Photonic Materials 3_Waveguides

8/8/2019 Photonic Materials 3_Waveguides

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PhotonicMaterials

Frank GellDaniel Navarro

1


2/92

TABLE OF CONTENTS

3. Fundamentals of waveguides 1D and 2D waveguides: fundamentals andexperimental characterization.

Reference: Silicon photonics: an introductionGraham T. Reed andAndrew P. Knights


3/92

n2

n1

2211 sinnsinn =

1

2n2

n1

Ei Er

Et

Lightraysrefractedandreflectedattheinterfaceoftwomedia

1

c

TheRay Optics Approach to Describing Planar Waveguides


4/92

n1

n2

211 nsinn =

1

2c

nnsin =

Totalinternalreflectionattwointerfacesdemonstrating

theconceptofawaveguide


5/92

Reflection Coefficients

irE.rE =

where r is a complex reflection coefficient, which ispolarisation dependant

The Transverse Electric (TE) condition is defined as the

condition when the electric fields of the waves are

perpendicular to the plane of incidence.

Correspondingly, the Transverse Magnetic (TM) conditionoccurs when the magnetic fields are perpendicular to the

plane of incidence.

n2

n1

Ei Er

Et

Circlesindicatethattheelectricfieldsarevertical(i.e.comingoutof

screen)

OrientationofelectricfieldsforTEincidenceattheinterface

between2media.

interface


6/92

The reflection coefficients rTE and rTM, are described by the

Fresnel formulae

For TE polarisation:

2211

2211TE

cosncosn

cosncosnr

+

=

For TM polarisation:

2112

2112TM

cosncosn

cosncosnr

+=

Using Snells Law:

1

22

1

2

211

1

22

1

2

211TE

sinnncosn

sinnncosnr

+

=

And

1

22

1

2

211

2

2

1

22

1

2

211

2

2

TM

sinnnncosn

sinnnncosnr

+

=


7/92

If1isgreaterthanc

( ) 2j

j j

j

a jb er e e

a jb e

and

= = = =

+

1r =

Where TE and TM are given by:

and

1

2

1

21

2

1

TEcos

n

nsin

tan2

=

1

1

2

1

2

2

2

2

1

1

TM

cosn

n

1sinn

n

tan2

=

Which arenegative indeed


8/92

Butrrelatesreflectedfields. Powerisdescribedby

thePoynting Vector

2

m

m2 EEZ

1S

==

And

reflected

power

is

related

by

reflectance

R:

2

2

i

2

r

i

r rE

E

S

SR ===

where E is electric field, m is the permittivity of themedium, m is the permeability of the medium, and Z is theimpedance of the medium


9/92

Phase of a propagating wave and its wavevector

Let

and

Where z is the direction of propagation

)]tkz(jexp[EE0 =

)]tkz(jexp[HH 0 =

Therefore,phase

is

tkz =

The phase varies with time (t), and with distance (z). These

variations are quantified by taking the time derivative and the

spatial derivatives:

where is angular frequency (rads/sec), and f is frequency (Hz).

f2t

==

kz

=


10/92

k is the wavevector (propagation constant) in the direction

of the wavefront. It is related to wavelength, , by:

=2

k

In free space k = k0, and

Hence in free space,

0nkk=

0

0

2

k

=


11/92

Modes of a planar waveguide

Thewavevector inaplanarwaveguide

n1

n2

n3

y

x

z

hk0n1

We can decompose the wavevector k, into two components,

in the y and z directions.

1k=n1k0

kz=n1k0sin1

ky=n1k0cos1

Therelationshipbetweenpropagationconstantsinthey,z,

andwavenormal directions


12/92

We require self consistency condition. As the wave reflects

twice it reproduces itself. The total phase shift must be a

multiple of 2, hence:

12 cos AC AB h =

= m2coshnk2 u110 l

, since we have seen that reflection coefficients give a phase

change upon reflection. We have refered to these phase

shifts as u, and l respectively.


13/92

Theplanarwaveguide

n1

n2

n3

y

x

z

h

0 1 12 cos 2uk n h m =l

Thuslightpropagatesindiscretemodesdescribedbythe

polarisationandthemodenumber.

E.g.TE0,TE1,TM0,etc

Eachmodewillhaveauniquepropagationconstantinthe

yandzdirections

Thenumberofmodesislimitedbysatisfactionofthe

requirementsoftotalinternalreflection.


