光电子学与光子学课后答案4-6章

Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 4.8 Fabry-Perot optical resonator a. Separation of the modes is, m = f = c2L = (3 10 8 m s-1) 2(0.5 m) = 310 8 Hz = 300 MHz. The finesse is F = piR 1/ 2 1 R = pi 0.991 / 2 1 0.99 = 312.6 and each mode width, spectral width, is m = fF = 3 10 8 312.6 = 9.610 5 Hz = 960 kHz. b Cavity mode nearest to the emission wavelength is m = 2L / n = 2(200 10 6 ) (1300 109 )/ 3.7 =1138.46 i.e. m = 1138. Separation of the modes is, m = f = c /n2L = (3 10 8 m s-1 ) / 3.7 2(200 106 m) = 2.0310 11 Hz The finesse is F = piR 1/ 2 1 R = pi 0.81/ 2 1 0.8 = 14.05 and each mode width, spectral width, is m = fF = 2.03 10 13 14.05 = 1.410 10 Hz. 4.10 Threshold current and power output from a laser diode a If Nph is the coherent radiation photon concentration, then only half of the photons, 1/2Nph, in the cavity would be moving towards the output face of the crystal at any instant. It takes t = nL/c seconds for photons to cross the laser cavity length L. Po = Energy flow per unit time in cavity towards face Transmittance = hc 1 2 Nph( )dWL( ) t Transmittance = hc 1 2 Nph( )dWL( ) Ln c 1 R( ) = hc 2 N phdW 2n 1 R( ) and I = Optical PowerArea = PodW = hc 2 N ph 2n 1 R( ) Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 where R is the reflectance of the crystal face. b Consider one round trip through the cavity. The length L is traversed twice and there is one reflection at each end. The overall attenuation of the coherent radiation after one-round trip is RRexp(2L) where R is the reflectance of the crystal end. Equivalently we can represent this reduction by an effective or a total loss coefficient t such that after one round trip, the reduction factor is expt(2L) Equating the two, RRexp(2L) = expt(2L) and rearranging, t = + 12L ln 1R2 c The reflectance is R = n 1n +1 2 = 3.5 13.5 +1 2 = 0.309 The total loss coefficient is t = + 12L ln 1R2 =1000 m1 + 160 106 m1 ln 10.3092 = 2.06104 m-1. ph = nc t = 3.5(3 108 m s1)(2.06 108 m1) = 5.710-13 s (0.57 ps) Coherent radiation is lost from the cavity after, on average, 0.57 ps. For the above device, threshold current density Jth 500 A cm-2 and sp 10 ps, d 0.25 m, From Jth = nthed sp we have, nth Jth sped (500 10 4 A m-2)(10 109 s) (1.6 1019C)(0.25 106 m) 1.2510 24 m-3 or 1.21018 cm-3 Now, the current density corresponding to I = 30 mA is J = I/(WL) = (0.05 A)/106010-610-6 m2) = 833104 A m-2. And, N ph phed J Jth( ) (5.7 10 13s) (1.6 1019C)(0.25 106 m) (833 500) 10 4 A m-2 4.71019 photons m-3 The optical power is Po = hc 2N phdW 2n 1R( ) = (6.62 10 34 J s)(3108 m s-1)2(4.7 1019 m-3)(0.25 106 m)(10 106 m) 2(3.5)(1310 109 m) 1 0.309( ) Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 0.00053 W or 0.53 mW. Intensity = Optical Power / Area = Po / (dW) = (0.00053)/0.251010-310-3 mm2) = 223 W mm-2. This intensity is right at the crystal face over the optical cavity cross section. As the beam diverges, the intensity decreases away from the laser diode. 