RECEPTORES ÓPTICOS

1. Comunicações Ópticas RECEPTORES ÓPTICOS Prof. Dr. Vitaly F. Rodríguez-Esquerre

2. Diagrama básico de um link óptico Transmissor  Receptor Sinal óptico: gerado por um LED ou laser. Meio de propagação: ar ou guia de ondas (fibra óptica). Receptor: fotodetector  converte sinal óptico em uma corrente elétrica.

3. Signal Path through an Optical Link

4. Fundamental Receiver Operation • The first receiver element is a pin or an avalanche photodiode, which produces an electric current proportional to the received power level. • Since this electric current typically is very weak, a front-end amplifierboosts it to a level that can be used by the following electronics. • After being amplified, the signal passes through a low-pass filter to reduce the noise that is outside of the signal bandwidth. • The also filter can reshape (equalize) the pulses that have become distorted as they traveled through the fiber. • Together with a clock (timing) recovery circuit, a decision circuit decides whether a 1 or 0 pulse was received,

5. Fotodetetor Dispositivos que convertem a luz incidente em uma corrente elétrica Requisitos: Boa sensibilidade (responsividade) no comprimento de onda de operação e nula em outros comprimentos de onda -> Seletivo Resposta temporal rápida -> largura de banda grande Dimensões físicas compatíveis com as fibras ópticas Baixo ruído Insensível a variações de temperatura Vida útil longa e baixos custos

6. Photodiodes • Due to above requirements, only photodiodes are used as photo detectors in optical communication systems • Positive-Intrinsic-Negative (pin) photodiode • No internal gain • Avalanche Photo Diode (APD) • An internal gain of M due to self multiplication • Photodiodes are sufficiently reverse biased during normal operation  no current flow, the intrinsic region is fully depleted of carriers

8. Physical Principles of Photodiodes • As a photon flux Φ penetrates into a semiconductor, it will be absorbed as it progresses through the material. • If αs(λ) is the photon absorption coefficient at a wavelength λ, the power level at a distance x into the material is Absorbed photons trigger photocurrent Ip in the external circuitry Photocurrent Incident Light Power

9. Examples of Photon Absorption

13. V out I ph h v > Eg Photodetectors Principle of the p-n junction Photodiode R V SiO2 r • Schematic diagram of a reverse biased p-n junction photodiode Electrode – + e h • Photocurrent is depend on number of EHP and drift velocity. • The electrode do not inject carriers but allow excess carriers in the sample to leave and become collected by the battery. n p+ E AR coating Electrode W Depletion region r net • Net space charge across the diode in thedepletion region. Nd and Naare the donor and acceptorconcentrations in the p and n sides. eN d x – eN a E ( x ) • The field in the depletion region. x E max

14. Principle of pn junction photodiode • (a) Reversed biased p+n junction photodiode. • Annular electrode to allow photon to enter the device. • Antireflection coating (Si3N4) to reduce the reflection. • The p+-side thickness < 1 μm. • (b) Net space charge distribution, within SCL. • (c) The E field across depletion region.

15. Photodetectors Principle of the p-n junction Photodiode • Operation of a p-i-n photodiode. (b) Energy band diagram under reverse bias. (a) Cross-section view of a p-i-n photodiode. (c) Carrier absorption characteristics.

16. Photodetectors Principle of the p-n junction Photodiode • A generic photodiode.

17. Photodetectors Principle of the p-n junction Photodiode • Variation of photon flux with distance. A physical diagram showing the depletion region. • A plot of the the flux as a function of distance. • There is a loss due to Fresnel reflection at the surface,followed by the decaying exponential loss due to absorption. • The photon penetration depth x0 is defined as thedepth at which the photon flux is reduced to e-1 of its surface value.

18. E Photodetectors RAMO’s Theorem and External Photocurrent • An EHP is photogenerated at x = l. The electron and the hole drift in oppositedirections with drift velocities vhand ve. • The electron arrives at time telectron= (L-l )/veandthe hole arrives at time thole= l/vh. V Iphoto(t) 0 iphoto(t) Semiconductor Area = Charge = e telectron e- + h velectron vhole thole L - l l t e v /L e v /L 0 0 l L h e x i (t) + h – e ielectron(t) telectron photocurrent thole thole ihole(t) t t t

19.  Photodetectors RAMO’s Theorem and External Photocurrent • As the electron and hole drift, each generates ielectron(t) and ihole(t). • The total photocurrent is the sum of hole and electron photocurrents each lasting a duration th and te respectively. Transit time Photocurrent The collected charge is not 2e but just “one electron”. If a charge q is being drifted with a velocity vd(t) by a field between two biased electrodes separated by L, the motion of q generates an external current given by Ramo’s Theorem

20.  Photodetectors Absorption Coefficient and Photodiode Materials • Absorbed Photon create Electron-Hole Pair. Cut-off wavelength vs. Energy bandgap • Incident photons become absorbed as they travel in the semiconductor and light intensity decays exponentially with distance into the semiconductor. Absorption coefficient

