ECEN 5645 Introduc0on to Optoelectronics Class Mee0ng 25. Non- PIN Solid State Detectors

Transcription

1 ECEN 5645 Introduc0on to Optoelectronics Class Mee0ng 25 Non- PIN Solid State Detectors

2 Today s Topics Avalanche Photodiodes Problem 5.6 APD Numerics and Examples Heterojunc0on Detectors Problem 5.10 Quantum Well Detectors SchoOkey Diodes Phototransistors Photoconduc0ve Detectors

3 Avalanche Photodiode (a) A schematic illustration of the structure of an avalanche photodiode (APD) biased for avalanche gain. (b) The net space charge density across the photodiode. (c) The field across the diode and the identification of absorption and multiplication regions.

4 Avalanche Photodiode (a) A pictorial view of impact ionization processes releasing EHPs and the resulting avalanche multiplication. (b) Impact of an energetic conduction electron with crystal vibrations transfers the electron's kinetic energy to a valence electron and thereby excites it to the conduction band

5 Avalanche Photodiode Gain or Multiplication M Multiplied photocurrent M = = Primary unmultiplied photocurrent M = 1 1 V V r br m I I ph pho

6 Avalanche Photodiode Typical multiplication (gain) M vs. reverse bias characteristics for a typical commercial Si APD, and the effect of temperature. (M measured for a photocurrent generated at 650 nm of illumination)

7 Avalanche Photodiode (a) A Si APD structure without a guard ring. (b) A schematic illustration of the structure of a more practical Si APD. Note: SiO 2 is silicon dioxide and serves as an insulating passivation layer.

8 Photodiode Comparison Photodiode λ range λ peak R at λ peak Gain I d For 1 mm 2 Features nm nm A/W GaP pin <1 1 nm UV detection a GaAsP pn < na UV to visible, covering the human eye, low I d. GaAs pin < na High speed and low I d Si pn < na Inexpensive, general purpose, low I d Si pin < na Faster than pn Si APD b na c High gains and fast Ge pin < µa IR detection, fast. Ge APD b µa c IR detection, fast InGaAs pin < na Telecom, high speed, low I d InGaAs APD b µa c Telecom, high speed and gain. InAs pn µm µm <1 >100 µa Photovoltaic mode. Normally cooled InSb pn µm 5 µm 3 <1 Large Photovoltaic mode. Normally cooled NOTE: c FGAP71 (Thorlabs); a At M = 1; c At operating multiplication.

9 Problem 5.6 Solu0on by Imbert Wang

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11 Avalanche Photodiode Gain or Multiplication M Ionization coefficient ratio is the ratio of hole to electron ionization k = h e α e = Aexp(-B/E) Chyoweth's law

12 Avalanche Photodiode Gain or Multiplication M Electrons only M = exp(α e w) Ionization coefficient Electrons and holes 1 k M = exp[ (1 k) α e w] k k = α h / α e

13 APD Characteristics Typical current and gain (M) vs. reverse bias voltage for a commercial InGaAs reach-through APD. I d and I ph are the dark current and photocurrent respectively. The input optical power is ~100 nw. The gain M is 1 when the diode has attained reach-through and then increases with the applied voltage. (The data extracted selectively from Voxtel Catalog, Voxtel, Beaverton, OR 97006)

14 EXAMPLE: InGaAs APD Responsivity An InGaAs APD has a quantum efficiency (QE, η e ) of 60 % at 1.55 µm in the absence of mul0plica0on (M = 1). It is biased to operate with a mul0plica0on of 12. Calculate the photocurrent if the incident op0cal power is 20 nw. What is the responsivity when the mul0plica0on is 12?

15 EXAMPLE: InGaAs APD Responsivity An InGaAs APD has a quantum efficiency (QE, η e ) of 60 % at 1.55 µm in the absence of mul0plica0on (M = 1). It is biased to operate with a mul0plica0on of 12. Calculate the photocurrent if the incident op0cal power is 20 nw. What is the responsivity when the mul0plica0on is 12? Solution The responsivity at M = 1 in terms of the quantum efficiency is = 0.75 A W eλ ( C)( m) R = ηe = (0.6) ( J s)( m s -1 hc ) If I pho is the primary photocurrent (unmul0plied) and P o is the incident op0cal power then by defini0on, R = I pho /P o so that I pho = RP o = (0.75 A W - 1 )( W) = A or 15 na. The photocurrent I ph in the APD will be I pho mul0plied by M, I ph = MI pho = (12)( A) = A or 180 na. The responsivity at M = 12 is Rʹ = I ph /P o = MR = (12) / (0.75) = 9.0 A W - 1

16 EXAMPLE: Silicon APD A Si APD has a QE of 70 % at 830 nm in the absence of mul0plica0on, that is M = 1. The APD is biased to operate with a mul0plica0on of 100. If the incident op0cal power is 10 nw what is the photocurrent?

