Milster, Socha, Brooker SPIE- SC707 1 Basics of Optical Imaging in Microlithography: A "Hands-on" Approach Tom D. Milster (University of Arizona) Robert

Milster, Socha, BrookerSPIE- 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

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 This To This

fc fcf1 f1 f2

f2 2fcam 2fcam

Condenser

Grating(Mask)

Lens 1

Stop

Lens 2

Image Plane (Aerial Image of Mask)

CCD Camera (AIMS)

SourceAperture

Tom D. Milster

f1 = 91 mmf2 = 245 mmfc = 50 mmlambda = 525 nm


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

Introduction• What is photo lithography ?

Etymology: Photolithography = Light Stone Writing

Photoresist Development

Negative Photoresist Positive Photoresist

Object: reticle or mask

Optics

Aerial ImagePhotoresist

Wafer + films Latent Image

• Optical image is recorded in the resist via changes in concentrations of species.

• Concentration level controls development

z

X

yResist Crosssections


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

0.25m 5-BAR Structures Focus=0.0m, NA=0.57 NA=0.6, 248nm

Image CrossSection

Resist CrossSection(not top down!)

Z

Z


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

Increase NA

Introduction• Imaging Resolution and Lord Rayleigh

– Q: When can you resolve the image of 2 distance stars?– A: When the 1st Intensity min of one lines up with peak of other

Decrease

Large NA

Web Top Optics, 1999

Resolution 0.61NA

Resolution 0.61

NA

Small NA

Large

From the math of the Airy function


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.


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• 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


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• You again grasshopper…• Master…• …look at the following data


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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!


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• 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!!


fc fcf1 f1 f2

f2 2fcam 2fcam

LEDSource

Condenser

Grating(Mask)

Lens 1

Stop

Lens 2


CCD Camera (AIMS)

SourceAperture

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.


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That is what your image looks like Draw the light pattern at the stop here.


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Basic Illumination and Imaging• Kohler Illumination

Condenser Imaging Lens

Mask PlaneSource

Aperture

Aerial Image

StopImage 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


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Basic Illumination and Imaging• Definition of Coherence Factor ‘Sigma’

CondenserImaging Lens

Mask Plane

Source

Stop Diameter

Source Image

Diameter

View of Entrance Pupil with blank mask

Pupil Edge (the “NA”)

Source Image

Diameter of Source Image in Stop

σDiameter of Stop


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Simple Binary Mask• Model a Cr on quartz grating

mask as an infinitely thin grating

Position

E

SiO2

Cr

E-Field

P

0+1

+2

+3

-1

-2

-3Diffraction

Orders

Lens/Pupil

Grating Equation:

0th-1st +1st

Note: For 1:1 lines and spaces, P= 2 * LW

LW = Line Width

Pitch

)sin(


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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.

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.


19


20

Binary Mask and Diffraction Orders

NAP

min min 2

LWNA

• Must have more than 1 order in pupil to have image modulation

Pupil (stop)

o

+1

+3

-3

-1

pupil

o

+1

-1

pupil

o

+1

-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.

Pmin is the minimum pitch that is at the limit of resolution.

Max

NA=sin(Max)

Max

1Photoresist CD kNA

1Photoresist CD k

NA

k1=1/2

Strong Image Modulation


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Coffee Break


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• 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!!!


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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


CCD Camera (AIMS)

SourceAperture

fc fc f1 f1 f2f2 2fcam 2fcam


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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.


Condenser

Grating(Mask)

Lens 1

Stop

Lens 2


CCD Camera (AIMS)

SourceAperture


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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


Condenser

Grating(Mask)

Lens 1

Stop

Lens 2


CCD Camera (AIMS)

SourceAperture


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• Comedy writer’s strike…

• No more multiple choice answers

• Let’s continue to cement the concepts associated with the bench


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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


CCD Camera (AIMS)

SourceAperture



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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


CCD Camera (AIMS)

SourceAperture



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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.

f1 f1 f2 f2

Grating(Mask)

Lens 1

Stop

Lens 2



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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.

