Photodetectors. Photodiodes

Transkript

1 Photodetectors Photodiodes

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5 Photo-detectors: Principle of the P-N junction photo-diode Schematic diagram of a reverse biased p-n junction photodiode Electrode SiO 2 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. Net space charge across the diode in the depletion region. N d and N a are the donor and acceptor concentrations in the p and n sides. hv > E g AR coating en d ρ net p + h + V r E e W Depletion region I ph n R V out Electrode x en a 5

6 Principle of pn junction photodiode (a) Reversed biased pn junction photodiode. Annular electrode to allow photon to enter the device. Anti-reflection coating (Si 3 N 4 ) to reduce the reflection. The p + -side thickness < 1 μ m. (b) Net space charge distribution, within SCL. 6

7 Photo-detectors: Principle of the p-n junction Photo-diode Operation of a photo-diode (b) Energy band diagram under reverse bias. (a) Cross-section view of a photo-diode (c) Carrier absorption characteristics. 7

8 Photodetectors: Principle of the p-n junction Photo-diode A generic photo-diode 8

9 Photodetectors: Principle of the p-n junction Photo-diode 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 x 0 is defined as the depth at which the photon flux is reduced to e -1 of its surface value. 9

10 Photo-detectors: RAMO s Theorem and External Photo-current An EHP is photogenerated at x = l. The electron and the hole drift in opposite directions with drift velocities v h and v e. The electron arrives at time t electron = (L-l )/v e and the hole arrives at time t hole = l/v h. v hole h+ V Semiconductor e - v electron E I photo (t) t electron t hole 0 e v L h e v h e v + L L Area = Charge = e e i photo (t) l L l 0 l L h + e x t electron t 0 evh/l eve/l i (t) i electron (t) t hole t t t hole t photocurrent i hole (t) 10

11 Photo-detectors: RAMO s Theorem and External Photo-current As the electron and hole drift, each generates i electron (t) and i hole (t). The total photocurrent is the sum of hole and electron photocurrents each lasting a duration t h and t e respectively. t e L l l = h = Transit time v v () t and t () t Work e done = e E dx = V i e h ()dt t V E = v e L = dx dt e v i e < L e () t = ; t te e v i h < L h () t = ; t th t 0 0 () t dt + i () t dt e e t h Qcollected = ie h = Photocurrent The collected charge is not 2e but just one electron. If a charge q is being drifted with a velocity v d (t) by a field between two biased electrodes separated by L, the motion of q generates an external current given by ( ) e v L () t d i t = ; t < t transit Ramo s Theorem 11

12 Photo-detectors: Absorption Coefficient & Photo-diode Materials Absorbed Photon create Electron-Hole Pair. λg [ μm] = 1.24 E [ ev ] g 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. I α x ( x) = I 0 e Absorption coefficient 12

13 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 depth δ ) α (m -1 ) Si 3 2 a-si:h Ge 1 Photon energy (ev) In 0.7 Ga 0.3 As 0.64 P 0.36 In 0.53 Ga 0.47 As GaAs InP Wavelength (μm) 13

14 Photo-detectors: Absorption Coefficient & Photo-diode Materials Absorption Photon energy (ev) The indirect-gap materials are shown with a broken line Ge In 0.7 Ga 0.3 As 0.64 P 0.36 Absorption Coefficient α (m -1 ) Si a-si:h GaAs InP In 0.53 Ga 0.47 As Wavelength (mm) 14

15 Absorption Coefficient Direct bandgap semiconductors (GaAs, InAs, InP, GaSb, InGaAs, GaAsSb), the photon absorption does not 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 ħ k), since photon momentum is very small. hk hk CB VB = photon momentum 0 k CB E c Direct Bandgap E g Photon E v VB E (a) GaAs (Direct bandgap) k Absorption coefficient α for direct band-gap semiconductors rise sharply with decreasing wavelength from λ g (GaAs and InP). 15

16 Absorption Coefficient Indirect band-gap semiconductors (Si and Ge), the photon absorption requires assistant from lattice vibrations (phonon). If K is wave vector of lattice wave, then ħ K represents the momentum associated with lattice vibration K is a phonon momentum. h k hk = phonon momentum = CB VB k E Indirect Bandgap, E g CB E c Photon E v VB Phonon hk k (b) Si (Indirect bandgap) Thus the probability of photon absorption is not as high as in a direct transition and the λ g is not as sharp as for direct band-gap semiconductors

17 Photo-detectors: Absorption Coefficient & Photo-diode Materials Photon absorption in a direct bandgap semiconductor. E Photon absorption in an indirect bandgap semiconductor E CB Direct Bandgap E C E g E V Photon Photon CB Indirect Bandgap E C E g VB VB E V Phonons k k k k 17

18 Photo-detectors: Quantum Efficiency and Responsivity External Quantum Efficiency η = Number of EHP Number of geberated and collected incidnet photons = I P 0 ph e hν Responsivity R = Photocurrent (A) Incident Optical Power (W) = I P ph 0 R e eλ η = η hν h c = Spectral Responsivity 18

19 Photo-detectors Responsivity vs. wavelength for a typical Si photo-diode Responsivity (A/W) Ideal Photodiode QE = 100% ( η = 1) Si Photodiode λ g Wavelength (nm) 19

20 The pin Photo-diode Intrinsic layer has less doping and wider region (5 50 μm)

21 Photo-detectors: PIN Photo-diode Reverse-biased p-i-n photodiode pin energy-band diagram pin photodiode circuit 21

22 Photo-detectors: PIN Photo-diode Schematic diagram of pin photodiode SiO 2 E(x) Electrode Electrode p+ In contrast to pn junction built-in-field is uniform x i-si n+ E 0 W ρ net hυ > E g E en d h + e x I ph R V out ena V r Small depletion layer capacitance gives high modulation frequencies. High Quantum efficiency. 22

