Comparative study of asymmetric versus symmetric planar slab dielectric optical waveguides

Indian J. Phys. 84 (7) 831-846 (2010)

© 2010 IACS*Presently working at Department of Electronics Engineering, Indian School of Mines,Dhanbad-826 004, Jharkhand, India

Comparative study of asymmetric versus symmetric planarslab dielectric optical waveguides

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Abstract : In this paper we have studied the asymmetric versus symmetric planar waveguide in terms oftheir usefulness in optical fiber communication systems. We have explored the thin waveguide versus thickwaveguide first. Later on usefulness of asymmetric versus symmetric waveguide is carried out to target forWDM optical network application. All kinds of optical network components are fabricated on Si substrate with thepoint of view of their application. Here asymmetric planar structure may be more useful compared to symmetricwaveguide in terms of their non-uniform power confinement properties. However, the symmetric waveguidestructure may be more useful for their high power confinement properties. It is well known that the thin symmetricwaveguide supports at least one mode. However the thick waveguide may support many even as well as oddmodes. We study the power confinement properties for symmetric as well as asymmetric waveguide structure.We conclude that higher order modes show the nonlinear power variations. Mode field profile for various casesis discussed as well. Comparative study between asymmetric versus symmetric waveguide has a lot of significancein optical network area. It has been shown through analysis that in asymmetric waveguide, the power flowsmore through film region in the case of fundamental mode. Power confinement properties for asymmetricwaveguide versus symmetric waveguide have been studied.

Keywords : Optical waveguide structure, formal electromagnetic theory concepts, waveguide theory.

PACS Nos. : 42.79.Dj, 42.79.-e

1. IntroductionThe paper presents a comparison of symmetric and asymmetric planar slab dielectricwaveguides. In this paper, we establish the fundamental concepts of guided waves. Itis perhaps one of the most important concepts, as almost all the other importantconcepts build on these. This paper includes many numerical calculations. Due to thetranscendental nature of the eigenvalue equations, appropriate software is required to

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make the computations feasible. One can strongly be advised to develop a set ofstandard programs that can quickly evaluate the basic waveguide parameters of a givenstructure, based on the material in this paper. These programs can be useful forsomeone exploring different types of waveguide, mode coupling and device construction.The index of refraction of the guiding slab nf is higher than that of the cover material,nc or the substrate material ns in order that total internal reflection occurs at theinterfaces. If the cover and substrate materials have the same index of refraction, thewaveguide is called “symmetric” otherwise the waveguide is called “asymmetric”. Wewill always choose the direction of propagation to be along the z-axis. The slabwaveguide is clearly an idealization of real waveguides, because real waveguides arenot infinite in width. However, the one-dimensional analysis is directly applicable tomany real problems, and the techniques form the foundation for further understanding.The fabrication and performance of such waveguides, known as dielectric clad planarwaveguides, now form an essential component most of the present-day integratedcircuits. In order to understand the performance of a inhomogeneous waveguidestructure, it is necessary to understand the simplest planar slab waveguide structure.In this paper, we derive wave equations which determine the modes and are valid forsimple planar waveguide having a step variation of the dielectric constant. We will thendevelop formal mode concepts such as orthogonality, completeness and modal expansion.We will also see that a waveguide structure can support only a discrete number ofguided modes. In the coming section, we will discuss the wave guiding action in planarwaveguides with continuous dielectric constant variation.

2. General theoretical considerationConsider the waveguide structure shown in Figure 1. The three indices are chosensuch that f s cn n n� � , and the guiding layer has a thickness h. The choice of thecoordinate system is critical in making the problem as simple as possible [2]. Theappropriate coordinate system for this planar problem is a rectilinear Cartesian system,because the three components of the field, Ex, Ey and Ez are not coupled byreflections. For example, an electric field polarized in the y-direction, Ey will still be anEy directed field upon reflections at either interface; the reflection does not couple anyof the vector field into the x or z directions.

The Maxwell equation for the TE mode is given by [1],

� �yi y

Ek n E

x

22 2 202 0.�

��

�(1)

So depending on the value of �, the solution can be either oscillatory or exponentiallydecaying. If ik n0� � we define an attenuation coefficient, �

ik n2 2 20� �� (2)

��

and describe the field as xyE x E e0 .

0( ) .�� If ik n0 0� � then we define a transversewave vector, �

ik n2 2 20� �� . (3)

so, j xyE x E e0 .

0( ) �� . We know that guided wave must satisfy the condition

s fk n k n0 0�� (4)

where; it is assumed that c sn n [1]. This is a universal condition for any dielectricwaveguide, regardless of geometry.