14/92

The Symmetrical planar waveguide

Inthesymmetricalplanarwaveguide,n2 =n3,andhence

u=l. ThereforeforTEpolarisation,theequationbecomes:

This can be rearranged as :

=

m2cos

n

nsin

tan4coshnk2

1

2

1

21

2

1

110

=

1

2

1

21

2

110

cosn

nsin

2mcoshnktan

Laterwewillsolvethisequationforangle1


15/92

Wecanfindtheapproximatenumberofmodessupportedby

thewaveguideasfollows:

The minimum value that 1 can take is c. i.e.

1

2c

n

nsin =

Hence the right hand side of the previous equation reduces

to zero and the equation becomes:

02

mcoshnk maxc10 =

rearranging for m, the mode number,

= c10max

coshnkm

Numberofmodes=[mmax]int +1,sincethelowestorder

mode(usuallycalledthefundamentalmode),hasamode

numberm=0.Notethatthesymmetricalwaveguideis

nevercut

off.


16/92

The Asymmetrical planar waveguide

n2 n3,andu l ,hence

Propagationin

an

asymmetric

planar

waveguide

n1

n2

n3

y

x

z

h

[ ]

+

=

1

2

1

31

2

1

1

2

1

21

2

1

110cos

n

nsin

tancos

n

nsin

tanmcoshnk

Notethatthereisnotalwaysasolutionform=0,hencethe

asymmetricalguidemaybecutoff.


17/92

Solving the eigenvalue equation for symmetrical and

asymmetrical waveguides

Let n1 = 1.5, n2 = 1.49, n3 = 1.40, 0= 1.3m, and h = 0.3m,

and TE polarisation.

Notethattheasymmetricalwaveguideiscutoff,whereasthe

symmetricalwaveguideisnot

0 0.2 0.4 0.6 0.8 1 1.2 1.40

1

2

3

f( )

2( )

1( )

3( )

4( )

u l

k0n1hcos1

l

2u

u +l u

Phase

Change

(radians)

Fundamentalmode

angle

1 (radians)

Solutionoftheeigenvalue equationform=0

Solutionfor1

20 0.2 0.4 0.6 0.8 1 1.2 1.4 /2

3

2

1

0


18/92

Now consider a silicon waveguide:

n1 = 3.5 (silicon), n2 = 1.5 (silicon dioxide), n3 = 1.0 (air),

0= 1.3m, and h = 0.15m.

Notethatthesymmetricalandasymmetricalwaveguidesare

similar

0 0.2 0.4 0.6 0.8 1 1.2 1.40

1

2

3

f( )

2( )

1( )

3( )

4( )

k0n1hcos1

2l

u +l

ul

Fundamentalmodeangle1 (radians)

u l

Phase

Change

(radians)

Solutionoftheeigenvalue equationform=0

(silicononinsulator)

20 0.2 0.4 0.6 0.8 1 1.2 1.4 /2

3

2

1

0


19/92

Monomode Conditions

Consider the TE polarisation eigenvalue equation for a

symmetrical waveguide:

=

1

2

1

21

2

110

cos

n

nsin

2

mcoshnktan

Forthesecondmode,m=1,andtheequationbecomes:

i.e.

02

coshnktan c10 =

hn2hnkcos

1

0

10

c

=

=


20/92

Hence for monomode conditions


21/92

Effective Index

Wehaveseenthatpropagationconstantsinthezandy

directionsare:

and

101z sinknk =

101ycosknk =

kz is also known as , the propagation constant in thedirection of the waveguide

Now define a parameter N, called the effective index of the

mode, such that:

11 sinnN =

i.e.

0zNkk ==


22/92

The lower bound on is determined by the larger of thecritical angles of the waveguide, usually at the lower

interface:

2001nksinkn =

l

The upper bound on is governed by the maximum value of,

which is 90o. In this case = k =n1k0. Hence:

2010 nknk

Substituting for = k =Nk0:

21 nNn


23/92

M-line technique

n1

ns

n2

Sidetector

2

1

0 n0


24/92

M-line technique

1.12 1.16 1.20 1.24 1.28 1.32 1.36 1.403

4

5

Intensityinthedetector(a.u.)

effective refractive index

Detector

=633nm


25/92

0 2000 4000 6000 8000 10000 12000

1.0

1.2

1.4

drift=1.3%n

mat

thickness (nm)

no

ne

1.2391.255

1.15

0.5

1.0

TM exp

TM sim

Normalisedintensity

1.15 1.20 1.25 1.30

0.3

0.6

0.9

1.2

TEexp

TE sim

neff

Nice agreement between experiments and

simulations

M-line technique


26/92

Just a little electromagnetic theory

StartingwithMaxwellsequations,ifweassumealoss

less,nonconductingmedium,limitourselvesto

propagationinthezdirection,andconsiderone

polarisationatatime(TEorTM),we canderiveascalar

equationdescribingwavepropagationinourplanar

waveguide:

2

x

2

mm2

x

2

2

x

2

t

E

z

E

y

E

=

+

where m is thepermittivity of the waveguide, m is thepermeability of the waveguide, and in this case the electric

field polarised in the x direction corresponds to TE

polarisation. This is called the scalar wave equation.