17 EXAMPLE: Silicon APD A Si APD has a QE of 70 % at 830 nm in the absence of mul0plica0on, that is M = 1. The APD is biased to operate with a mul0plica0on of 100. If the incident op0cal power is 10 nw what is the photocurrent? Solution The unmul0plied responsivity is given by = 0.47 A W - 1 R eλ = ηe = hc 19 9 ( C)( m) (0.70) ( J s)( m s -1 ) The unmul0plied primary photocurrent from the defini0on of R is I pho = RP o = (0.47 A W - 1 )( W) = 4.7 na The mul0plied photocurrent is I ph = MI pho = (100)(4.67 na ) = 470 na or 0.47 µa

18 EXAMPLE: Avalanche multiplication in Si APDs The electron and hole ioniza0on coefficients α e and α h in silicon are approximately given by Eq. (5.6.4) with A cm - 1, B V cm - 1 for electrons (α e ) and A cm - 1 and B V cm - 1 for holes (α h ). Suppose that the width w of the avalanche region is 0.5 µm. Find the mul0plica0on gain M when the applied field in this region reaches V cm - 1, V cm - 1 and V cm - 1. What is your conclusion? What is the point of this Exercise?

19 EXAMPLE: Avalanche multiplication in Si APDs The electron and hole ioniza0on coefficients α e and α h in silicon are approximately given by Eq. (5.6.4) with A cm - 1, B V cm - 1 for electrons (α e ) and A cm - 1 and B V cm - 1 for holes (α h ). Suppose that the width w of the avalanche region is 0.5 µm. Find the mul0plica0on gain M when the applied field in this region reaches V cm - 1, V cm - 1 and V cm - 1. What is your conclusion? What is the point of this Exercise? Breakdown e w = log(k) 1 k M = 1 k exp[ (1 k) α w] e k Stability improves when k is small

20 EXAMPLE: Avalanche multiplication in Si APDs The electron and hole ioniza0on coefficients α e and α h in silicon are approximately given by Eq. (5.6.4) with A cm - 1, B V cm - 1 for electrons (α e ) and A cm - 1 and B V cm - 1 for holes (α h ). Suppose that the width w of the avalanche region is 0.5 µm. Find the mul0plica0on gain M when the applied field in this region reaches V cm - 1, V cm - 1 and V cm - 1. What is your conclusion? Solution At the field of E = V cm - 1, from Eq. (5.6.4) α e = Aexp(-B/E) = ( cm - 1 )exp[-( V cm - 1 )/( V cm - 1 )] = cm - 1. Similarly using Eq. (5.6.4) for holes, α h = cm - 1. Thus k = α h /α e = Using this k and α e above in Eq. (5.6.6) with w = cm, M = 4 exp[ ( )( cm)( cm -1 )] = 11.8 Note that if we had only electron avalanche without holes ionizing, then the mul0plica0on would be M e = exp (α e w) = exp[( cm - 1 )( cm)] = 7.65

21 EXAMPLE: Avalanche multiplication in Si APDs Solution (contiued) We can now repeat the calcula4ons for E = V cm - 1 and again for E = V cm - 1. The results are summarized in Table 5.3 for both M and M e. No0ce how quickly M builds up with the field and how a very small change at high fields causes an enormous change in M that eventually leads to a breakdown. (M running away to infinity as V r increases.) No4ce also that in the presence of only electron- ini4ated ioniza4on, M e simply increases without a sharp run- away to breakdown. E (V cm -1 ) a e (cm -1 ) a h (cm -1 ) k M M e Comment M and Me not too different at low E % increase in E, large difference between M and M e % increase in E

22 Heterojunction Photodiodes: SAM Simplified schematic diagram of a separate absorption and multiplication (SAM) APD using a heterostructure based on InGaAs-InP. P and N refer to p and n -type wider-bandgap semiconductor.

23 Heterojunction Photodiodes: SAM (a) Energy band diagrams for a SAM detector with a step junction between InP and InGaAs. There is a valence band step ΔE v from InGaAs to InP that slows hole entry into the InP layer. (b)an interposing grading layer (InGaAsP) with an intermediate bandgap breaks ΔE v and makes it easier for the hole to pass to the InP layer for a detector with a graded junction between InP and InGaAs. This is the SAGM structure.

24 Heterojunction Photodiodes: SAM Simplified schematic diagram of a more practical mesa-etched SAGM layered APD

25 Solu0on by Ian Barry Problem 5.10

26 HW Set 10 Problem 2 Kasap 5.10 P 0 = Pulse Energy / Pulse Length E V r / W = x 10 5 V/m v d = 8 x 10 4 m/s I ph = eλp 0 η e / hc = A t drik = W / v d = ns

27 A Different Solu0on J =en e v d Z I = J da = e N e W v d ê z N e = N ph = " pulse " photon The current is constant for ns. This is the substance of the Shockley- Ramo theorem. (The genie is out of the boole theorem). Find the magnitude and length of the current pulse.

28 Superlattice APD Multiple Quantum Well Detectors (a) Energy band diagram of a MQW superlance APD. (b) Energy band diagram with an applied field and impact ioniza0on.

