Presentation on theme: "LED Characteristics EBB 424E Lecture 4– LED 3 Dr Zainovia Lockman."— Presentation transcript:
LED Characteristics EBB 424E Lecture 4– LED 3 Dr Zainovia Lockman
Last Lectures.. Applications of LED LED Configuration How does LED Works? P-n diode, radiative transmission Materials Requirements Candidate Materials Designing efficient LED Epitaxial growth Right E g UV, VIS,IR LED Band-gap engineering LED I&2 GaAsP GaAsP:N Definition of LED Group III-V Group II-V
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 )
Why n + -p? e = D e n p /L e D e n p /L p + D h p n /L h Or e = D h p n L e D e n n L p e = e n n L p 1+ h p n L e If we use Einstein’s equation to substitute this equation. Then, III-V compounds, e >> h e needs to be close to unity. This can be circumvent by doping n with more electrons (n >> p) 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 (n n >> p n ) = 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 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
Recap- Total Internal Reflection Incident beam TIR
Why do we need the dome? n+n+ Electrodes Pn junction Plastic dome Semiconductor material is shaped like a hemisphere to reduce TIR… p
How to solve TIR problem GaAs-air interface, the C = 16 o 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 n i = 3.6 and n 2 = 1, what is the critical angle, C ?. C = sin -1 (n 2 /n 1 )
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:
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 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.
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. 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
Band-gap and refractive index engineering. Heterostructured LED Avoiding losses in LED Carrier confinement Photon Confinement
Double heterostructure Burrus type LED Shown bonded to a fiber with index-matching epoxy. Double Heterojunction LED (important) n + GaAs p Al GaAs p GaAs (active region) n AlGaAs Metal contact Epoxy Fiber Optics
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 Source of holes Active region (micron in thickness) 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 E p < E g ). 2eV 1.4eV n-AlGaAsp-GaAs Active region – Photons will not be absorbed by the n- AlGaAs since the band gap is much higher than GaAs p-AlGaAs n + GaAs p Al GaAs p GaAs (active region) n AlGaAs Metal contact Epoxy
Carrier confinement p + -AlGaAsn + -AlGaAsp-GaAs holes electrons Simplified band diagram of the ‘sandwich’ top show carrier confinement
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 Active layer n- GaAlAs N GaAlAs N+- GaAlAs GaAs(n) substrate Metal contact P GaAlAs P+ GaAlAs n- GaAlAs 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 Active Layer Substrate Epi growth