1. Milster, Socha, Brooker SPIE- SC707 1 Basics of Optical Imaging in Microlithography: A "Hands-on" Approach Tom D. Milster (University of Arizona) Robert Socha (ASML) Peter Brooker (SYNOPSYS) Thanks to: Del Hansen Phat Lu Warren Bletscher

2. Milster, Socha, Brooker SPIE- SC707 2 What we want to do with this course Take a complicated optical system, like a lithographic projection camera used to make computer chips, and simplify it to a working model that demonstrates basic principles. Use a simple optical system for the student to work with “hands on” and observe the results. Demonstrate the relationship of the simple system to a real lithographic system through a commercial simulator. Have fun and demonstrate our unparalled acting abilities From ThisTo This fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture

3. Milster, Socha, Brooker SPIE- SC707 3 OUTLINE Intro –Basic Imaging – What we do in lithography –The goal of making a small image –What limits the size of the image? Basic Illumination and Imaging –Koehler Illumination –Definition of coherence factor “sigma” Binary Mask –Contrast versus pitch for sigma ~ 0 –Contrast versus pitch for sigma > 0 –2-Beam and 3-Beam Imaging –Focus behavior Phase Mask –Contrast versus pitch –Focus behavior Off-Axis Illumination –Contrast versus pitch –Focus behavior Summary

4. Milster, Socha, Brooker SPIE- SC707 4 Introduction What is photo lithography ? Etymology: Photolithography = Light Stone Writing Photoresist Development Negative PhotoresistPositive Photoresist Object: reticle or mask Optics Aerial ImagePhotoresist Wafer + filmsLatent Image Optical image is recorded in the resist via changes in concentrations of species. Concentration level controls development z X y Resist Cross sections

5. Milster, Socha, Brooker SPIE- SC707 5 Introduction 1st approximation is that Aerial image propagates into photoresist normal to the wafer plane, creating a latent image Reality is more complicated; you need to calculate E fields in photoresist at many propagation angles 0.25  m 5-BAR Structures Focus=0.0  m, NA=0.57 NA=0.6, 248nm Image Cross Section Resist Cross Section (not top down!) Z Z

6. Milster, Socha, Brooker SPIE- SC707 6 Introduction The goal of making a small image –Transfer image into a photosensitive material, i.e., photoresist, for subsequent processing that results in a desired pattern to be used as a “stencil” photoresist

7. Milster, Socha, Brooker SPIE- SC707 7 Increase NA Introduction Imaging Resolution and Lord Rayleigh –Q: When can you resolve the image of 2 distance stars? –A: When the 1 st Intensity min of one lines up with peak of other Decrease Large NA Web Top Optics, 1999 Small NA Large From the math of the Airy function

8. Milster, Socha, Brooker SPIE- SC707 8 Oh Master-Litho…..what limits the size of the photoresist pattern? Grasshopper, there are three paths to improve resolution: Reduce Wavelength (Lambda) Increase numerical aperture (NA) Decrease k1 : “Process” knob –Includes off-axis illumination, complex masks, high contrast photoresist, acid diffusion, etc… ….now go away Grasshopper I am busy.

9. Milster, Socha, Brooker SPIE- SC707 9 What is it now Grasshopper… Master, what affects the contrast of the image? The answer is found in the values of NA CD and Pitch Partial Coherence or illumination (s) – s=0: Coherent Limit – s=1: Incoherent Limit

10. Milster, Socha, Brooker SPIE- SC707 10 You again grasshopper… Master… …look at the following data

11. Milster, Socha, Brooker SPIE- SC707 11 100nm L/S 150nm L/S Effect of Varying  =193nm, NA=0.75 Dense Lines vs.  (circular) Master, how come in one case increasing sigma is good (100nm L/S) and in the other case, increasing sigma is bad (200nm L/S)? It depends on the amount of diffraction orders that are being collected by the lens…now go away!

12. Milster, Socha, Brooker SPIE- SC707 12 Master, I am sure that your answers are correct but… …yes Grasshopper… But I find these facts confusing. What is sigma? How can in some cases a larger sigma be good and in other cases a larger sigma be bad? And what the heck is k1? Master…I do not want only the answers…I want to understand…please help me understand master… Grasshopper… you are finally asking the right question Go to the optical bench now… It holds the answer to your questions!!

