Week 3: Light Emitting Diode

Question 1

What is a double-heterostructure design and why would you use it?

Answer 1

A double-heteostructure design is when two different materials are used and you do so to try to confine the electrons and holes so that the probability of radiative recombination is higher which increases the efficiency of the devices.

Question 2

Describe the three stages of the formation of p-n junction in which electrons and holes recombine and emit light.

Answer 2

In the first stage, two separate p and n type semiconductors exist with different fermi levels. When they’re brought together, the fermi levels do not form a straight line. The third stage is at thermal equilibrium.

Question 3

What happens to the pn-junction under thermal equilibrium and how does that enable us to produce current?

Answer 3

Under thermal equilibrium, the energy bands bend to allow the fermi levels of the p-type and n-type materials form a straight line.

The first current producing mechanism is fusion, the motion of carriers in response to a concentration gradient generated by the bending and fusion of the materials.

The second is drift, motion of carriers in response to an applied electric field. The band bending generates an electric field that produces a force on the carriers. In the vicinity of the junction, all the free holes go over to the electron side and all the electrons go over to the hole side, creating a separation of charge.

Question 4

How do you calculate the built in potential for the pn-junction?

Answer 4

V_bi = (k_bT/q) * ln((n_n0p_p0)/n_i^2)

Question 5

What is the equation for current density in a pn junction?

Answer 5

The current density, J, is defined:

J = J_p + J_n = J_sexp(qV/k_bT)

where J_s is the saturation current.

J_s = (qD_pp_n0/L_p) + (qD_nn_p0/L_n)

Question 6

What happens to the potential of the junction with reverse bias? How does minority carrier concentration compare to thermal equilibrium concentration with reverse bias?

Answer 6

With reverse bias you increase the potential for the junction. You’ve increased the electric field which imbalances the drift and diffusion currents at thermal equilibrium. So under reverse bias, you make the electric field stronger until you get current from drift.

The minority carrier concentrations dip below the thermal equilibrium concentrations.

Question 7

What happens to the potential of the junction with forward bias? How does minority carrier concentration compare to thermal equilibrium concentration with forward bias?

Answer 7

With forward bias, the built in potential is reduced and when it’s reduced, it reduces the barrier for the electrons in the holes to diffuse. This means you get exponentially increasing current from diffusion.

With forward bias the minority carrier concentrations exceed their thermal equilibrium concentrations in the vicinity of the junction since most of the current is dominated by diffusion from the concentration gradient.

Question 8

Under what condition will a diode emit light?

Answer 8

We can use the case of the ideal diode where minority carriers are recombining with majority carriers. There’s an exponentially decaying carrier concentration and if the recombination is dominantly radiative, then the diode will emit light.

Question 9

What is the internal quantum efficiency and how do you calculate it?

Answer 9

The internal quantum efficiency is the percentage of the recombining carriers that will emit light. It can be calculated:

Internal quantum efficiency = # of photons emitted/s / # of carriers injected/s

The internal quantum efficiency is also equal to the radiative efficiency of the device so you can calculate this by:

1 / (1 + tau_R/tau_NR)

Question 10

What three mechanisms prevent light from escaping an LED?

Answer 10

Reabsorption by semiconductor, particularly by the substrate material.

Reflection at the semiconductor surface

Total internal reflection

Question 11

What causes an LED with a Gallium Arsenide substrate to lose 85% of the light it generates?

Answer 11

Gallium Arsenide substrate ends up absorbing 85 percent of the light generated. Reabsorption is the problem here.

Question 12

How would you reduce the losses from reabsorption by the substrate material? Which Gallium based substrate loses ~25% of the light?

Answer 12

You would have to switch to a material for the substrate with a larger band gap. Gallium Phosphide has a higher band gap than Gallium Arsenide and loses about 25% of the light.

Question 13

What is Fresnel loss?

Answer 13

Fresnel loss is light loss that occurs at the air to semiconductor interface because not all of the light can get out.

Question 14

What is the expression for the power reflected at normal incidence (also called the reflection coefficient sometimes)?

Answer 14

This expression is:

R = ((n_2 - n_1) / (n_2 + n_1))^2

where n_2 is the index of refraction of air and n_1 is the index of refraction of the semiconductor.

Question 15

What assumption can we make for the reflection coefficient when the semiconductor has a large index of refraction? Give an example.

Answer 15

When the semiconductor material has a large index of refraction, the average reflection coefficient from summing over all angles of incidence is approximately equal to the reflection coefficient at normal incidence (perpendicular). Gallium Phosphide has a high index of refraction, n = 3.3.

Question 16

What must you do when calculating reflection coefficients for materials with smaller index of refraction?

Answer 16

You have to use the average R, the average coefficient of reflection taken from summing over all angles of incidence. You cannot use the assumption that the coefficient of reflection for normal incidence is sufficient.

Question 17

What is the critical angle and what does it mean for total internal reflection? What is the critical angle for Gallium Arsenide? Gallium Phosphide?

Answer 17

Total internal reflection can be thought of as light that gets trapped within the semiconductor and cannot cross the air interface.

The critical angle is determined from Snell’s law and it indicates the angle at which all angles above it will have total internal reflection. The critical angle defines a cone in which generated light may be coupled out to reach the viewers.

theta_1 = theta_c = sin^-1(n_2/n_1)
This occurs when theta_2 = 90 degrees.

Gallium Arsenide critical angle = 16 degrees. Gallium phosphide is 17 degrees.

Question 18

How would you use the critical angle to determine the reduction in generated light?

Answer 18

You can find that the total generated light is reduced by:

1/2)(1-cos(theta_critical)

Question 19

What are two alternatives to a planar LED that help combat total internal reflection?