21. Absorption Coefficient • Absorption coefficient α is a material property. • Most of the photon absorption (63%) occurs over a distance 1/α (it is called penetration depthδ)

22. Photodetectors Absorption Coefficient and Photodiode Materials • Absorption The indirect-gap materials are shown with a brokenline. Photon energy (eV) 5 4 3 2 1 0.9 0.8 0.7 8 ´ 1 10 Ge 7 In0.7Ga0.3As0.64P0.36 ´ 1 10 In0.53Ga0.47As Si 6 ´ 1 10 GaAs InP 5 ´ Absorption Coefficient a (m-1) 1 10 a-Si:H 4 ´ 1 10 3 ´ 1 10 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Wavelength (mm)

23. Absorption Coefficient • Direct bandgap semiconductors (GaAs, InAs, InP, GaSb, InGaAs, GaAsSb), the photon absorption doesnot require assistant from lattice vibrations. The photon is absorbed and the electron is excited directly from the VB to CB without a change in its k-vector (crystal momentum ħk), since photon momentum is very small. • Absorption coefficient α for direct bandgap semiconductors rise sharply with decreasing wavelength from λg (GaAs and InP).

24. Absorption Coefficient • Indirect bandgap semiconductors (Si and Ge), the photon absorption requires assistant from lattice vibrations (phonon). If K is wave vector of lattice wave, then ħK represents the momentum associated with lattice vibration  ħK is a phonon momentum. • Thus the probability of photon absorption is not as high as in a direct transition and the λg is not as sharp as for direct bandgap semiconductors.

25. Photodetectors Absorption Coefficient and Photodiode Materials Photon absorption in a direct bandgap semiconductor. Photon absorptionin an indirect bandgap semiconductor E E C B Indirect Bandgap EC C B Eg Direct Bandgap Eg Photon EC EV Photon EV V B V B Phonons – k k k – k

26.  Photodetectors Quantum Efficiency and Responsivity • External Quantum Efficiency • Responsivity Spectral Responsivity

27. 1 0.9 0.8 Ideal Photodiode 0.7 QE = 100% ( h = 1) 0.6 Responsivity (A/W) 0.5 lg 0.4 0.3 Si Photodiode 0.2 0.1 0 0 200 400 600 800 1000 1200 Wavelength (nm) Photodetectors • Responsivity vs. wavelength for a typical Si photodiode.

28. The pin Photodiode • The pn junction photodiode has two drawbacks: • Depletion layer capacitance is not sufficiently small to allow photodetection at high modulation frequencies (RC time constant limitation). • Narrow SCL (at most a few microns)  long wavelengths incident photons are absorbed outside SCL  low QE • The pin photodiode can significantly reduce these problems. • Intrinsic layer has less doping and wider region (5 – 50 μm).

29. Photodetectors The pin Photodiode • Reverse-biased p-i-n photodiode • pin energy-band diagram • pinphotodiode circuit

30. E Photodetectors The pin Photodiode • Schematic diagram of pin photodiode In contrast to pn junction built-in-field is uniform SiO2 E(x) Electrode Electrode + p x E0 + i-Si n W rnet h u > Eg eN d e– h+ x Iph R Vout – eN a Vr • Small depletion layer capacitance gives high modulation frequencies. • High Quantum efficiency.

31. + p i -Si E Diffusion u h > E – e g Drift + h W l V r Photodetectors The pin Photodiode • A reverse biased pin photodiode is illuminated with a short wavelength photon that is absorbed very near the surface. • The photogenerated electron has to diffuse to the depletion region where it is swept into the i-layer and drifted across.

32. Photodetectors The pin Photodiode p-i-ndiode (a) The structure; (b) equilibrium energy band diagram; (c) energy band diagram underreverse bias.

33. Photodetectors The pin Photodiode • The responsivity of pin photodiodes

34. Photodetectors Photoconductive Detectors and Photoconductive gain • Quantum efficiency versus wavelength for various photodetectors.

35.  Photodetectors The pin Photodiode • Junction capacitance of pin • Small capacitance: High modulation frequency • RCdeptime constant is50 psec. • Response time • Electric field of biased pin • The speed of pin photodiodes are invariably limited by the transit time of photogenerated carriers across the i-Si layer. • For i-Si layer of width 10 m, the drift time is about is about 0.1 nsec.

