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LIGHT EMITTING DIODE – Materials Issues and Selection

LIGHT EMITTING DIODE – Materials Issues and Selection. EBB 424E Lecture 3– LED 2 Dr Zainovia Lockman. At the end of this lecture you would be able to…. Cite semiconductor materials suitable for LED of different colours (red, yellow, green, blue, white)

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LIGHT EMITTING DIODE – Materials Issues and Selection

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  1. LIGHT EMITTING DIODE – Materials Issues and Selection EBB 424E Lecture 3– LED 2 Dr Zainovia Lockman

  2. At the end of this lecture you would be able to… • Cite semiconductor materials suitable for LED of different colours (red, yellow, green, blue, white) • Describe the GaAsP system as an example of ternary compounds • Use the knowledge of band gap engineering to design LED material to emit suitable coloured lights • Discuss the current phenomenon in LED research activities

  3. Ga P As GaAs(1+x) Px

  4. What is GaAs(1+x) Px? • GaAs(1+x) Px is a ternary compound based on GaAs and GaP • GaAs is a direct gap semiconductor and GaP is indirect semiconductor • When alloyed, there is cross over point where GaAs(1+x) Px will transformed from being direct gap material to indirect gap material • Red, yellow and orange coloured LED can be made with GaAs(1+x) Px

  5. Band Diagram Direct band gap 100%GaAs Indirect to Direct transition 50% GaP Indirect band gap 100% GaP Composition of GaP % Red photon Green photon Doped with nitrogen efficiency increases

  6. GaN+GaP = GaAs (1+x)Px At the transition, the band gap correspond to  from near IR to the orange-red part of the vis-spectrum Spectral response of human eye eV  GaP = 2.26eV 1.997eV GaAs = 1.42eV GaP= indirect but when alloyed with GaAs, the band gap will become direct at x = 0.45 Indirect ----------- > Direct GaAs (1+x) Px

  7. GaAs(1+x) Px system • x = 0.45  indirect to direct transition

  8. GaAs(1+x) Px system doped with N • Indirect  no radiative transition • Indirect GaAs(1+x) Px can have radiative transition. • HOW? • By adding nitrogen to the system • When N added to GaAs(1+x) Px: • The quantum efficiency increases ~ 100x • The emission wavelength increases • Quantum efficiencies = rate of emission of photons Rate of electron supply How efficient the e-h pair can recombine

  9. Isoelectronic Doping and Heisenberg Uncertainty Principles (N +GaAsP) • N has the same valancy as that of P and As • N can enter the As or P site in the GaAsP crystal structure. • N and P has similar number of valance electrons but different core shell structure • N produces a perturbance in the electronic confinement • Electronics confinement changes and acts as a ‘trap’ • Electron trapped at a level just below a conduction band. • Hole can be captured to produce electron-hole pair (exciton) • The carriers are localised, the momentum and the wavenumber are diffuse due to Heisenberg uncertainties principle

  10. Typical Exam Question!!!! The figure below shows the quantum efficiencies of GaAsP based LED as a function of alloy composition with and without nitrogen doping. Explain why the additional of nitrogen leads to such dramatic changes in the quantum efficiencies of the device. Why is this phenomenon important from a practical point of view? (100 marks +5 bonus)

  11. The figure: Quantum efficiencies

  12. N substitution to GaAsP CB No N VB e N produces perturbances CB N doping can dramatically increases the radiative efficiency of GaP (indirect), the doping changes the emission wavelength to longer wavelength because the energy of the transition is now reduced to Eg-Ed VB e falls inside the ‘trap’ producing excitons CB e ED VB

  13. Heisenberg Uncertainty Principle – the uncertainties of the doped electrons position and momentum Px = h P = hk/2 P = momentum Px = h x = 2/k Set x = 2/a (a= lattice parameter) The position of electron is uncertain, when electron is at k=0 then recombination occurs, if not then no recombination. The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. E h K

  14. Question 2. • GaP and GaAs can be mixed to produce a direct gap semiconductor that produce red-light, explain this statement.

  15. Question 3. List down al of the possible application of IR LED

  16. Band Gap Engineering A process of varying the elemental components of the semiconductor alloy in a controlled way to achieve a desired band gap that can emit a desired wavelength of radiation.