23 Photo-detectors: PIN Photo-diode 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. hυ > E g p + Diffusion e h + i-si E Drift l W V r 23

24 Photo-detectors: PIN Photo-diode p-i-n diode (a) The structure; (b) equilibrium energy band diagram; (c) energy band diagram under reverse bias. 24

25 Photo-detectors: PIN Photo-diode The responsivity of PIN photodiodes 25

26 Photo-detectors: Photo-conductive Detectors and Gain Quantum efficiency versus wavelength for various photo-detectors 26

27 Photo-detectors: PIN Photo-diode Junction capacitance of pin C dep = ε 0 ε r A W Small capacitance: High modulation frequency RC dep time constant is 50 psec. Electric field of biased pin Response time E E = 0 + Vr W Vr W t drift W = v vd = μ d d E 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 μm, the drift time is about is about 0.1 nsec. 27

28 Photo-detectors: PIN Photo-diode Drift velocity vs. electric field for holes and electrons in Silicon. Drift velocity (m sec -1 ) Electron Hole Electric field (V m -1 )

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32 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 cm 2 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, λ = 600 nm, we need E ph = hυ = E g so that, E g = hc/λ = ( J s)( m s -1 )/( m) = 2.07 ev (b) Area = cm 2 and I light = W/cm 2. The received power is P = Area I light = ( cm 2 )( W/cm 2 ) = 10-3 W N ph = number of photons arriving per second = P/E ph = (10-3 W)/( J/eV) = photons s -1 = EHP s

33 Example Bandgap and Photodetection (c) From the known energy gap of the semiconductor GaAs (E g = 1.42 ev), calculate the primary wavelength of photons emitted from this crystal as a result of electronhole 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, E g = 1.42 ev and the corresponding wavelength is λ = hc/ E g = ( J s)( m s -1 )/(1.42 ev J/eV) = 873 nm (invisible IR) The wavelength of emitted radiation due to EHP recombination is 873 nm. (d) For Si, E g = 1.1 ev and the corresponding cut-off wavelength is λ g = hc/ E g = ( J s)( m s -1 )/(1.1 ev 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 E g, 1.1 ev, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation) 33

34 Example Absorption coefficient (a)if d is the thickness of a photodetector material, I o is the intensity of the incoming radiation, the number of photons absorbed per unit volume of sample is Solution [ 1 exp( α d ] I0 ) n ph = d hυ (a) If I 0 is the intensity of incoming radiation (energy flowing per unit area per second), I 0 exp( α d ) is the transmitted intensity through the specimen with thickness d and thus I 0 exp( α d ) is the absorbed intensity 34

35 Example (b) What is the thickness of a Ge and In 0.53 Ga 0.47 As crystal layer that is needed for absorbing 90% of the incident radiation at 1.5 μm? For Ge, α m -1 at 1.5 μm incident radiation. For In 0.53 Ga 0.47 As, α m -1 at 1.5 μm incident radiation. (b) For Ge, α m -1 at 1.5 μm incident radiation. d 1 exp( α d ) = = ln α = ln m 5 = = μm For In 0.53 Ga 0.47 As, α m -1 at 1.5 μm incident radiation. d = ln m 5 = = μm 35

36 Example InGaAs pin Photodiodes Consider a commercial InGaAs pin photodiode whose responsivity is shown below. Its dark current is 5 na. (a) (b) What optical power at a wavelength of 1.55 μm would give a photocurrent that is twice the dark current? What is the QE of the photodetector at 1.55 μm? What would be the photocurrent if the incident power in (a) was at 1.3 μm? What is the QE at 1.3 μm operation? Responsivity (A/W) Wavelength (nm) The responsivity of an InGaAs pin photodiode 36

37 Solution (a) At λ = m, from the responsivity vs. wavelength curve we have R 0.87 A/W. From the definition of responsivity, η R = Photocurrent ( A) Incident Optical Power = ( W ) 9 I ph 2Idark ( A) we have P0 = = = = 11.5 nw R R 0.87 A / W ) From the definitions of quantum efficiency η and responsivity, e eλ R = η = η h hc hcr eλ υ 34 8 ( J sec)(3 10 m / s)(0.87a / W ) = = 0.70 (70 %) 19 6 ( coul)( m) = 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. I P ph 0 37

38 Solution (b) At λ = m, from the responsivity vs. wavelength curve, R = 0.82 A/W. Since P o is the same and 11.5 nw as in (a), I ph = R P = (0.82 A / W )(1.15 nw ) = na The QE at λ = 1.3 μm is 34 8 hcr ( J sec)(3 10 m / s)(0.82a / W η = = 19 6 eλ ( coul)( m) ) 0.78 (78 %) 38

39 Photo-detectors: Avalanche Photo-diode (APD) Electrode SiO 2 I photo R Impact ionization processes resulting avalanche multiplication hυ > E g n+ p e h+ π p+ E E h + e ρ net Electrode x n + p Avalanche region e - π E c E(x) h + E v Absorption region Avalanche region x Impact of an energetic electron's kinetic energy excites VB electron to the CV. 39

40 Photo detectors: Avalanche Photo-diode (APD) Schematic diagram of typical Si APD. Anti-reflection coating Electrode SiO 2 n + p Guard ring n n + p n π Avalanche breakdown π p + Substrate Electrode p + Substrate Electrode Si APD structure without a guard ring More practical Si APD Breakdown voltage around periphery is higher and avalanche is confined more to illuminated region (n + p junction). 40