70 Chapter 17 17.1 We wish to use a photodiode as a detector for a signal of 9000 Å wavelength. Which would be the best choice of material for the photodiode, a semiconductor of bandgap = 0.5 ev, bandgap = 2 ev, or bandgap = 1 ev? Why? (Assume all three are direct gap and are equivalent in impurity content, etc.) λ 1 = 1.24 = 2.48 µm 0.5 λ 2 = 1.24 = 6200 Å 2 λ 3 = 1.24 = 1.24 µm 1 λ 2 cutoff at too short a wavelength λ 1 cutoff at to long a wavelength (increases bandwidth for noise reduces S/N). 17.2 To improve the signal-to-noise ratio of the diode in Problem 17.1, we wish to use a semiconductor low-pass filte which has the following absorption properties at room temperature: for 9000 Å radiation α = 0.2cm 1 for 7000 Å radiation α = 10 3 cm 1. How thick must the filte be to attenuate 7000 Å background noise by a factor of 10 4? By what factor is the signal (at 9000 Å) attenuated by a filte of this thickness? Neglect reflectio at the surfaces. I e αz I 0 1 10 4 = e 103 z z = 9.21 10 3 cm I = e 2 10 1 9.21 10 3 = 0.999 I 0 17.3 If the minimum useful photocurrent of the diode in Problem 17.1 is 1 µa (peak pulse value), what is the minimum signal light intensity (peak pulse value) which must fall on the detector? Assume an internal yield or quantum efficien y η q = 0.8 and sensitive area = 10 mm 2.
71 10 6 coul/sec 1.25 photons/carrier = 7.8 10 12 1.6 10 19 photons/sec. coul/carrier 7.8 10 12 photons/sec 1.38 ev/photon 1.60 10 19 joules/ev 10 1 cm 2 = 1.72 10 5 watts/cm 2 17.4 Below is a list of semiconductor materials and their bandgaps. E g [ev] Si 1.1 GaAs 1.4 GaSb 0.81 GaP 2.3 InAs 0.36 a) Based solely on bandgap energy, which of these materials could possibly be used to make a detector for the light from a GaAs laser? b) Which of the following ternary compounds could possibly be used to make a detector for the light from a GaAs laser? GaAs (1 x) Sb x GaAs (1 x) P x Ga (1 x) In x As c) If a reverse biased Si photodiode with a quantum efficien y of η = 80% and an area of 1 cm 2 is uniformly illuminated with the light from a GaAs laser to an intensity of 10 mw/cm 2 what is the photocurrent which f ows? a) For eff cient detection you need E g < E photon and for a GaAs laser E phot 1.4eV Si, GaSb, InAs GaP will work will not work b) GaAs 1 x Sb x and Ga 1 x In x As would work because GaSb and InAs have bandgaps less than that of GaAs. GaAs 1 x P x would not work because the bandgap of GaP exceeds that of GaAs. c) 10 10 3 W/cm 2 1cm 2 = 10 2 W of optical power, each photon has 1.24/0.9 1.6 10 19 = 2.20 10 19 joules of optical power the number of photons striking the detector is 10 2 2.2 10 19 = 4.55 1016 photons/sec at 80% quantum efficien y the current is 4.55 10 16 0.8 1.6 10 19 = 5.82 ma
72 17.5 a) Determine the maximum value of the energy gap which a semiconductor, used as a photodetector, can have if it is to be sensitive to light of wavelength λ 0 = 0.600 µm. b) A photodetector whose area is 5 10 6 m 2 is irradiated with light whose wavelength is λ 0 = 0.600 µm and intensity is 20 W/m 2. Assuming each photon generates one electron-hole pair, calculated the number of pairs generated per second. c) By what factor does the answer to (b) change if the intensity is reduced by a factor of 1/2? d) By what factor does the answer to (b) change if the wavelength is reduced by a factor of 1/2? a) E = hν = hc λ = 6.6 10 34 3 10 8 6000 10 10 = 3.3 10 19 joule 3.3 10 19 = = 2.06 ev 1.6 10 19 b) 20 Watts m 2 5 10 6 m 2 = 10 4 watts pairs generated sec = 10 4 joule/sec 3.3 10 19 joules/pair = 3 1014 pairs/sec c) Intensity = 20 Watts/cm 2 1 2 = 10 Watts/cm2 pairs/sec reduced by factor of 1 2 6000 10 10 d) If λ = = 3000 10 10 2 Energy per photon = E = hc λ = 6.6 10 19 joule 10 4 joules/sec 6.6 10 19 also a factor of 1 joule 2 17.6 We wish to design a waveguide photodiode in GaAs, with the geometry as shown in Fig. 17.5, for operation at λ 0 = 0.900 µm wavelength. a) If the photodiode has a waveguide thickness and depletion width W = 3 µm at an applied reverse bias voltage of 40.5 V, what is the magnitude of the change in effective bandgap due to the electric f eld? (Assume m = 0.067m 0 ). b) What length L is required to produce a quantum efficien y of 0.99? (Assume that scattering loss and free carrier absorption are negligible).