27/92

Since there is only electric field in the x direction the general

solution of this field as:

tjzj

xx ey)e(EE

=

This means that there is a field directed (polarised) in the x

direction, with a variation in the y direction yet to be

determined, propagating in the z direction with propagation

constant , with sinusoidal (ejt) time dependence. Thewaveguide configuration is:

Propagationinanasymmetricplanarwaveguide

n1

n2

n3

y x

z

h

k0n1y=h/2

y=h/2

y=0


28/92


29/92

Solution of the wave equation provides us withexpressions of the electric fields in the core and

claddings:

Intheuppercladding:

For y(h/2))2hy(k

ux

yu

eEy)(E

=In the core we will have

For -(h/2) y (h/2)yjk

cx

yc

eEy)(E

=In the lower cladding we will have:

For y -(h/2))

2

hy(k

x

y

eEy)(E+

=l

l

Where 22i

22

yinkk

0=

Note that n and ky are written ni and kyi because they can

now represent any of the three media (core, upper cladding,

or lower cladding), by letting i=1, 2, or 3.


30/92

Propagation constants again

In solving the wave equation we assumed field solutions

with propagation constants in the core, and the upper and

lower claddings, but we didnt discuss why the propagationconstants took the form they did.

Our general solution was of the form:

tjzjyk

cx eeeEEy =

However, in the three media, core, upper cladding and lower

cladding, the propagation constant ky took different forms.

In the claddings ky was a real number, whereas in the core

ky was an imaginary number

Totalinternalreflectionoccurs

fielddecaysincladding

partofthefieldpropagatesinthecladding


31/92

Mode Profiles

Now that we have field solutions for modes in the planar

waveguide, we can plot the field distribution, Ex(y) or the

intensity distribution, Ex(y)2. Consider the following

waveguide:

n1 = 3.5, n2 = 1.5, n3 = 1.0, 0= 1.3m, and h = 1.0m.

For even functions, the field in the core is a cosine

function, and in the claddings, an exponential decay.

Therefore, solutions for m=0, and m=2 are even functions,

and are plotted below. The field distribution is plotted form = 0, and the intensity distribution is plotted for m = 2.


32/92

Electricfieldprofileofthefundamentalmode(m=0)

1 106

5 107

0 5 107

1 106

0

0.2

0.4

0.6

0.8

1

EC0( )y

EU0( )y2

EL0 ( )y1

,,y y2 y1

Normalised

Electric

fieldEx(y)

Distanceintheydirection(m)

1 0.5 0 0.5 1

1

0.8

0.6

0.4

0.2

0


33/92


34/92

1106

5107

0 5107

11060

0.2

0.4

0.6

0.8

1

IC2( )y

IU2( )y2

IL2( )y1

,,yy2y1

Normalised

Intensity

Ex(y)2

Distanceintheydirection

(m)

Intensityprofileofthe2nd evenmode(m=2)

1 0.5 0 0.5 1

1

0.8

0.6

0.4

0.2

0


35/92

Confinement factor

Howmuchofthepowerpropagatesinsidethecore?

Wecan

define

aconfinement

factor

:

=

dy)y(E

dy)y(E

2

x

2/h

2/h

2

x


36/92

3.0 2D waevguides


37/92

Silicon on Insulator Waveguides

SiO2

Surfaceguidinglayer

siliconsubstrate

SiOSiO22(n(n22))

rh

w

n1

n3


38/92

Modes of two dimensional waveguides

xq,pE

Modesaredesignated

yq,pE

or

q,pHE q,pEHPmaximainthexdirection,qmaximaintheydirection

HenceFundamentalmode:

x1,1E

or y1,1E

Sometimeslabelled

x 0,0E y 0,0E


39/92

The Effective Index Method of analysis

n1

n3y

x

h

w

Firstsolve:

n3

n1

n2

h

Thensolve

n3 neff n3

w


40/92

The Effective Index Method of analysis

EXAMPLE

nn22=1.5=1.5

3m

5m

3.5m

n1 =3.5=3.5

n3 =1.0=1.0

FindeffectiveindexNwg?


