Electromagnetic Wave Propagation Lecture 10: Multilayer ...

Electromagnetic Wave PropagationLecture 10: Multilayer structures I

Daniel Sjoberg

Department of Electrical and Information Technology

October 4, 2012

Outline

1 Introduction

2 Multiple dielectric slabs

3 Antireflection coatings

4 Dielectric mirrors

5 Propagation bandgaps

6 Narrow-band transmission filters (Fabry-Perot resonators)

7 Conclusions

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Outline

1 Introduction






7 Conclusions

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Key questions

I How can we analyze multilayer structures?

I What can we build with them?

I How can we design the structures?

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Outline

1 Introduction






7 Conclusions

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Scattering from multilayer structure

(Fig. 6.1.1 in Orfanidis)(E1+

E1−

)=

1

2

(1 ηa1 −ηa

)P1 · · ·PM

(1 11ηb− 1ηb

)(E′M+1,+

0

)

Pi =

(cos(kiì) jηi sin(kiì)

jη−1i sin(kiì) cos(kiì)

)6 / 50

Scattering from multilayer structure


E1−

)= P′1 · · ·P′M

1

τM+1

(1 ρM+1

ρM+1 1

)(E′M+1,+

0

)

P′i =1

τi

(ejkiì ρie

−jkiì

ρiejkiì e−jkiì

)7 / 50

Scattering parameters

The total transfer matrix relation is(E1+

E1−

)=

(T11 T12T21 T22

)(E′M+1,+

0

)=

(T11E′M+1,+

T21E′M+1,+

)which implies

T =E′M+1,+

E1+=

1

T11

Γ1 =E1−E1+

=T21E′M+1,+

E1+=T21T11

Thus, having computed the total transfer matrix, the reflectionand transmission coefficients correspond to simple rearrangements.

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Layer recursion for reflection

With propagation matrices(Ei,+Ei,−

)=

1

τi

(ejkiì ρie

−jkiì

ρiejkiì e−jkiì

)(Ei+1,+

Ei+1,−

)(EiHi

)=

(cos(kiì) jηi sin(kiì)

jη−1i sin(kiì) cos(kiì)

)(Ei+1

Hi+1

)the reflection coefficient at interface i can be found by recursion

Ei−Ei+

= Γi =ρi + Γi+1e

−2jkiì

1 + ρiΓi+1e−2jkiì, ΓM+1 = ρM+1

and the impedance at interface i in the same way

EiHi

= Zi = ηiZi+1 + jηi tan(kiì)

ηi + jZi+1 tan(kiì), ZM+1 = ηb

These are equivalent. Thus, the reflection properties can be foundfrom a one-pass calculation, iterating from M + 1 to 1.

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Computation of transmission

In general, to find also the transmission coefficient, the fullcascading technique must be employed. However, in the importantcase of no losses energy conservation gives us the result

1− |Γ |2

2ηa=|T |2

2ηb

Thus, for lossless structures the number 1− |Γ |2 represents thetransmittance.

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Dielectric structures

From now on, we assume the medium is non-magnetic, that is,

k = ω√εµ =

ω

c0

√εr = k0n

and

η =

√µ

ε=

√µ0ε0

1√εr

=η0n

Thus, a slab is characterized solely by its physical length ` and itsrefractive index n =

√εr.

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Orfanidis’ matlab codes: multidiel.m

% multidiel.m - reflection response of isotropic or birefringent multilayer structure

%

% na | n1 | n2 | ... | nM | nb

% left medium | L1 | L2 | ... | LM | right medium

% interface 1 2 3 M M+1

%

% Usage: [Gamma,Z] = multidiel(n,L,lambda,theta,pol)

% [Gamma,Z] = multidiel(n,L,lambda,theta) (equivalent to pol=’te’)

% [Gamma,Z] = multidiel(n,L,lambda) (equivalent to theta=0)

%

% n = isotropic 1x(M+2), uniaxial 2x(M+2), or biaxial 3x(M+2), matrix of refractive indices

% L = vector of optical lengths of layers, in units of lambda_0

% lambda = vector of free-space wavelengths at which to evaluate the reflection response

% theta = incidence angle from left medium (in degrees)

% pol = for ’tm’ or ’te’, parallel or perpendicular, p or s, polarizations

%

% Gamma = reflection response at interface-1 into left medium evaluated at lambda

% Z = transverse wave impedance at interface-1 in units of eta_a (left medium)

%

% notes: M is the number of layers (M >= 0)

% n = [na, n1, n2, ..., nM, nb] = 1x(M+2) row vector of isotropic indices

%

% [ na1 n11 n12 ... n1M nb1 ] 3x(M+2) matrix of birefringent indices,

% n = [ na2 n21 n22 ... n2M nb2 ] = if 2x(M+2), it is extended to 3x(M+2)

% [ na3 n31 n32 ... n3M nb3 ] by repeating the top row

%

% optical lengths are in units of a reference free-space wavelength lambda_0:

% for i=1,2,...,M, L(i) = n(1,i) * l(i), for TM,

% L(i) = n(2,i) * l(i), for TE,

% TM and TE L(i) are the same in isotropic case. If M=0, use L=[].