  17. 2 critical considerations • The wavelength of the radiation emitted • The lattice parameters of the compounds The wavelength  visible, UV or IR The lattice parameter  for epitaxial growth Why? How? A good device requires a defect free semiconductor films. Defect free  good crystallographic orientation of the grains of the semiconductor materials, low defect Thin film To achieve defect free semiconductor thin film, adopt a so-call epitaxial growth of the film on a substrate  growth process where the deposited films will ‘follow’ the surface structure of a substrate. Substrate

  18. P-dopant Epitaxial growth p P-n junction n Substrate Semiconductor materials need to be deposited onto a textured substrate (thin film technology) Substrate must have similar lattice parameter to that of the semiconductor thin film to avoid lattice mismatch (strain at the interface will induce crack) and to allow epitaxial growth The semiconductor then need to be doped to achieve both p and n type  require p-n junction

  19. IR & Red LED • GaAs  direct band gap, p-n junctions are readily formed with high radiative efficiency. High radiative efficiency can be induced by doping GaAs with Zn or Si. Si doped GaAs is now the industry standard near IR LEDs. • GaAsP  direct – in direct transition • GaInAsP  Grown on InP substrate and band gap can be varied to get wavelength from 919nm to 1600nm. A true story of band gap engineering.

  20. Band Gap and Lattice Constant The band gap energy can be tailored to get desired visible light radiation Substrates must have similar lattice parameter to the semiconductor films, GaAs, GaN and InP are often used as substrates.

  21. LED + band gap engineering “LEDs are specialized semiconductor devices that can potentially convert electricity to light, without the wasteful creation of heat. The color emitted is controlled in large part by the energy gap of the semiconductor and in advanced structures by the “photonic band gap,” a range of wavelengths that cannot travel through that particular substance. By suppressing certain wavelengths and enhancing others, the band gap determines the color.” One of the pioneers in the field of LED; Fred Schubert

  22. Examples of Substrate/semiconductor p-n diode/visible light produces

  23. Cross section of a typical epitaxial layers

  24. Calculation. InGaAs on InP substrate (Kasap)

  25. The nitrides and blue LED • Difficulties: • to find suitable substrates for the nitrides. • to get p-type nitrides • But with constant R&D works, better materials are produced • GaN, InGaN, AlGaN  high efficiency LEDs emitting blue/green part of the spectrum. • First blue LED 1994 Shuji & Nakamura (10 000 hours lifetime) • SiC can also be used as blue LED- SiC on GaN substrate

  26. The device • Applications: • Flat panel displays (display requires, R,G,B now B is found, all LED displays can be made. • High resolution printers • Light source for communications • Microwave transistors (electrons have high mobility)

  27. UV-LED Apart blue LED, UV LED can also be made using nitrides. UV-LED can be used as UV calibration devices, UV detector etc.

  28. The Blue-Violet LED + Phosphor and White LED White LEDs are slightly more efficient than a 100W incandescent bulb and three times more efficient than a 7W night light type bulb. The lifetime of white LED could reach >10 000 hours while incandescent filament (100watt) normally reaches about 750-1500 hours.

  29. Phosphor

  30. Another typical exam question Draw a table to list down some examples of possible materials for visible LEDs. In your table state also the visible wavelength your LED will emit as well as some applications of a given visible LED. Explain why group III-V materials have been selected as an LED emitter for use in an optical fiber network. (100 marks)

  31. The Selenide • Group II-V is also important (ZnSe especially even though ZnO has being a contender as well) • ZnSe can be made into LED, emitting blue and green lights. • Problem with finding suitable template (substrate) for growth. • GaAs and GaN can be used as the substrate for selenide. The lattice parameter for GaAs = 5.6Å and ZnSe = 5.5Å • ZnSe has been used as blue/green laser (study later). • The selenide degrade more rapidly hence shorter working life-time

  32. The selenides - E gap vs lattice parameter ZnSe can be made ternary allow with ZnTe to produce ZnSeTe  blue-green

  33. Homework question A diagram given to you shows the energy gap versus lattice constant of some group III-V semiconductors. Explain the importance of band gap engineering in designing an LED and expand your answer to include some examples of materials used in an IR-LED.

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