13. Milster, Socha, Brooker SPIE- SC707 fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam LED Source Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture

14. First Light – Get An Image Let’s do an experiment: –Set up the bench with: Pinhole Source Aperture Stop of 6.35 mm (1/4 in) diameter. −Put in the L (25.2µm) pitch mask and observe the aerial image. −The grating simulates a mask. −The aerial image simulates what is used to expose the resist. −In our system, the aerial image is reimaged onto a CCD camera, which is like an Aerial Image Measurement System (AIMS). −Draw picture of the light pattern at the stop. Milster, Socha, Brooker SPIE- SC707 14 That is what your image looks likeDraw the light pattern at the stop here.

15. Milster, Socha, Brooker SPIE- SC707 15 Basic Illumination and Imaging Kohler Illumination Condenser Imaging Lens Mask Plane Source Aperture Aerial Image Stop Image of source Field Stop of Imaging Lens is Aperture stop condenser and vice versa Lithographic systems use Koehler illumination where the illumination source aperture is imaged into the stop of the imaging lens. Lens 1 Lens 2

16. Milster, Socha, Brooker SPIE- SC707 16 Basic Illumination and Imaging Definition of Coherence Factor ‘Sigma’ Condenser Imaging Lens Mask Plane Source Stop Diameter Source Image Diameter View of Entrance Pupil with blank mask Pupil Edge (the “NA”) Source Image

17. Milster, Socha, Brooker SPIE- SC707 17 Simple Binary Mask Model a Cr on quartz grating mask as an infinitely thin grating Position E SiO 2 Cr E-Field P  0 +1 +2 +3 -2 -3 Diffraction Orders Lens/Pupil Grating Equation: 0 th -1 st +1 st Note: For 1:1 lines and spaces, P= 2 * LW LW = Line Width

18. Milster, Socha, Brooker SPIE- SC707 18 Effect of Varying Pitch Let’s do an experiment –Set up the bench with: Pinhole Source Aperture Stop of 6.35 mm (1/4 in) diameter. –Use the S(8.4µm), M(12.6µm) and L(25.2µm) pitches of the mask and observe the effects in the image plane and at the stop. –Draw the light pattern at the stop on the next page. –What is the relationship between the light pattern at the stop and the image? –What is the smallest pitch for which we can obtain an image? –This system is very similar to what would be observed if an on-axis laser beam was used to illuminate the mask. Therefore, we call this case coherent imaging. –Notice that the lines in the image are either completely resolved, or they are not. There is no ‘partially resolved’ case.

19. Effect of Varying Pitch Draw the light pattern at the stop for the S(8.4 µm) grating. Draw the light pattern at the stop for the M(12.6 µm) grating. Draw the light pattern at the stop for the L(25.2 µm) grating. Milster, Socha, Brooker SPIE- SC707 19

20. Milster, Socha, Brooker SPIE- SC707 20 Binary Mask and Diffraction Orders Must have more than 1 order in pupil to have image modulation Pupil (stop) o +1 +3 -3 pupil o +1 pupil o +1 Coherent limit No Image just constant Irradiance For 1:1 grating We see diffraction orders emanating from the mask that are necessary for imaging. P min is the minimum pitch that is at the limit of resolution.  Max NA=sin(  Max )  Max k 1 =1/2 Strong Image Modulation

21. Milster, Socha, Brooker SPIE- SC707 21 Coffee Break

22. Milster, Socha, Brooker SPIE- SC707 22 Time for the Late Shows new and exciting quiz game sensation. Do you want to play: –Know your “Current events”? –Know your “Cuts of Beef’? –Know your “Optics Bench Basics”? Know your Bench Basics! Excellent choice!!!

23. Milster, Socha, Brooker SPIE- SC707 23 Bench basics: Where is the Source Aperture relative to the condenser lens? Is it: A: at minus infinity B: it refuses to reveal its location C: The source aperture is located at the front focus of the condenser lens Answer is C: The source aperture (effective source for the system) is located at the focus of the condenser lens. Collimated light from the LED illuminates the grating. Light from every part of the source aperture illuminates each point on the grating. Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam

24. Milster, Socha, Brooker SPIE- SC707 24 Bench basics Q: Where does the image of the Source Aperture appear? Does it appear … A: only in the Borg space time continuum B: at the grating C: in the plane of the “Stop”. Correct answer is C: The image of the Source Aperture appears in the plane of the stop. fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture

25. Milster, Socha, Brooker SPIE- SC707 25 Bench basics Q: Collimated light from the Source Aperture illuminates the Grating. This is because…. A: The grating is not worthy of the sources “focused” attention B: The source is the grating…question is irrelevant C: Kohler Illumination of the grating averages out non uniformities in the source. Answer is C fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture

26. Milster, Socha, Brooker SPIE- SC707 26 Comedy writer’s strike… No more multiple choice answers Let’s continue to cement the concepts associated with the bench

27. Milster, Socha, Brooker SPIE- SC707 27 Bench Basics Q: Where is the grating located with respect to Lens1? A: The grating is located at the focus of lens 1. Q: Where does the image of the grating appear? A: The image of the grating appears at the “Image plane” Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam

28. Milster, Socha, Brooker SPIE- SC707 28 Bench Basics: Q: If the Image occurs at the image plane, why is the microscope needed? A: The image of the source at the image plane cannot be seen with the eye. The microscope is needed to magnify the image so it can be seen by your eye. Condenser Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) Source Aperture fcfc fcfc f1f1 f1f1 f2f2 f2f2 2f cam

29. Milster, Socha, Brooker SPIE- SC707 29 Bench Basics: Grating off axis point Q: Look at the above picture. Estimate the vertical magnification? ~3.7 How can the vertical magnification be decreased? Decrease f2 but keep “Stop” at focus of Lens2. f1f1 f1f1 f2f2 f2f2 Grating (Mask) Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask)

30. Milster, Socha, Brooker SPIE- SC707 30 Connection back to real Scanner Optics Q: Where is the mask plane and image of the mask? A: First plane on the left and last plane on right. Q: Can you find the stop in the lens column? A: On the right side of center. Q: What is the magnification? A: 4x demagnification.

31. Milster, Socha, Brooker SPIE- SC707 31 Effect of Varying Sigma Let’s do an experiment –Set up the bench with: Pinhole Source Aperture Stop of 6.35 mm diameter. S(8.4µm) grating –Use the PH, 3.18mm (1/8 in) and 6.35mm (1/4in) diameter sources and observe the effect at the stop and at the image plane. Estimate  for each source. –Draw the light pattern at the stop on the next page. –Is there a point where we can resolve the lines in the image? –By changing , we are allowing more light through the stop that can interfere to form an image. –Not all of the light that is passed through the stop can interfere, thus giving us background light that reduces our contrast. The amount of background light is a function of the pitch, therefore the contrast is a function of the pitch. –This case is called partially coherent imaging, because of the dependence of the contrast on pitch.

32. Effect of Varying Sigma Milster, Socha, Brooker SPIE- SC707 32 Draw the light pattern at the stop for the PH light source. Draw the light pattern at the stop for the 3.18mm diameter light source. Draw the light pattern at the stop for the 6.45mm diameter light source.

33. Milster, Socha, Brooker SPIE- SC707 33 Contrast Curves versus Pitch & Sigma Sigma=0.05 ---Coherent Sigma=0.5 -----Partially Coherent Sigma=1 ---Incoherent limit

34. Milster, Socha, Brooker SPIE- SC707 34 Modulation Transfer Function (MTF) Optics types love this plot!!!! Can you find the Coherent frequency cut off?

35. Milster, Socha, Brooker SPIE- SC707 35 Binary Mask: Influence of Sigma Pupil diagrams with Partial Coherence : No imaging Imaging!! σ NA’ Each source point is projected by the diffraction orders from the mask –These will interfere with each other for a given source point –need more than 1 for interference and hence image modulation We must have at least 2 conjugate sources points in the pupil to form an image.

36. Milster, Socha, Brooker SPIE- SC707 36 Binary Mask: Sigma < 1 No grating - just blank mask 0th order  -1st +1st Grating period at cut-off frequency Grating period resolution limit at given  0th order -1st +1st Resolution limit with 0<  <1 for a circular source

37. Milster, Socha, Brooker SPIE- SC707 37 Binary Mask: Sigma = 1 0th order No grating - just blank mask -1st +1st Grating period at cut-off frequency 0th order -1st +1st Grating period corresponds to incoherent cut-off Resolution limit with  1 for a circular source