29 Schottky Junction Photodiodes Schottky kunction type metalsemiconductor-metal (MSM) type photodetectors. (Courtesy of Hamamatsu) GaAsP Schottky junction photodiode for nm detection, from UV to red (Courtesy of Hamamatsu) GaP Schottky junction photodiode for 190 nm to 550 nm detection. (Courtesy of Hamamatsu) AlGaN Scottky junction photodiode for UV detection (Courtesy of sglux, Germany)

30 Schottky Junction (a) Metal and an n-type semiconductor before contact. The metal work function Φ m is greater than that of the n-type semiconductor (b) A Schottky junction forms between the metal and the semiconductor. There is a depletion region in the semiconductor next to the metal and a built-in field E o (c) Typical I vs. V characteristics of a Schottky contact device.

31 Schottky Junction Reverse biased Schottky junction and the dark current due to the injection of electrons from the metal into the semiconductor over the barrier Φ B.

32 Schottky Junction LEFT: Photogeneration in the depletion region and the resulting photocurrent. RIGHT: The Schottky junction photodetector

33 Schottky Junction Photodiodes Schottky junction based photodetectors and some of their features. τ R and τ F are the rise and fall times of the output of the photodetector for an optical pulse input. The rise and fall times represent the times required for the output to rise from 10% to 90% of its final steady state value and to fall from 90% to 10% of its value before the optical pulse is turned off. Schottky junction λ range nm R peak (at peak) (A/W) J dark per mm 2 Features with typical values GaAsP (610 nm) 5 pa UV to red, τ R = 3.5 µs. (G1126 series a ) GaP (440 nm) 5 pa UV to green, τ R = 5 µs. (G1961 a ) AlGaN (350 nm) 1 pa Measurement of UV; blind to visible light. (AG38S b ) GaAs (830 nm) ~ 1 na InGaAs MSM (1300 nm) 5 µa GaAs MSM (850 nm) 0.1 na Wide bandwidth > 10 GHz, τ R < 30 ps. (UPD-30- VSG-P c ) Optical high speed measurements, τ R = 80 ps, τ F = 160 ps. (G7096 a ) Optical high speed measurements, τ R = 30 ps, τ F = 30 ps. (G4176 a ) a Hamamatsu (Japan); b sglux (Germany); c Alphalas

34 Schottky Junction Photodiodes LEFT: The metal electrodes are on the surface of the semiconductor crystal (which is grown on a suitable substrate). RIGHT: The electrodes are configured to be interdigital and on the surface of the crystal.

35 Schottky Junction Photodiodes LEFT: Two neighboring Schottky junctions are connected end-to-end, but in opposite directions as shown for A and B. The energy band diagram without any bias is symmetrical. The grey areas represent the SCL 1 and SCL 2 at A and B. RIGHT: Under a sufficiently large bias, the SCL 1 from A extends and meets that from B so that the whole semiconductor between the electrodes is depleted. There is a large field in this region, and the photogenerated EHPs become separated and then drifted, which results in a photocurrent.

36 Phototransistor Transistor action I E exp(ev BE /k B T) Gain I ph βi pho

37 Photoconductive Detectors PbS (lead sulfide) photoconductive detectors for the detection of IR radiation up to 2.9 µm. They are typically used in such applications as radiation thermometers, flame monitors, water content and food ingredient analyzers, spectrophotometers etc.. (P9217 series) (Courtesy of Hamamatsu.)

38 Photoconductive Detectors A semiconductor slab of length l, width w and depth d is illuminated with light of wavelength λ

39 Photoconductive Detectors A photoconductor with ohmic contacts (contacts not limiting carrier entry) can exhibit gain. As the slow hole drifts through the photoconductors, many fast electrons enter and drift through the photoconductor because, at any instant, the photoconductor must be neutral. Electrons drift faster which means as one leaves, another must enter.

40 Photoconduc0vity Δσ and Photocurrent Density J ph Photon flux = Φ ph Photogeneration rate I ηi ηi AΦ ph hv ηiiλ g ph = = = Ad d hcd η i = Internal quantum efficiency Steady state illumination dδn dt = Δn gph = 0 τ Photoconductivity Δσ = eµ e Δn + eµ h Δp = eδn(µ e + µ h ) Δσ = eη Iλτ ( µ e + µ h hcd i ) J ph V = Δσ = ΔσE l

41 Photoconductive Gain Photon flux = Φ ph Rate of electron flow = I e = wdj ph ph µ e η ii wλτ ( µ e + h) E = hc Rate of electron generation = (Volume) gph = ( wdl ) gph = η I wl hc i λ Photoconductive gain G G = Rate of electron flow in external circuit Rate of electron generation by light absorption τ ( µ e + µ h)e = l

42 Photon flux = Φ ph Photoconductive Gain Rate of electron flow in external circuit G = Rate of electron generation by light absorption τ ( µ e + µ h)e = l Electron and hole transit times (time to cross the semiconductor) are t e = l / (µ e E) t h = l / (µ h E) Electron Hole G Photoconductive gain G τ τ τ µ = + = 1 + h t t t e h e µ e