4.12 Laser diode efficiency a The external quantum efficiency EQE, of a laser diode is defined as EQE = Number of output photons from the diode per unit second( )Number of injected electrons into diode per unit second( ) EQE = Optical Power / hDiode Current / e = Po /EgI / e = ePoIE g The external differential quantum efficiency, EDQE, of a laser diode is defined as EDQE = Increase in number of output photons from diode per unit second( )Number of injected electrons into diode per unit second( ) EDQE = (Change in Optical Power) / h(Change Diode Current) / e = Po /EgI / e = eE g dPo dI The external power efficiency, EPE, of the laser diode is defined by EPE = Optical ouput powerElectical input power = PoIV = PoIV eEgeE g = ePoIE g EgeV EPE = ePoIE g EgeV = EQE EgeV b 670 nm laser diode Eg hc/ = (6.62610-34)(3108)/(67010-9)(1.610-19) = 1.85 eV, so that EQE = (1.6 10 19 C)(2 103 Js1) (80 103 A)(1.85 eV 1.6 1019 eV/J) = 0.0135 or 1.35% EDQE = eE g dPo dI = (1.6 1019 C) (1.85 eV 1.6 1019 C) 3 103 2 103 Js1 82 103 80 103 A = 0.27 or 27%. EPE = PoIV = 2 10 3 W (80 103 A )(2.3 V) = 0.011 or 1.1% c 1310 nm laser diode Eg hc/ = (6.62610-34)(3108)/(131010-9)(1.610-19) = 0.9464 eV, so that EQE = (1.6 10 19 C)(3 103 Js1 ) (40 103 A)(0.9464 eV 1.6 1019 J/eV) = 0.079 or 7.9% Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 EDQE = eE g dPo dI = (1.6 1019 C) (0.9464 eV 1.6 1019 C) 4 103 3 103 Js1 45 103 40 103 A = 0.21 or 21%. and EPE = PoIV = 3 10 3 W (40 103 A )(1.4 V) = 0.054 or 5.4% 4.15 The SQW laser The lowest energy levels with respect to the CB edge Ec in InGaAs are determined by the energy of an electron in a one-dimensional potential energy well n = h 2n2 8me*d2 where n is a quantum number 1, 2, , n is the electron energy with respect to Ec in InGaAs, or n = En Ec. Using d = 1010-9 m, me* = 0.04me and n = 1 and 2, we find the following electron energy levels n =1 1=n = h 2n2 8me*d2 = (6.626 1034 )2 (1)2 8(0.04 9.11 1031)(10 109 )2 = 1.5110 -20 J = 0.094 eV n = 2 2 = 0.376 eV Using d = 1010-9 m, mh* = 0.44me and n = 1, the hole energy levels below Ev is n =1 n = h 2n2 8mh*d2 = (6.626 1034 )2 (1)2 8(0.44 9.11 1031)(10 109 )2 = 1.3710 -21 J = 0.00855 eV The wavelength light emission from the QW laser with Eg = 0.70 eV is QW = hcE g + 1 + 1 = (6.626 10 34)(3108 ) (0.70 + 0.094 + 0.00855)(1.602 1019) = 154510 -9 m (1545 nm) The wavelength of emission from bulk InGaAs with Eg = 0.70 eV is g = hcE g = (6.626 10 34 )(3 108) (0.70)(1.602 1019) = 177110 -9 m (1771 nm) The difference is g QW = 1771 - 1545 = 226 nm. 4.16 A GaAs quantum well The lowest energy levels with respect to the CB edge Ec in GaAs are determined by the energy of an electron in a one-dimensional potential energy well n = h 2n2 8me*d2 where n is a quantum number 1, 2, , n is the electron energy with respect to Ec in GaAs, or n = En Ec. Thus, n =1 1=n = h 2n2 8me*d2 = (6.626 1034 )2 (1)2 8(0.07 9.111031)(8 109 )2 = 0.0839 eV n = 2 2 = 0336 eV n = 3 3 = 0.755 eV Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 Note: Whether 3 is allowed depends on the depth of the QW and hence on the bandgap of the sandwiching semiconductor. The hole energy levels below Ev is n = h 2n2 8mh*d2 = (6.626 1034)2(1)2 8(0.47 9.111031)(8 109)2 = 0.0125 eV The wavelength of emission from bulk GaAs with Eg = 1.42 eV is g = hcE g = (6.626 10 34 )(3 108 ) (1.42)(1.602 1019) = 87410 -9 m or 874 nm. Whereas from the GaAs QW, the wavelength is, QW = hcE g + 1 + 1 = (6.626 10 34)(3108 ) (1.42 + 0.0839 + 0.0125)(1.602 1019) = 81810 -9 m or 818 