Led Powerpoint

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

Last Lectures.. Definition of LED

Candidate Materials Group III-V Group II-V

Applications of LED

LED I&2

GaAsP GaAsP:N

LED Configuration

How does LED Works? P-n diode, adiative transmission

Band-gap engineering

Designing efficient LED Epitaxial growth Right Eg Materials Requirements

UV, VIS,IR LED

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 pside efficiently. ◘ Consider the fraction of the total diode current that is carried by electrons being injected into the p-side of the junction (η e)

Why n -p? +

Denp/Le ∀η e= Denp/Lp + Dhpn/Lh Or η

e

=

DhpnLe

DennLp η

e

=

µ e nn Lp

If we use Einstein’s equation to substitute this equation. Then,

III-V compounds, µ e >> µ h then, η e needs to be close to unity. This can be circumvent by doping n with more electrons (n >> p)

1+ µ hpnLe If (nn >> pn) = one sided junction n+ -p diode

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)

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. 

Calculation I. LED Output spectrum (Kasap)

Calculation II. Output wavelength variations (Kasap)

Calculation III. InGaAs on InP substrate (Kasap)

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

ηexternal =

Poutput (optical ) IV

x100 %

Recap- Total Internal Reflection

TIR

Incident beam

Why do we need the dome? Semiconductor material is shaped like a hemisphere

Plastic dome p

Pn junction

n+

Electrodes

Electrodes

to reduce TIR…

How to solve TIR problem GaAs-air interface, the θ C = 16o which means that much of the light suffers TIR. • To solve the problem we could: 1. Shape the surface of the semiconductor into a dome or hemisphere so that light rays strike the surface angles < θ C therefore does not experience TIR. But expensive and not practical to shape p-n junction with dome-like structure. 2. Encapsulation of the semiconductor junction within a dome-shaped transparent plastic medium (an epoxy) that has higher refractive index than air. •

Calculation IV. Calculating θ C If we take a GaAs/air interface where ni = 3.6 and n2 = 1, what is the critical angle, θ C?. θ

C

= sin-1 (n2/n1)

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:

Calculation V. Fraction of light being emitted

LED Structure

Basic Layer by Layer Structure

LED 

1. Surface emitter 2. Edge emitter

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

Homo- and HetroJunction • 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.

Why Homojunction is bad? 1. 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. 1. 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

Avoiding losses in LED Carrier confinement

Photon Confinement

Band-gap and refractive index engineering.

Heterostructured LED

Double Heterojunction LED (important) Fiber Optics

Epoxy Metal contact n AlGaAs p GaAs (active region) p Al GaAs n+ GaAs Metal contact

 Double heterostructure  Burrus type LED  Shown bonded to a fiber with index-matching

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 1µ m wide.

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.

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

n-AlGaAs p-GaAs

2eV

p-AlGaAs

Epoxy Metal contact n AlGaAs p GaAs (active region) p Al GaAs n+ GaAs Metal contact

Active region – Photons will not be absorbed by the nAlGaAs since the band gap is much higher than GaAs

Carrier confinement electro ns

n+-AlGaAs

p-GaAs

holes p+-AlGaAs

Simplified band diagram of the ‘sandwich’ top show carrier confinement

Burrus-Type LED

Communication LED

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)

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.

Edge emitter  using double heterostructure Metal contact

GaAs(n) substrate N+- GaAlAs N GaAlAs Active layer nGaAlAs P GaAlAs P+ GaAlAs n- GaAlAs Metal contact

Light emits from the edge

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)

Contact Layer Cladding Layer

Epi growth

Active Layer

Cladding Layer Substrat e

Another Example of Edge Emitter