Tom D. Milster

f1 = 91 mmf2 = 245 mmfc = 50 mmlambda = 525 nm


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.

Effect of Varying Sigma


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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

Contrast Curves versus Pitch & Sigma

• Sigma=0.05 ---Coherent• Sigma=0.5 -----Partially Coherent• Sigma=1 ---Incoherent limit


34

Modulation Transfer Function (MTF)

• Optics types love this plot!!!!• Can you find the Coherent frequency cut off?


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

Binary Mask: Sigma < 1

• No grating - just blank mask

0th order

0th order

-1st +1st

• Grating period at cut-off frequency

PNA

• Grating period resolution limit at given

min (1 )P

NA

0th order-1st +1st

limit 2 (1 )NA

Resolution limit with 0<<1 for a circular source

min

For 1:1 grating,

2 (1 )LW

NA

min 2LW

NA


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Binary Mask: Sigma = 1

0th order• No grating - just

blank mask

-1st +1st


PNA

0th order-1st +1st

limitmin

4 2NAP

• Grating period corresponds to incoherent cut-off

min 2P

NA

Resolution limit with 1 for a circular source

min

For 1:1 grating,

4LW

NA


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Binary Mask, =248nm, NA=0.63


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=0.05 & = 0.7 for k1=0.5


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 = 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

Binary Mask: Round and Annular Illumination

Larger and k1<0.5Small and k1>0.5

Conventional or Circular Source

Annular Source

• All power is inside pupil (for 0th and 1st 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

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?

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 +/- 1st 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.


43


44

Binary Mask, =248nm, NA=0.63, 300nm L/S

3-Beam Imaging


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.

Missing Orders

L pitch and PH source Block zero order Block ± 1 order


46


47

Binary Mask, =248nm, NA=0.63, 250nm L/S


48

Phase MaskFor phase:

2( 1)d

n

E

E-Field

+1+3

+5

-1-3

-5Diffraction Orders

Lens/Pupil

Grating Equation:

-1st +1st

Cr

SiO2

Etched depth

d

Position

+1

-1

)(2

)sin(PitchCr

P


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Pure Phase – Chromeless

E

E-Field

Diffraction Orders

Lens/Pupil

Position

+1

-1

SiO2

Etched depth

d

For phase: 2( 1)

dn

+1+3

+5

-1-3

-5

Grating Equation:

-1st +1st

Pitch

)sin(

P


50

Phase Mask• The phase mask produces no

zero order Pupil (stop)

+1

+3

-3

-1

pupil +1

-1

pupil

+1

-1

Coherent limit


Strong Image Modulation

No zero order is emitted from the phase mask.

pmin is the minimum Cr pitch that is at the limit of resolution.

min 4LW

NA

For alternating phase shift grating

NA=sin(Max)

Max

Max

NAP

2min

1Photoresist CD kNA

1Photoresist CD k

NA

k1=1/4


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 +/-1st 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?

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.


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Draw a sketch of image. Draw a sketch of light pattern at the stop.


53

=248nm, NA=0.63, sigma = 0.3


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 Quadrupole 0 Quadrupole y- dipole x- dipole


55

Coherent Off-Axis Illumination and a Binary Mask• Orders shift relative to pupil

Image Modulation

pupil

0

+1

+3

-3

-1

0 Orderth

“Incoherent” limitpupil

0

-1

0 Orderth

pupil

0

-1


min 4LW

NA

For 1:1 grating

Max

Max

NA=sin(Max)

NAP

2min

1Photoresist CD kNA

1Photoresist CD k

NA

k1=1/4

Pmin is the minimum pitch that is at the limit of resolution.


56

Binary Mask with Annular Illumination

min

For 1:1 grating,

2 (1 )outer

LWNA

• No grating - just blank mask

-1st +1st


PNA

• Grating period resolution limit at given

min (1 )outer

PNA

-1st +1st

limit 2 (1 )outerNA

Resolution limit with 0<<1 for a circular source

inner

center

0th order

outer

0th order

0th order


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.