nm. The difference is g QW = 874 818 =56 nm. 5.1 Bandgap and photodetection a Given, = 600 nm, we need Eph = h = Eg so that, Eg = hc/ = (6.62610-34 J s)(3108 m s-1)/(60010-9 m) = 2.07 eV b A = 510-2 cm2 and Ilight = 2010-3 W/cm2. The received power is P = AIlight = (510-2 cm2)(2010-3 W/cm2) = 10-3 W Nph = number of photons arriving per second = P/Eph = (10-3 W)/(2.071.60218 10-19 J/eV) = 2.97871015 Photons s-1. = 2.97871015 EHP s-1. c For GaAs, Eg = 1.42 eV and the corresponding wavelength is = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.42 eV 1.6 10-19 J/eV)= 873 nm The wavelength of emitted radiation due to EHP recombination is therefore 873 nm. It is not in the visible region (it is in the IR). d For Si, Eg = 1.1 eV and the corresponding cut-off wavelength is, g = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.1 eV 1.6 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 Eg, 1.1 eV, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation) 5.2 Absorption coefficient a If Io is the intensity of incoming radiation (energy flowing per unit area per second), Ioexp(d) is the transmitted intensity through the specimen with thickness d (Figure 5.17) and thus Io1 exp(d) is the “absorbed” intensity. If ph is the number of photons arriving per unit area per unit second (the Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 photon flux), then ph = Io/h where h is the energy per photon. Thus the number of photons absorbed per unit volume per unit second of sample is nph = AphAd = Io 1 exp(d) h d = Io 1 exp(d) dh b For Ge, 5.2 105 m-1 at 1.5 m incident radiation (from figure 5.3). 1 exp(d) = 0.9 d = 1 ln 11 0.9 = 15.2 105 ln 11 0.9 = 4.428 106 m = 4.428 m. For In0.53Ga0.47As, 7.5 105 m-1 at 1.5 m incident radiation (Figure 5.3). d = 17.5 105 ln 11 0.9 = 3.07 106 m = 3.07 m. c The quantum efficiency is unity. Therefore the collected electrons per unit area per unit second is equal to the absorbed photons per unit area per unit second. So, the current density (current per unit area), hc deh deJ ph )exp(1)exp(1 = oo II Given, Io = 100 W mm-2 = 100 10-6 106 W m-2 = 100 W m-2, 81.108100.310626.6 9.0105.11001060218.1 834 619 = = phJ A m -2 =10.881 mA/cm2. NOTE: We neglected any light reflection from the surface of the semiconductor crystal (100% efficient AR coating assumed). 5.3 Ge Photodiode a At = 85010-9 m, from the responsivity vs. wavelength curve we have R = 0.25 A/W. From the definitions of quantum efficiency (QE) and responsivity we have, %36.5)m10850()C1060218.1( )A/W25.0()ms103(Js)10626.6( 919 1834 = = e hcR Similarly, we can calculate quantum efficiency at other wavelengths. The results are summarized in the following table. Wavelength, (nm) 850 1300 1550 Responsivity R, (A/W) 0.25 0.57 0.73 Quantum efficiency , (%) 36.5 54.3 58.4 b Given, photocurrent Iph = Id = 0.3 A = 0.310-6 A and area, A = 810-9 m2, the incident optical power, Po = Iph/R = (0.310-6 A)/(0.73 A W-1) = 4.1096 10-7 W Light intensity, Io = P0/A = (4.1096 10-7 W)/(810-9 m2) = 51.4 W m-2 or 5.14 mW cm-2. c From semiconductor data under Selected Semiconductors, for most semiconductors dEg/dT is negative, Eg increases with decreasing temperature. Stated differently, vs curve shifts towards shorter with decreasing T. The change