Waveguide Coupler I Waveguide Coupler waveguide to other waveguides

Transcription

Waveguide Coupler I
Class: Integrated Photonic Devices
Time: Fri. 8:00am ~ 11:00am.
Classroom: 資電206
Lecturer: Prof. 李明昌(Ming-Chang Lee)
Ming-Chang Lee, Integrated Photonic Devices
Waveguide Coupler
n1 > n0
Waveguide 1
n0
n1
n0
Waveguide 2
n1
n0
• How to switch the power from one
waveguide to other waveguides
1
Two Types of Directional Couplers
I. Planar Waveguide Coupler (Out-of-Plane)
(Cross Section)
II. Dual-Channel Waveguide Coupler (In-Plane)
Coupled Mode Theory for Directional Coupler
monochromatic waveguide modes
E ( x, y, z, t ) = AL( z )ψ ( x, y ) exp(− jωt )
Axial (Longitudinal)
Part
Transverse
Part
Time harmonic
L( z ) is a complex amplitude including phase term exp( j β z )
ψ ( x, y ) represents the field distribution of a single waveguide
2
N1(x,y)
N2(x,y)
We define the eigen modes in each optical waveguide satisfying Maxwell’s equations
 ∇ × E p = jωµ0 H p

2
∇ × H p = − jωε 0 N p E p
(p = 1,2)
Np(x,y): refractive index
The electromagnetic fields of the coupled waveguide is summation of two eigen modes
 E = A1 ( z )E1 + A2 ( z )E 2

H = A1 ( z )H1 + A2 ( z )H 2
A1 and A2 is the amplitude of
optical fields
The summed fields should also satisfy Maxwell’s equation
 ∇ × E = jωµ0 H

2
∇ × H = − jωε 0 N E
The vector formula
(1)
N(x,y) denote the entire refractive
index of coupled waveguide
∇ × ( AE) = A∇ × E + ∇A × E = A∇ × E +
dA
uz × E
dz
(the amplitude only varies at z-direction)
(2)
Perturbation!
Combine (1) and (2)
(u z × E1 )
(u z × H1 )
dA1
dA
+ (u z × E2 ) 2 = 0
dz
dz
(3)
dA1
dA
+ jωε 0 ( N 2 − N12 ) A1 E1 + (u z × H 2 ) 2 + jωε 0 ( N 2 − N 2 2 ) A2 E2 = 0 (4)
dz
dz
3
∞
∫ ∫
∞
−∞ −∞
∞
∫ ∫
E1* ⋅ (4) − H1* ⋅ (3)  dxdy = 0


(5)
E2* ⋅ (4) − H 2* ⋅ (3)  dxdy = 0


(6)
∞
−∞ −∞
From Eq. (5)
∞
∞
*
*
dA1 dA2 ∫−∞ ∫−∞ u z ⋅ (E1 × H 2 + E 2 × H1 )dxdy
+
⋅ ∞ ∞
*
*
dz
dz
u ⋅ (E1 × H1 + E1 × H1 )dxdy
∫ ∫
ωε ∫ ∫ ( N − N )E ⋅ E dxdy
+ jA
∫ ∫ u ⋅ (E × H + E × H )dxdy
ωε ∫ ∫ ( N − N )E ⋅ E dxdy
=0
+ jA
∫ ∫ u ⋅ (E × H + E × H )dxdy
z
−∞ −∞
∞
∞
1
∞
0 −∞ −∞
∞
z
−∞ −∞
∞
0
∞
2
∞
2
*
1
∞
1
z
*
1
*
1
2
2
*
1
1
1
2
−∞ −∞
−∞ −∞
*
1
2
1
1
2
*
1
1
(7)
From Eq. (6)
∞
∞
*
*
∫ ∫ u ⋅ (E × H + E × H )dxdy
∫ ∫ u ⋅ (E × H + E × H )dxdy
ωε ∫ ∫ ( N − N )E ⋅ E dxdy
+ jA
∫ ∫ u ⋅ (E × H + E × H )dxdy
ωε ∫ ∫ ( N − N )E ⋅ E dxdy
=0
+ jA
(
)
⋅
×
+
×
dxdy
u
E
H
E
H
∫ ∫
dA2 dA1
+
⋅
dz
dz
−∞ −∞
∞ ∞
z
2
−∞ −∞
z
2
∞
2
∞
2
∞
1
∞
∞
−∞ −∞
2
2
2
2
*
*
2
2
2
2
*
*
2
z
∞
0
1
*
0 −∞ −∞
∞
−∞ −∞
1
∞
2
2
2
*
2
2
1
−∞ −∞
2
1
*
z
2
*
2
2
(8)
2
Here we separate the transverse and axial dependencies of electromagnetic fields
 E p = E p exp( j β p z )

