1. EBB 424E Lecture 4– LED 3 Dr Zainovia Lockman
LED Characteristics
EBB 424E
Lecture 4– LED 3
Dr Zainovia Lockman

2. radiative transmission
Last Lectures..
Candidate Materials
Definition of LED
Group III-V
Group II-V
Applications of LED
LED I&2
GaAsP
GaAsP:N
LED Configuration
Band-gap
engineering
Designing efficient
LED
How does LED
Works?
P-n diode,
radiative transmission
Epitaxial growth
Right Eg
Materials
Requirements
UV,
VIS,IR LED

3. LED Construction – Aim – 100% light emitting efficiency
Important consideration - radiative recombination must take place from the side of the junction nearest to the surface to reduce reabsorption.
Carrier from n must be injected into the p-side efficiently.
Consider the fraction of the total diode current that is carried by electrons being injected into the p-side of the junction (e)

4. If we use Einstein’s equation to substitute this equation. Then,
Why n+-p?
If we use Einstein’s equation to substitute this equation. Then,
e = Denp/Le
Denp/Lp + Dhpn/Lh
Or e = DhpnLe
DennLp
e = ennLp
1+ hpnLe
III-V compounds, e >> h then, e needs to be close to unity. This can be circumvent by doping n with more electrons (n >> p)
If (nn >> pn) = one sided junction n+ -p diode

5. Typical exam question
Justify the reason why in a typical construction of an LED, the n side in the p-n diode must be made heavily doped.
(50 marks)

6. LED Characteristic
The energy of an emitted photon = to the size of the band gap
BUT this is a simplified statement.
The energy of an emitted photon from LED is distributed appropriately according to the energy distribution of electrons on the conduction band and holes in the valance band.
You need to know the distribution of electrons and holes in the CB and VB respectively.

7. Calculation I. LED Output spectrum (Kasap)

8. Calculation II. Output wavelength variations (Kasap)

9. Calculation III. InGaAs on InP substrate (Kasap)

10. The quantum efficiency
Internal quantum efficiency can of some LED approaches 100% but the external efficiencies are much lower. This is due to reabsorption and TIR.
III-V materials have small critical angles therefore the radiation emitted suffers from TIR

11. Recap- Total Internal Reflection
TIR
Incident beam

12. Why do we need the dome? to reduce TIR…
Semiconductor material is shaped like a hemisphere
Electrodes
Pn junction
Plastic dome n+
Electrodes p
to reduce TIR…

13. How to solve TIR problem
GaAs-air interface, the C = 16o which means that much of the light suffers TIR.
To solve the problem we could:
Shape the surface of the semiconductor into a dome or hemisphere so that light rays strike the surface angles < C therefore does not experience TIR. But expensive and not practical to shape p-n junction with dome-like structure.
Encapsulation of the semiconductor junction within a dome-shaped transparent plastic medium (an epoxy) that has higher refractive index than air.

14. Calculation IV. Calculating C
If we take a GaAs/air interface where ni = 3.6 and n2 = 1, what is the critical angle, C?.
C = sin-1 (n2/n1)

15. Fraction of light being emitted, F
If light is isotropically generated in a medium then the fraction transmitted to the outside world is given by:

16. Calculation V. Fraction of light being emitted

17. LED Structure

18. Basic Layer by Layer Structure
LED  1. Surface emitter
2. Edge emitter

19. 1. Surface Emitter
In surface emitter the emitting area is defined by oxide isolation, with the metal contact area a circle of diameter ~ 10m-15 m.
The surface layer is kept as thin as possible (10-15 m) to minimise reabsorbtion

20. Homo- and Hetro-Junction
Homojunction = a p-n junction made out of two differently doped semiconductors that are of the same material (i.e having the same band gap).
Heterojunction = junction formed between two different band gaps semiconductors.
Heterostructure device = semiconductor device structure that has junctions between different bandgap materials.

21. Why Homojunction is bad?
Shallow p-region  narrow to allow photons to escape without reabsorption.
If the p-region is too shallow, electrons can escape the p-region by diffusion and recombine through crystal defect in the surface of the layer.
This recombination is non-radiative and decreases the efficiency of the LED.
Thick p–region  then reabsoprtion will be the main problem as the photons will have a long way to go before can be successful emitted.
Create a heterojunction instead since heterojunction solves:
Reabsoption problem (photon confinement)
Also carrier confinement

22. Band-gap and refractive index engineering.
Avoiding losses in LED
Carrier
confinement
Photon
Confinement
Band-gap and refractive index engineering.
Heterostructured LED

23. Double Heterojunction LED (important)
Fiber Optics
n+ GaAs
p Al GaAs
p GaAs (active region)
n AlGaAs
Metal contact
Epoxy
Double heterostructure
Burrus type LED
Shown bonded to a fiber with index-matching epoxy.

24. Double Heterostructure
The double heterostructure is invariably used for optical sources for communication as seen in the figure in the pervious slide.
Heterostucture can be used to increase:
Efficiency by carrier confinement (band gap engineering)
Efficiency by photon confinement (refractive index)
The double heterostructure enables the source radiation to be much better defined, but further, the optical power generated per unit volume is much greater as well. If the central layer of a double heterostructure, the narrow band-gap region is made no more than 1m wide.

25. Photon confinement - Reabsorption problem
Source of electrons
Active region (micron in thickness)
Source of holes
Active region (thin layer of GaAs) has smaller band gap, energy of photons emitted is smaller then the band gap of the P and N-GaAlAs hence could not be reabsorbed.

26. Reabsorption Problem
In order to prevent reabsorption, the upper layer (one that is above the active region) needs to have higher band gap therefore the emitted photons will not be absorbed by the upper layer (photons will be absorbed when Ep < Eg).
1.4eV
2eV
n-AlGaAs
p-GaAs
p-AlGaAs
n+ GaAs
p Al GaAs
p GaAs (active region)
n AlGaAs
Metal contact
Epoxy
Active region – Photons will not be absorbed by the n-AlGaAs since the band gap is much higher than GaAs

27. Simplified band diagram of the ‘sandwich’ top show carrier confinement
p+-AlGaAs
n+-AlGaAs
p-GaAs
holes
electrons
Simplified band diagram of the ‘sandwich’ top show carrier confinement

28. Burrus-Type LED

29. Communication LED

30. Typical Exam Question
Sketch a typical surface emitted LED that can emit red light. Explain why for such configuration, the light source can be suitable for optical communication. Start your answer with the reasons why photon and carrier confinement are needed.
(80 marks)

31. 2. Edge Emitter
In edge emitter a double heterostructure band gap engineering is used to achieve carrier confinement and recombination in an active layer but in addition layers of relatively low refractive index are included to produce optical guide. A large fraction of the photons are therefore confined between two ‘plates’ of material and emerge at the edge of the device as highly directional flux compatible with coupling to a fibre optic cable.

32. Edge emitter  using double heterostructure
Metal contact
GaAs(n) substrate
N+- GaAlAs
N GaAlAs
Active layer n- GaAlAs
P GaAlAs
P+ GaAlAs
Light emits from the edge
n- GaAlAs
Metal contact

33. The waveguide
We can use refractive index engineering to create a multilayer structure in which light can be trapped within the central layers. This layer act as waveguide. (TIR is used in Edge Emitter)

34. Contact Layer
Cladding Layer
Active Layer
Substrate
Epi growth

35. Another Example of Edge Emitter
Dr Zainovia Lockman L3