1. Photodetectors

5. Principle of the p-n junction Photodiode  Schematic diagram of a reverse biased p-n junction photodiode SiO 2 Electrode  net –eN a d x x E (x) R E max e – h + I ph hv  >EgEg W E n Depletion region AR coating V r Electrode V out  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.  The field in the depletion region. p+p+  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.

6. 6 Principle of pn junction photodiode (a) Reversed biased p + n junction photodiode. Annular electrode to allow photon to enter the device. Antireflection coating (Si 3 N 4 ) to reduce the reflection. The p + -side thickness < 1 μm. (b) Net space charge distribution, within SCL. (c) The E field across depletion region.

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

8.  A generic photodiode. Photodetectors Principle of the p-n junction Photodiode

9.  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. Photodetectors Principle of the p-n junction Photodiode

10. Photodetectors RAMO’s Theorem and External Photocurrent tt e-e- h + I photo (t) Semiconductor V x l L  l t v hole 0 L l e – h + 0 t 0 Area = Charge = e e v h /Le v e  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. photocurrent i electron (t) E t hole t electron t hole i hole (t) i (t) t electron t hole v electron i photo (t)

11.  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. Photodetectors RAMO’s Theorem and External Photocurrent Transit time Ramo’s Theorem 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 

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

13. 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)

14. Photodetectors Absorption Coefficient and Photodiode Materials 0.20.40.60.81.21.41.61.8 Wavelength (mm) Ge Si In 0.7 Ga 0.3 As 0.64 P 0.36 InP GaAs a-Si:H 12345 0.90.80.7 1  10 3 1  4 1  5 1  6 1  7 1  8 Photon energy (eV) Absorption Coefficient  (m -1 ) 1.0 In 0.53 Ga 0.47 As  Absorption The indirect-gap materials are shown with a broken line.

15. 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 ħk), since photon momentum is very small. Absorption coefficient α for direct bandgap semiconductors rise sharply with decreasing wavelength from λ g (GaAs and InP).

16. 16 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 λ g is not as sharp as for direct bandgap semiconductors.

17. Photodetectors Absorption Coefficient and Photodiode Materials E CB VB k –k Direct Bandgap EgEg E k –k VB CB Indirect Bandgap Photon Phonons Photon absorption in an indirect bandgap semiconductor EgEg ECEC EVEV ECEC EVEV Photon absorption in a direct bandgap semiconductor. Photon

18. Photodetectors Quantum Efficiency and Responsivity  External Quantum Efficiency  Responsivity Spectral Responsivity 

19.  Responsivity vs. wavelength for a typical Si photodiode. Photodetectors

20. 20 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).

21.  Reverse-biased p-i-n photodiode Photodetectors The pin Photodiode  pin energy-band diagram  pin photodiode circuit

22. SiO 2  Schematic diagram of pin photodiode p + i-Si n + Electrode  net –eN a d x x E(x) R E0E0 e–e– h+h+ I ph h  >EgEg W VrVr V out Electrode E Photodetectors The pin Photodiode  Small depletion layer capacitance gives high modulation frequencies.  High Quantum efficiency. In contrast to pn junction built-in-field is uniform

23.  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. h  >E g p + i-Si e – h + W Drift Diffusion V r E Photodetectors The pin Photodiode

24. p-i-n diode (a) The structure; (b) equilibrium energy band diagram; (c) energy band diagram under reverse bias. Photodetectors The pin Photodiode

25. Photodetectors The pin Photodiode  The responsivity of pin photodiodes

26.  Quantum efficiency versus wavelength for various photodetectors. Photodetectors Photoconductive Detectors and Photoconductive gain

27.  Electric field of biased pin  Junction capacitance of pin Photodetectors The pin Photodiode  Small capacitance: High modulation frequency  RC dep time constant is  50 psec.  Response time  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. 