38. Milster, Socha, Brooker SPIE- SC707 38 Binary Mask, =248nm, NA=0.63

39. Milster, Socha, Brooker SPIE- SC707 39  =0.05 &  = 0.7 for k1=0.5

40. Milster, Socha, Brooker SPIE- SC707 40 Different cases for on axis, k1=0.5 Assume circular, on axis illumination Assume dense L/S k1=0.5 –  Center of n=1 diffraction orders are at edge of lens –  CD = LW = 0.5*Lambda/NA For 248nm illumination, NA=0.63 –CD = 0.5*248nm/0.63 = 197nm  200nm L/S give k1=0.5 For 193nm illumination, NA=0.93 –CD = 0.5*193nm/0.93 = 104nm  104nm L/S give k1=0.5 For 193nm illumination, NA = 1.2 –CD = 0.5*193/1.2 = 80.4nm  80nm L/S gives k1 = 0.5 Results above are only good for on axis illumination. The usual off-axis case is different.

41. Milster, Socha, Brooker SPIE- SC707 41 Binary Mask: Round and Annular Illumination Larger  and k 1 <0.5 Small  and k 1 >0.5 Conventional or Circular Source Annular Source All power is inside pupil (for 0 th and  1 st orders) Coherent source points have 3- beam interaction Some power is inside pupil (center of  1st orders is outside) Coherent source points have 2- beam interaction

42. Milster, Socha, Brooker SPIE- SC707 42 Binary Mask: 3-Beam Imaging Let’s do an experiment –Set up the bench with: L(25µm) Pitch grating PH Source –Observe the behavior (position and contrast) of the image as the observation plane is moved from the perfect focus. Write down your observations. –What happens as the observation plane is moved beyond the point of zero contrast?

43. Binary Mask: 3-Beam Imaging –Do you see reversed-contrast lines? –This type of focus behavior is indicative of three-beam imaging, where all of the power from the 0 and +/- 1 st diffraction orders passes the stop. –Every point in the image is derived from three conjugate source points in the pupil. –Three-beam imaging has the characteristic that reversed-contrast planes can occur if the focus is too far or the resist is too thick. Milster, Socha, Brooker SPIE- SC707 43

44. Milster, Socha, Brooker SPIE- SC707 44 Binary Mask, =248nm, NA=0.63, 300nm L/S 3-Beam Imaging

45. Milster, Socha, Brooker SPIE- SC707 45 Missing Orders Let’s do an experiment –Set the bench with L(25 µm) pitch PH source –Draw a sketch of the image on the next page. –Block the zero diffraction order at the stop. –Draw a sketch of the image on the next page. –Does the pitch of the image change? –This type of focus behavior is indicative of two-beam imaging. –Every point in the image is derived from two conjugate source points in the pupil, which are widely separated and lead to a double-frequency image. –Now change the system to block either the +1 or -1 order, but let the zero order pass the stop. –Draw a sketch of the image on the next page. –Observe the image pitch and defocus behavior. Write down your observations.

46. Missing Orders L pitch and PH sourceBlock zero orderBlock ± 1 order Milster, Socha, Brooker SPIE- SC707 46

47. Milster, Socha, Brooker SPIE- SC707 47 Binary Mask, =248nm, NA=0.63, 250nm L/S

48. Milster, Socha, Brooker SPIE- SC707 48 Phase Mask E E-Field +1 +3 +5 -3 -5 Diffraction Orders Lens/Pupil Grating Equation: -1 st +1 st Cr SiO 2 Etched depth d Position +1  P

49. Milster, Socha, Brooker SPIE- SC707 49 Pure Phase – Chromeless E E-Field Diffraction Orders Lens/Pupil Position +1 SiO 2 Etched depth d +1 +3 +5 -3 -5  Grating Equation: -1 st +1 st P

50. Milster, Socha, Brooker SPIE- SC707 50 Phase Mask The phase mask produces no zero order Pupil (stop) +1 +3 -3 pupil +1 pupil +1 Coherent limit No Image just constant Irradiance Strong Image Modulation No zero order is emitted from the phase mask. p min is the minimum Cr pitch that is at the limit of resolution. For alternating phase shift grating NA=sin(  Max )  Max k 1 =1/4

51. Milster, Socha, Brooker SPIE- SC707 51 Phase Mask Let’s do an experiment –Set the bench with: . 12.5µm Pitch Phase Mask 14.25mm Diameter stop (No Magnet) 3mm Diameter Source (  ~ 0.3) –Observe the light pattern at the stop. How many diffraction orders do you see? –Draw a sketch of image and the light pattern at the stop on the next page. – Note the relative brightness of the zero order and the +/-1 st orders. If needed, remove the grating to identify where the zero order occurs. –Observe the line pattern at the observation plane. (Block the zero order if present) –How does the image pitch compare to using a simple grating mask?

