光电子学与光子学的原理及应用sokasap课后答案

Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 1.4 Antireflection coatingggg For light traveling in medium 1 incident on the 1-2 interface at normal incidence, r12 = n1 n2n 1 + n2 = n1 n1n3n 1 + n1n3 = 1 n3n 1 1+ n3n 1 For light traveling in medium 2 incident on the 2-3 interface at normal incidence, r23 = n2 n3n 2 + n3 = n1n3 n3n 1n3 + n3 = n1 n3 1 n1 n3 +1 = 1 n3n 1 1 + n3n 1 thus, r23 = r12 Significance? For an efficient antireflection effect, waves A (reflected at 1-2) and B (reflected at 2-3) in Figure 1Q4 below should interfere with near “total destruction”. That means they should have the same magnitude and that requires that the reflection coefficient between 1 and 2 should be the same as that between 2 and 3; r12 = r23. Thus, the layer 2 can act as an antireflection coating if its index n2 = (n1n3)1/2. This can be achieved by r12 = r23. The best antireflection coating has to have a refractive index n2 such that n2 = (n1n3)1/2 = (1)(3.5)1/2 = 1.87. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3, both are close . The phase change for wave B going through the coating of thickness d is 2k2d where k2 = n2ko and ko = wavevector in free space = 2pi/. This should be 180 or pi. Thus we need 2n2(2pi/)d = pi or For SiO2 d = 4n 2 = 900 10 9 m 4(1.5) = 0.15 m For TiO2 d = 4n 2 = 900 10 9 m 4(2.3) = 0.10 m 1.8 Thin film coating and multiple reflections: Assume that n1 1 is Areflected A0 k = t1t2r 1 r1r2e j 2( )k (3) so that the reflection coefficient is r = AreflectedA 0 = r1 t1t2r 1 r1r2e j2( )k k =1 Since the Eq. (2) is a geometric series with terms given by Eq. (3), the summation is simple, r = r1 t1t2r 1 r1r2e j 2( ) 1 r1r2e j 2( ) = r1 + t1t2 r2e j2 1+ r1r2e j 2 (4) Using Eq. (1), Eq. (4), r = r1 1 + r1r2e j2( )+ 1 r 1 2( )r 2e j 2 1+ r1r2e j 2 i.e. r = r1 + r2e j 2 1+ r1r2e j 2 (5) The amplitude of the transmitted beam is Ctransmitted = C1 + C2 + C3 + . i.e. Ctransmitted /A0 = t1t23ej t1t23r1r2ej3 + t1t23r12r22ej5 + . (6) so that the k-th term is Ctransmitted A0 k = t1t23e j r1r2 r1r2e j2( )k (7) so that the transmission coefficient is t = CtransmittedA 0 = t1t23e j r1r2 r1r2e j 2( )k k=1 = t1t23e j r1r2 r1r2e j 2 1+ r1r2e j 2 (8) Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 i.e. t = t1t23e j 1+ r1r2e j 2 1.9 Antireflection coating: Consider the transmission coefficient obtained in Question 1.8, t = t1t2e j 1+ r1r2e j 2 r1 and r2 are positive numbers. To maximize t we need exp(j2) = 1. which means that exp(j2) = cos(2) + jsin(2) = 1. This will be so when 2 = mpi, where m is an odd-integer, or when = 2pin2d = m 12 pi leading to d = m4n 2 In addition we need r1r2 = 1. Consider choosing n2 = (n1n3)1/2. For light traveling in medium 1 incident on the 1-2 interface at normal incidence, r1 = r12 = n1 n2n 1 + n2 = n1 n1n3n 1 + n1n3 = 1 n3n 1 1 + n3n 1 For light traveling in medium 2 incident on the 2-3 interface at normal incidence, r2 = r23 = n2 n3n 2 + n3 = n1n3 n3n 1n3 + n3 = n1 n3 1 n1 n3 +1 = 1 n3n 1 1+ n3n 1 thus, r2 = r1 which confirms that we need n2 = (n1n3)1/2. The reflection coefficient from Question 1.8 is r = r1 + r2e j 2 1+ r1r2e j 2 This is zero (no reflection) when the numerator