2.1. The symmetric waveguide :Figure 1 shows a symmetric waveguide, if a guiding film with index nf and thicknessh is surrounded on both sides by an index ns. It is convenient to place the coordinatesystem in the middle of this waveguide since the fields will reflect the symmetry of thestructure.

� �x hyE Ae 2�� for x h 2�

� �yE C xcos �� or � �D xsin � for h x h2 2�

� �x hyE Be 2� � � � for x h 2 � (5)

nc

nf

ns

h

x

z

y

Figure 1. The planar slab waveguide consists of three materials.

Since there is no Ex component to the field, we get an explicit equation for thetangential component of the magnetic field, Hz [1,2],

yz

EiHx0��

��

�

Since � and � are defined in all the media, the continuity of is guaranteed if� �yE x� � is made continuous across the interface. Hence we can find amplitudes A,B, C and D using the electric field description only. At the x h 2� � interface, thecondition that Ey is continuous requires that

��

hA C Bcos2

�� .

We can show that the general field description of a TE mode within this symmetricstructure is

� �x hyE A e / 2. �� for x h / 2�

yxE A

hcos.

cos / 2�

�� or

xAh

sinsin / 2

�

� for h x h/ 2 / 2� (6)

� �x hyE A e / 2. � �� for x h / 2 � .

The magnetic amplitude of the TM mode can be similarly described. There are twochoices for the description of the field in the guiding layer, depending on whether asymmetric (cosine) or anti-symmetric (sine) mode is excited. The fact that the modescan be uniquely characterized in terms of even or odd groups is a natural consequenceof the symmetry of the index structure. Applying the continuity of yE x� � at theinterfaces leads to the characteristics eigenvalue equation for the TE mode in asymmetric waveguide given by

htan / 2 /� � �� for even modes

/� �� for odd modes. (7)

A unique feature of the symmetric waveguide is that it will always support at least onemode. Consider the graphical solution for a symmetric waveguide described by a filmof index fn 1.49,� and cladding index equal to sn 1.485� . The difference in indexbetween the two layers is very small. Let the wavelength be 0.8 �m.

We can use the graphical solution to find �, as it demonstrates why thesymmetric waveguide will always support at least one mode. Two thicknesses areexamined for h = 3 �m and h = 15 �m. There are only two variables in this problemviz., � and �. To find the eigenvalues of the TE modes we must solve eq. (7).Graphically, the functions htan 2,� �� and –�� are plotted on the same as afunction of �. These are plotted against � for the case where h = 3 �m in Figure 2.The top curve, which corresponds to the even mode, begins at �� and terminateswith a value of 0. Notice that the htan 2,� starts at zero and increases. Crossing ofthe two curves is unavoidable; so there must always be at least one mode, no matterhow thin the waveguide is. As the waveguide is made thicker, more modes appear.Consider the graphical plot of the equations for the case when the waveguide slab is15 �m thick, as shown in Figure 3. Note that as the transverse wavevector (�)increases, the allowed modes alternate between even and odd structure. The spatialprofile of these allowed modes alternate between even and odd structure.

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Figure 2. For the thin waveguide, there is only one allowed mode, which occurs near 16000 cm� �� .

Figure 3. The thick waveguide supports both even and odd modes.

��

h = 3 �m

Tan �h/2

–�/�

10

8

6

4

2

0

–2

–4

–6

–8

–100 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

�

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000�

10

8

6

4

2

0

–2

–4

–6

–8

–10

��

h = 15 �m� = 0.8 �m

� �

tan �h/2

��

2.2. Power associated with a mode :Now, we will calculate the power associated with the TE mode. The power flow is givenby

� �S E H *1 Re2

� ��

Now, for the TE mode we have through Maxwell’s equation, y xS S0, 0� �� and

� �z y x yS E H E* 2

0

1 Re | |2 2

�

�� .

The power associated with the mode (per unit length in the y-direction) is given by

yP E dx2

0

1 | |2

�

��

��

��

� � . (8)

We consider the symmetric mode eq. (6) for which � �h2cos 2 1� � , where we have

used eq. (7),

hh x

h

AP xdx A e e dxh

/ 222 2 2

20 0 / 2

12 cos2 cos 2

� ��

��

��

� ��

A h h h2

2

0

1 sin 1 cos / 22 2 2� �

��

� ��

h hA hP h

22

0

2 sin cos1 22 2 (1 sin )4 2

� ��

��

� ��

h hA hh

2

0

2 sin cos1 2 2 2 tan4 2

� ��

� ��

� ��

AP h2

total0

1 24�

��

� �� for both even and odd modes. (9)

The power carried by the fundamental mode and the spatial profile of these allowedmodes are shown in Figures 4 and 5 respectively. Power in the guiding layer is foundby integrating the Poynting vector over the area of the waveguide structure. The fractionof the power contained in the core is given by

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Figure 4. The power carried by the TE01 mode for the case of thin symmetric waveguide structure (A = 1).

