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EE 232: Lightwave Devices Lecture #15 – Optical waveguides Instructor: Seth A. Fortuna Dept. of Electrical Engineering and Computer Sciences University of California, Berkeley 3/14/2019
EE 232: Lightwave Devices
Lecture #15 – Optical waveguides
Instructor: Seth A. Fortuna
Dept. of Electrical Engineering and Computer Sciences
University of California, Berkeley
3/14/2019
2Fortuna – E3S Seminar
Slab waveguide
1
2
2
2d
2d−
z
x
y
Slab waveguide consists of a slab of high-index material surroundedby low-index material ( ). The waveguide is assumed to beinfinitely large in the y and z-directions.
We wish to find confined electromagnetic modes that propagatein the +z direction and solve the source-free time-harmonic wave equation
1 2
( )2 2 0 + =E
3Fortuna – E3S Seminar
Wave equation
In general, solution for the electric-field can be written as
( , , ) ( , , ) ( ,ˆ ˆ ˆ( , , )) ,y zxx y z xE yx y z x y xzE zE z y+= +E
If we plug back into the wave equation,
( ) ( )2 2 2 2ˆ ˆ ˆ ˆˆ ˆx y z x y zxE yE zE xE yE zE + ++ = + + +E E
We separate into three equations
2 2
2 2
2 2
0
0
0
x x
y y
z z
E
E
E
E
EE
=
+ =
+
+ =
Let’s expand equation (1)
2 2 22 2 2
2 2 20x x x
x x xy
E EE E E
Ey z
+ = =
+ +
+
(1)
(2)
4Fortuna – E3S Seminar
Wave equation
Using separation of variables, we assume
( , , ) ( ) ( ) ( )xE x y z f x g y h z=
If we plug back into equation (2)
2 2 22
2 2 2
1 1 1d f d g d h
f dx g dy h dz+ + = −
The sum of the terms can equal a constant only if each individual term is a constant
2 2 22 2 2
2 2 2
1 1 1 x y z
d f d g d h
f dx g dy f dz = = =
Typical solutions for these differential equations:
21
1 2
1
( )
( ) cos( ) sin( )
( )
x x
x
x x
x
i i
x x
i
f x Ae A e
f x B x B x
f x C e
−
−
= +
= +
=
(travelling wave)
(standing wave)
(evanescent wave)
5Fortuna – E3S Seminar
Slab waveguide
1
2
2
We look for a solution that is transverse electric (TE),“bound” to the core of the waveguide, and travelling
in the z-direction such that
( , ) ( , ) ( ) ( )ˆy y zyE x z E x z f x h→ ==E
where we assume there is no dependence on ygiven the slab is translationally invariant in they-direction.
Along the z-direction we expect a traveling wave solution
21( ) z zi z i zh z C e C e −= +
2
(| | 2)
1 2
2( )
sin( ) 2
x d
x
B e xf
dx
B
d
x dx
−− = − 2
(| | 2)
1 2
2( )
cos( ) 2
x d
x
Ae xf
dx
A
d
x dx
−− = −
Along the x-direction, we expect a standing wave solution in the waveguide core and evanescent solution in the cladding.
Even solution Odd solution
Slab waveguide
Plug into wave equation
1
2 2 2
1
2 2
2 2
2
x z
z
+ =
− + =
Apply boundary conditions at interfacebetween core and cladding.Tangential component of electricand magnetic field are equal acrossinterface.