41/92

Solution

Firstly decompose the rib structure into vertical and horizontal

planar waveguides This means we must first solve the

horizontal planar waveguide. Since the polarisation is TE, weuse the asymmetrical TE eigenvalue equation:

[ ]

+

=

1

2

1

31

2

1

1

2

1

21

2

1

110cos

n

nsin

tancos

n

nsin

tanmcoshnk

Thisgivesafundamentalmodepropagationangleof

87.92o (1.53456radians),andneffg =n1sin1 =3.4977

Firstplanarwaveguideofthedecomposedribstructure

n1=3.5

n2 =1.5

n3=1.0

h=5m


42/92

We now need to solve the second planar waveguide of the

decomposed rib structure:

(i) First find the effective index of the planar regionseither side of the core.

(ii) Solve the asymmetrical TE eigenvalue equation

again, for r = 3m.

Thisyields

apropagation

angle

of

86.595

o

,and

an

effective

indexfortheplanarregionofneffp =3.4938.

Secondplanarwaveguideofthedecomposedribstructure

neffg=

3.4977neffp =3.4938 neffp =3.4938

w=3.5m


43/92

We can now solve the second decomposed planar

waveguide of the decomposed rib structure, using the

effective indices just calculated. This time the symmetrical

TM eigenvalue equation should be used:

=

wg

effg

effp

wg

effp

effg

1

wgeffg0

cos

n

n

12sin

2

n

n

tan2coswnk

Hence wg is be found. The solution is wg= 88.285o, which

corresponds to an effective index of Nwg= 3.496. Knowing

the effective index we can evaluate all of the propagation

constants of the waveguide. For example we are

particularly interested in :

1

wg0m897.16Nk ==

Note: Highconfinementinsilicon

Differentdegrees

of

confinement

horizontally

&

vertically


44/92

Large single mode rib waveguides

5m planarwaveguide

n1=3.5

n2 =1.5

n3=1.0

h=5m

Firstlyconsidera5mplanarwaveguide:

Fromsection1,theapproximatenumberofmodescanbe

foundas:

= c10maxcoshnk

m

Since

o1

c 4.255.3

5.1sin ==

and hence 25 modes are supported (including the m=0 mode).

24cos10x5x5.3x2

]m[0

c

6

intmax =

=


45/92

Alternativelyconsiderthethicknessrequiredtomakesucha

waveguidesinglemode. Fromsection1:


46/92

Itisthereforealittlesurprisingthatribwaveguideswith

crosssectionaldimensionsofseveralmicronsare

routinelydescribedassinglemodewaveguides.

The answer to this apparent paradox is that, if the

waveguide is correctly designed, higher order modes leak

out of the waveguide over a very short distance, as

demonstrated by Soref et al, in 1991.

Firstly the authors defined a rib waveguide in terms of

some normalising parameters:

nn22

Rib Waveguide definitionsRib Waveguide definitions

2br2b n1

n0

2a


47/92

The authors limited their analysis to waveguides in which

0.5 r < 1.0, because for r 0.5, the effective index ofvertical modes in the planar region either side of the rib,

becomes higher than the effective index of all vertical

modes in the rib, otherthan the fundamental.

They then used an effective index approach to define a

parameter related to the aspect ratio of the rib waveguide,a/b. They then found the limiting condition such that the

EH01and HE01 just failed to be guided. This resulted in a

condition for the aspect ratio:

2r1

r3.0

b

a

+

Inordertodemonstratethistheysimulatedexcitationofhigher

ordermodesandwatchedthemleakoutofthewaveguide.


48/92


49/92

Pogossian et al took the experimental data of Rickman &

Reed, and fitted an equation of the form:

c being approximately 0, resulting in a modified equation:

2

r1

rc

b

a

+

2r1

r

b

a

Thesinglemode condition

a/b

r


50/92

Refractive index and loss coefficient in optical waveguides

Complex refractive index can be defined as:

IRjnn'n +=

Andwehaveexpressedapropagatingfieldas:

)tkz(j

0eEE=

Substitutingforpropagationconstantandcomplex

refractiveindex:

tjznkznjk0

)tz'nk(j0 e.e.e.EeEE I0R00 ==

The term exp(-k0nIz) is often re-designated .