%

% lambda is in units of lambda_0, that is, lambda/lambda_0 = f_0/f

%

% reflectance = |Gamma|^2, input impedance = Z = (1+Gamma)./(1-Gamma)

%

% delta(i) = 2*pi*[n(1,i) * l(i) * sqrt(1 - (Na*sin(theta))^2 ./ n(3,i).^2))]/lambda, for TM

% delta(i) = 2*pi*[n(2,i) * l(i) * sqrt(1 - (Na*sin(theta))^2 ./ n(2,i).^2))]/lambda, for TE

%

% if n(3,i)=n(3,i+1)=Na, then will get NaN’s at theta=90 because of 0/0, (see also FRESNEL)

%

% it uses SQRTE, which is a modified version of SQRT approriate for evanescent waves

%

% see also MULTIDIEL1, MULTIDIEL2

% Sophocles J. Orfanidis - 1999-2008 - www.ece.rutgers.edu/~orfanidi/ewa

function [Gamma,Z] = multidiel(n,L,lambda,theta,pol)

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Outline

1 Introduction






7 Conclusions

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Application: antireflection coating

nb = 1.5

na = 1 Er = 0Ei

nb = 1.5 nb = 1.5

n2 = 1.38

na = 1 ErEi na = 1 Er = 0Ei

n1 = 1.22 n3 = 2.45

An anti-reflection coating can be realized by one or many layers.

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Materials

Some materials in use for optical anti-reflection coatings are foundon page 190 in Orfanidis:

Not all refractive indices are available, meaning the design needs toeither be tolerant to deviations or adapted to the availablematerials.

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How can the parameters of the coating be chosen?

Each slab is characterized by two parameters, thickness `i andrefractive index ni.

We may fix the parameter vector a = [{`i}Ni=1, {ni}Ni=1], andcompute the response Γ (a). Setting up an optimization loop, wetypically minimize a penalty function like

p(a) =

∫ f2

f1

|Γ (a, f)− Γ0(f)|nw(f) df

where Γ0(f) is the target frequency response, and the weightfunction w(f) may emphasize different parts of the spectrum.

I This is straightforward, but as the number of layers increases,it becomes important to have a good start design, andunderstand what can and cannot be done.

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Demo

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Explicit design of a two-layer structure

We wish to match refractive index nb = 1.5 to na = 1 using twoslabs. The reflection coefficients are

Γ1 =ρ1 + Γ2e

−2jk1`1

1 + ρ1Γ2e−2jk1`1, Γ2 =

ρ2 + ρ3e−2jk2`2

1 + ρ2ρ3e−2jk2`2

Zero reflection requires Γ1 = 0, or e2jk1`1 = −Γ2/ρ1.

|Γ2|2 =ρ22 + ρ23 + 2ρ2ρ3 cos(2k2`2)

1 + ρ22ρ23 + 2ρ2ρ3 cos(2k2`2)

= ρ21

which can be transformed to

cos2(k2`2) =ρ21(1− ρ2ρ3)2 − (ρ2 − ρ3)2

4ρ2ρ3(1− ρ21)

sin2(k2`2) =(ρ2 + ρ3)

2 − ρ21(1 + ρ2ρ3)2

4ρ2ρ3(1− ρ21)

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Determining materials

Assuming n1 < n2 and n2 > nb, we have ρ2 < 0 and ρ3 > 0. Thefollowing conditions must be satisfied for the equations to have asolution: ∣∣∣∣ ρ3 + ρ2

1 + ρ2ρ3

∣∣∣∣2 < ρ21 <

∣∣∣∣ ρ3 − ρ21− ρ2ρ3

∣∣∣∣2which requires

n1 ∈ [√nb, nb] = [1.22, 1.5] and n2 6∈ [

√nb, n1

√nb] = [1.22, 1.69]︸︷︷︸

using n1 = 1.38

Materials:I Magnesium flouride: n1 = 1.38I Bismuth oxide: n2 = 2.45

Lengths (target wavelength λ0 = 550 nm)

k1`1 = 2.0696 n1`1 = 0.3294λ0 `1 = 131 nm

k2`2 = 0.2848 n2`2 = 0.0453λ0 `2 = 10.2 nm

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Results

nb = 1.5

na = 1 Er = 0Ei

nb = 1.5 nb = 1.5

n2 = 1.38

na = 1 ErEi na = 1 Er = 0Ei

n1 = 1.22 n3 = 2.45

Some drawbacks:

I Smaller bandwidth

I Complicated procedure

I Small dimensions

(Fig. 6.2.1 in Orfanidis)

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Special case: quarter and half wavelength slabs

For a quarter wavelength slab, niì = λ0/4, we have

Zi = ηiZi+1 + jηi tan(kiì)

ηi + jZi+1 tan(kiì)=

η2iZi+1

and for a half wavelength slab, niì = λ0/2, we have

Zi = ηiZi+1 + jηi tan(kiì)

ηi + jZi+1 tan(kiì)= Zi+1

Thus, quarter wavelength slabs inverts the impedance, whereashalf wavelength slabs preserves impedance.

I These rules can be used to make simple designs at targetwavelength.

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An impedance approach to antireflection

The goal is to transform the impedance ηb to ηa. Two possibilities:


First possibility (quarter-quarter):

ηa =η21Z2

=η21

η22/ηb= ηb

η21η22

⇒ nanb

=n21n22

Second possibility (quarter-half-quarter):

ηa =η21Z2

=η21

η23/ηb= ηb

η21η23

⇒ nanb

=n21n23

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Resulting bandwidth

The design was made for the center wavelength λ = 550 nm, theresult for any wavelength is computed with multidiel.m:


Note there was no requirement on the half-wavelength slab! Thefunction of this slab is to increase the bandwidth.

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How does this look in the complex plane?

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

-0.15 -0.1 -0.05 0 0.05

Mismatched quarter-quarter, matched quarter, matchedquarter-half-quarter, mismatched quarter-half-quarter.The effect of the half-wavelength slab is to add the twist for thered curve, since Γ` = Γ0e

−2jk`.

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Outline

1 Introduction






7 Conclusions

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Dielectric mirrors

Instead of anti-reflection, we can design a multilayered structureaimed at significant reflection.

Useful to avoid losses in metal structures, which may absorb a fewpercent of power at optical frequencies. Typical design issues:thickness and bandwidth.

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Design at center wavelength

Alternating quarter wavelength slabs of high (nH) and low (nL)refractive index (short hand notation AH(LH)4G):


The impedance at interface 2 is

Z2 =η2LZ3

=η2L

η2H/Z4=

(nHnL

)2

Z4 =

(nHnL

)4

Z6 = · · · =(nHnL

)8

ηb

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Reflection

The impedance at interface 1 after N bilayers and a final nH layeris

Z1 =η2HZ2

=η20/n

2H(

nHnL

)2Nηb

which implies the reflection coefficient

Γ1 =Z1 − ηaZ1 + ηa

= · · · =1−

(nHnL

)2N n2H

nanb

1 +(nHnL

)2N n2H

nanb

→ −1, N →∞

Thus, for many layers, we get high reflection. Without the final nHlayer, the reflection would have the limit Γ → +1.

nH nL N 1 2 4 8 16

2.32 1.38 |Γ1|2 0.7685 0.9111 0.9884 0.9998 1.0000

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Propagation on other wavelengths


E1−

)=

1

τ1

(ejδH ρ1e

−jδH

ρ1ejδH e−jδH

)FN

1

τ2

(1 ρ2ρ2 1

)(E′M+1,+

0

)F =

1

1 + ρ

(ejδL ρe−jδL

ρejδL e−jδL

)1

1− ρ

(ejδH −ρe−jδH−ρejδH e−jδH

)δH = kH`H, δL = kL`L, ρ =

nH − nLnH + nL

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Results for different number of layers


Computed using multidiel.m with the parameters na = 1,nb = 1.52, nH = 2.32, nL = 1.38. Bandwidth is relativelyinsensitive to the number of layers, ca 600−430

500 = 0.34.

For the infinite structure, the bandwidth is (will be shown later)

∆λ

λ0=π

2

(1

arccos(ρ)− 1

arccos(−ρ)

)= 0.336, ρ =

nH − nLnH + nL

= 0.254

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Different optical thicknesses


Computed using multidiel.m with the parameters na = 1,nb = 1.48, nH = 4.6, nL = 1.6. The optical thicknesses arenH`H = 0.2944λ0 and nL`L = 0.2112λ0, and N = 4.

I Note that nH`H + nL`L = 0.5λ0, this is the important period.I Higher bandwidth (∆λ/λ0 = 0.73) due to higher contrast

(ρ = 0.48).

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Multiband reflectors

Short-hand notation for the dielectric mirror: AH(LH)8G, whereA denotes the air medium, H the high-index medium, L thelow-index medium, and G is the glass. Multiband reflectors can bemade by inserting extra multiples of one or both of H and L slabs:


Note fundamental period of ∆f/f0 = 2.