Figure 5. Mode field pattern for the first three even TE modes (h = 15 �m).

��

TE01 mode

× 10–7

14

13

12

11

10

9

8

7

6

5

40.5 1 1.5 2

�(�m)

Pow

er (A

rbitr

ary

Uni

t)

TE0

TE0

TE0

4000

3000

2000

1000

0–15 –10 –5 0 5 10 15

Position (�m)

–15 –10 –5 0 5 10 15Position (�m)

–15 –10 –5 0 5 10 15Position (�m)

2000

1000

0

–1000

–2000

1000

500

0

–500

–1000

E(V

/cm

)E

(V/c

m)

E(V

/cm

)

��

h

y xh

y x

E x H x dxPP

E x H x dx

/ 2

core / 2

total

( ) ( ).

( ) ( )

�

��

�

��

��

�(10)

Hence

� �

� �

h hA A xP xdx dxh h

h hA h Ahh h

/ 2 / 22 22

core 20 00 0

2 2

0 0

1 2 1 cos 22 cos2 1 cos 2cos 2

sinsin .2 1 cos 2 1 cos

� � ��

��

� ��

��

� � ��

� ��

� �

.

Therefore, we have

� �

� �

h hPP h

core

total

sin2

� � �

� �

��

� (even mode) (11)

� �

� �

h hPP h

core

total

sin2

� � �

� �

��

� (odd mode). (12)

In general, higher order modes are less confined than their lower order counterparts.Application of eq. (12) on mode 0 shows that it has a power confinement of 99.47%,while mode 2 has only 85.9% confinement. Mode confinement is an important propertyfor waveguide designs. A mode that is loosely confined will be more affected by bendsand more strongly coupled evanescently to neighboring structures than what a tightlybound mode will. It is apparent from the Figures 6 and 7 that higher order modes showthe nonlinear variation with wavelength. Figure 8 shows the fraction power carried bythe few modes in the core (film) region, it is apparent that the fundamental mode willcarry more power.

3. Eigenvalues for the asymmetric slab waveguideTo find the values of � that lead to allowed solutions to the wave equation, we mustapply the boundary conditions to the general solutions developed in eq. (1). Assumethat � satisfies eq. (4). Then the transverse portions of the electric field amplitudes inthe three regions are

c xyE x Ae( ) �� x0 �

� � � �y f fE x B co x C x( ) . . sin� �� h x 0� � �

� �y sE x D e x h( ) . �� x h� � (14)

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Figure 6. Variation of power for three even and two odd TE modes (h = 15 �m).

TE01

TE02

TE03

��

2.05

2

1.95

1.9

1.85

1.8

1.75

1.7

1.65

1.6

Pow

er (A

rbitr

ary

Uni

t)

Pow

er (A

rbitr

ary

Uni

t)

1.9

1.85

1.8

1.75

1.7

1.65

1.6

TE01

TE02

��

06. 0.8 1 1.2 1.4 1.6 1.8 2� (�m)

06. 0.8 1 1.2 1.4 1.6 1.8 2� (�m)

Figure 7. Comparison of the above cases.

��

��

��

��

��

��

1.9

1.85

1.8

1.75

1.7

1.65

1.606. 0.8 1 1.2 1.4 1.6 1.8 2

� (�m)

x10–6

x10–6 x10–6

where A, B, C and D are amplitude coefficients to be determined from the boundaryconditions, �c and �s refer to the attenuation coefficients in the cover and substrate,respectively and �f is the transverse component of k in the guiding film. First considerthe continuity conditions for the tangential electromagnetic field components at the

Pow

er (A

rbitr

ary

Uni

t)

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interface x = 0. Since Ey is transverse to the interface, the first boundary condition isstraightforward to apply. What about the condition for continuity of magnetic field, H?Indeed, there is usually a simpler way to derive expressions for the magnetic field. Ifwe assume that the fields are harmonic, then we can describe the magnetic intensityin terms of the electric intensity and derive a simple boundary condition for themagnetic terms. The tangential component of H, Hz is thus defined in terms of electricfield quantities. Since � and �� are identical in all the media, the continuity of thetangential magnetic field is guaranteed if yE x.� � is made continuous across theinterface. Hence we can now find the amplitudes, A, B, C and D using only theelectric field description. At the x = 0 interface, the condition that Ey is continuousrequires that

� � � �cf fA e B C0 cos 0 sin 0�

� �� . (15)

This is satisfied only if A = B. Making the magnetic field continuous at x = 0 requiresthat the first derivative, yE x. . ,� � be continuous at x = 0 so that

� � � �cc f f f fA e B C.0. . sin 0 . cos 0�

� � � � ��

or, c fA C. .� �� (16)

yielding

c

fC A

�

�� (17)

Figure 8. The fraction of the power contained in the core for few cases.