The term is called the loss coefficient. The factor of is

included in the definition above because is an intensity

loss coefficient. Therefore we can write :

)2

1exp(

z

0eII=

Thebenchmarkforacceptablewaveguidelossisoftheorder

of1dB/cm. TypicallossesforSOIwaveguidesareintherange

0.1 0.5dB/cm


51/92

The contributions to loss in an optical waveguide

Scattering

Losses result from scattering, absorption, and radiation:

Scattering in an optical waveguide can result from two

sources: volume scattering and interface scattering

Volumescatteringfollowseithera3 dependence,ora1 dependence,asaconsequenceofthetypeand

concentrationof

scattering

centres.

Interfacescatteringhasbeenmodelledbymanyauthors,

butareasonablyaccurateandattractivelysimplemodel

wasproducedbyTien:

where u is the rms roughness for the upper waveguideinterface, lis the rms roughness for the lower waveguideinterface, kyu is the decay constant in the upper cladding, kyl is

decay constant in the lower cladding, and h is the waveguidethickness

++

+

=

l

l

yyu

2

0

2

1

22

u1

3

s

k

1

k

1h

1)(n4

sin2

cos


52/92

Exampleofinterfacescattering

Consider a planar waveguide:

n1 = 3.5; n2 = 1.5; n3 = 1.0; h = 1.0m, and the operatingwavelength 0 = 1.3m. Let us compare the scattering loss of

two different modes of the waveguide, say the TE0 and the TE2

modes.

The propagation angles, 1 of these two are 80.8o (TE0), and

60.7o (TE2). Thus the decay constants in the claddings are

given by:2

i

2

0

22

yi nkk =

Therefore we can evaluate the decay constants for each mode are

as follows:

TE0 TE2

m-1 m-1

m-1

m-1

98.15nkk2

3

2

0

2

yu == 94.13nkk2

3

2

0

2

yu ==

04.15nkk2

2

2

0

2

y ==l 85.12nkk 22202y ==l


53/92

If we now let both u and l equal 1nm, we can evaluate thescattering loss for each mode, for each nanometre of rms

roughness at each interface.

For the TE0 mode,

This is equivalent to a loss of 0.18 dB/cm.

1

yyu

2

0

2

122

u1

3

s cm04.0

k

1

k

1h

1)(n4

sin2

cos =

++

+

=

l

l

For the TE2 mode

This is equivalent to a loss of 5.79 dB/cm.

1

yyu

2

0

2

1

22

u1

3

s cm33.1

k1

k1h

1)(n4

sin2

cos =

++

+

=

l

l


54/92

Absorption

The two main potential sources of absorption loss for

semiconductor waveguides are band edge absorption andfree carrier absorption. If we operate at a wavelength well

away from the band edge, the former is negligible.

Changes in free carrier absorption can be described by

Drude-Lorenz equation:

where e is the electronic charge; c is the velocity of light in

vacuum; e is the electron mobility; h is the holemobility; is the effective mass of electrons; is

the effective mass of holes; Ne is the free electron

concentration; ; Nh is the free hole concentration; 0 is the

permittivity of free space; and 0 is the free space

wavelength.

+

=

2*

chh

h

2*

cee

e

0

32

2

0

3

)m(

N

)m(

N

nc4

e

*

cem*

chm


55/92

Additionallossofsiliconduetofreecarriers

Silicon

=1.3m1018 cm3 electrons

andholesintroduces

anadditionallossof~

2.5cm1. i.e 10.86

dB/cm!


56/92

Radiation Losses in optical waveguides

Ideallynegligible.

Possibility

of

radiation

via

leaky

modesorcurvatureattoofastarate.


57/92

Coupling to the Optical circuit

Couplinglighttoawaveguideisnontrivial. The

mainmethods

are:

(d)endfirecoupling

waveguide

Input

beam

(a)prismcoupling

Inputbeam

waveguide

(b)grating

coupling

(c)buttcoupling

waveguide

Optical

fibre

waveguide

lens

Input

bea

m

Fourtechniquesforcouplinglighttoopticalwaveguides


58/92

Grating couplers

Gratingcouplersallowindividualmodeselection.

Inordertocouplelighttothewaveguide,the

propagationconstants

of

the

exciting

beam

and

the

waveguidemustbematched.