H p = H p exp( j β p z )
(p = 1,2)
4
dA1
dA
+ c12 2 exp[ j ( β 2 − β1 ) z ] + j χ1 A1 + jκ12 A2 exp[ j ( β 2 − β1 ) z ] = 0
dz
dz
(9) (from (7))
dA2
dA
+ c21 1 exp[− j ( β 2 − β1 ) z ] + j χ 2 A2 + jκ 21 A1 exp[− j ( β 2 − β1 ) z ] = 0 (10) (from (8))
dz
dz
where
κ pq =
ωε 0 ∫
∞
∞
∫ ∫
∞
c pq
∞
−∞ −∞
∞ ∞
−∞ −∞
χp =
∞
∞
( N 2 − N q 2 )E p* ⋅ Eq dxdy
u z ⋅ (E p* × H q + Eq × H p* )dxdy
u z ⋅ (E p* × H p + E p × H p* )dxdy
ωε 0 ∫
∫ ∫
∫
u z ⋅ (E p* × H p + E p × H p* )dxdy
−∞ −∞
∫ ∫
=
∫ ∫
∞
−∞ −∞
∞
−∞ −∞
∞
∫
∞
−∞ −∞
( q perturb p)
( q perturb p)
( N 2 − N p 2 )E p* ⋅ E p dxdy
u z ⋅ (E p* × H p + E p × H p* )dxdy
N1 ( x, y )
N 2 ( x, y )
Waveguide I
κ12
•
Waveguide II
c12
χ1
χ is much smaller than κ
5
(a). Consider
χp
χ p is real number
χ p = χ p*
(b). Consider c pq
Recall
1
Pp = ∫ ∫
Re(E p × H p* ) ⋅ u z dxdy
−∞ −∞ 2
∞
∞
∫ ∫
∞
−∞ −∞
∞
(E p* × H p + E p × H p* ) ⋅ u z dxdy = 4 Pp = 4
)
(
(
*
1
E exp(− jωt ) + E exp( jωt )
2
*
1
H = H exp(− jωt ) + H exp( jωt )
2
E=
)
If Ep, Hp is the eigen mode
∞
c pq
∫ ∫
=
∫ ∫
∞
−∞ −∞
∞ ∞
−∞ −∞
u z ⋅ (E p* × H q + Eq × H p* )dxdy
c21 = c12*
u z ⋅ (E p × H p + E p × H p )dxdy
*
*
(c). Consider κ pq
Suppose there is no coupling loss, the total power should be constant
P=∫
∞
∫
∞
−∞ −∞
*
1
Re(E × H ) ⋅ u z dxdy
2
and
dP
=0
dz
jA1* A2 (κ 21* − κ12 − 2δ c12 ) exp(− j 2δ z ) − jA1 A2* (κ 21 − κ12* − 2δ c12* ) exp( j 2δ z ) = 0
where
δ=
( β 2 − β1 )
2
for every z
κ 21 = κ12* + 2δ c12*
• The coupling coefficients are complicate
conjugated when
– Two modes are in the same waveguide or
– Two waveguides are phase matching
δ=
β 2 − β1
2
=0
6
General expression of coupled mode theory
dA1
= jκ a A2 exp( j 2δ z ) + jγ a A1
dz
[(9) − (10) × c12 exp( j 2δ z ) = 0]
dA2
= jκ b A1 exp(− j 2δ z ) + jγ b A2
dz
[(10) − (9) × c21 exp(− j 2δ z ) = 0]
κa =
where
κb =
κ12 − c12 χ 2
1 − c12
γa =
2
and
κ 21 − c12* χ1
1 − c12
γb =
2
κ 21c12 − χ1
1 − c12
2
κ12 c12* − χ 2
1 − c12
2
In most cases, we suppose two waveguides are sufficient separated
χ p ,q ∼ 0
Then
and
C pq
1
κ12 = κ 21 and γ a = γ b = 0
The general coupled mode theory is simplified by
dA1
= jκ12 A2 exp[ j ( β 2 − β1 ) z ]
dz
dA2
= jκ 21 A1 exp[− j ( β 2 − β1 ) z ]
dz
7
The coupling between mode is given by the coupled mode
equations for the amplitudes of the two modes (guide 1 and guide 2)
 d A1 ( z )
= j β1 A1 ( z ) + jκ12 A2 ( z )

