光电子学与光子学的原理及应用s.o.kasa 课后答案.pdf

solutions for optoelectronics and photonics principles and practices chapter 1 2 3 7 1 4 antireflection coating g g g for light traveling in medium 1 incident on the 1 2 interface at normal incidence r12 n1 n2 n1 n2 n1 n1n3 n1 n1n3 1 n3 n1 1 n3 n1 for light traveling in medium 2 incident on the 2 3 interface at normal incidence r23 n2 n3 n2 n3 n1n3 n3 n1n3 n3 n1 n3 1 n1 n3 1 1 n3 n1 1 n3 n1 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 2 this should be 180 or thus we need 2n2 2 d or for sio2 d 4n2 900 10 9 m 4 1 5 0 15 m for tio2 d 4n2 900 10 9 m 4 2 3 0 10 m 1 8 thin film coating and multiple reflections assume that n1 n2 1 is areflected a0 k t1t2 r1 r1r2e j 2 k 3 so that the reflection coefficient is r areflected a0 r1 t1t2 r1 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 t1t2 r1 r1r2e j2 1 r1r2e j2 r1 t1t2r2e j2 1 r1r2e j2 4 using eq 1 eq 4 r r11 r1r2e j2 1 r1 2 r2e j2 1 r1r2e j2 i e r r1 r2e j2 1 r1r2e j2 5 the amplitude of the transmitted beam is ctransmitted c1 c2 c3 i e ctransmitted a0 t1t23e j t1t23r1r2e j3 t1t23r12r22e j5 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 ctransmitted a0 t1t23e j r1r2 r 1r2e j2 k k 1 t1t23e j r1r2 r1r2e j2 1 r1r2e j2 8 solutions for optoelectronics and photonics principles and practices chapter 1 2 3 7 i e t t1t23e j 1 r1r2e j2 1 9 antireflection coating consider the transmission coefficient obtained in question 1 8 t t1t2e j 1 r1r2e j2 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 m where m is an odd integer or when 2 n2d m 1 2 leading to d m 4n2 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 n2 n1 n2 n1 n1n3 n1 n1n3 1 n3 n1 1 n3 n1 for light traveling in medium 2 incident on the 2 3 interface at normal incidence r2 r23 n2 n3 n2 n3 n1n3 n3 n1n3 n3 n1 n3 1 n1 n3 1 1 n3 n1 1 n3 n1 thus r2 r1 which confirms that we need n2 n1n3 1 2 the reflection coefficient from question 1 8 is r r1 r2e j2 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 m where m is an odd integer or when solutions for optoelectronics and photonics principles and practices chapter 1 2 3 7 2 n2d m 1 2 leading to d m 4n2 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 10 2 3 6 10 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 40 10 2 m 3 6 10 5 rad 1 44 10 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 2 the phase difference between the rays meeting at c is kac ka c k1ac k1accos 2 k1ac 1 cos 2 k1ac 1 cos 2 k1 a y cos 1 2cos2 1 k1 a y cos 2cos2 2k1 a y cos given 2 2a n1 cos m m m cos m m m 2 n1 2a m m k1 2a then m 2k1 a y cos m m 2k1 a y m m k1 2a m m 1 y a m m m m y a m m m m y m y a m 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 2 10 7 m or a 0 1 m for only a single mode we need v 2 a n1 2 n2 2 1 2 2 a n1 2 n2 2 1 2 2 2 0 1 m 3 66 2 3 402 1 2 2 0 542 m the cut off wavelength is 542 nm when 870 nm v 2 1 m 3 66 2 3 402 1 2 0 870 m 0 979 2 therefore 870 nm is a single mode operation for a rectangular waveguide the fundamental mode has a mode field distance 2wo mfd 2a v 1 v 0 2 m 0 979 1 0 979 0 404 m the decay constant of the evanescent wave is given by v a 0 979 0 1 m 9 79 m 1 or 9 79 106 m 1 the penetration depth 1 1 9 79 m 1 0 102 m the penetration depth is half the core thickness 2 8 a multimode fiber given n1 1 475 n2 1 455 2a 100 10 6 m or a 50 m and 0 850 m the v number is v 2 a n1 2 n2 2 1 2 2 50 m 1 4752 1 4552 1 2 0 850 m 89 47 number of modes m m v2 2 89 472 2 4002 the fiber becomes monomode when v 2 a n1 2 n2 2 1 2 2 a n1 2 n2 2 1 2 2 405 2 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 n1 2 n2 2 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 na no arcsin 0 242 1 14 modal dispersion is given by intermode l n1 n2 c 1 475 1 455 3 108m s 1 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 km 1 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 total 2 intermode 2 material 2 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 8 10 6 m or a 4 m and 1 3 m the v number is v 2 a n1 2 n2 2 1 2 2 4 m 1 4682 1 4642 1 2 1 3 m 2 094 b since v 2 405 or 2 a n1 2 n2 2 1 2 2 405 2 4 m 1 4682 1 4642 1 2 2 405 1 13 m for wavelengths shorter than 1 13 m the fiber is a multi mode waveguide c the numerical aperture na is na n1 2 n2 2 1 2 1 4682 1 4642 1 2 0 108 d if max is the maximum acceptance angle then max arcsin na no arcsin 0 108 1 6 2 so that the total acceptance angle is 12 4 e at 1 3 m from the figure dm 7 5 ps km 1 nm 1 dw 5 ps km 1 nm 1 1 2 l dm dw 1 2 solutions for optoelectronics and photonics principles and practices chapter 1 2 3 7 7 5 5 ps km 1 nm 1 2 nm 15 ps km 1 10 ps km 1 0 025 ns km 1 obviously materials dispersion is 15 ps km 1 and waveguide dispersion is 10 ps km 1 the maximum bit rate distance product is then bl 0 59l 1 2 0 59 0 025 ns km 1 23 6 gb s 1 km material and waveguide dispersion coefficients in an optical fiber with a core sio2 13 5 geo2 for a 2 5 to 