73 a) The fiel in the depletion layer is E = 40.5 3 10 4 V/cm = 1.35 105 V/cm or 1.35 10 7 V/m From (9.19) E = 3 2 (m ) 1/3 (qeh) 2/3 = 3 2 (9.1 10 31 0.067 kg) 1/3 [(1.6 10 19 1.35 10 7 V/m) (1.05 10 34 Js)] 2/3 = 3 2 (3.72 10 31 )m kg 4/3 /s 2 (3.94 10 11 ) kg 1/3 = 1.42 10 20 J 1.42 10 20 = ev 1.6 10 19 = 0.089 ev b) From Fig.9.2, α = 10 4 cm 1 η q = 1 e αl 0.99 = 1 e 104 L 10 4 L = ln 0.01 L = 4.6. µm L = 4.6. 10 4 cm
80 Chapter 18 18.1 Explain the difference between a multiple quantum well structure and a superlattice. See text Sect. 16.1. 18.2 For a deep quantum well of GaAs with a thickness L z = 70 Å, L x = L y = 100 µm, calculate the magnitude of the steps in the conduction band cumulative density of states function and the energies at which they occur relative to the bottom of the well for the n = 1, 2 and 3 levels. (Assume that the well has infinitel high sidewalls.) ϱ(e) m e πh 2 = 4m eπ h 2 = 4(0.08)(9.1 10 31 )π (6.625 10 34 ) 2 = 2.08 10 36 K y = K x = nπ 10 4 n = 1, 2, 3 E n = h2 2m e ( ) nπ 2 + h2 L z E n = (1.054 10 34 ) 2 2(0.08)9.1 10 31 (Kx 2 2m + K y 2 ) n = 1, 2, 3 e ( ) nπ 2 7 10 9 + (1.504 10 34 ) 2 ( ( nπ ) 2 ( nπ ) ) 2 + 2(0.08)9.1 10 31 10 4 10 4 ( ) nπ 2 E n = 7.36 10 38 7 10 9 + 7.63 10 38 (2) for n = 1 ( nπ 10 4 ) 2 E 1 = 7.63 10 38 (2.01 10 17 ) + 7.63 10 38 (2)(9.87 10 8 ) = 1.53 10 20 + 1.51 10 28 = 1.53 10 20 Joule = 0.0956 ev E 2 = 1.53 10 20 (4) + 1.51 10 28 (9) = 6.12 10 20 =0.383 ev E 3 = 1.53 10 20 (9) + 1.51 10 28 (9) = 1.38 10 19 =0.860 ev 18.3 If the quantum well of Problem 18.2 were incorporated into a diode laser, what would be the emission wavelength for electron transitions between the n = 1 level in the conduction band and the n = 1 level in the valence band? (Assume 300 K operation.) From problem 16.2 the energy of a conduction band electron in n = 1 level E 1 = 0.0956 ev
For a hole in the valence band n = 1 level, the equation is the same except that m e = 0.08 m 0 is replaced by m h = 0.5 m 0, thus for the hole E 1 = 0.0956 0.08 0.5 = 0.0153 ev 81 0.0956 ev E 1 E c E g = 1.38 ev 0.0153 ev The photon energy is E ph = 0.0956 + 1.38 + 0.0152 = 1.49 ev The emission wavelength is λ = 1.24 = 1.24 = 0.832 µm E ph 1.49 E v E 1 18.4 a) Explain how a single p-i-n MQW diode can function as a self-electro-opticeffect device (SEED). b) Why are SEEDs envisioned as being possible digital logic elements in See text Sect. 16.5. 18.5 A single-quantum well of GaAs with a thickness of 60 Å is incorporated into a GaAlAs heterojunction laser. The laser has a stripe geometry, being 200 µm long and 10 µm wide. a.) What is the emission wavelength if the energy bandgap is E g = 1.52 ev in the emitting region and electron transitions occur between the n = 1 level in the conduction band and the n = 1 level in the valence band? b). If the gain coefficien of the laser in part (a) is g = 80 cm 1, what would be the gain coefficien of a multiple quantum well laser with 20 identical wells? ( ) a.) E n = 2 nπ 2 + 2 (k 2 x 2m e L z 2m + k2 y ) e where k x = nπ/l x and k y = nπ/l y taking n = 1 and realizing that k 2 x and k2 y are negligible since L x, L y L z ( ) E 1 = 2 π 2 (1.055 10 34 ) 2 π 2 = 2m e L z 2 0.08 9.1 10 31 (6 10 9 = 2.123 10 2 ) 2 2.123 10 20 E 1 = = 0.13 ev 1.60 10 19
