光电子学与光子学 课后答案 46章.pdf

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 c 2L 3 108 m s 1 2 0 5 m 3 108 Hz 300 MHz The finesse is F R1 2 1 R 0 991 2 1 0 99 312 6 and each mode width spectral width is m f F 3 108 312 6 9 6 105 Hz 960 kHz b Cavity mode nearest to the emission wavelength is m 2L n 2 200 10 6 1300 10 9 3 7 1138 46 i e m 1138 Separation of the modes is m f c n 2L 3 108 m s 1 3 7 2 200 10 6 m 2 03 1011 Hz The finesse is F R1 2 1 R 0 81 2 1 0 8 14 05 and each mode width spectral width is m f F 2 03 1013 14 05 1 4 1010 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 P o Energy flow per unit time in cavity towards face Transmittance hc 1 2Nph dWL t Transmittance hc 1 2Nph dWL Ln c 1 R hc2NphdW 2n 1 R and I Optical Power Area P o dW hc2Nph 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 exp t 2L Equating the two RRexp 2L exp t 2L and rearranging t 1 2L ln 1 R2 c The reflectance is R n 1 n 1 2 3 5 1 3 5 1 2 0 309 The total loss coefficient is t 1 2L ln 1 R2 1000 m 1 1 60 10 6 m 1 ln 1 0 3092 2 06 10 4 m 1 ph n c t 3 5 3 108 m s 1 2 06 108 m 1 5 7 10 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 sp ed 500 104 A m 2 10 10 9s 1 6 10 19C 0 25 10 6m 1 25 1024 m 3 or 1 2 102 102 102 1018 181818 cm 3 Now the current density corresponding to I 30 mA is J I WL 0 05 A 10 60 10 610 6 m2 833 104 A m 2 And Nph ph ed J Jth 5 7 10 13s 1 6 10 19C 0 25 10 6m 833 500 104 A m 2 4 7 1 1 1 10 0 0 019 191919 photons m 3 The optical power is Po hc 2N phdW 2n 1 R 6 62 10 34 J s 3 108 m s 1 2 4 7 1019 m 3 0 25 10 6 m 10 10 6 m 2 3 5 1310 10 9 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 25 10 10 310 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 h Diode Current e P o Eg I e ePo IEg 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 P o Eg I e e Eg dP o dI The external power efficiency EPE of the laser diode is defined by EPE Optical ouput power Electical input power P o IV P o IV eEg eEg eP o IEg Eg eV EPE ePo IEg Eg eV EQE Eg eV b 670 nm laser diode Eg hc 6 626 10 34 3 108 670 10 9 1 6 10 19 1 85 eV so that EQE 1 6 10 19 C 2 10 3 Js 1 80 10 3 A 1 85 eV 1 6 10 19 eV J 0 0135 or 1 35 EDQE e Eg dP o dI 1 6 10 19 C 1 85 eV 1 6 10 19 C 3 10 3 2 10 3 Js 1 82 10 3 80 10 3 A 0 27 or 27 EPE Po IV 2 10 3W 80 10 3A 2 3 V 0 011 or 1 1 c 1310 nm laser diode Eg hc 6 626 10 34 3 108 1310 10 9 1 6 10 19 0 9464 eV so that EQE 1 6 10 19 C 3 10 3 Js 1 40 10 3 A 0 9464 eV 1 6 10 19 J eV 0 079 or 7 9 Solutions for Optoelectronics and Photonics Principles and Practices Chapter 4 5 6 EDQE e Eg dP o dI 1 6 10 19 C 0 9464 eV 1 6 10 19 C 4 10 3 3 10 3 Js 1 45 10 3 40 10 3 A 0 21 or 21 and EPE Po IV 3 10 3W 40 10 3A 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 h2n2 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 10 10 9 m me 0 04me and n 1 and 2 we find the following electron energy levels n 1 1 n h2n2 8me d2 6 626 10 34 2 1 2 8 0 04 9 11 10 31 10 10 9 2 1 51 10 20 J 0 094 eV n 2 2 0 376 eV Using d 10 10 9 m mh 0 44me and n 1 the hole energy levels below Ev is n 1 n h2n2 8mh d2 6 626 10 34 2 1 2 8 0 44 9 11 10 31 10 10 9 2 1 37 10 21 J 0 00855 eV The wavelength light emission from the QW laser with Eg 0 70 eV is QW hc Eg 1 1 6 626 10 34 3 108 0 70 0 094 0 00855 1 602 10 19 1545 10 9 m 1545 nm The wavelength of emission from bulk InGaAs with Eg 0 70 eV is g hc Eg 6 626 10 34 3 108 0 70 1 602 10 19 1771 10 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 h2n2 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 h2n2 8me d2 6 626 10 34 2 1 2 8 0 07 9 11 10 31 8 10 9 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 h2n2 8mh d2 6 626 10 34 2 1 2 8 0 47 9 11 10 31 8 10 9 2 0 0125 eV The wavelength of emission from bulk GaAs with Eg 1 42 eV is g hc Eg 6 626 10 34 3 108 1 42 1 602 10 19 874 10 9 m or 874 nm Whereas from the GaAs QW the wavelength is QW hc Eg 1 1 6 626 10 34 3 108 1 42 0 0839 0 0125 1 602 10 19 818 10 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 6 626 10 34 J s 3 108 m s 1 600 10 9 m 2 07 eV b A 5 10 2 cm2 and Ilight 20 10 3 W cm2 The received power is P AIlight 5 10 2 cm2 20 10 3 W cm2 10 3 W Nph number of photons arriving per second P Eph 10 3 W 2 07 1 60218 10 19 J eV 2 9787 1015 Photons s 1 2 9787 1015 EHP s 1 c For GaAs Eg 1 42 eV and the corresponding wavelength is hc Eg 6 626 10 34 J s 3 108 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 6 626 10 34 J s 3 108 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 Io 1 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 A ph Ad Io1 exp d h d Io1 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 