Consideralightrayincidentuponawaveguide

surface:

Lightincidentuponthesurfaceofawaveguide

n1

n2

n3a z

Inmediumn3 thepropagationconstantisk0n3. Thez

directedpropagationconstantinmediumn3 willbe:

a30z sinnkk =


59/92

Therefore the phase-match condition will be

where is the waveguide propagation constant

a30zsinnkk ==

But k0n3. Thereforetheconditioncanneverbemet,sincesina willbelessthanunity. Thisiswhyagratingisrequiredtocouplelightintothewaveguide.

The periodic nature of the grating causes a periodic

modulation of the effective index of the waveguide. For an

optical mode with propagation constant W when the grating

is not present, the modulation results in a series of possible

propagation constants, p given by

where is the period of the grating, and p = 1, 2, 3, etc.

+= p2Wp


60/92

Only the negative values of p can result in a phase match.

It is usual to fabricate the grating such that only p = -1

results in a phase match with a waveguide mode.

Therefore the waveguide propagation constant becomes:

=2

Wp

And the phase match condition becomes:

a30W sinnk2

=

Writing W in terms of the effective index, N,

a300

sinnk2

Nk =

and substituting for k0

a3 sinnN

=


61/92

If medium n3 is air, n3 = 1 and

asinN

=

Because the refractive index of silicon is large, the period of

grating couplers in silicon is of the order of 400nm. The highest

coupling efficiency from gratings in silicon to date has been

reported by Ang et al., who reported an output coupling efficiency

of approximately 70% for rectangular gratings and 84% for a non

symmetrical profile. One of their devices is shown below:

WaveguidecouplerfabricatedinanSOIwaveguide


62/92

Butt coupling and End fire coupling

The efficiency with which the light is coupled into the

waveguide is a function of (i) how well the fields of theexcitation and the waveguide modes match; (ii) the degree

of reflection from the waveguide facet; (iii) the quality of

the waveguide endface; (iv) and the spatial misalignment of

the excitation and waveguide fields.

Overlap of excitation and waveguide fields

The overlap integral of two fields E and , is given by

2

1

22 dxdy.dxEdy

dx..Edy

=

The factor lies between 0 and 1, and therefore representsthe range between 0 coupling and 100% coupling due to

field overlap.


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Theproblemisoftensimplifiedtotheoverlapoftwo

gaussian functions.Let

This represents a waveguide field with 1/e widths in thex

and y direction of 2x and 2y respectively. Similarly let

+

=

2y

2

2x

2 yxexpE

This represents a circularly symmetrical input beam.

( )

+=

2

0

22 yxexp

Using the mathematical identity for a definite integral:

[ ]r2

dx.xrexp0

22 =

The overlap integral reduces to:

2

1

2

0

2

y

2

1

2

0

2

x

2

1

yx0

1111

12

+

+

=


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Since the overlap integral describes the coupling efficiency of

the field profiles, the power coupling efficiency is given by:

+

+

=

2

0

2

y

2

0

2

x

yx

2

2

1111

14

0

Afewexamplesshowhowcouplingefficiencyvaries:

0 x y 2 Loss due to

2 (dBs)

5m 5m 5m 1.0 1.0 0

10m 10m 5m 0.894 0.8 0.97

20m 16m 3m 0.535 0.286 5.4

20m 1m 1m 0.1 0.01 20

5m 5m 3m 0.939 0.882 0.55

5m 4.8m 4.9m 0.999 0.999 0.004


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Reflection from the waveguide facet

UsingtheFresnelequationsfromsection1.

ReflectioncoefficientforTEpolarisationwas:

2211

2211TE

cosncosn

cosncosnr

+

=

Similarly, the reflection coefficient rTM was:

2112

2112TM

cosncosncosncosnr

+=

Using Snells Law rTE reduces to:

)sin(

)sin(r

21

21TE

+

=

Hence the reflectivity R is given by:

)(sin

)(sinrR

21

2

21

22

TE TE +

==


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Similarly RTM can be found to be

)(tan

)(tanrR

212

21

22

TMTM

+

==

The two functions are plotted below for an air/silicon

interface (i.e. n1= 1.0, n2 =3.5).

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

RTE( )1

RTM( )1

1

Incidentangle1 (radians)

Reflectance

TM

TE

Reflectionatanair/siliconinterface

1

0.8

0.6

0.4

0.2

00 0.2 0.4 0.6 0.8 1.0 1.2 1.4


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At normal incidence (1=0), the reflection of both TE and TM

polarisations is the same. Furthermore, end-fire coupling

introduces light at near normal incidence. Consequently the

approximation is usually made that the Fresnel reflection at the

waveguide facets is that due to normal incidence. In this case,

reflectivity is:

2

21

21

nn

nnR

+

=

For a silicon/air interface this reflection is approximately

31%, which introduces an additional loss of 1.6dBs. A lossof 1.6dBs for each facet of the waveguide is considerable,

and is reduced in commercial devices by the use of anti-

reflection coatings.