82 for holes m changes from 0.08 m e to 0.5 m e then E 1 = 0.13 0.08 0.5 The photon energy is = 0.02 ev E phot = w = E g + E 1C + E 1V = 1.52 + 0.13 + 0.02 = 1.67 ev The emission wavelength is λ 0 = 1.24/1.67 = 0.743 µm b). 20 80 = 1600 cm 1
83 Chapter 19 19.1 If a silicon rod of 1 mm diameter is exposed to a tensional force by supporting a 3 g weight, a. What axial stress does the rod experience? b. What is the axial strain? c. What is the longitudinal strain? a. G = F/π(θ/2) 2 (1lb/in 2 = 703.1kg/m 2 ) = 3 10 3 kg π(10 3 /2) m 2 = 3819 kg/m2 = 5.43 lb/in 2 b. G = εe εe = G E = 5.43 lb/in 2 190 GPa 1.45 10 5( lb /GPa ) in 2 = 1.97 10 7 tension c. ν = ε l /ε a ε l = ν ε a = ( 0.28 1.97 10 7 ) = 5.5 10 8 19.2 A square membrane of silicon, 7 mm on a side and 150 µm thick, is exposed uniformly over its surface to a pressure of 1 10 4 kg/m 2. a. What is the deflectio at the center point of the membrane? b. What is the maximum longitudinal stress? c. What is the maximum transverse stress? d. What is the resonant frequency of the membrane?
84 a. D = EA 3 /[12(1 ν 2 )] = 190 GPa (150 10 6 ) 3 [12(1 0.28 2 )] = 6.96 10 10 GPa m 2 w max = 0.001265 Pa 4 /D = = 6.41 10 10 (1 0.28 2 ) 0.001265 kg/m 2 (7 10 3 ) 4 m 4 6.96 10 10 GPa m 2 1.02 10 8 (kg/m 2 )/GPa = 6.41 10 10.9216 = 4.28 10 11 P meters = 4.28 10 5 P µm with P in kg/m 2 take P = 10 4 kg/m 2 w max = 4.28 10 1 µm = 0.428 µm b. G l = 0.3081 P(a/t) 2 = 0.3081 10 4 (7/0.15) 2 = 6.71 10 6 kg/m 2 c. G t = νg l = 0.28 6.71 10 6 = 1.88 10 6 kg/m 2 d. F 0 = (1.65 t/a 2 )[E/e(1 ν 2 )] 1/2 1.654 150 10 6 (7 10 3 ) 2 [ 190 1.02 10 8 2.3 10 3 (1 0.28) = 5.063 9.14 10 6 = 4.63 10 7 = 46.3MHz ] 1/2 19.3 A silicon cantilever beam, 1.5 mm long, 500 µm wide and 250 µm thick, is loaded with a uniformly distributed force per unit width of 100 kg/m 2. a. What is the deflectio at the point 0.5 mm from the free end? b. What is the maximum stress? c. What is the frequency of the fundamental vibrational mode?
85 a. W(P, x) = (Px 2 /24 EI)(6L 2 4Lx + x 2 ) I = at 3 /12 = 500 10 6 (250 10 6 ) 3 = 6.51 10 16 12 W(P, 1 10 3 P 1 10 6 ) = 24 190 1.02 10 +8 6.51 10 16 ( 6(1.5 10 3 ) 2 42x x 2 ) = P 3.30 10 3 (1.35 10 5 4 1.5 10 3 1 10 3 1 10 6 ) = P 3.30 10 3 (6.5 10 6 ) = P 2.1 10 8 let P = 100 kg/m w = 2.1 10 6 m = 2.1 µm b. G max = PL2 t = 100 (1.5 10 3 ) 2 (250 10 6 ) 4I 4 6.51 10 16 = 2.16 10 7 kg/m 2 c. F 0 = 0.161(t/L 2 )(E/C) 1/2 ( 250 10 6 )( 190 1.02 10 8 = 0.161 (1.5 10 3 ) 2 2.3 10 3 = 3.23 10 5 = 323 khz ) 1/2