is zero, that is r1 = r2exp(j2) The magnitude of exp(j2) is unity and since r1 and r2 are positive quantities, we must have two conditions to obtain zero in the numerator: Condition 1: r2 = r1. This requires n2 = (n1n3)1/2 as derived above. Condition 2: exp(j2) = cos(2) + jsin(2) = 1 which will be so when 2 = mpi, where m is an odd-integer, or when Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 = 2pin2d = m 12 pi leading to d = m4n 2 1.16 Diffraction by a lens The angular position of the first dark ring is determined by the diameter D of the aperture and the wavelength , and is given by sin =1.22 D Since is small, = sin =1.22 D =1.22 590 10 9 2 102 = 3.610 -5 rad. From the Rayleigh criterion this is also the resolving power min of the lens. If f = focal length of the lens, the radius r of the central Airy disk is determined by = r/f r = f = (4010-2 m)( 3.610-5 rad) = 1.4410-5 m or 14.4 m. For nearly all practical purposes, this 29 m diameter spot at the focal plane is a point. 2.1 Dielectric slab waveguide From the geometry we have the following: (a y)/AC = cos and C/AC = cos(pi 2) The phase difference between the rays meeting at C is = kAC kAC = k1AC k1ACcos(pi 2) = k1AC1 cos(pi 2) = k1AC1 + cos(2) = k1(a y)/cos 1 + 2cos2 1 = k1 (a y)/cos2cos2 = 2k1(a y)cos Given, 2pi(2a)n1 cosm m = mpi cosm = (mpi + m )2pin 1(2a) = mpi +mk 1(2a) Then m = 2k1(a y) cosm m = 2k1(a y) mpi +mk 1(2a) m m = (1 ya)(mpi +m) m = mpi ya (mpi +m) m = m(y) = mpi ya (mpi + m ) 2.5 Dielectric slab waveguide Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 Given n1 = 3.66 (AlGaAs), n2 = 3.4 (AlGaAs), 2a = 210-7 m or a = 0.1 m, for only a single mode we need V = 2pia n12 n22( )1/ 2 2pia n1 2 n 2 2( )1/ 2 pi 2 = 2pi(0.1 m) 3.66 2 3.402( )1/ 2 pi 2 = 0. 542 m. The cut-off wavelength is 542 nm. When = 870 nm, V = 2pi(1 m) 3.66 2 3.402( )1/ 2 (0.870 m) = 0.979 2pia n1 2 n 2 2( )1/ 2 2.405 = 2pi(50 m) 1.4752 1.4552( )1/ 2 2.405 = 31.6 m For wavelengths longer than 31.6 m, the fiber is a single mode waveguide. The numerical aperture NA is NA = n12 n22( )1 / 2 = 1.4752 1.4552( )1/ 2 = 0.242 Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 If max is the maximum acceptance angle, then, max = arcsin NAn o = arcsin(0.242/1) = 14 Modal dispersion is given by intermodeL = n1 n2c = 1.475 1.4553 108 m s1 = 66.7 ps m-1 or 66.7 ns per km Given that 0.29, maximum bit-rate is BL = 0.25L total 0.25L intermode = 0.25(0.29)(66.7 ns km1) = 13 Mb s-1 km (only an estimate!) We neglected material dispersion at this wavelength which would further decrease BL. Material dispersion and modal dispersion must be combined by total2 = intermode2 + material2 For example, assuming an LED with a spectral rms deviation of about 20 nm, and a Dm 200 ps km-1 nm-1 (at about 850 nm)we would find m = (200 ps km-1 nm-1)(20 nm)(1 km) 4000 ps km-1 or 4 ns km-1, which is substantially smaller than the intermode dispersion and can be neglected. 