36. Photodetectors The pin Photodiode • Drift velocity vs. electric field for holes and electrons in Silicon. 5 10 Electron 4 10 Hole Drift velocity (m sec-1) 3 10 2 10 4 6 7 5 10 10 10 10 Electric field (V m-1)

38. Photodetectors Absorption Coefficient and Photodiode Materials • Absorption The indirect-gap materials are shown with a brokenline. Photon energy (eV) 5 4 3 2 1 0.9 0.8 0.7 8 ´ 1 10 Ge 7 In0.7Ga0.3As0.64P0.36 ´ 1 10 In0.53Ga0.47As Si 6 ´ 1 10 GaAs InP 5 ´ Absorption Coefficient a (m-1) 1 10 a-Si:H 4 ´ 1 10 3 ´ 1 10 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Wavelength (mm)

41. Example Bandgap and photodetection (a) Determine the maximum value of the energy gap which a semiconductor, used as a photoconductor, can have if it is to be sensitive to yellow light (600 nm). (b) A photodetector whose area is 510-2 cm2 is irradiated with yellow light whose intensity is 20 mW cm-2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs generated per second. Solution (a) Given, l = 600 nm, we need Eph = hu = Eg so that, Eg = hc/l = (6.62610-34 J s)(3108 m s-1)/(60010-9 m)= 2.07 eV (b)Area = 510-2 cm2 and Ilight = 2010-3 W/cm2. The received power is P = Area Ilight = (510-2 cm2)(2010-3 W/cm2) = 10-3 W Nph = number of photons arriving per second = P/Eph = (10-3 W)/(2.0591.6021810-19 J/eV) = 2.97871015photons s-1 = 2.97871015 EHP s-1.

42. Example Bandgap and Photodetection (c) From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primary wavelength of photons emitted from this crystal as a result of electron-hole recombination. Is this wavelength in the visible? (d) Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why? Solution (c) For GaAs, Eg = 1.42 eV and the corresponding wavelength is l = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.42 eV1.610-19 J/eV) = 873 nm (invisible IR) The wavelength of emitted radiation due to EHP recombination is 873 nm. (d) For Si, Eg = 1.1 eV and the corresponding cut-off wavelength is, lg = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.1 eV1.610-19 J/eV) = 1120 nm Since the 873 nm wavelength is shorter than the cut-off wavelength of 1120 nm, the Si photodetector can detect the 873 nm radiation (Put differently, the photon energy corresponding to 873 nm, 1.42 eV, is larger than the Eg, 1.1 eV, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation)

43. Example Absorption coefficient (a) If d is the thickness of a photodetector material, Io is the intensity of the incoming radiation, the number of photons absorbed per unit volume of sample is Solution (a) If I0 is the intensity of incoming radiation (energy flowing per unit area per second), I0exp(-d) is the transmitted intensity through the specimen with thickness d and thus I0exp(-d) is the “absorbed” intensity

44. Example (b) What is the thickness of a Ge and In0.53Ga0.47As crystal layer that is needed for absorbing 90% of the incident radiation at 1.5 mm? For Ge, 5.2  105 m-1 at 1.5 m incident radiation. For In0.53Ga0.47As,  7.5  105 m-1 at 1.5 m incident radiation. For Ge, 5.2  105 m-1 at 1.5 m incident radiation. (b) For In0.53Ga0.47As,  7.5  105 m-1 at 1.5 m incident radiation.

45. Example Responsivity(A/W) 1 0.8 0.6 0.4 0.2 0 800 1000 1200 1400 1600 1800 Wavelength(nm) • InGaAs pin Photodiodes • Consider a commercial InGaAs pin photodiode whose responsivity is shown in fig. Its dark current is 5 nA. • What optical power at a wavelength of 1.55 mm would give a photocurrent that is twice the dark current? What is the QE of the photodetector at 1.55 mm? • What would be the photocurrent if the incident power in a was at 1.3 mm? What is the QE at 1.3 mm operation? • The responsivity of an InGaAs pin photodiode

46. Solution (a)Atl= 1.55´10-6 m, from the responsivity vs. wavelength curve we have R 0.87 A/W. From the definition of responsivity, we have From the definitions of quantum efficiency h and responsivity,  Note the following dimensional identities: A = C s-1 and W = J s-1 so that A W-1 = C J-1. Thus, responsivity in terms of photocurrent per unit incident optical power is also charge collected per unit incident energy.

47. Solution (b)Atl= 1.310-6 m, from the responsivity vs. wavelength curve, R= 0.82 A/W. Since Po is the same and 11.5 nW as in (a), The QE at l = 1.3 mm is

48. Comparisons of pin Photodiodes NOTE: The values were derived from various vendor data sheets and from performance numbers reported in the literature. They are guidelines for comparison purposes. Detailed values on specific devices for particular applications can be obtained from photodetector and receiver module suppliers.

49. Avalanche Photodiode (APD) • APD has an internal gain obtained by having a high electric field that energizes photo-generated electrons and holes • These electrons and holes ionize bound electrons in the valence band upon colliding with them • This mechanism is known as impact ionization • The newly generated electrons and holes are also accelerated by the high electric field and they gain enough energy to cause further impact ionization • This phenomena is called the avalanche effect

50. Photodetectors Avalanche Photodiode (APD) • Impact ionization processes resulting avalanche multiplication R Iphoto SiO2 Electrode E h u > Eg + – h e + + p  n p + h E – e r Electrode n e t + n p  Avalanche region e- E x c E v E ( x ) + h x Absorption region • Impact of an energeticelectron's kinetic energy excites VB electron to the CV. Avalanche region