in with T means that the amount of optical power absorbed in the depletion region and hence the quantum efficiency will change with temperature. The peak Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 responsivity will shift to lower wavelengths with decreasing temperature. If maximum photogeneration requires a certain absorption depth and hence a certain max, then the same max will occur at a lower wavelength at lower temperatures. In Figure 5Q3, maximum responsivity corresponds to max which occurs at max at high T and at max at lower T. Figure 5Q3. Absortion coefficient = 1/ Low T High T maxmax max The absorption coefficient depends on the temperature d Dark current ( exp(Eg/kT) will be drastically reduced if we decrease the temperature. Reduction of dark current improves SNR. e The RC time constant = 100 (410-12) = 0.4 ns. The RC time constant is comparable to the rise time, 0.5 ns. Therefore, the speed of response depends on both the rise time and RC time constant. (It is not simply 0.4 ns + 0.5 ns.) 5.4 Si pin Photodiodes a For type A, responsivity RA = 0.2 A/W at 450 nm wavelength light. Given, intensity Io = 1 W cm-2 = 10-8 W mm-2 and area A = 0.125 mm2. Power, P0 = IoA = (10-8 W mm-2) (12.5 mm2) = 1.25 10-7 W. Photocurrent, Iph = RAP0 = (0.2 A/W) (1.25 10-7 W) = 2.5 10-8 A = 25 nA. Quantum efficiency, %1.55)m10450()C1060218.1( )A/W2.0()ms103(Js)10626.6( 919 1834 = = e hc A AR For type B, responsivity RB = 0.12 A/W at 450 nm wavelength light. Photocurrent, Iph = RBP0 = (0.12 A/W) (1.25 10-9 W) = 1.5 10-10 A = 0.15 nA. Quantum efficiency, %1.33)m10450()C1060218.1( )A/W12.0()ms103(Js)10626.6( 919 1834 = = e hc B BR b, c Photocurrent and quantum efficiency can be calculated for other wavelengths in the same way. The values are summarized in the following table. Table 5Q4 Summarized values for photocurrent and quantum efficiency. Wavelength, nm Type A Type B Responsivi ty, A/W Photocurrent, nA Quantum efficiency, % Responsivity , A/W Photocurrent, nA Quantum efficiency, % 450 0.20 25 55.1 0.12 15 33.1 700 0.46 57 81.5 0.46 57 81.5 1000 0.15 19 18.6 0.40 50 50 d Quantum efficiency depends on the wavelength and also on the device structure. Both devices use a Si crystal , but, at a given wavelength such as 450 nm, for A, QE = 55.1% and for B, QE = 33.1%. Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 maximum = hcRe = (6.62 10 34 J s)(3 108 m s-1)(0.78 A W-1) (1.6 1019 C)(1210 109 m) 0.80 5.6 Maximum QE The relationship between the responsivity R and the quantum efficiency (QE) is, R = ehc or = hce R The QE is maximum when d/d = 0, thus differentiating the above expression with respect to we have, d d = hc e dR d + hcR e d d 1 = 0 hc e dR d hcR e 1 2 = 0 dR d R = 0 i.e. dR d = R 0 0.2 0.4 0.6 0.8 1 800 1000 1200 1400 1600 1800 Wavelength(nm) The responsivity of an InGaAs pin photodiode Responsivity(A/W) 6004002000 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.5 1 1.5 2 Wavelength(m) The responsivity of a commercial Ge pn junction photodiode Responsivity(A/W) Tangent through origin 00 0.1 0.2 0.3 0.4 0.5 0.6 200 400 600 800 1000 1200 Wavelength(nm) A B The responsivity of two commercial Si pin photodiodes Responsivity(A/W) 0 Figure 5Q6 We can find the maximum QE by drawing a tangent