 dz

 d A2 ( z )
 dz = j β 2 A2 ( z ) + jκ 21 A1 ( z )
A1 = A1 exp( j β1 z )
A2 = A2 exp( j β 2 z )
 β1 : the propagation constant of guide 1

 β 2 : the propagation constant of guide 2
κ12 : the coupling coefficient from 2 to 1

 κ 21 : the coupling coefficient from 1 to 2
In general, κ12 and κ21 are not equal. We suppose the coupling
coefficients are approximated and real
κ 21 = κ12 = κ
Solutions of Directional Coupler



δ
κ
A1 ( z ) =  cos( gz ) + j sin( gz )  A1 (0) − j sin( gz ) A2 (0)  exp( j ( β1 + δ ) z )
g
g



 κ



δ
A2 ( z ) =  − j sin( gz ) A1 (0) + cos( gz ) + j sin( gz )  A2 (0)  exp( j ( β 2 − δ ) z )
g


 g

Where g ≡ κ + δ
2
2
2
and δ ≡
β 2 − β1
2
If the initial condition A1(0) = 1 and A2(0) = 0, the solution is


δ
A1 ( z ) = cos( gz ) − j sin( gz )  exp( j ( β1 + δ ) z )
g


A2 ( z ) = j
κ
g
sin( gz ) exp( j ( β 2 − δ ) z )
• The propagation constant of each eigen mode becomes the
average of two individual propagation constants in the
coupling region
8
Transferred Power
Phase Mismatching
Phase Matching
Transferred Power
Transferred Power
P1
P2
Normalized Distance (gz)
δ =0
P1
P2
Normalized Distance (gz)
δ = 2κ
2

*
2
2 sin ( gz )
 P1 ( z ) = A1 ( z ) ⋅ A1 ( z ) = cos ( gz ) + δ
g2

β − β1
2
2
2

and δ ≡ 2
2
Where g ≡ κ + δ
 P ( z ) = A ( z ) ⋅ A ( z )* = κ sin 2 ( gz )
2
2
2
 2
g2
• With a phase difference δ, the power transfer is incomplete
Directional Coupler with Phase Matching
Suppose the two waveguides are identical
β1 = β 2 = β
and
Real
κ 21 = κ12 = κ
The coupled mode equation becomes
 d A1 ( z )
= j β A1 ( z ) + jκ A2 ( z )

 dz

 d A2 ( z )
 dz = j β A2 ( z ) + jκ A1 ( z )
If the initial condition A1(0) = 1 and A2(0) = 0, the solution is
 A1 ( z ) = cos(κ z ) exp( j β z )

 A2 ( z ) = j sin(κ z ) exp( j β z )
9
Directional Coupler with Phase Matching and
Loss
Intensity
If we consider the loss α
α

 A1 ( z ) = cos(κ z ) exp( j β z ) exp(− 2 z )

 A2 ( z ) = j sin(κ z ) exp( j β z ) exp(− α z )

2
P1(z)
 P1 ( z ) = A1 ( z ) A1* ( z ) = cos 2 (κ z ) exp(−α z )

*
2
 P2 ( z ) = A2 ( z ) A2 ( z ) = sin (κ z ) exp( −α z )
P2(z)
Propagation Distance
The optical power is completely transferred as propagation length (L)
L=
(2m + 1)π
, m = 0,1,2 …
2κ
What is the coupling coefficient?
Suppose two identical slab waveguides
n2
E1
w
n1
s
n2
E2
n1
x
n2
h,γ
∞
κ=
ωε 0 ∫ (n12 − n0 2 )E1* ⋅ E2 dx
−∞
∞
∫ uˆ
−∞
z
⋅ (E1* × H1 + E1 × H1* )dx
2h 2γ exp(−γ s )
κ=
β dTE (h 2 + γ 2 )
κ≈
k0n1
γ h2
β
z
2β 2 + γ 2 − h2
exp(−γ s )
k2 2
β dTM (h + γ )
2
2
⋅
• The coupling coefficient is an exponential function of gap s.
10
Transfer Matrix Expression
Suppose a lossless coupler
Ai1
Ao1
L
Waveguide 1
Waveguide 2
Ao 2
Ai 2
recall
2
 Ao1   1 − c

