EE 5337 Computational Electromagnetics

Transcription

1 Instructor Dr. Ramon Rumpf (915) EE 5337 Computational Electromagnetics Lecture #23 RCWA Extras Lecture 23 These notes ma contain coprighte material obtaine uner fair use rules. Distribution of these materials is strictl prohibite Slie 1 Outline One spatial harmonic: P=Q=1 Simulation of 1D Gratings with 3D RCWA Formulation of 2D RCWA with fast Fourier factorization Danger of RCWA an convergence RCWA an curve structures Strategicall truncating the set of spatial harmonics RCWA for generalize smmetries Moeling hexagonal gratings with rectangular RCWA Enhance transmittance matrix approach Lecture 23 Slie 2 1

2 One Spatial Harmonic P = Q = 1 Lecture 23 Slie 3 Anatom of the Convolution Matrix Device r x, Convolution Matrix r Higher orer Fourier coefficients Higher orer Fourier coefficients escribe perioic variations in r (x,) 0 orer Fourier coefficient is the average value Lecture 23 Slie 4 2

3 One Spatial Harmonic (P=Q=1) When onl one spatial harmonic is use, RCWA reuces to the 1D transfer matrix metho, but uses the average value for the material properties in each laer. This is first orer effective meium theor an can make use of fast Fourier factorization so it is more sophisticate than the stanar TMM r Phsical Device Effective Meium Approximation in 1D Lecture 23 Slie 5 Simulation of 1D Gratings with 3D RCWA Lecture 23 Slie 6 3

4 Grating Terminolog 1D grating Rule grating 2D grating Crosse grating Requires a 2D simulation Requires a 3D simulation Lecture 23 Slie 7 3D RCWA for 1D Gratings Three imensional RCWA simulates all polarizations at the same time. For 1D Gratings, Maxwell s equations ecouple into the E moe an the H moe. It is possible to reformulate RCWA specificall for 1D Gratings so that it will onl simulate either the E moe or H moe, but not both. This approach will be several times faster ue to smaller matrices. It is m experience that 3D RCWA is fast enough that few applications warrant formulating a 2D RCWA coe. Exceptions inclue when fast Fourier factorization is important or for running optimizations that require man thousans of simulations to be iterate. There are some tricks that can be use when using RCWA to moel 1D gratings to maximize the spee an efficienc. Lecture 23 Slie 8 4

5 Number of Spatial Harmonics There is no contrast in this irection. Onl one spatial harmonic is neee. Q 1 N 1 You must use some number of spatial harmonics in this irection. P 7 N 500 x Lecture 23 Slie 9 E an H Moes with 3D RCWA The fiel for both E an H moes are thought of completel in terms of the polarization vector for the electric fiel. If ou have implemente our coes following these lectures, no changes to our coe are neee outsie of the ashboar. k inc k inc P H P H P H P E P E P E x Grating irection P > 1 k inc Uniform irection Q = 1 Source Wave Vector inc 0 inc sin 0 cos The concept of E an H moes onl appl when: (1) LHI or iagonall anisotropic materials, (2) evice is uniform in irection, an (3) propagation is restricte to x z plane. Lecture 23 Slie 10 k k n Source Polarization Vector 0 P E 1 E Moe 0 cos P H 0 H Moe sin 5

6 FDFD Vs. RCWA Representations FDFD represents a evice as the sie view of a single unit cell. In contrast, RCWA represents a evice as a top view of each laer of a single unit cell. Lecture 23 Slie 11 Formulation of 2D RCWA with FFF (1D Gratings) Lecture 23 Slie 12 6

7 Starting Point for Derivation We will start with the semi analtical matrix form of Maxwell s equations in Fourier space. These are rigorous an vali even for 3D evices. jku z u rsx z ux jk xuz rs z Ku Ku s Ks z s r ux z sx jk xsz r u z Ks Ks u x x j r j z j x x r z Lecture 23 Slie 13 Reuction to Two Dimensions When evices are uniform in the irection an no wave propagation occurs in this irection, we have K 0 Our matrix equations reuce to u rsx z ux jk xuz rs z Ku j s x r z s rux z sx jk xszru z Ks j u x r z Lecture 23 Slie 14 7

