band gap:
alpha = A_2 (h * nu - E_g +/- E_phonon)^2

A_1 and A_2 are constants.
h * nu: energy of the photon

Question 12

What is the joint density of states and how does it relate to absorption in direct band gap semiconductors?

Answer 12

For direct band gap semiconductors the absorption is proportional to the joint density of states, which means it’s proportional to the the fact that we find an electron in the valence band and a hole in the conduction band.

Question 13

How do you calculate the energy that a photon needs to make the transition in a direct band gap semiconductor?

Answer 13

The energy that the photon needs to make the transition is:

h_barw_c,v = E_g + (h_bar^2 * k^2 / 2)(1/m_e + 1/m_h*)

Question 14

What is the expression for the joint density of states in a direct band gap semiconductor?

Answer 14

Joint density of states in a direct band gap semiconductor:

g(h_bar * w) = { 0, when E_photon < E_g
(1/ 2pi)(m_em_h/(m_e+m_h))(2^(3/2)/h_bar^3)(h_barw - E_g)^(1/2), when E_photon > E_g }

w: lowercase omega

Question 15

What is minority carrier lifetime?

Answer 15

The minority carrier lifetime is the average length of time before radiative recombination with a majority carrier.

Question 16

How does phonon emission and absorption vary with temperature?

Answer 16

When you plot the sqrt(absorpotion coefficient) vs the energy of a photon, h*v, you’ll see that at low temperatures, it’s an almost vertical line corresponding to phonon emission only. At high temperatures, you’ll see a diagonal line with positive slop in the lower sqrt(absorption coefficient) range that becomes a vertical line at a certain point along the y axis. The vertical portion is phonon emission and the positive slope diagonal portion is phonon absorption.

Recall that in the absorption coefficient relationship there is a plus/minus for phonon energy that corresponds to emission or absorption and which one depends on the temperature.

Question 17

What is radiative recombination?

Answer 17

Radiative recombination is a radiative transition of an electron to an empty state.

Question 18

What is the radiative recombination rate in thermal equilibrium? Describe the equations used to derive it.

Answer 18

The radiative recombination rate in thermal equilibrium is:

R_0(v)dv = P(v) * rho(v) * dv

v: greek letter nu
P(v): probability/time of photon absorption
P(v) = alpha(v) * c / n

rho(v): density of photon modes
rho(v) = [(8piv^3hn^3)/c^3][(dv/exp(hv/k_bT) - 1]

Question 19

What determines the radiative recombination rate in lasers and LEDS?

Answer 19

Lasers and LEDs have radiative recombination that occurs in non-thermal-equilibrium conditions. The state is determined by the carrier injection or optical pumping.

Question 20

How do you determine the radiative recombination rate for non-equilibrium conditions?

Answer 20

R = Bnp where n is the electron density, p is the hole density, and B is the probability of radiative recombination known as the Einstein coefficient.

We have the R_0 equation for thermal equilibrium and we can write that R_0 = Bn_0p_0 or B*n_i^2 so we can solve to get an equation for R that doesn’t have B:

R = R_0 * n * p / n_i^2

Question 21

How does the radiative recombination rate change when you have excess carriers? How would you express the total rate?

Answer 21

With excess carriers the recombination rate, R, becomes:

R = R_0 + delta(R) 
where delta(R) is from the excess carriers.

To write the total rate we have:

R_0 + delta(R) = (R_0 * (n_p0 + delta(n))*(p_p0 + delta(p)) / n_i^2

Question 22

What is the ratio delta(R)/R_0 for a p-type material and why? What determines the recombination rate?

Answer 22

Delta(R)/R_0 ~= delta(n)/n_p0 for a p-type material because the number of electrons in the p-type material, n_p0, is significantly less than the number of holes in the p type material, p_p0.

The recombination rate is determined by the minority carrier density.

Question 23

How would you define a minority carrier lifetime?

Answer 23

The minority carrier lifetime is the average length of time before radiative recombination with a majority carrier.

There are several ways to write this. The minority carrier lifetime of the electrons can be determined by:

delta(R) = delta(n) / tao_n

where tao_n = n_p0 / R_0 = n_p0 / B*n_i^2 = 1 / B * P_p0

or

d(n(t)) / dt = -delta(R)

or

delta(n(t)) = delta(n(t=0))*exp(-t/ tau_n)

Question 24

What is the low level injection assumption?

Answer 24

The low level injection assumption means you’re assuming the concentration of excess minority carriers has to be less than that of thermal equilibrium so d(n(t)) and d(p(t)) have to be less than n_0 and p_0.

Question 25

If a material is p-type, what can you assume and how do you use this assumption to simplify the differential equation d(d(n(t))/dt = -alpha_r*d(n(T))[(n_0 + p_0) + d(n(t))]?

Answer 25

If the material is p-type we can assume that there are more holes than electrons, so p_0 > n_0.

Our simplification is:

d(n(t)) = d(n(t=0))* exp(-t/tau_n0)

tau_n0: excess minority carrier lifetime
tau_n0 = 1 / alpha_r * p_0

Question 26

Why do we say that d(n(t)) = d(p(t))?

Answer 26

We can assume this because electrons and holes are generated in pairs.

Question 27

What is the recombination rate just from the excess minority electrons?

Answer 27

This is known as R_n’.

R_n’ = d(n(t)) / tau_n0

Question 28

Problem: If holes are the minority carrier with a lifetime of 10 seconds and we have 10,000 holes, at what rate will they disappear?

Answer 28

The holes will disappear at a rate of 10,000 / 10 seconds so roughly 1000 holes disappearing per second.

Question 29

How would you determine the radiative efficiency? What types of recombination contribute to light output?

Answer 29

To answer this, we have to understand that there are both radiative and non-radiative recombinations. Only radiative recombination contribute to light output. Therefore, if we want the radiative efficiency, we need the ratio of the radiative minority carrier lifetime to the total minority carrier lifetime.

We have that:

(1 / tau_total) = (1/tau_R) + (1/tau_NR)

so the radiative efficiency is just:

tau_R^-1 / (tau_R^-1 + tau_NR^-1)

Question 30

Practice Q1

Enter equation to calculate electron density if you are given the Fermi level. Do not assume non-degeneracy. Your answer can be in terms of the following variables:

g (representing g(E), density of states in conduction band as a function of energy)
Ec (representing Ec, the conduction band energy)
b, (representing ∫Ec∞dE
f (representing f(E), Fermi level as function of energy)

You can assume parabolic bands.

Answer 30

ANS: bfg

Mathematically:
n(E)=g(E)f(E)

n=∫Ec∞g(E)f(E)dEn = \int_{E_c}^\infty g(E) f(E) dEn=∫Ec​∞​g(E)f(E)dE

Question 31

Practice Q2

Write expression for nnn, the electron density, without assuming non-degeneracy in terms of:

A=kbT where kb is Boltzman constant and T is temperature.
a=12π2
b=∫Ec∞dE
c=(2×me(h2π)2)1.5
d=1/(1+exp((E−EF)/kbT)
R=(E−Ec)0.5 where E (energy), EF (Fermi energy), me (effective mass), h (Planck’s constant), kb (Boltzman constant),T (temperature), and Ec (conduction band energy).

Answer 31

bacRd

Question 32

Practice Q3

What is the expression for the probability of electron occupation in the non-degenerate case? Your answer should be in terms of E (Energy), E_F (Fermi energy), k_b (Boltzmann constant) and T (temperature).

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

Answer 32

exp(-((E-E_F)/(k_b*T)))