to the R vs. curve that passes through the origin as in the three examples below. The actual graphical values are listed in Table 5Q6. For example for the InGaAs pin photodiode, the maximum QE is maximum = hcRe = (6.62 10 34 J s)(3 108 m s-1)(0.78 A W-1) (1.6 1019 C)(1210 109 m) 0.80 Table 5Q6 InGaAs pin Si-pin-A Si-pin-B Ge photodiode (nm) 1210 700 810 1500 R (A/W) 0.78 0.46 0.57 0.71 Maximum QE 0.80 0.81 0.87 0.57 Maximum QE % 80% 81% 87% 57% Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 5.12 The APD and excess avalanche noise a We can find the value of x by plotting F vs. M on a log-log plot, which is shown in Figure 5Q12. From the plot, the index x = 0.857. The fit shows that, F = 1.095M.8571 which agrees well with the equation, F Mx. 1 10 1 10 Ex ce ss no ise fa cto r Multiplication F = (1.095)M0.857 Excess noise factor F vs. M for a GE APD; from Scansen and Kasap 1992. Figure 5Q12 b Given, Ido = 0.5 A, M = 6, B=500 MHz and x = 0.857. From Equation (12) 5.10, the SNR can be written as, SNR= Signal PowerNoise Power= M 2I pho 2e Ido + Ipho( )M2 +x B M2Ipho2 2eM2 + xB SNR( ) Ipho 2eM2+ x B SNR( )Ido = 0 (2) Solving this quadratic Equation (2) with a SNR = 1 for Ipho we find, Ipho = 1.966510-8 A or 19.665 nA If Po is the incident optical power, then by the definition of responsivity, R = Ipho/Po, Po = Ipho/R = (1.966510-8 A)/(0.8 A/W) = 2.45810-8 W or 24.58 nW. c Solving this quadratic Equation (2) with a SNR = 10 for Ipho we find, Ipho = 6.483210-8 A or 64.665 nA The incident optical power, Po = Ipho/R = (6.483210-8 A)/(0.8 A/W) = 8.10410-8 W or 81.04 nW. Note: Although the SNR has gone up by a factor of 10, the required increase in the incident optical power is only a factor of 3.3. 6.3 Solar cell driving a load a The solar cell is used under an illumination of 1 kW m-2. The short circuit current has to be scale up by 1000/600 = 1.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 1.67 along the current axis. The load line for R = 20 and its intersection with the solar cell IV characteristics at P which is the operating point P. Thus, I 22.5 mA and V 0.45 V The power delivered to the load is Pout = V = (22.510-3)(0.45V) = 0.0101W, or 10.1 mW. This is not the maximum power available from the solar cell. The input sun-light power is Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6 Pin = (Light Intensity)(Surface Area) = (1000 W m-2)(4 cm2 10-4 m2/cm2) = 0.4 W The efficiency is = 100 PoutP in = 100 0.0100.4 = 2.50 0 which is poor. b Point M on Figure 6Q3-2 is probably close to the maximum efficiency point, I 23.5 mA and V 0.44 V. The load should be R = 18.7 , close to the 20 load. At 600 W m-2 illumination, the load has to be about 30 as in Figure 6.8 (b). Thus, the maximum efficiency requires the load R to be decreased as the light intensity is increased. The fill factor is FF = ImVmI scVoc = (23.5 mA)(0.44 V)(27 mA)(0.50 V) 0.78 c The solar cell is used under an illumination of 400 W m-2. The short circuit current has to be scale up by 400/600 = 0.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 0.67 along the current axis. Suppose we have N identical cells in series, and the voltage across the calculator is Vcalculator. The current taken by the calculator is 3 mA in the voltage