8 Two Inepenent Moes We see that Maxwell s equations have ecouple into two inepenent moes. u rsx z ux jk xuz rs z Ku j s x r z s rux z sx jk xsz ru z Ks j u x r z E Moe ux jk xuz rs z s rux z Ks j u x r z H Moe sx jk xszru z u rsx z Ku j s x r z Lecture 23 Slie 15 Orientation of the Fiel Components For 1D gratings, the orientation of the fiel components relative to the interfaces is fixe. In this case, it is straightforwar to incorporate fast Fourier factorization into the formulation. S S x x U U x S z U z Within the laer S x is alwas perpenicular to interfaces. S is alwas parallel to interfaces. S z is alwas parallel to interfaces. z Within the laer U x is alwas perpenicular to interfaces. U is alwas parallel to interfaces. U z is alwas parallel to interfaces. Lecture 23 Slie 16 8

9 Incorporating Fast Fourier Factorization For 1D gratings, it is ver straightforwar to incorporate fast Fourier factorization rules. This will improve convergence rates. E Moe ux jk xuz rs z s z Ks u 1 j 1 r ux x r z H Moe sx jk xsz ru z 1 u s z Ku s 1 j r x x r z s is alwas parallel to interfaces so is the stanar convolution matrix. r u is alwas parallel to interfaces so is the stanar convolution matrix. r u x is alwas perpenicular so FFF rules are use to construct 1 1 r. u z is alwas parallel to interfaces so is the stanar convolution matrix. s x is alwas perpenicular so FFF rules are use to construct 1 1 r. r r s z is alwas parallel to interfaces so Is the stanar convolution matrix. Lecture 23 Note: FFF in E moe with non magnetic materials makes no ifference. Slie 17 Eliminate Longituinal Components We solve the thir equation for u z an the sixth equation for s z to obtain u 1 j K s z r x 1 s j K u z r x We substitute these expressions into the remaining Maxwell s equations. E Moe 1 u s K K s z 1 s 1 r ux z x r x r x H Moe 1 s u K K u z 1 u 1 r sx z x r x r x Lecture 23 Slie 18 9

10 Stanar P an Q Form E Moe H Moe s z u z x Pu Qs x u z s z x Psx Qu P 1 1 r 1 Q K K r x r x P 1 1 r 1 Q K K r x r x We see that FFF is incorporate solel into the matrix P. Lecture 23 Slie 19 Matrix Wave Equations Now that we have our equations in stanar P an Q form, we erive the wave equations in the same manner as before. E Moe H Moe z s Ω s 0 z u Ω u 0 Ω 2 PQ Ω 2 PQ an now ou know the rest of the stor. Lecture 23 Slie 20 10

11 Convergence Stu for 1D Gratings r = 1.0, r = 3.0 Orinar FFF H Moe r = 1.0, r = 40.0 Orinar FFF Lecture 23 Slie 21 Convergence Stu for 1D Curve Structures r = 1.0, r = 3.0 Orinar FFF H Moe r = 1.0, r = 40.0? Orinar FFF Lecture 23 Slie 22 11

12 Danger of RCWA an Convergence Lecture 23 Slie 23 Danger of RCWA In real space, poor gri resolution le to fluctuations in conservation of power an other ver recognizable signs that things are wrong. The anger of RCWA is that results can look correct even with ver few spatial harmonics. Conservation of power will alwas be obee in RCWA even using just one spatial harmonic. It must become habit to look for convergence, as there are few other signs that more spatial harmonics are neee. Lecture 23 Slie 24 12

13 Tpical Convergence Plot Device from last f 0 = 47.6 GHz Convergence convergence 250 harmonics 1515 harmonics Lecture 23 Slie 25 RCWA an Curve Structures Lecture 23 Slie 26 13

14 EBG Material Lecture 23 Slie 27 Divie into Thin Laers Laers 1 to 20 Staircase Approximation Lecture 23 Slie 28 14

15 Strategicall Truncating the Set of Spatial Harmonics Lecture 23 Slie 29 Notes on Truncating the Set of Spatial Harmonics The choice of which spatial harmonics to inclue in the expansion is arbitrar. Improper choice can lea to slow convergence an inaccurate results. We chose irections consistent with the phsics of iffraction an a rectangular Fourier space gri for simplicit. The number of harmonics retaine in a particular irection etermines the spatial resolution of structures with contrast in that irection. It seems optimal to keep the number of spatial harmonics uniform in all irections. Lecture 23 Slie 30 15