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Anantireflectioncoatinghasathicknessof/4,reducingoreliminatingthereflection.Fornormalincidence,thenet

reflectivityRisgivenby:

where nar is the refractive index of the anti-reflection

coating. R will be zero if:

2

2

ar21

2

ar21

nnn

nnnR

+

=

For a silicon/air interface, nar needs to be approximately

1.87.

For silicon nitride (Si3N4), n = 2.05.

For silicon oxynitride (SiOxNy), n ranges from 1.46 (SiO2)

to 2.05 (Si3N4)

2

ar21 nnn =


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The quality of the waveguide endface

Three main options are available for endface preparation of

semiconductor waveguides: cleaving; polishing; etching.

Polishing is probably the most common method of

preparing a waveguide facet. The sample endface is lapped

with abrasive materials with sequentially decreasing grit

sizes, and can result in an excellent surface finish, although

rounding of the endface is a common failing.

Endfaces can be prepared by chemical or dry etching. The

details are beyond the scope of this course, but suffice to say

it is a technique that can be developed to a sufficiently high

level for commercial applications.


70/92

polished edge

clived edge

The edge problem


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Spatial misalignment of the excitation and waveguide

fields

Alignment is of critical importance due to the very small

dimensions involved. Recall the overlap of exciting and

mode fields:

where

and

2

1

22 dxdy.dxEdy

dx..Edy

=

+

=

2

y

2

2

x

2 yxexpE

( )

+

=2

0

22yx

exp

Letusintroduceandoffset,A, intooneofthefieldequations:

i.e.

( )

+= 2

0

22 )Ay(xexp


72/92

This expression can be re-written as:

( )

++

= 20

222)Ay2Ay(x

exp

Subsequently, the overlap integral equation can be

manipulated into the form:

2

1

22

2

0

2

y

2I

dxdy.dxEdy

dx.Edy.Aexp

+=

i.e.

+= 202

y

2I A

exp

Therefore we can evaluate the term

+=

2

0

2

y

2Aexpoffset


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Offset2 is also evaluated as it represents to additional loss

when considering power transmission (coupling).

The terms are evaluated below for a range of offsets A,

when y = 0 = 5m

0 2 .10 6 4 .10 6 6 .10 6 8 .10 6 1 .10 5 1.2 .10 5 1.4 .10 50

0.2

0.4

0.6

0.8

11

0

C A( )

D A( )

15 100 A

0 2 4 6 8 10 12 14

OffsetA(m)

TheeffectofanoffsetAontheelectricfield

overlapandthepowercouplingefficiency

Offset

andOffset

2

OffsetOffset

2

AnoffsetofA=2mproducesoffset2 of0.85 0.7dB


74/92

Measurement of propagation loss is integrated

optical waveguides

There is often confusion between insertion loss and

propagation loss

Insertion loss and propagation loss

The insertion loss of a device, is the total loss associatedwith introducing that element into a system, and includes

the inherent loss and the coupling losses.

Alternatively, the propagation loss is the loss associated

with propagation in the waveguide alone i.e.

measurement of loss coefficient, .

There are three main experimental techniques associated

with waveguide measurement. These are (i) the cutback

method; (ii) the Fabry-Perot resonancemethod; and (iii)

scattered light measurement.


75/92

1( ) 10log 10log

prop prop

L L

out o in

incoupling

o

I I e I Ce

IdB

I C

= = = =

Origin of the coupling losses

Dimension fiber-waveguide (modal mismatch)

Numerical Aperture

Fresnel Reflection

Misalignement

Defects on the facets

Scattering

Insertion losses in a waveguide:

Intrinsic

Extrinsic


76/92

1 110log ( )

prop propL L

out o in

incoupling

o

I I e I Ce

IdB

C I C

= =

= =

Origin of the propagation losses

Absorption

Scattering

Radiation

Insertion losses in a waveguide:


77/92

Positioning systemFor the fiber.

Nanopositioning system with Piezoelectric actuators

(precision 1nm)

For the sample

Micropositioning system


78/92

The cut-back method

The cut back method is conceptually simple. A waveguide

of length L1 is excited by one of the coupling methods

mentioned, and the output power from the waveguide, I1,

and the input power to the waveguide, I0 are recorded. The

waveguide is then shortened to another length, L2, and the

measurement repeated to determine I2. Hence:

i.e.