=


 Ao 2   jc
 A1 ( z ) = cos(κ z ) exp( j β z )

 A2 ( z ) = j sin(κ z ) exp( j β z )
A1(0)=1, A2(0)=0
2
where c = sin (κ L)
 A 
  i1 
1 − c   Ai 2 
jc
2
c 2 : power coupling ratio
Varying L affects the power coupling ratio
Experimental Measurement
3µm
3µm
GaAs
3µm
L = 2.1 mm (100% coupler)
L = 1 mm (3dB coupler)
11
Summary of Waveguide Coupling
When you design a waveguide coupler, you have to
• Consider the coupling coefficient (κ)
• Consider the phase matching (∆β)
• Consider the coupling length (L)
2

2
2 sin ( gz )
exp(−α z )
 P1 ( z ) = cos ( gz ) exp(−α z ) + δ
g2


2
 P ( z ) = κ sin 2 ( gz ) exp(−α z )
2

g2
g2 ≡ κ 2 +δ 2
Supermodes
Actually, the power oscillating between two waveguides can be thought
as the beating between two supermodes.
For example, suppose the two waveguides are identical,
 A1 ( z ) = A0 cos(κ z ) exp( j β z )

 A2 ( z ) = A0 j sin(κ z ) exp( j β z )
if
 A1 (0) = A0

 A2 (0) = 0
E ( z ) = A1 ( z )ψ 1 ( x, y ) + A2 ( z )ψ 2 ( x, y )
= A0 [ψ 1 ( x, y ) cos(κ z ) exp( j β z ) + jψ 2 ( x, y )sin(κ z ) exp( j β z )]
1
1

= A0  ψ 1 [ exp( jκ z ) + exp(− jκ z )] exp( j β z ) + ψ 2 [ exp( jκ z ) − exp(− jκ z ) ] exp( j β z ) 
2
2

1
1

= A0  (ψ 1 −ψ 2 ) exp [ j ( β − κ ) z ] + (ψ 1 +ψ 2 ) exp [ j ( β + κ ) z ]
2
2

Ea ( z )
Eb ( z )
12
Supermodes
1
1

E ( z ) = A0  (ψ 1 −ψ 2 ) exp [ j ( β − κ ) z ] + (ψ 1 +ψ 2 ) exp [ j ( β + κ ) z ]
2
2

= A0ψ a exp( j β a z ) + A0ψ b exp( j βb z ) = A0 Ea ( z ) + A0 Eb ( z )
Where
βb
ψb
1

ψ a = 2 (ψ 1 −ψ 2 )

ψ = 1 (ψ +ψ )
2
 b 2 1
ψ1
Waveguide II
β a = β − κ

 βb = β + κ
βa
ψa
ψ2
Waveguide I
and
ψ1
Waveguide I
(Symmetric)
ψ2
Waveguide II
(Antisymmetric)
Supermodes
βb
βa
+
ψb
ψa
( βb − β a ) L = 2mπ , m = 0,1, 2,...
βa
βb
ψb
+
ψa
( βb − β a ) L = 2(m + 1)π , m = 0,1, 2,...
Lπ =
π
π
=
βb − β a 2κ
13
Polarization Dependent
Because the coupling coefficients of TE modes and TM
modes are different, we can design a polarization splitter.
TE
Suppose two identical waveguides
TE+TM
ls
ls =
Nπ
κTE
(2 N ± 1)π
=
2κTM
TM
where N :integer and κ TE κ TM are coupling coefficient for TE and TM
Symmetric Curved Waveguide Coupling
Symmetric Curved Waveguide
Parallel Waveguide
A1(z)
A1(s)
Waveguide 1
Waveguide 1
z
A2(z)
s
s
Waveguide 2
Waveguide 2
A2(s)
d A1 ( z ) = j β1 A1 ( z )dz + jκ12 A2 ( z )dz