52. Phase Mask –Change the observation plane location. How sensitive is the observation– plane location to focus changes? –The phase mask has no zero order, and it produces a double-frequency pitch in the aerial image compared to a binary mask. –The minimum pitch in the image is half the minimum pitch of a simple grating mask. –The phase-mask image is relatively insensitive to focus changes, due to the missing zero order. Milster, Socha, Brooker SPIE- SC707 52 Draw a sketch of image.Draw a sketch of light pattern at the stop.

53. Milster, Socha, Brooker SPIE- SC707 53 =248nm, NA=0.63, sigma = 0.3

54. Milster, Socha, Brooker SPIE- SC707 54 Off-Axis Illumination Illumination source shapes that do not have axial intensity as usually known as off-axis sources –Examples are annular, quadrupole, and dipole Off-axis illumination helps to enable k1<0.5 with binary masks –Reduction of on axis source reduces “DC” terms and enhances contrast A conventional on axis small source Some Off-axis sources Annular 45  Quadrupole0  Quadrupole y- dipole x- dipole

55. Milster, Socha, Brooker SPIE- SC707 55 Coherent Off-Axis Illumination and a Binary Mask Orders shift relative to pupil Image Modulation pupil 0 +1 +3 -3 “Incoherent” limit pupil 0 pupil 0 No Image just constant Irradiance For 1:1 grating  Max NA=sin(  Max ) k 1 =1/4 P min is the minimum pitch that is at the limit of resolution.

56. Milster, Socha, Brooker SPIE- SC707 56 Binary Mask with Annular Illumination No grating - just blank mask -1st+1st Grating period at cut-off frequency Grating period resolution limit at given  -1st+1st Resolution limit with 0<  <1 for a circular source  inner  center 0th order  outer 0th order

57. Milster, Socha, Brooker SPIE- SC707 57 Off-Axis Illumination with a Binary Mask Let’s do an experiment –Set up the bench with: system for minimum . S(8.2µm) pitch mask PH Source centered on axis –Observe the pattern at the stop. Draw the light pattern at the stop on the next page. –Do you see an image? Sketch the camera output on the next page. –Move the source until at least two orders pass through the stop. Draw the pattern at the stop and the image on the next page.

58. Off-Axis Illumination with a Binary Mask Draw a sketch with the centered sourceDraw a sketch with decentered source Camera Output Milster, Socha, Brooker SPIE- SC707 58

59. Milster, Socha, Brooker SPIE- SC707 59 =248nm, NA=0.63, sigma = 0.3

60. Milster, Socha, Brooker SPIE- SC707 60 Different cases for off axis, k1=0.25 Assume off axis illumination Assume dense L/S k1=0.25 –  Center of n=0 and n=1 diff. orders are at edge of lens –  CD = LW = 0.25*Lambda/NA For 248nm illumination, NA=0.63 –CD = 0.5*248nm/0.63 = 98nm  100nm L/S give k1=0.25 For 193nm illumination, NA=0.93 –CD = 0.25*193nm/0.93 = 52nm  50nm L/S give k1=0.25 For 193nm illumination, NA = 1.2 –CD = 0.25*193/1.2 = 40.2nm  40nm L/S gives k1 = 0.25 Current off-axis results. Actually might want whole orders inside with sigma=0.3

61. Milster, Socha, Brooker SPIE- SC707 61 Summary What have we learned? –The basic optical components of a lithography system are the source, condenser and imaging lens. –The size and shape of the source influence properties of the aerial image. –The stop of the system determines the maximum angle of diffraction orders that can pass to the image. –It takes at least two diffraction orders passing the stop to form a line-space image. –By increasing , we can change from coherent-like illumination to partially-coherent illumination. –Partially coherent illumination can allow higher pitch in the image at the expense of reduced contrast. –2-Beam and 3-Beam geometries have different focus characteristics. –By using a phase-shift mask, the zero order is eliminated and the first diffraction orders move closer to the center. –Off-axis illumination can produce a half-pitch image, but the contrast is lower than with a phase-shift mask.