4 m 0 0 10 20 1 21 31 41 51 6 0 m dm dw sio2 13 5 geo2 2 5 3 0 3 5 4 0 a m dispersion coefficient ps km 1 nm 1 figure 2q9 2 18 microbending loss radius of curvature v1 v2 kt e 0 02586 v where so ajso aeni2 dh lhnd de lena given ni 1 8 1012 m 3 a 0 1 10 6 m2 thus iso 0 1 10 6 m2 1 602 10 19 c 1 8 10 12 m 3 2 0 000803 m2 s 1 1 056 10 5 m 1 10 22 m 3 0 1 10 6 m2 1 602 10 19 c 1 8 10 12 m 3 2 0 01813 m2 s 1 5 02 10 5 m 1 10 22 m 3 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 10 17ndopant 5 10 7 1 2 10 17 1015 cm 3 490 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 kt e e 0 02585 v 250 cm2 v 1 s 1 6 463 cm2 s 1 and dh kt h e 0 02585 v 450 cm2 v 1 s 1 11 63 cm2 s 1 the diffusion lengths are then le de e 6 463 cm2 s 1 23 81 10 9 s 3 92 10 4 cm or 3 92 m and lh dh h 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 so exp 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 1 10 2 cm2 1 60 10 19 c 1 45 10 10 cm 3 2 11 63 cm2 s 1 2 39 10 3 cm 1 10 15 cm 3 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 enand 1 2 2 vo v end 1 2 w 2 11 9 8 854 10 12 f m 1 0 755 v 0 6 v 1 60 10 19 c 10 21 m 3 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 aeniwn 2 h aeniwp 2 e aeniw 2 h 1 10 2 cm2 1 602 10 19 c 1 45 10 10 cm 3 0 451 10 4 cm 2 490 10 9 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 10 17ndopant 5 10 7 1 2 10 17 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 kt e e 0 02585 v 250 cm2 v 1 s 1 6 463 cm2 s 1 and dh kt h e 0 02585 v 130 cm2 v 1 s 1 3 361 cm2 s 1 diffusion lengths are le de e 6 463 cm2 s 1 23 81 10 9 s 3 923 10 4 cm or 3 92 m and lh dh h 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 iso exp ev kt 1 isoexp ev kt for v kt e 0 02585 v where so ajso aeni2 dh lhnd de lena iso 1 10 2 cm2 1 602 10 19 c 1 45 10 10 cm 3 2 3 361 cm2 s 1 2 829 10 4 cm 10 18 cm 3 6 463 cm2 s 1 3 923 10 4 cm 10 18 cm 3 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 15 10 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 45 1010 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 enand 1 2 4 vo v end 1 2 w 4 11 9 8 854 10 12 f m 1 0 933 v 0 6 v 1 602 10 19 c 10 24 cm 3 1 2 w 2 96 10 8 m or 0 0296 m the recombination current pre exponential term is iro aewni 2 r 1 10 2 cm2 1 602 10 19 c 0 0296 10 4 cm 1 45 10 10 cm 3 2 23 81 10 9 s iro 1 44 10 9 a the forward current due to recombination is irecom iroexp ev 2kt 1 44 10 9 a exp 0 6 v 2 0 02585 v irecom 1 58 10 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 d deph hc eph 2 we can represent small changes or intervals or by differentials e g eph d deph then hc eph 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 3kbt hc or 2 3kbt hc 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 0100200300400 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 eph hc 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 ch e peak kbt e and at peak 820 109 m taking t 25 273k eg 3 108 6 626 10 34 1 6 10 19 820 10 9 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 power input electrical power p o iv 25 10 6 w 40 10 3 a 1 5 v 0 000417 0 0417 3 7 external conversion efficiency ext p o iv 2 5 10 3 w 50 10 3 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 h h h hh h h h 0 20 40 60 80 100 120 140 160 0100000020000003000000 algaas algaas algaas gaas gaas ingaasp ingaasp ingaasp 1 2 nm 2 nm 2 the best line forced through zero is 1 2 6 57 10 5 nm 1 2 and slope 6 57 10 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 p o iv 200 10 6 w 75 10 3 a 1 5 v 1 8 10 3 0 18 b i overall p o iv 48 10 6 w 120 10 3 a 1 3 v 0 0307 solutions for optoelectronics and photonics principles and practices chapter 1 2 3 7 ii overall p o iv 7 10 6 w 120 10 3 a 1 3 v 0 0045 7 3 wire grid polarizer ex is along the wires and drives the conduction electrons along the length of the wires and thereby generates a current the collisions of these driven electrons with lattice vibrations leads to joule loss i e i2r thus energy is absorbed from the ex field and the wires heat up a little negligible amount just as energy would be absorbed from a battery driving a current through a wire as the wires are very thin ey field cannot drive the electrons too far or ex at the grid location is changing sinusoidally with time which means its associated magnetic field by is also changing this induces a voltage across the wires by virtue of the faraday effect because the magnetic field by cuts the wires as it increases or decreases and the wires are at right angles to by the induced voltage drives a current along the wires and hence leads to a joule loss i2r or power absorption from the ex wave reduction in by means a reduction in ex on the other hand the magnetic field bx associated with ey is parallel to the wires and does not induce a voltage hence it is not absorbed 7 7 birefringence the waves in the calcite plate propagate as o wave and e wave with fields