1 1 0 9 1 5 2 105 ln 1 1 0 9 4 428 10 6 m 4 428 m For In0 53Ga0 47As 7 5 105 m 1 at 1 5 m incident radiation Figure 5 3 d 1 7 5 105 ln 1 1 0 9 3 07 10 6m 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 de h de J ph exp 1 exp 1 oo II Given Io 100 W mm 2 100 10 6 106 W m 2 100 W m 2 81 108 100 310626 6 9 0105 11001060218 1 834 619 ph J 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 850 10 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 3 10 6 A and area A 8 10 9 m2 the incident optical power Po Iph R 0 3 10 6 A 0 73 A W 1 4 1096 10 7 W Light intensity Io P0 A 4 1096 10 7 W 8 10 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 max max 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 4 10 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 A R 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 B R 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 hcR e 6 62 10 34 J s 3 108 m s 1 0 78 A W 1 1 6 10 19 C 1210 10 9 m 0 80 5 6 Maximum QE The relationship between the responsivity R and the quantum efficiency QE is R e hc or hc e 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 0 511 52 Wavelength m The responsivity of a commercial Ge pn junction photodiode Responsivity A W Tangent through origin 0 0 0 1 0 2 0 3 0 4 0 5 0 6 2004006008001000 1200 Wavelength nm AB 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 hcR e 6 62 10 34 J s 3 108 m s 1 0 78 A W 1 1 6 10 19 C 1210 10 9 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 10 110 Excess noise factor 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 Power Noise Power M 2I pho 2e Ido Ipho M 2 xB M2Ipho 2 2eM2 xB SNR Ipho 2eM2 xB SNR Ido 0 2 Solving this quadratic Equation 2 with a SNR 1 for Ipho we find Ipho 1 9665 10 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 9665 10 8 A 0 8 A W 2 458 10 8 W or 24 58 nW c Solving this quadratic Equation 2 with a SNR 10 for Ipho we find Ipho 6 4832 10 8 A or 64 665 nA The incident optical power Po Ipho R 6 4832 10 8 A 0 8 A W 8 104 10 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 I V 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 5 10 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 Pout Pin 100 0 010 0 4 2 500 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 ImVm IscVoc 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 range 3 to 4 V and the calculator stops working when Vcalculator nVT Solutions for Optoelectronics and Photonics Principles and Practices Chapter 4 5 6 Rearranging we find the following 2 Cells in series V 2nVTln I Iph Io 2R sI 1 Cell V nVTln I Iph Io RsI Figure 6Q6 1 shows the I V characteristics for 1 cell and 2 cells in series Notice that the OC voltage for the series cells is twice the OC voltage for one cell 1 cell Iph 10 mA 2 cells in series Current in mA Voltage Figure 6Q6 1 Since Power IV we can plot the power delivered as a function of current as shown in Figure 6Q6 2 Maximum power corresponds to 1 cell I 7 7 mA V 0 3 V P 2 3 W 2 cell I 7 7 mA V 0 58 V P 4 5 W Power in mW Current in mA 2 cells in series 1 cell 4 5 mW 2 3 mW 7 7 mA Figure 6Q6 2 6 7 Series connected solar cells A Iph1 V Id1 Rs2 RL I Two different solar cells in series Iph2 Id2 Rs1 Iph2 Iph2 V1 V2 I Figure 6Q7 1 Solutions for Optoelectronics and Photonics Principles and Practices Chapter 4 5 6 The equivalent circuit is shown in Figure 6Q7 1 The current through both the devices has to be the same Thus for cell 1 I Iph1 Io1exp Vd1 n1VT 1 Iph1 Io1exp V1 IRs1 n1VT 1 V1 IRs1 n1VT ln I Iph1 Io1 1 V1 n1VTln I Iph1 Io1 1 IRs1 For cell 2 I Iph1 Io2exp Vd 2 n2VT 1 Iph2 Io2exp V2 IRs2 n2VT 1 V2 n2VTln I Iph2 Io2 1 IRs2 But V V1 V2 V n1VTln I Iph1 Io1 1 IRs1 n2V Tln I Iph2 Io2 1 IR s2 We can now substitute Io1 25 10 6 mA n1 1 5 Rs1 10 Io2 1 10 7 mA n2 1 Rs2 50 and then plot V vs I rather I vs V since we can calculate V from I for each cell and the two cells in series as in Figure 6Q7 2 Notice that the short circuit is determined by the smallest Isc cell The total open circuit voltage is the sum of the two Cell 1 Voltage Current mA Cell 2 Cell 1 in series with 2 Figure 6Q7 2 6 11 Maximum Power and Fill Factor The current through a solar cell is I Iph Ioexp V nVT so that the power is P IV IphV IoV exp V nVT At maximum power dP dV 0 when V Vm dP dV Iph Ioexp V nVT Io V nVT exp V nVT dP dV Iph Ioexp Vm nVT 1 Vm nVT 0 Solutions for Optoelectronics and Photonics Principles and Practices Chapter 4 5 6 1 Vm nVT exp Vm nVT Iph Io and since Vm nVT Vm nVT exp Vm nVT Iph Io 1 Consider the current at maximum power Im Iph Ioexp Vm nVT Substitute for the exponential term Im