Off-Axis Illumination with a Binary Mask

Draw a sketch with the centered source Draw a sketch with decentered source

Camera Output Camera Output


58


59

=248nm, NA=0.63, sigma = 0.3


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 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

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

ReferencesIntroductory Articles

SPIE Proceedings for Microlithography

Journal of Microlithography, Microfabrication, and Microsystems (JM3) – SPIE Press

Industry MagazinesMicrolithography World

www.pennwell.com

BooksIntro to Fourier Optics and Statistical Optics

by J. GoodmanResolution Enhancement Techniques and Optical Imaging in Projection Microlithography

Alfred Wong, SPIE PressMicrolithography: Science and Technology

Ed: James Sheats and Bruce SmithPub: Marcel Dekker

Linear Systemsby J. Gaskill


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 PapersU of A Dissertation by Doug Goodman (1979), Stationary Optical ProjectorsPapers by H.H. Hopkins for partial coherent imaging, Richards and Wolf for high NA


64

Thank You for Taking This Course!


65

Backup Slides


66

Basic Illumination and Imaging

sin mNA n

NAh h

NA

• 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

m


67

Basic Illumination and Imaging

• Let’s do an experiment– Calculate NA at the image

plane for rs = _______ .

– Calculate the coherent

resolution limit in terms of pitch in the aerial image.

1

2

sin sin tan sm

rNA

f

rs

’m

2

1

________

525nm

n

f

sin mNA n

Imaging Lens

minpNA


68

Optimum DOF and Modulation for Annular (and dipole)

• Optimum when phase differences between 0th and 1st orders are minimum

center

/Pitch

1 1 1 1

2 4center Pitch NA LW NA

2 centerNAPitch

+1-1 0


69

Optimum DOF and Modulation for Quadrupole

• Optimum when phase differences between 0th and 1st orders are minimum

center

/Pitch

2 1 2 1

2 4center Pitch NA LW NA

22

centerNA

Pitch

+1-1 0


70

Periodic features benefit most from QUASAR illumination

Optimum illumination is specific to reticle features

conventional

annular

QUASARafter IMEC 1997

Resolution [/NA]2.01.51.00.50.0

0.0

0.5

1.0

1.5QUASAR hor. / vert.

QUASAR 45° lines

annular

conventional

DO

F [

µm

]

Dense Lines @ 60% contrast

Off-axis Illumination PrinciplesEffects of different illumination modes


71

More Facts: Aerial Image Cross Section

0.10

0.20

0.30

0.40

0.50

-100 -80 -60 -40 -20 0 20 40 60 80 100

Image Intensity

Horizontal Position (nm)

0.1

0.3

0.5

0.7

0.9

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

-100 0 100

Image Intensity

Horizontal Position (nm)

0.1

0.3

0.5

0.7

0.9

100nm L/S150nm 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

Lithography Imaging Laws• What limits the size of the photoresist pattern ?

– Three paths to improve resolution:

• Wavelength ()

• Numerical Aperture (NA)

• k1 : “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??

1Photoresist CD kNA

1Photoresist CD k

NA

1Optical Resolution

2(1 ) NA

1Optical Resolution

2(1 ) NA

• Note: resolution is often written as Linewidth (LW) or critical dimension (CD) in the context with photoresist


73

Basic Illumination and Imaging• Definition of Coherence Factor ‘Sigma’

CondenserImaging Lens

Mask Plane

Source

Stop Diameter

Source Image

Diameter

View of Entrance Pupil with blank mask

Pupil Edge (the “NA”)

Source Image

Diameter of Source Image in Stop

σDiameter of Stop

If pupil diameter = NA, then source size = NA • (pupil or NA units)

Documents

Milster, Socha, Brooker SPIE- SC707 1 Basics of Optical Imaging in Microlithography: A "Hands-on" Approach Tom D. Milster (University of Arizona) Robert