(a) (b) (c)�

��

��

��

��

��

��

��

��

��

Pco

re/P

T

�

��

��

��

��

��

��

��

��

��

��

Pco

re/P

T

�

��

��

��

��

��

��P

core

/PT

�� (�m)

�� (�m)

�� (�m)

��

��

��

h = 15 �m h = 15 �m h = 15 �m

��

All coefficients are written in terms of A. Using these coefficients and applying thecondition that Ey be continuous at x = –h yields

� � � � � �s h hcf f

fA h h D ecos . sin . ��

� ��

��

. (18)

This can be solved for D as

� � � �cf f

fD A h h. cos sin

��

�

� ��

(19)

Putting all the terms together,

c xyE A e �� x 0�

� � � �cy f f

fE A x x. cos sin

��

�

� ��

h x 0� � �

� � � � � �s x hcy f f

fE A h h e .. cos sin . ��

� ��

��

x h� � (20)

where A is the amplitude at the x = 0 interface. Eq. (20) describes the amplitude ofthe electric field in all regions of the problem. The propagation and decay constants�s, �c and �f, all depend on �, which is still undefined. The fourth and final boundarycondition, namely the continuity of yE x. .� � at x h,� � gives an equation for �,

� � � �yx h f f c f

EA h h

x.

| . sin . cos.

� � � ��

��

� � � �cf f s

fA h h. cos sin .

��

�

� ��

(21)

Dividing both sides of the equation by � �f hcos � we get the eigenvalue equation

� � c sf

c sf

f

h

2

tan .1

� ��

� ��

�

��

� ��

(22)

This is a transcendental equation that must be solved numerically or graphically. Allterms s c f( , , )� � � depend on the value of �. Eq. (22) is called the characteristicsequation for TE modes of the slab waveguides. Solution of this equation will yield theeigenvalues, �TE that correspond to allowed TE modes in the waveguide. Everywaveguide structure, no matter what shape or symmetry, will have a characteristicsequation that must be solved to find the eigenvalues of the modes. The transcendentalequation can be solved numerically on a computer or it can be solved graphically.

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Consider the planar dielectric structure shown in Figure 1. The guiding index has value1.50, the substrate index is 1.45 and the cover index is 1.40. Now this is anasymmetric waveguide structure. The thickness of the guiding layer is a 15 �m. Wewant to determine the allowed value of � using eq. (22) for this structure. Assume thatlight with wavelength of 0.8 �m is used to excite the waveguide.

The power carried by the asymmetric waveguide structure obtained by using eqs.(8) and (20) is given by

cx cf f

fh h

P A e dx A x x dx20

22 2

0 / 2

1 cos sin2

� ��

��

��

�

� � � ��

� �sh

x hcf f

fA h h e dx

222 cos sin ��

� ��

��

��

� ��

� �AP M N O

2

0

14�

�� (23)

where

cf f

c f ff

h hM N h h2

2sin 2 sin 21 ;

2 2��

� � ��

� � � ��

� �c cf f f

sf fh O h h

2

2 211 cos 2 ; cos sin

2� �

� � ��

� �� . (24)

For the sake of comparison we have used same simulation parameters as symmetricwaveguide case. It is apparent from the Figure 9 that, asymmetric waveguide structuresupports many modes. Figure 10 shows the comparison of transverse wave-vectorwhich is less for symmetric waveguide case. It means that asymmetric waveguide maycarry more power in film. Figure 11 shows that in case of asymmetric waveguide, thepower varies more rapidly. Initially the power carried by asymmetric waveguide is lessupto � ! 1.2 �m, thereafter it increases rapidly. Hence for � � 1.2 �m, theasymmetric waveguide carries more power, which is apparent for the higher mode also.Figure 12 shows that the asymmetric waveguide carried more power in the film region,hence may be more advantageous for specific application (low loss region � ! 1.5�m). Figure 13 shows the comparison between fundamental mode spot sizes forfundamental mode case, which revals that power level is high for asymmetric case;however, the profile is offset due to asymmetric structure.

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Figure 9. The thick asymmetric waveguide supports both even and odd modes.