2 2
2 2
, ,
, ,
d dx xay core y cl d
z core z cla dx xdd
E E
H H
= =
= =
=
=
1
2
(even)
(odd)
2
1
ta2
nx x
d
=
2
1
cot2
x x
dB
= −
Slab waveguide
After rearranging3 ( )2 2 2
1
2
1 2 22 2 2
x
d d d + −
=
(even)
(odd)
2
1
tan2 22
x x
d d d
=
2
1
cot2 22
x x
d d dB
= −
Slab waveguide
2
2x
d
2
d
1 1 2 22
d −
0TE 1TE 2TE
Boundsolution
Solve graphically4
Cutoff conditionIn the example on the previous slide we see that the TE1 mode would nothave a solution and would be “cutoff” if the radius of the circle is less than π/2
The cutoff condition for each mode can be generalized as
1 1 2 2( m=0,1,2,3...)2 2
dm
− = (Cutoff condition for TEm mode)
The waveguide will be single-mode if all modes except the fundamentalmode are cutoff.
1 1 2 22
2
( )d
− (Single mode condition)
Effective index
Effective index0
zeffn
= 0
0
2
=
High-frequency limit R aad ius s
→
→
→
2 2 2
1 1
2
1 1
z x
= −1 1
1 1 0
00 0
for zeffn n
= == =
Low-frequency limit as 0
0
Radius 0
→
→
→
2
2 2
2
2
2
2
2z
= +2
2
0
2
2
0 00
for zeffn n
= == =
Effective index
eff
n
1n
2n
0TE
1TE
2TE
1
2
2
1
2
2
Low ω
High ω
Effective index is a measure of howconfined the mode is to the core
Optical confinement factor
( )
( )
*
* ˆ
1
2Power in core
1Total power in mode
2
ˆRe
Re
core
total
zdx
zdx
= =
E H
E H
Weak guidance limit (mode is mostly within cladding)
( )2
2 2
1 2
0
2 nd
n
−
For largest possible 𝜞
(1) Thick core(2) Small wavelength(3) Large index contrast
TM modes( , ) ( , ) ( ) ( )ˆ
y y zyH x z H x z f x h→ ==H
21( ) z zi z i zh z C e C e −= +
2
(| | 2)
1 2
2( )
sin( ) 2
x d
x
B e xf
dx
B
d
x dx
−− = −
2
(| | 2)
1 2
2( )
cos( ) 2
x d
x
Ae xf
dx
A
d
x dx
−− = −
Even solution Odd solution
Along the z-direction we expect a traveling wave solution
Along the x-direction, we expect a standing wave solution in the waveguide core and evanescent solution in the cladding.
Eigenequations: ( )2 2 2
2
2 2 1 12 2 2
x
d d d + −
=
2
1
tan2 2 2
x x
d d d
=
2
1
cot2 22
x xBd d d
= −
Rectangular waveguides
Rectangular waveguides have dielectric contrast in two-directions
y
3
x1
2
Rectangular waveguides do not support pure TE or TM modes!Instead they support hybrid modes.
Rectangular waveguides
Intensity patterns
00 00or EHHE 01 01or EHHE 10 10or EHHE
y
x
Hybrid modes
pqHE ,x yH E are the dominant components (quasi-TE)
pqEH ,x yE H are the dominant components (quasi-TM)
number of nodes in the x-directionp →
number of nodes in the y-directionq →
Effective index method
We estimate the propagation constant of the HE00 mode with the effective index method. We essentially break the 2D problem into a 1D slab waveguide problem.
To simplify this problem we assume that the waveguide is completelysurrounded by the same index. More sophisticated examples are foundin the book.
y
2
x1
w
d
Step 1
y
2
x1
1
2
2
Solve for the TE mode of the slab waveguide with core of permittivity ϵ1and cladding with permittivity ϵ2
Calculate the effective index 𝑛𝑒𝑓𝑓,1and modal distribution 𝐹(𝑥)
w
d
d
Step 2
2
,1effn22
Solve for the TM mode slab waveguide with core of permittivity 𝑛𝑒𝑓𝑓2
and cladding with permittivity ϵ2. Calculate the propagation constant 𝛽𝑧and modal distribution G(𝑦).
y2
x12
w
The overall propagation constant of the 2D waveguide is then 𝛽𝑧and the modal distribution of the 2D waveguide is given by
( , ) ( ) ( )yE x y F x G y=
