EE 529 Semiconductor Optoelectronics
– Optical Waveguides
EE529 Semiconductor
Optoelectronics
Optical Waveguides
1. Modes in planar waveguides
2. Ray-optics approach
3. EM-wave approach
4. Modes in channel waveguides
“Photonic Devices,” Jia-Ming Liu, Chapter 2
“Theory of Optical Waveguides,” by H. Kogelnik, in “Guided-wave Optoelectronics,” T. Tamir,
ed., Chapter 2, Springer Verlag
Lih Y. Lin
EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Why study waveguides for integrated
optoelectronics?
Cleaved reflecting surface
W
Semiconductor laser
Photonic integrated circuits
L
Stripe electrode
Oxide insulator
p-GaAs (Contacting layer)
p-Al xGa1-xAs (Confining layer)
p-GaAs (Active layer)
n-Al xGa1-xAs (Confining layer)
n-GaAs (Substrate)
2
1
Current
paths
Substrate
3
Substrate
Electrode
Cleaved reflecting surface
Elliptical
laser
beam
Active region where
(Emission region)
Modulator
Light: Science and Applications (2012)
J > J t h.
Directional coupler
In
V(t)
W aveguides
A
Electrode
C
In
B
B
A
A
Waveguide
LiNbO
3
V(t)
Out
B
Lo
E le c tr o d e
D
F ib e r s
L iN b 3O
EOSubstrate
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
E-M Field in a Planar Waveguide
Warm-up question: What kind of structure can be a waveguide?
Assuming a monochromatic wave propagating in z-direction:
j ωt
− j β z j ωt
E(r, t ) = E(r )e
= E( x , y ) e
e
∇ 2 E(r ) + k 2 n 2 (r )E(r ) = 0
Region I:
Region II:
Region III:
∂2
2 2
2
E
(
x
,
y
)
+
(
k
n
−
β
) E ( x, y ) = 0
1
2
∂x
∂2
2 2
2
E
x
y
k
n
(
,
)
(
) E ( x, y ) = 0
+
−
β
2
2
∂x
∂2
2 2
2
E
(
x
,
y
)
+
(
k
n
−
β
) E ( x, y ) = 0
3
∂x 2
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Modes in a Planar Waveguide
Modal solutions are sinusoidal or exponential, depending on the sign of
Boundary conditions:
E ( x, y ) and
Assuming
x
(k 2 ni2 − β 2 )
∂E ( x, y )
must be continuous at the interface between layers.
∂x
n2 > n3 > n1 , let’s draw possible waveguide modes:
0
kn1
kn3
kn2
β
n1
n2
n3
(The technique you learned from solving optical waveguide modes can be applied
to the design of many photonic components.)
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Guided Modes in a Planar Waveguide
Examples of guided modes in a symmetrical waveguide
m: Mode order
Q: How to define the mode order?
Q: Can we obtain an infinite number of solutions to β with
continuous values?
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Experimental Observation of
Waveguide Modes
Q1: How to choose the laser wavelength?
Q2: How to create different modes?
Q3: How to tell which side is air, which side is the substrate?
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Do things in simple ways first.
→ Geometrical optics.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Ray Patterns in the
Three-Layer Planar Waveguide
Remember that only discrete values of β are allowed. How to solve for allowable β?
Step 1:
Determine the relation between β and the angle of the optical ray. Different modes
have different angles.
In the guided region,
E ∝ sin(hx + γ )
β 2 + h 2 = k 2 n22
Overall
propagation
constant
→Propagation constant in x
For the m-th mode,
→Propagation constant in z
 h 
θ m = tan −1  
 βm 