62. Milster, Socha, Brooker SPIE- SC707 62 References Introductory Articles SPIE Proceedings for Microlithography Journal of Microlithography, Microfabrication, and Microsystems (JM 3 ) – SPIE Press Industry Magazines Microlithography World www.pennwell.com Books Intro to Fourier Optics and Statistical Optics by J. Goodman Resolution Enhancement Techniques and Optical Imaging in Projection Microlithography Alfred Wong, SPIE Press Microlithography: Science and Technology Ed: James Sheats and Bruce Smith Pub: Marcel Dekker Linear Systems by J. Gaskill

63. Milster, Socha, Brooker SPIE- SC707 63 References Intro Papers “Using location of diffraction orders to predict performance of future scanners”, Peter Brooker Publication: Proc. SPIE Vol. 5256, p. 973-984, 23rd BACUS (2003) “Roles of NA, sigma, and lambda in low-k1 aerial image formation”, Peter D. Brooker Publication: Proc. SPIE Vol. 4346, p. 1575-1586, (2001) Advanced Papers U of A Dissertation by Doug Goodman (1979), Stationary Optical Projectors Papers by H.H. Hopkins for partial coherent imaging, Richards and Wolf for high NA

64. Milster, Socha, Brooker SPIE- SC707 64 Thank You for Taking This Course!

65. Milster, Socha, Brooker SPIE- SC707 65 Backup Slides

66. Milster, Socha, Brooker SPIE- SC707 66 Basic Illumination and Imaging Pupil or “the aperture stop”: Physical Limiting aperture of system –Location and size defined by Chief Ray and Marginal Ray Chief Ray: Starts at edge of object (field) goes through center of pupil Marginal Ray: Starts at axial object and goes through edge of pupil Pupil Aerial Image Object h’ h n’ = image side refractive index n = object side refractive index Chief ray Marginal ray NA: numerical aperture –defined by marginal ray –maximum angle accepted by system

67. Milster, Socha, Brooker SPIE- SC707 67 Basic Illumination and Imaging Let’s do an experiment –Calculate NA at the image plane for r s = _______. –Calculate the coherent resolution limit in terms of pitch in the aerial image. rsrs ’m’m Imaging Lens

68. Milster, Socha, Brooker SPIE- SC707 68 Optimum DOF and Modulation for Annular (and dipole) Optimum when phase differences between 0th and 1st orders are minimum  center /Pitch +1 0

69. Milster, Socha, Brooker SPIE- SC707 69 Optimum DOF and Modulation for Quadrupole Optimum when phase differences between 0th and 1st orders are minimum  center /Pitch +1 0

70. Milster, Socha, Brooker SPIE- SC707 70 n Periodic features benefit most from QUASAR illumination n Optimum illumination is specific to reticle features conventional annular QUASAR after IMEC 1997 Resolution [ /NA] 2.01.51.00.50.0 0.5 1.0 1.5 QUASAR hor. / vert. QUASAR 45° lines annular conventional Dense Lines @ 60% contrast Off-axis Illumination Principles Effects of different illumination modes

71. Milster, Socha, Brooker SPIE- SC707 71 More Facts: Aerial Image Cross Section 100nm L/S 150nm L/S Varying  =193nm, NA=0.75 Dense Lines vs.  (circular) Increase sigma and contrast goes up (100nm L/S) Increase sigma and contrast goes down (200 nm L/S) Very confusing!!! What is going on??

72. Milster, Socha, Brooker SPIE- SC707 72 Lithography Imaging Laws What limits the size of the photoresist pattern ? –Three paths to improve resolution: Wavelength ( ) Numerical Aperture (NA) k 1 : “Process” knob –Includes off-axis illumination, complex masks, high contrast photoresist, acid diffusion, etc… What limits the size of the optical (and/or aerial) image? (Assuming circular illumination source and binary reticle) NA Partial Coherence or illumination (  ) –  =0: Coherent Limit –  =1: Incoherent Limit Fine…but where do these come from?? Note: resolution is often written as Linewidth (LW) or critical dimension (CD) in the context with photoresist

73. Milster, Socha, Brooker SPIE- SC707 73 Basic Illumination and Imaging Definition of Coherence Factor ‘Sigma’ Condenser Imaging Lens Mask Plane Source Stop Diameter Source Image Diameter View of Entrance Pupil with blank mask Pupil Edge (the “NA”) Source Image If pupil diameter = NA, then source size = NA  (pupil or NA units)