2.9 A single mode fiber a Given n1 = 1.475, n2 = 1.455, 2a = 810-6 m or a = 4 m and =1.3 m. The V-number is, V = 2pia n12 n22( )1/ 2 = 2pi(4 m) 1.468 2 1.4642( )1/ 2 (1.3 m) = 2.094 b Since V 2.405 or kT/e ( 0.02586 V) where so = AJso = Aeni2(Dh/(LhNd) + (De/(LeNa) Given ni = 1.81012 m-3, A = 0.110-6 m2, Thus Iso = 0.110 6 m2( )1.602 1019 C( )1.8 1012 m3( )2 0.000803 m2 s1( ) 1.056 105 m( )11022 m3( ) + 0.110 6 m2( )1.602 1019 C( )1.8 1012 m3( )2 0.01813 m2 s1( ) 5.02 105 m( )11022 m3( ) Iso = 2.27 10-21 A. The forward current due to diffusion is Idiff = Isoexp(eV/(kT) = (2.27 10-21 A)exp(1 V)/(0.0259 V) Idiff = 0.00013 A or 0.13 mA 3.4 The Si pn junction Consider temperature T = 300 K. kT/e = 0.02586 V. a This is a p+n diode: Nd = 1015 cm-3. Hole lifetime h in the n-side is h = 5 10 7 1 + 2 1017 Ndopant( )= 5 107 1 + 2 1017 1015 cm3( )= 490.2 ns and, using the same equation with Ndopant = 1018 cm-3, electron lifetime in the p-side is e = 23.81 ns. I. Diffusion component of diode current Given Na = 1018 cm-3, e 250 cm2 V-1 s-1, and with Nd = 1015 cm-3, h 450 cm2 V-1 s-1. Thus: De = kTe/e (0.02585 V)(250 cm2 V-1 s-1) = 6.463 cm2 s-1 and Dh = kTh/e (0.02585 V)(450 cm2 V-1 s-1) = 11.63 cm2 s-1 The diffusion lengths are then: Le = Dee = 6.463 cm2 s-1)(23.81 10-9 s) = 3.92 10-4 cm, or 3.92 m and Lh = Dhh = (11.63 cm2 s-1)(490.2 10-9 s) = 2.39 10-3 cm, or 23.9 m The diffusion component of the current is I = Idiff = soexp(eV/(kT) 1 soexp(eV/(kT) for V kT/e ( 0.02585 V) Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 where so = AJso = Aeni2(Dh/(LhNd) + (De/(LeNa) Aeni2Dh/(LhNd) as Na Nd. In other words, the current is mainly due to the diffusion of holes in the n-region. Thus: Iso = 110 2 cm2( )1.60 1019 C( )1.45 1010 cm3( )2 11.63 cm2 s1( ) 2.39 103 cm( )11015 cm3( ) Iso = 1.64 10-12 A or 1.64 pA The forward current due to diffusion is Idiff = Isoexp(eV/(kT) = (1.64 10-12 A)exp(0.6 V)/(0.02585 V) Idiff = 0.020 A or 20 mA II. Recombination component The built-in potential is (where ni is the intrinsic concentration, found in the table in the inside front cover): Vo = (kT/e)ln(NdNa/ni2) = (0.02585 V)ln(1018 cm-3 1015 cm-3)/(1.45 1010 cm-3)2 Vo = 0.755 V The depletion region width W is mainly on the n-side (r of Si is 11.9 from Table 5.1). W = 2 Na + Nd( )Vo V( )eN a Nd 1 / 2 2 Vo V( )eN d 1/ 2 W = 2 11.9( ) 8.854 10 12 F m1( ) 0.755 V 0.6 V( ) 1.60 1019 C( )1021 m3( ) 1/ 2 i.e. W 0.451 10-6 m or 0.451 m Recombination in the wider depletion region in the n-side exceeds the recombination in the narrow depletion region in the p-side. Further the width of the depletion region on the n-side Wn W. The recombination time r here is not necessarily h but let us assume that is so (very roughly). Then, Iro = AeniWn2 h + AeniW p2 e AeniW2 h = 110 2 cm2( )1.602 1019 C( )1.45 1010 cm3( ) 0.451104 cm( ) 2 490 109 s( ) Iro = 1.070 10-9 A The forward current due to recombination is Irecom = Iroexp(eV/2kT) = (1.070 10-9 A)exp(0.6 V)/(2 0.02586 V) Irecom = 1.17 10-4 A or 0.117 mA Clearly, the diffusion component dominates the recombination component. b This is a symmetrical pn diode: Nd = Na = 1018 cm-3. Hole lifetime h in the n-side is h = e = 5 10 7 1 + 2 1017 Ndopant( )= 5 107 1+ 2 1017 1018 cm 3( ) = 23.81 10 -9 s Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 I. Diffusion component We are given, e 250 cm2 V-1 s-1, and with Nd = 1018 cm-3, h 130 cm2 V-1 s-1. Thus: Thus De = kTe/e (0.02585 V)(250 