))LL(exp(I

I21

2

1 =

=1

2

21 I

Iln

LL

1


79/92

I = I0 e-L

L1

Laser in

Fotodiode

Waveguide

I

LL1

Cut-back technique


80/92

I

LL1L2

I = I0 e-L

L2

Laser in

Fotodiode

Waveguid

e

Cut-back technique


81/92

I

LL1L2L3

I = I0 e-L

L3

Laserin

Fotodi

ode

Waveg

uide

Cut-back technique


82/92

I

LL1L2L3L4

I = I0 e-L

L

4

Laser in

Fotodio

de

Waveg

uide

ln(I)

LL1L2L3L4

ln (I)= ln(I0) - L

Cut-back technique

L=0

ln(I)=ln(I0)=ln(Iin)+ln(C)

MeasurableEstimaton of

the coupling losses


83/92

The accuracy of the technique can be improved by taking

multiple measurements and plotting a graph

Optical

Loss

(dBs)

Propagationlength

(cm)

Note that the data presented is now insertion loss for eachdevice length, not propagation loss. Hence the loss at zero

propagation loss represents coupling loss.

A useful variation of the cut-back method, is to carry out an

insertion loss measurement for a single waveguide length,

and calculate the coupling. Whilst this technique is less, it

has the enormous advantage of being non-destructive


84/92

Similar technique

Equivalent to cut-back but without the coupling lossesproblems

Losses from a curve

Propagation losses


85/92

The Fabry-Perot resonance method

An optical waveguide with polished end faces (facets), is

similar in structure to the cavity of a laser, and may be

regarded as a potentially resonant cavity. Such a cavity is

called a Fabry-Perot cavity. The optical intensity

transmitted through such a cavity, It, is related to the

incident light intensity, I0, by the well known equation:

where R is the facet reflectivity, L is the waveguide

length, is the loss coefficient, and is the phasedifference between successive waves in the cavity

)

2

(sinRe4)Re1(

e)R1(

I

I

2L2L

L2

0

t

+

=


86/92

This transfer function has a maximum value when =0 (or

multiples of 2), and a minimum value when =. i.e. :

2L

L2

L2L

L2

0

min

)Re1(

e)R1(

Re4)Re1(

e)R1(

I

I

+

=+

=

2L

L2

0

max

)Re1(

e)R1(

I

I

=

Which we can rearrange as:

+

= 11

R

1lnL

1

Ifweknowthereflectivity,R,ifwecanmeasure,thelosscoefficientcanbeevaluated.

Thereforeifwecansweepthroughafewcyclesof2,canbemeasured. Suchcyclingcanbeachieved

thermally,or

by

varying

the

wavelength

of

the

light

source.

Therefore the ratio of the maximum intensity to minimum

intensity, , is:

2L

2L

min

max

)Re1(

)Re1(

I

I

+

==


87/92

0 5 10 15 20 250

0.2

0.4

0.6

0.8

11

0

FP1 ( )

FP2 ( )

FP3 ( )

8 0

Plot of the Fabry-Perot transfer function for

three different mirror reflectivities

R=0.5

R=0.31

R=0.1

It/I0

(radians)

IminImax


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1550.00 1550.05 1550.10 1550.15 1550.20 1550.25

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

0.65

intensity(arb.units)

wavelength(nm)

FabryPerotscanofasinglemode

waveguide


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FabryPerotscanofamultimode

waveguide

1550.00 1550.05 1550.10 1550.15 1550.20 1550.25

0.4

0.5

0.6

0.7

0.8

0.9

intensity(arb.units)

wavelength(nm)


90/92

Scattered Light Measurement

The measurement of scattered light from the surface of a

waveguide can be used to determine the loss. The assumption is

that the amount of light scattered is proportional to the

propagating light. Therefore, the rate of decay of scattered light

with length will mimic the rate of decay of light in the waveguide.

However, it is clear that light is only scattered significantly if the

loss of the waveguide is high, so this approach has limited uses.


91/92

Scattering light collection

0 5 10 15 20 25 30 35

= 0.1 0.07 dB/cm

Si3N4 channel =780nm

Ln

Intensity

[a.u.]

Length [mm]

The sample has to be quite lossy


92/92

Examplesofmicromachining:

Photonic Materials 3_Waveguides

Documents

Transcript of Photonic Materials 3_Waveguides