Answer 19

You could use a hemispherical shape LED or a parabolic shape LED instead of a planar LED and these would help reduce the total internal reflection.

Question 20

How do you determine the extraction efficiency?

Answer 20

Extraction efficiency is equal to:

n_extraction = (# of photons emitted into free space / s) / (# of photons emitted from semiconductor / s)

Question 21

How do you define the external quantum efficiency?

Answer 21

The external quantum efficiency is the number of photons that are emitted into free space divided by the number of electrons that are injected.

n_quantumext = (# photons emitted/s) / (# of electrons injected/s)

Question 22

How do calculate the power efficiency?

Answer 22

The power efficiency can be calculate:

power_effic = P / I * V

Question 23

What determines the emission spectra of LEDs? Where are most of the emissions taking place?

Answer 23

The energy band structure determines the emission spectra of LEDs. The electrons and holes are concentrated to the bottom and top of the energy bands so most emissions take place close to the band gap energy.

Question 24

What is the energy of the photon, h*nu, equal to?

Answer 24

The energy of the photon, hnu = E_g + (h_bar^2 * k^2)/(2m_r) where m_r is the reduced mass.

Reduced mass = (1/m_e) + (1/m_h)

Question 25

How do you get the joint density of states, g_cv(E)?

Answer 25

g_cv(E) = (1/2pi^2)((2m_r/h_bar^2)^3/2)sqrt(E - E_g)

Question 26

What is emission intensity proportional to? What assumptions did you make to get this?

Answer 26

We make two assumptions to get the emission intensity relation: low level injection and assume a Boltzmann distribution which allows us to use classical statistics.

Emission intensity is proportional, I(E), to: sqrt(E - E_g)exp(-E / k_bT)

Question 27

How do you get the peak energy? How would you get the full width half max, delta(E)?

Answer 27

The peak energy can be derived by:

E = E_g + (k_b*T/2)

The FWHM or delta(E) is:
delta(E) = 1.8k_bT. We can also write this in terms of wavelength:

delta(lambda) = 1.8k_bTlambda^2 / hc

Question 28

Why would it be useful to use the wavelength version of the FWHM calculation?

Answer 28

You can use that wavelength version to determine the wavelength spread of your LED.

Question 29

How does the carrier temperature compare to the temperature of the lattice?

Answer 29

The carrier temperature is much hotter than the temperature of the lattice due to the high rate of injection in these structures.

Question 30

What does junction temperature represent?

Answer 30

Junction temperature represents the temperature of the semiconductor lattice.

Question 31

What causes heating of the semiconductor lattice? What happens to the band gap as the temperature increases?

Answer 31

The semiconductor lattice temperature, known as the junction temperature, is changed by the large current and dual heating which causes the temperature in the lattice to rise. As the lattice temperature goes up, the band gap energy goes down, leading to a color change.

Question 32

What happens to the peak wavelength as a function of current?

Answer 32

The peak wavelength increases as the current increases. The current increase causes temperature change which increases the peak wavelength.

Question 33

What is the wavelength range that LEDs exist within?

Answer 33

LED wavelengths cover a range from 400 nanometers to 1300 nanometers.

Question 34

What material would you use to get a violet LED? What about for infrared LEDs?

Answer 34

You could use gallium nitride for violet. For infrared, you could use indium gallium arsenide phosphide.

Question 35

What is a lumen?

Answer 35

A lumen is the standard unit for luminous flux. This is the part of the power that the eye can perceive.

Lumen = radiant_power[W] * 683 lumens/Watt * lumen_efficacy_source

683 lumens/Watt is the value at the peak responsivity of the human eye, 555 nanometers.

Question 36

What is the luminous efficacy?

Answer 36

Luminous efficacy is a measure of how well the light source produces visible light.

Question 37

What are some drawbacks of fluorescent and incandescent lamps?

Answer 37

limited efficiency, environmental problem, incompatibility with microfrabrication technology

Question 38

What are the pros of LEDs?

Answer 38

Pros: high efficiency fast response, long device life

Question 39

What color(s) in the visible spectrum does the relative luminosity of LEDs peak at?

Answer 39

between yellow and green

Question 40

Practice Q4:

As the p doping is increased in a pn junction, what do you expect to happen to the depletion region?

Answer 40

Decrease. As the doping increases, the depletion region shrinks. In a metal, the depletion region can be approximated as a delta function.

Question 41

Practice Q5

Write down equation for built in voltage in a pn junction. Your answer can be in terms of

NA (acceptor doping on p side)
ND (donor doping on n side)
ni (intrinsic carrier concentration)
V_T in Volts, representing VT=kbT/q (thermal voltage which is product of kb (Boltzmann constant), T (temperature) and q charge.

Answer 41

V_Tln(NAND/ni^2)

Question 42

Practice Q3

You are given a pn junction. Explain what the depletion region is.

Answer 42

A region around the immediate vicinity of the junction with no free carriers (electrons or holes), only positively and negatively charged ions.

Question 43

Practice Q2

Explain what happens at the critical angle.

Answer 43

The wave that would exit the material is evanescent (there is no power in that direction) and all light stays in the material.

If you calculate the angle of refraction from Snell’s law, you will notice that you are at 90 degrees for the wave that should exit the material. This is the point at which transmitted wave goes evanescent.

Snell’s law is below:

n1sin(θ1)=n2sin(θ2) where n1 and θ1 represent the medium that the light comes from, and n2​ and θ2 represent the medium that the light is going to.

Question 44

Practice Q1

Explain why total internal reflection occurs.

Answer 44

The wave that would exit the material is evanescent (there is no power in that direction) and all light stays in the material.