Transcript LED Characteristics - USM :: Universiti Sains Malaysia

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,
radiative transmission
Designing efficient
LED
Materials
Requirements
Band-gap
engineering
Epitaxial growth
Right Eg
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 (e)
Why n+-p?
If we use Einstein’s equation
to substitute this equation.
Then,
• e
=
Denp/Le
Denp/Lp + Dhpn/Lh
Or e
=
e
=
DhpnLe
DennLp
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
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
external 
Poutput (opt ical)
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 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 < 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 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)
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 ~ 10m-15 m.
• The surface layer is kept as thin as possible (10-15 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 pregion 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.
2. 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
epoxy.
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.
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
electrons
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 n- GaAlAs
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
Epi growth
Cladding
Layer
Active Layer
Cladding
Layer
Substrate
Another Example of
Edge Emitter