cm2 V-1 s-1) = 6.463 cm2 s-1 and Dh = kTh/e (0.02585 V)(130 cm2 V-1 s-1) = 3.361 cm2 s-1 Diffusion lengths are Le = Dee = 6.463 cm2 s-1)(23.81 10-9 s) = 3.923 10-4 cm or 3.92 m and Lh = Dhh = (3.361 cm2 s-1)(23.81 10-9 s) = 2.829 10-4 cm or 2.829 m The diffusion component of the current is Idiff = I = Isoexp(eV/kT) 1 Isoexp(eV/kT) for V kT/e ( 0.02585 V) where so = AJso = Aeni2(Dh/(LhNd) + De/(LeNa) Iso = 1102 cm2( )1.602 1019 C( )1.45 1010 cm3( )2 3.361 cm 2 s1( ) 2.829 104 cm( )1018 cm3( )+ 6.463 cm2 s1( ) 3.923104 cm( )1018 cm3( ) Iso = 9.554 10-15 A The forward current due to diffusion is Idiff = Isoexp(eV/kT) = (9.554 10-15 A)exp(0.6 V)/(0.02585 V) Idiff = 1.1510-4 A or 0.115 mA II. Recombination component The built-in potential is Vo = (kT/e)ln(NdNa/ni2) = (0.02585 V)ln(1018 cm-3 1018 cm-3)/(1.451010 cm-3)2 Vo = 0.933 V The depletion region width is symmetrical about the junction as Na = Nd. W = 2 Na + Nd( ) Vo V( )eN aNd 1/ 2 = 4 Vo V( )eN d 1/ 2 W = 4 11.9( ) 8.854 10 12 F m1( ) 0.933 V 0.6 V( ) 1.602 1019 C( )1024 cm3( ) 1/ 2 . W 2.96 10-8 m or 0.0296 m The recombination current pre-exponential term is Iro = AeWni2 r = 110 2 cm2( )1.602 1019 C( ) 0.0296 104 cm( )1.45 1010 cm3( ) 2 23.81109 s( ) Iro = 1.44 10-9 A The forward current due to recombination is Irecom = Iroexp(eV/2kT) = (1.4410-9 A)exp(0.6 V)/(2 0.02585 V) Irecom = 1.5810-4 A or 0.158 mA Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 3.5 AlGaAs LED emitter a We note that the emitted wavelength is related to the photon energy Eph by = c/ = hc/Eph. If we differentiate with respect to photon energy Eph we get ddE ph = hcE ph 2 We can represent small changes or intervals (or ) by differentials, e.g. /Eph |d/dEph|, then hcE ph 2 Eph We are given the energy width of the output spectrum, Eph = (h) 3kBT. Then, using the latter and substituting for Eph in terms of we find, 2 3kBThc or 2 3kBThc Temperature Parameter 40 C 25 C 85 C Comment peak (nm) 804 820 837 (nm) (Measured) 30 40 48 (nm) (Calculated); Eph = 2.5 kT 26.2 34.8 43.6 (nm) (Calculated); Eph = 3 kT 31.4 41.7 52.3 Very close B B B 0 10000 20000 30000 40000 50000 60000 70000 0 100 200 300 400 /(2) Temperature (K) Best line forced through zero is /(2) = 1956T ; R2 = 0.9932374 The theory predicts that /2 vs T should be a straight line because, 2 Ephhc so that 2 = mkBT hc where Eph = mkBT and m is a numerical constant that represents the ratio (h)/(kBT) and is determined from the slope of the /2 vs. T plot. Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 The three points plotted in the figure seems to follow this behavior. The best line forced through zero has a slope that indicates m = 2.8. b The bandgap decreases with temperature c There are two factors to consider. (i) Spectral intensity means intensity per unit wavelength, that is, dI/d. The integration of the spectral curve gives the total intensity, the total number of photons emitted per unit area per unit time. As the spectrum broadens with temperature we would naturally expect the peak to decrease with temperature. (ii) Higher the temperature, the stronger are the lattice vibrations (there are more phonons). Indirect or radiationless transitions, those