10

8

6

4

2

0

–2

–4

–6

–8

–10

tan �h

h = 15 �m� = 0.8 �m

0 5000 10000 15000�

Figure 10. The transverse wave-vector versus wavelength for asymmetric versus symmetric waveguide.

TE01AsymetricTE01Symetric

2000

1950

1900

1850

1800

1750

1700

1650

1600

1550

15000.5 1 1.5 2

�(�m)

�

��

Figure 11. The variation of power for asymmetric versus symmetric waveguides for few TE modes.

3.5

3

2.5

2

1.5

1

0.5

x 10–8

Pow

er (A

rbitr

ary

unit)

TE01(Asy)

TE01(Sym)

TE01(Asy)

TE02(Sym)

0.6 0.8 1 1.2 1.4 1.6 1.8 2� (�m)

0.6 0.8 1 1.2 1.4 1.6� (�m)

3

2.5

2

1.5

1

0.5

x 10–8(a) (a)

Pow

er (A

rbitr

ary

unit)

Figure 12. The fraction of the power contained in the core for the asymmetric waveguide case.

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0

P cor

e�P T

P cor

e�P T

0.25

0.2

0.15

0.1

0.05

00.6 0.8 1 1.2 1.4 1.6

� (�m)0.6 0.8 1 1.2 1.4 1.6

� (�m)

�� !"�

�� !"�

(a) (b)

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4. ConclusionThis paper is an outcome of comparison between asymmetric waveguide structureversus symmetric waveguide structure. Formal electromagnetic wave theory has beenapplied for the various cases. The results show that for a specific application theasymmetric waveguide may be more useful due to its high power confinementproperties at film region. However, at lower wavelength region the symmetric waveguideis more useful which has been discussed in detail. But at low loss wavelength regionthe asymmetric waveguide shows significant improvement. It has been found that powerlevels for various modes show the nonlinear variation over the band, which may beconsidered undesirable for many applications. It has shown also that for the same filmthickness the asymmetric waveguide structure may allow the large number of modes.

References[1] A Yariv IEEE J. Quantum Electron. 9 919 (1973)

[2] S K Raghuwanshi and T Srinivas ICMO 3 (2007)

[3] M S Sodha and A K Ghatak Inhomogeneous Optical Waveguide (New York : Plenum) (1977)

[4] W H Louisell Coupled Mode and Parametric Electronics (New York : John Wiley and Sons) (1960)

[5] A W Snyder J. Opt. Soc. Amer. 62 1267 (1972)

[6] A Syahriar, R B Yahya, Bukhari and J Dewanto KOMMIT B-38 (2000)

[7] C R Pollock and M Lipson Integrated Photonics (Boston : Kluwer Academic Publishers) (2003)

Figure 13. Comparison of mode field pattern for fundamental mode.

��

� � #��$��%��

& �� #��$��%�

4500

4000

3500

3000

2500

2000

1500

1000

500

0–30 –20 –10 0 10 20 30

Position (�m)–30 –20 –10 0 10 20 30

Position (�m)

6000

5000

4000

3000

2000

1000

0

E(V�

cm)

E(V�

cm)

(a) (b)

��

[8] A W Snyder and J D Love Optical Waveguide Theory (London : Chapman and Hall) (1983)

[9] D Marcuse Theory of Dielectric Optical Waveguides (New York : Academic Press) (1974)

[10] A Hardy and W Streifer IEEE J. Lightwave Technol. 5 1135 (1985)

[11] S L Chuang IEEE J. Lightwave Technol. 6 294 (1988)

[12] W Streifer, M Osinski and A Hardy IEEE J. Lightwave Technol. 5 1 (1987)

[13] H A Haus, W P Huang, S Kawakami and N A Whitaker IEEE J. Lightwave Technol. 5 16 (1987)

[14] H A Haus and W Huang Proc. IEEE 79 1505 (1991)

[15] A A Hardy IEEE J. Quantum Electron. 34 1109 (1998)

[16] A W Snyder, Y Chen and A Ankiewicz IEEE J. Lightwave Technol. 7 1400 (1989)

[17] K Yasumoto Opt. Lett. 7 503 (1993)

[18] W P Huang, S T Chu and S K Chaudhuri IEEE J. Quantum Electron. 28 184 (1992)

[19] W P Huang J. Opt. Soc. Amer. 3 963 (1994)

[20] H A Haus, W P Huang and A W Snyder Opt. Lett. 14 1222 (1989)

[21] A N Kireev and T Graf IEEE J. Quantum Electron. 39 866 (2003)

Comparative study of asymmetric versus symmetric planar slab dielectric optical waveguides

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Transcript of Comparative study of asymmetric versus symmetric planar slab dielectric optical waveguides