d A2 ( z ) = j β 2 A2 ( z )dz + jκ 21 A1 ( z )dz
d A1 ( s) = j β1 A1 ( s)ds + jκ12 ( s ) A2 ( s)ds

d A2 ( s ) = j β 2 A2 ( s )ds + jκ 21 ( s) A1 ( s)ds
• Line coordinate vs. curve coordinate
• The amplitude increment can be modelled in point-to-point
correspondence
14
Asymmetric Curved Waveguide Coupling
Waveguide 2
Y
Waveguide 2
Y
Tangent
θ1= 900
R
l
θ2= 900
X
Waveguide 1
θ1
θ2
Y=0
D
X=0
X: Axial coordinate for waveguide 2
Y: Axial coordinate for waveguide 1
X =Y
l
Waveguide 1
X
X
Y
If 2 R + D = tan( 2 R )
then θ1 = θ 2
• If we can define a function to link X,Y curve coordinates such
that θ1 = θ2, the asymmetric curved waveguide coupling can be
treated as a symmetric curved waveguide coupling.
The asymmetric curved waveguide coupling can be expressed by
coupled mode theory.
 d A1(x) = j β1 ( x) A1 ( x)dx + jκ 2 ( y ) A2 ( y )dy

d A2 (y) = j β 2 ( y ) A2 ( y )dy + jκ1 ( x) A1 ( x)dx
where
X
Y
= tan( )
2R + D
2R
β1 ( x)
and
1/ 2

 dy 
κ1 = κ 0 ⋅ cos(π − 2θ ) ⋅  
 dx 


1/ 2
 dx 

=
⋅
−
⋅
κ
κ
cos(
π
2
θ
)
 
0
 2
 dy 

l
Coupling coefficient
due to gap
The angle of
β1 ( x), β 2 ( y)
π − 2θ
β 2 ( y)
15
Consider the coupler is lossless; that is, the coupling
power is conservative
κ1dx = κ 2 dy
The coupled mode theory becomes
 d A1(x)
= j β1 ( x) A1 ( x) + jκ1 ( x) A2 ( y )

 dx

 d A2 (y) = j β ( y ) A2 ( y ) + jκ ( y ) A1 ( x)
2
2
 dy
 d A1(x) = j β1 ( x) A1 ( x)dx + jκ1 ( x) A2 ( y )dx

d A2 (y) = j β 2 ( y ) A2 ( y )dy + jκ 2 ( y ) A1 ( x)dy
A2 ( y0 )
A2 ( y )
A1 ( x)
A1 ( x0 )
 A1 ( x)   C11 ( x, y ) C12 ( x, y )   A1 ( x0 ) 
 A ( y )  =  C ( x, y ) C ( x, y )  ⋅  A ( y ) 
22
 2   21
  2 0 
2
2
2
2
C11 ( x, y ) + C21 ( x, y ) = C12 ( x, y ) + C22 ( x, y ) = 1
C11 ( x, y ) ⋅ C12 ( x, y )* + C21 ( x, y ) ⋅ C22 ( x, y )* = 0
Due to power conservation
Example of Asymmetric Curved Waveguide
Coupling (Vertical Coupling)
Ring
R
w: width
Waveguide
d: thickness
s: gap
(i)
(ii)
Waveguide
microdisk
(i) Tangent Position
(cross section)
(ii) Deviated Position
(cross section)
16
Coupling Coefficient (m-1)
For a silicon curved waveguide with 20-um radius of curvature and 0.8 um
width, coupling coefficient (k0) is analyzed as follow:
Gap: 0.2 µm
Gap: 0.4 µm
Gap: 0.6 µm
Coupling Position (µm)
The effective coupling length is only from -5µm to 5µm
Because the coupling coefficient is dependent on the gap spacing, the
power coupling ratio is also a function of gap spacing.
Gap
Spacing
R: 20 µm
A2 ( y0 )
Transfer back to the
original waveguide
A1 ( x)
A1 ( x0 )
Power Coupling Ratio
A2 ( y )
Straight Waveguide
Gap Spacing (µm)
Curved Waveguide
17
Coupler Fabrication
Gold
18

Waveguide Coupler I Waveguide Coupler waveguide to other waveguides

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