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Short- and longpass reflectors

The frequency dependence can be manipulated by the outmostlayers. Original design is AH(LH)8G centered at 500nm.

Shortpass: A(0.5L)H(LH)8(0.5L)G

Longpass: A(0.5H)L(HL)8(0.5H)G


Note how the roles of H and L are dual in the two designs. Dottedcurve is original design centered at the corresponding λ0.

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Outline

1 Introduction






7 Conclusions

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Bandwidth for an infinite structure

Consider the limit case N →∞. We look for waves propagatingwith the effective wave number K, so that there should existsolutions (where δL = kL`L, δH = kH`H, and ` = `L + `H)(

EiHi

)=

(cos(δL) jηL sin(δL)

jη−1L sin(δL) cos(δL)

)·(

cos(δH) jηH sin(δH)

jη−1H sin(δH) cos(δH)

) (Ei+2

Hi+2

)︸︷︷︸=e−jK`(Ei

Hi)

This is an eigenvalue equation, which can be put in the form

cos(K`) =cos(δH + δL)− ρ2 cos(δH − δL)

1− ρ2

where ρ = nH−nLnH+nL

. K is called the Bloch wavenumber.

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Bandgaps

With δH = δL = δ, the equation is

cos(K`) =cos(2δ)− ρ2

1− ρ2

Propagating waves correspond to real K, which implies| cos(K`)| ≤ 1.

0 1 2 3 4 5−5

−4

−3

−2

−1

0

1

δ/δ0

nH/n

L=2

nH/n

L=4

nH/n

L=8

Inside the band gap, the waves are exponentially attenuated.36 / 50

Bandwidth

The band edges are given by the condition

−1 =cos(2δ)− ρ2

1− ρ2⇒ cos2 δ = ρ2

Since δ = π2λ0/λ, where λ0 is the center wavelength, the band

edges are

λ1 =λ0π/2

arccos(−ρ)λ2 =

λ0π/2

arccos(ρ)

and the bandwidth is

∆λ

λ0=π

2

(1

arccos(ρ)− 1

arccos(−ρ)

),

∆f

f0=

4

πarcsin(ρ)

The bandwidth is zero for ρ = 0, and ∆f/f0 = 2 for ρ = 1(λ2 →∞ and f2 → 0 when ρ→ 1), where

ρ =nH − nLnH + nL

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Bandgaps generalize to 3D structures

Manufactured and measured at FOI in Linkoping.

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General periodic structures

I Propagation bandgaps may appear in any periodic structure:I CrystalsI GratingsI Periodically loaded transmission lines

I Often analyzed by solving for the Bloch waves, propagatingwith factor e−jK·r, which leads to eigenvalue problems.

Looking for solutions [E(r),H(r)]e−jK·r where E(r) and H(r)are periodic functions, turns Maxwell’s equations into(

0 −(∇− jK)× I(∇− jK)× I 0

)(E(r)

H(r)

)= −jω

(ε(r) ξ(r)ζ(r) µ(r)

)(E(r)

H(r)

)with periodic boundary conditions. For some bands of ω, there areno solutions with real-valued K-vectors.

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Multidimensional bandgap structures

See http://ab-initio.mit.edu for free text book, lecture notes andcomputational software MPB (eigensolver) and Meep (FDTD).

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MIT website

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Outline

1 Introduction






7 Conclusions

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Application: Fabry-Perot resonator

Having designed a dielectric mirror, we can use it to construct aFabry-Perot resonator, which is a narrow-band transmission filter.

Since a half wavelength slab preserves impedance, we can eliminatesuch slabs at the design frequency. By inserting a low-indexmaterial between two bilayer stacks, we obtain

(HL)NL(HL)N = (HL)N−1HL L HL(HL)N−1

→ (HL)N−1HHL(HL)N−1

→ (HL)N−1L(HL)N−1 → · · · → L

Adding another outer layer L, the final structure(HL)NL(HL)NL→ 2L allows perfect transmission at the designwavelength.

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Example of a Fabry-Perot resonator design


Computed using multidiel.m with parameters na = nb = 1.52,nL = 1.4, and nH = 2.1.

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Two FPRs

G|(HL)N1L(HL)N1 |(HL)N2L(HL)N2 |G


The resonance bandwidth is controlled by the number of layers.

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Control of the design

The resonance bandwidth is controlled by the number of layers.


The isolation of the resonance is controlled by the contrast.

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Outline

1 Introduction






7 Conclusions

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Conclusions

I Multilayer structures are easily analyzed by cascadingtechniques.

I Typical designs are based on quarter wavelength and halfwavelength slabs.

I A stack of high/low index slabs can form a dielectric mirror.

I Insertion of extra layers can form a Fabry-Perot resonator.

I High bandwidth and high attenuation requires many layers.

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