16 Fourier Space Gri Notation The components of the wave vector expansion look like: k x m k * * 2 2 k m, n k mk m n * r r z x We visualize it this wa: A simpler view of our 2D Fourier space gri is Lecture 23 Slie 31 Simple Gri Truncation Scheme We will retain all spatial harmonics with inices that satisf the following equation: m M 2 2 n N Conventional n N an m M pincushion iamon circular barrel L. Li, New formulation of the Fourier moal metho for crosse surface relief gratings, J. Opt. Soc. Am. A 14(10), , Lecture 23 Slie 32 16

17 Implementation Step 1 Buil Unit Cell on High Resolution Gri Lecture 23 Slie 33 Implementation Step 2 Calculate Fourier Coefficients Using FFT Technique Lecture 23 Slie 34 17

18 18 Lecture 23 Slie 35 Implementation Step 3 Assemble Stanar Convolution Matrix From Fourier Coefficients r Lecture 23 Slie 36 Implementation Step 4 Buil Truncation Map % Construct Truncation Map TMAP = abs(m/m).^(2*p) + abs(n/n).^(2*p); TMAP = (TMAP <= 1);

19 19 Lecture 23 Slie 37 Implementation Step 5 Extract Inices of Spatial Harmonics to Retain % Construct Truncation Map TMAP = abs(m/m).^(2*p) + abs(n/n).^(2*p); TMAP = (TMAP <= 1); % Extract Arra Inices in = fin(tmap(:)); Lecture 23 Slie 38 Implementation Step 6 Truncate Convolution Matrix r r % Truncate Convolution Matrix ERCT = ERC(in,in);

20 Implementation Step 7 Perform RCWA The rest of the RCWA algorithm remains virtuall unchange. You will nee to consier our truncation again: 1. When ou calculate the source. 2. If ou calculate the fiels from the eigen moes. Lecture 23 Slie 39 RCWA for Generalize Smmetries Lecture 23 Slie 40 20

21 Revise Fourier Transforms The materials r a e r m n m, n jmt1nt2 r 1 jmt 1nT2r amn, rre A A unit cell r a e r m n m, n jmt1nt2 r 1 jmt 1nT2r bmn, rre A A unit cell The fiels jkm, nr Exr, z Sxm, n; ze m n jkm, nr Er, z Sm, n; ze m n jkm, nr Ezr, z Szm, n; ze m n jkm, n r H xr, z Uxm, n; ze m n jkm, n r H r, z Um, n; ze m n jkm, n r H zr, z Uzm, n; ze m n Wave vector expansion k m n k mt nt t, inc 1 2 kz m n k0 r r kt m n 2 * *,, 2 * m,, 2, 1,0,1, 2,, n,, 2, 1, 0,1, 2,, T, T reciprocal lattice vectors of the unit cell 1 2 Lecture 23 Slie 41 Convolution Matrices We cannot use the FFT to calculate the Fourier coefficients. Instea, we use numerical integration. 1 jmt 1nT2r amn, rre A A unit cell 1 jmt nt r 1 2 bmn, rre A A unit cell After this, we construct the convolution matrices an implement RCWA exactl the same wa with exactl the same coe. We woul onl nee to consier the smmetr again if we were reconstructing the fiels. Due to the numerical integrations, calculation of the convolution matrices is computational intensive an time consuming. Lecture 23 Slie 42 21

22 Moeling Hexagonal Gratings with Rectangular RCWA Lecture 23 Slie 43 Geometr of a Hexagon Lecture 23 Slie 44 22

23 Grating Vectors of Hexagonal Structures conventional unit cell Direct Lattice z a Reciprocal Lattice z 2 a c 2 c t 1 t 2 x a a 3 t ˆ ˆ 1 x 2 2 a a 3 t ˆ ˆ 2 x 2 2 t czˆ T 1 T 2 x 2 2 T ˆ ˆ 1 x a a T ˆ ˆ 2 x a a 3 T 2 c zˆ 3 3 Lecture 23 Slie 45 Rectangular Unit Cell in Hexagonal Arra We must ientif a rectangular unit cell that reconstructs a hexagonal arra. a 3 This implies that we will nee more spatial harmonics along the irection than x. Q = roun(p*s/sx); Lecture 23 Slie 46 a 23