Lower-order mode has smaller θm and larger βm.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Ray Patterns for Different Modes
β
n
φ2 = sin
≤ sin −1 1
n2
kn2
φ2
−1
n
n1
≤ sin φ2 ≤ 3
n2
n2
0
kn1
φ 2 ≥ sin −1
kn3
n3
n2
kn2
Higher-order
Lower-order
β
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Reflection at a Dielectric Interface
Step 2:
Determine phase changes at the interfaces.
E3 = rE1 , E2 = tE1
For TE wave:
rTE =
n1 cos θ1 − n2 cos θ2
n1 cos θ1 + n2 cos θ 2
tTE = 1 + rTE
For TM wave:
rTM =
n2 cos θ1 − n1 cos θ 2
n2 cos θ1 + n1 cos θ 2
tTM =
n1
(1 + rTM )
n2
rTE = | rTE | exp( jϕTE ), rTM = | rTM | exp( jϕTM )
→ Phase change accompanies reflection.
Ref: Pedrotti3, “Introduction to Optics,” Sec. 23.1-23.3
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Phase Change on Total Internal Reflection
sin 2 θ1 − sin 2 θc
n12 sin 2 θ1 − n22
φTE
tan
=
=
cos θ1
2
n1 cos θ1
sin 2 θ1 − sin 2 θc n12
φTM
tan
= 2
=
2
cos θ1 sin 2 θc
n2
n12 sin 2 θ1 − n22
n1 cos θ1
2
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Dispersion Equation
Step 3:
Define transverse resonance condition.
Transverse resonance condition:
2kn f h cos θ − 2φc − 2φ s = 2mπ
m
: mode number
kn f h cos θ
: phase shift for the transverse passage through the film
2φc (= φTE ,TM )
: phase shift due to total internal reflection from film/cover interface
2φ s (= φTE ,TM )
: phase shift due to total internal reflection from film/substrate interface
Dispersion equation (β vs. ω):
kn f h cos θ − φc − φ s = mπ → Solve for θ.
Effective guide index
N ≡β
k
= n f sin θ
ns < N < n f
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Graphical Solution
of the Dispersion Equation
Symmetrical waveguide, φs = φc
Asymmetrical waveguide, φs ≠ φc
For a symmetrical waveguide, there is always a solution (no cutoff) for fundamental mode (m = 0).
Increasing h (and/or decreasing λ) will support more modes.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Typical β – ω diagram
Cut-off
Higher-order
kns
Lower-order
knf
β
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Numerical Solution for Dispersion Equation (I)
Define:
Normalized frequency and film thickness
V ≡ kh n 2f − ns2
Normalized guide index
N 2 − ns2
b≡ 2
n f − ns2
b = 0 at cut-ooff (N = ns), and approaches 1 as N → nf.
Measure for the asymmetry
ns2 − nc2
a≡ 2
for TE,
n f − ns2
n 4f ns2 − nc2
for TM
a≡ 4 2
nc n f − ns2
a = 0 for perfect symmetry (ns = nc), and a approaches infinity for strong asymmetry (ns ≠ nc, ns ~ nf).
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Numerical Solution for Dispersion Equation (II)
For TE modes, dispersion relation
kn f h cos θ − φc − φ s = mπ →
V 1 − b = mπ + tan −1
ν=m
b
b+a
+ tan −1
1− b
1− b
: Mode number
(Normalized) cut-off frequency:
V0 = tan −1 a
Vm = V0 + mπ
# of guided modes allowed:
m=
2h 2
n f − ns2
λ
<Example>
AlGaAs/GaAs/AlGaAs double
heterostructure, n = 3.55/3.6/3.55. Determine
a waveguide thickness supporting 0th, 0th and
1st order modes for λ = 1.55 µm..
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
The Goos-Hänchen Shift
zs =
For TE modes
For TM modes
d ϕs
dβ
kz s = ( N 2 − ns2 ) −1 / 2 tan θ
( N 2 − ns2 ) −1 / 2 tan θ
kz s =
 N2 N2 