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

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 5  10 -2 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, = 600 nm, we need E ph = h  = E g so that, E g = hc/ = (6.626  10 -34 J s)(3  10 8 m s -1 )/(600  10 -9 m) = 2.07 eV (b)Area = 5  10 -2 cm 2 and I light = 20  10 -3 W/cm 2. The received power is P = Area  I light = (5  10 -2 cm 2 )(20  10 -3 W/cm 2 ) = 10 -3 W N ph = number of photons arriving per second = P/E ph = (10 -3 W)/(2.059  60218  10 -19 J/eV) = 2.9787   photons s -1 = 2.9787   EHP s -1.

33. (c) For GaAs, E g = 1.42 eV and the corresponding wavelength is = hc/ E g = (6.626  10 -34 J s)(3  10 8 m s -1 )  (1.42 eV  6  10 -19 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, g = hc/ E g = (6.626  10 -34 J s)(3  10 8 m s -1 )  (1.1 eV   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 E g, 1.1 eV, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation) 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 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

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 (a) If I 0 is the intensity of incoming radiation (energy flowing per unit area per second), I 0 exp(  d ) is the transmitted intensity through the specimen with thickness d and thus I 0 exp(  d ) is the “absorbed” intensity Solution

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  m? For Ge,   5.2  10 5 m -1 at 1.5  m incident radiation. For In 0.53 Ga 0.47 As,   7.5  10 5 m -1 at 1.5  m incident radiation. For Ge,   5.2  10 5 m -1 at 1.5  m incident radiation. For In 0.53 Ga 0.47 As,   7.5  10 5 m -1 at 1.5  m incident radiation. (b)

36. Example 0 0.2 0.4 0.6 0.8 1 80010001200140016001800 Wavelength (nm) Responsivity (A/W)  The responsivity of an InGaAs pin photodiode InGaAs pin Photodiodes Consider a commercial InGaAs pin photodiode whose responsivity is shown in fig. Its dark current is 5 nA. (a)What optical power at a wavelength of 1.55  m would give a photocurrent that is twice the dark current? What is the QE of the photodetector at 1.55  m? (b)What would be the photocurrent if the incident power in a was at 1.3  m? What is the QE at 1.3  m operation?

37. Solution (a)At  = 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  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.

38. Solution (b)At  = 1.3  10 -6 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), The QE at = 1.3  m is

39.  p + SiO 2 Electrode  net x x E (x) R h  >EgEg p I photo e – h + Absorption region Avalanche region Electrode n + E Photodetectors Avalanche Photodiode (APD) h +  n + p e – Avalanche region E e-e- h + E c E v  Impact ionization processes resulting avalanche multiplication  Impact of an energetic electron's kinetic energy excites VB electron to the CV.

40. SiO 2 Guard ring Electrode Antireflection coating n n n + p +  p Substrate Electrode n + p +  p Substrate Electrode Avalanche breakdown Si APD structure without a guard ring More practical Si APD Photodetectors Avalanche Photodiode (APD)  Schematic diagram of typical Si APD.  Breakdown voltage around periphery is higher and avalanche is confined more to illuminated region (n + p junction).

41. E Nn Electrode x E (x) R h  I ph Absorption region Avalanche region InP InGaAs h + e – InP P + n + V r V out E Photodetectors Heterojunction Photodiode Separate Absorption and Multiplication (SAM) APD InGaAs-InP heterostructure Separate Absorption and Multiplication APD P and N refer to p- and n-type wider-bandgap semiconductor.

42. InP InGaAs h + e – E c E v E c E v InP InGaAs E v E v h +  E v E Photodetectors Heterojunction Photodiode Separate Absorption and Multiplication (SAM) APD InGaAsP grading layer (a) Energy band diagram for a SAM heterojunction APD where there is a valence band step  E v from InGaAs to InP that slows hole entry into the InP layer. (b) An interposing grading layer (InGaAsP) with an intermediate bandgap breaks  E v and makes it easier for the hole to pass to the InP layer.

43. x R e – h + i ph h  >E g W V r Photogenerated electron concentration exp(  x) at timet = 0 B A v de E