that do not emit photons, require phonons (lattice vibrations) which encourage indirect transitions. Thus increasing the temperature increases indirect transitions at the expense of direct transitions and the light intensity decreases. If (ii) was totally absent then the areas under the curves for all the three spectra would be identical. d Use the peak emission wavelength to find Eg as follows: At -40 C (233 K) , peak = 804 nm. At 25 C (298 K), peak = 820 nm. At 85 C (358 K), peak = 837 nm. We first note that we need the required bandgap Eg at the wavelength of interest. The photon energy at peak emission is hc/peak = Eg + kBT. Then, Eg = che peak kBTe and at peak = 820109 m, taking T = 25 + 273K, Eg = (310 8 )(6.626 1034) (1.6 1019)(820 109) 0.0257 eV = 1.4863 eV e The bandgap Eg of the ternary alloys AlxGa1-xAs follows the empirical expression, Eg(eV) = 1.424+1.266x+0.266x2. Eg(eV) = 1.4863 = 1.424 + 1.266x + 0.266x2. Solving for x we find, x = 0.05. f From the definition of efficiency , = Output optical powerInput electrical power = PoIV = 25 10 6 W (40 103 A)(1.5 V) = 0.000417 = 0.0417% 3.7 External conversion efficiency ext = PoIV = 2.5 10 3 W (50 103 A)(1.6 V) = 0.03125 = 3.125 % 3.8 Linewidth of LEDs Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 1/2/3/7 a72 a72a72 a72 a72 a72 a72 a72 0 20 40 60 80 100 120 140 160 0 1000000 2000000 3000000 AlGaAs AlGaAsAlGaAs GaAsGaAs InGaAsP InGaAsP InGaAsP 1/2 (nm) 2 (nm)2 The best line forced through zero is 1/2 = 6.5710-5 (nm)-12 and Slope = 6.5710-5 (nm)-1. This slope is mkT/hc; thus m 3.15. Very close to the theoretically predicted value of m = 2.5 3. Table 2Q8-3 shows the calculated spectral widths 1/2 using m = 3. The actual observed widths are substantially larger than the expected 1/2 using m = 3, as we found above for the direct bandgap materials. If the recombination center were a discrete level, we would expect a spread in the photon energy that is controlled by the energy distribution of holes (electrons) in the valence (conduction) band, that is about 1.5kBT, or m = 1.5. The observed spread is much more than m = 1.5 and hence the energy level cannot be discrete. ( It is possible to give a semiquantitative plausible explanation as follows. A captured electron will have a wavefunction that is localized and hence a smaller uncertainty in its position than in the band; i.e. x will be small. That means the uncertainty p in its momentum will be higher and hence the uncertainly in its energy will also be higher. We would expect that the spread of photon energies will be more than from than in band to band recombinations.) Table 3Q8-3 Linewidth 1/2 between half points in the output spectrum (Intensity vs wavelength) of four various visible LEDs using SiC and GaAsP materials. Peak wavelength of emission () nm 468 565 583 600 635 1/2 nm 66 28 36 40 45 Expected 1/2 nm using m = 3 13.7 20.0 21.3 22.5 25.2 Color Blue Green Yellow Orange Red Material SiC (Al) GaP (N) GaAsP (N) GaAsP (N) GaAsP 3.10 LED-Fiber coupling Efficiency a overall = PoIV = 200 10 6 W (75 103 A)(1.5 V) = 1.810 -3 = 0.18 % b (i) overall = PoIV = 48 10 6 W (120 103 A)(1.3 V) = 0.0307% Solutions for Optoele