24 Enhance Transmittance Matrix Approach M. G. Moharam, Drew A. Pommet, Eric B. Grann, Stable implementation of the rigorous couple wave analsis for surface relief gratings: enhance transmittance matrix approach, J. Opt. Soc. Am. A, Vol. 12, No. 5, pp , Ma Lecture 23 Slie 47 Motivation The enhance transmittance matrix (ETM) metho involves less matrix manipulations so it is much faster than using scattering matrices. Mabe 10. It also provies easier computation of internal fiels. ETM works b first stepping backwar through each laer (backwar analsis) an then stepping forwar (forwar analsis). Intermeiate parameters must be store for each laer uring the backwar analsis that are recalle uring the forwar analsis. This leas to severe memor limitations when man laers are use. Conclusion Unless the evice is compose of prohibitivel large number of laers or if other features of scattering matrices are not neee, use ETM. Lecture 23 Slie 48 24

25 The Problem The source of the instabilit is the following matrix. X ikl 0 i i e Ω X e ikl 0 i i Ω Growing exponentials!!! The enhance transmittance matrix (ETM) metho was the first technique applie to RCWA that fixe the instabilit. ETM is much faster than scattering matrices, but is much less memor efficient because it requires parameters to be store for all laers at the same time. Scattering matrices can procee one laer at a time, forgetting everthing about previous laers. Lecture 23 Slie 49 Bounar Conitions I 0 First interface: 0 I 2 2 I 0 j KK x j K K z,i A sar F1 c1 I K z,i I K z,i 0 X1 2 2 j K x K z,i j KK x unknown I z,i K I K z,i I 0 Intermeiate interfaces: 0 I 2 2 j KK x j K K z,ii X i 0 I 0 B II K z,ii II K F z,ii i ci Fi 1 ci1 2 2 j K x K z,ii j KK 0 I 0 Xi 1 x II z,ii K II K z,ii pxδ0, pq p δ0, pq Last interface: j s kz,inc p k,inc pzδ0, pq I X N 0 F j N cn Bt kx,inc pz kz,inc pxδ0, pq 0 I I unknown Wi Wi ΩikL 0 i Fi Xi e Vi Vi Lecture 23 Slie 50 25

26 Work Backwar Through Laers (1 of 4) The goal is to solve for r an t without using X -1. We start b solving the equation at the last interface for c N. X 0 X 0 F c Bt c F 0 I 0 I 1 N N 1 N N N N Bt We write this as X 0a a 1 N N N 1 cn tn FN 0 I b N b N B Lecture 23 Slie 51 Work Backwar Through Laers (2 of 4) To eliminate the potentiall ill conitione matrix X -1, we introuce an intermeiate transmittance matrix parameter t N efine as t a X t 1 N N N t an t N remain unknown. Our equation for c N becomes 1 X N 0aN I cn tn cn 1 t 0 IbN bnanxn N Lecture 23 Slie 52 26

27 Work Backwar Through Laers (3 of 4) The bounar conition equation at the secon to last interface is X 0 I 0 F c F c N 1 N1 N1 N N 0 I 0 XN Substituting our expression for c N into the equation iels X 0 I 0 I F c F t N 1 N1 N1 N 1 N 0 I 0 XN bnanxn Solving this for c N-1 leas to c X 0 F F I 0 I t 1 N1 N 1 1 N1 N 1 0 I 0 XN bnanxn N 1 We have now worke backwar b one laer while avoiing X. N Lecture 23 Slie 53 Work Backwar Through Laers (4 of 4) This process continues through all the laers. c X 0 F F I 0 I t 1 N1 N 1 1 N1 N 1 0 I 0 XN bnanxn N 1 X N 1 0aN1 an1 1 I 0 I cn 1 tn FN 1FN 1 0 IbN1 bn1 0 XNbNaNXN I cn 1 tn 1 bn1an1x N1 t a X t 1 N N1 N1 N1 T N-1 an T N remain unknown. Lecture 23 Slie 54 27

28 Solve for Reflecte an Transmitte Fiels After working through all interfaces, we are left with I 0 I sar F t 0 X b a X We solve this matrix equation for r an t 1. r I 0 I 1 t1 0 X1b1a1 X1 A B 1 S BF1 t1 Now that we know t 1, we can work forwar through the laers to solve for t. 1 t2 a1 X1t1 1 t a X t t a X t t a X t 1 N N1 N1 N1 Note: we must store a an X for each laer. This leas to poor memor efficienc. 1 Lecture 23 N N N Slie 55 Calculating the Diffraction Efficiencies First, we calculate the longituinal fiel components.,ref r K K r K r t K K t K t 1 1 z z,ref x x z z x x Secon, we calculate the iffraction efficiencies. Re K R r r r r r Re k Re K z,trn r,trn T t t tx t tz Re k z,inc r,inc z,ref r,inc x z z,inc r,inc Lecture 23 Slie 56 28