+ 2 − 1
2
 ns

nf


The lateral ray shift indicates a penetration depth:
zs
xs =
tan θ
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Effective Waveguide Thickness
Effective thickness:
heff = h + xs + xc
Normalized effective thickness:
H ≡ kheff n 2f − ns2
For TE modes:
1
1
H =V +
+
b
b+a
Minimum H → Maximum confinement
Effective waveguide thickness cannot be zero,
even for symmetrical waveguide (a = 0).
Example:
Sputtered glass, ns = 1.515, nf = 1.62,
nc = 1, a = 3.9. Determine the minimum
effective waveguide thickness.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Ray-optic approach can solve for
the effective index,
but this is not good enough.
Why?
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Guided E-M Wave in a Planar Waveguide
Define:
κ c2 = nc2 k 2 − β 2 = − γ c2
κ 2f = n 2f k 2 − β 2
κ 2s = ns2 k 2 − β 2 = − γ 2s
Cover:
Film:
Substrate:
∂2
∂2
2 2
2
2
E
x
y
n
k
E
x
y
E
E=0
+
−
β
=
→
(
,
)
(
)
(
,
)
0
−
γ
c
c
2
2
∂x
∂x
2
∂
∂2
2 2
2
2
E
(
x
,
y
)
+
(
n
k
−
β
)
E
(
x
,
y
)
=
0
→
E
+
κ
f
fE =0
2
2
∂x
∂x
2
∂
∂2
2 2
2
2
E
x
y
n
k
E
x
y
E
+
−
β
=
→
(
,
)
(
)
(
,
)
0
−
γ
s
sE = 0
2
2
∂x
∂x
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
TE Modes
Modal solutions are sinusoidal or exponential, depending on the sign of ( k ni − β )
2 2
Boundary conditions:
For guided modes:
Cover:
Film:
Substrate:
2
The tangential components of E and H are continuous at the interface
between layers. → E y and ∂E y ∂x continuous at the interface.
∂2
2
−
γ
E
E y = 0 → E y = Ec exp[− γ c ( x − h)]
y
c
2
∂x
∂2
2
E
+
κ
y
f E y = 0 → E y = E f cos( κ f x − φ s )
2
∂x
∂2
2
−
γ
E
y
s E y = 0 → E y = Ec exp( γ s x )
2
∂x
Applying boundary conditions, we obtain:
tan φ s =
γs
γ
, tan φc = c
κf
κf
κ f h − φ s − φc = mπ
Ec, Ef, and Es can be determined by,
Optical power
P=
→ Dispersion relation
1
Re{E × H*} ⋅ zˆ dx
∫
2
h
Optical confinement factor
Γ=
1
Re{E × H*} ⋅ zˆ dx
∫
20
∞
1
Re{E × H*} ⋅ zˆ dx
∫
2 −∞
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
TM Modes
Cover:
Film:
Substrate:
∂2
H y − γ c2 H y = 0 → H y = H c exp[− γ c ( x − h)]
2
∂x
∂2
H y + κ 2f H y = 0 → H y = H f cos( κ f x − φ s )
2
∂x
∂2
H y − γ 2s H y = 0 → H y = H c exp( γ s x)
2
∂x
H y and E z continuous at the interface between the layers
1 ∂H y
→ H y and 2
continuous at the interface between the layers
n ∂x
Boundary conditions:
Applying boundary conditions, we obtain:
2
2
 nf  γ
 nf  γ
tan φ s =   s , tan φc =   c
 ns  κ f
 nc  κ f
κ f h − φ s − φc = mπ → Dispersion relation
Relation between the peak fields:
H 2f
(n 2f − N 2 )
n 2f
N
qs = 

 nf
2
= H s2 (n 2f − ns2 )
 N
 +   − 1,
  ns 
  
2
qs
2
2
2 qc
=
H
(
n
−
n
c
f
c) 2
2
ns
nc
N
qc = 
 nf

2
 N
 +   −1
  nc 
  
2
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Multilayer Matrix Theory
Focusing on TE modes first,
U ≡ Ey ,
V ≡ ωµH z
U = A exp(− jκx) + B exp( jκx)
V = κ[ A exp(− jκx) − B exp( jκx)]
At x = 0,
U 0 ≡ U (0),
V0 ≡ V (0)
j
 U 
U 0   cos( κx)
sin(
)
x
κ
 
κ
V  = 
 0   jκ sin( κx) cos( κx)  V 
U 
≡ M 
M: Characteristic matrix of the layer
V 

cos( κi hi )
Mi = 

 jκi sin( κi hi )
U 0 
U n 
=
M
V 
V 
 0
 n
j

sin( κi hi )
κi

cos( κi hi ) 
m12 
m
M ≡  11
= M1 ⋅ M 2 ⋅  M n

m21 m22 
But β, and therefore κi, is unknown …
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Dispersion Relation
for Multilayer Slab Waveguides
Q: What do we know about the fields in the substrate and cover?
Consider guided mode. For substrate and cover,
U = A exp( γx) + B exp(− γx)
V = jγ[ A exp( γx) − B exp(− γx)]
In the substrate,
U 0 = As ,
V0 = jγ s As
In the cover,
U n = Ac ,
Vn = − jγ c Ac
Using the multilayer stack matrix theory, we obtain:
j ( γ s m11 + γ c m22 ) = m21 − γ s γ c m12
→ Dispersion relation for multilayer slab waveguides
<Example> Four-layer waveguides
MathCAD programs
Effective index and modal
distribution in 3-layer and
4-layer waveguides.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Multilayer Matrix Theory for TM Modes
U ≡ Hy,
V ≡ ωε 0 E z
U = A exp(− jκx) + B exp( jκx)
V =−
κ
[ A exp(− jκx) − B exp( jκx)]
n2
TE Modes:
U = A exp(− jκx) + B exp( jκx)
V = κ[ A exp(− jκx) − B exp( jκx)]
Therefore,
TE → TM
 κ
κ → − 2 
n 
(except the phase terms)
Dispersion relation:
γs
γc
γ sγc
− j (m11 2 + m22 2 ) = m21 − 2 2 m12
ns
nc
ns nc
j ( γ s m11 + γ c m22 ) = m21 − γ s γ c m12
Characteristic matrix of the i-th layer:


Mi = 
−



ni2
cos( κi hi )
− j sin( κi hi )
κi

κi
j 2 sin( κi hi )
cos( κi hi ) 
ni


cos( κi hi )
Mi = 

 jκi sin( κi hi )
j

sin( κi hi )
κi

cos( κi hi ) 
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
What is Next?
What we learned cannot be
practical waveguides.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Channel Waveguide Structures
Q: Can you think of more?
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
The Method of Field Shadows (I)
Ignore the fields and refractive indices in the shaded field shadow regions.
→ Results in separable index profiles.
Works well as long as the fields are well confined in the high index (nf) region of the
waveguide.
→ Not applicable at cutoff.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
The Method of Field Shadows (II)
E ( x, y ) = X ( x)Y ( y )
Assuming a buried channel waveguide structure.
β 2 = β 2x + β 2y
N 2 = N x2 + N y2
Vx = kh n 2f − ns2
V y = kw n 2f − ns2
Obtain Nx and Ny, therefore N, by using
the dispersion relation chart and
N x2 − ns2 + n 2f / 2
bx =
n 2f − ns2
N y2 − ns2 + n 2f / 2
by =
n 2f − ns2
Or instead of solving for Nx and Ny, we can use
N 2 − ns2
b= 2
= bx + by − 1
2
n f − ns
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Method of Field Shadows Exercise
Ti:LiNbO3 buried channel
waveguide for λ = 0.8 µm
2µm
1µm nf = 2.234
ns = 2.214
Determine effective index and
modal distribution
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
The Effective-Index Method
(1) Determine the normalized thickness of the channel and lateral guides.
V f = kh n 2f − ns2 ,
Vl = kl n 2f − ns2
(2) Use the dispersion relation chart to determine the normalized guide indices bf and bl.
Determine the corresponding effective indices.
N 2f ,l = ns2 + b f ,l (n 2f − ns2 )
(3) Determine the normalized width. Veq = kw N 2f − N l2
Then determine the normalized guide index beq using the dispersion relation chart.
(4) The effective index of the waveguide can be determined from
N 2 − N l2
beq = 2
N f − N l2
→ N 2 = N l2 + beq ( N 2f − N l2 )
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Effective-Index Method Exercise
Ti:LiNbO3 rib waveguide for λ = 0.8
µm, n f = 2.234 , ns = 2.214 , nc = 1 ,
h = 1.8 µm, l = 1µm, w = 2 µm.
Determine the effective index.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Effective Index Parameters
for Channel Waveguides
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
Numerical Comparisons
Between Different Methods
Effective index method provides good
approximation even near cut-off.
Lih Y. Lin
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EE 529 Semiconductor Optoelectronics
– Optical Waveguides
To Complete the Story …
Q1: What about waveguides which have more than 3 layers in each region?
Air, nc=1
0.5 µm
GaAs, n2=3.6
0.5 µm
Al0.25Ga0.75As, n1=3.45
0.3 µm
x=0
Al0.35Ga0.65As, ns=3.4
x
y
y=0
A: Use the matrix method from Multilayer Stack Theory to determine Nf and
Nl, then continue on Step (3) in the Effective Index Method.
Q2: What about 2-D modal profile in the waveguide?
Ex ,i A exp(− j κ x ,i x) + B exp( j κ x ,i x)
=
2
(kni ) 2 − (kN f ,l ) 2
κ=
x ,i
Ey,=
C exp(− j κ y , f
f or l
(kN f
κ 2y , f =
or l
or l
or l
y ) + D exp( j κ y , f
or l
y)
) 2 − (kN ) 2
κ 2x +=
κ 2y (kni ) 2 − (kN ) 2
Lih Y. Lin
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