Lecture 26
Dielectric Slab Waveguides
In this lecture you will learn:
• Dielectric slab waveguides
•TE and TM guided modes in dielectric slab waveguides
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Guided Modes in Parallel-Plate Metal Waveguides
r r
E (r )
x >0
= yˆ Eo sin(k x x ) e − j k z z
x
Ei
Hi
r
ki
r
kr
Er
Hr
r
k i = − k x xˆ + k z zˆ
Ei
Hi
ε
r
ki
Ey
µo
z
r
k r = k x xˆ + k z zˆ
Guided TE modes are TE-waves bouncing back and fourth between two metal
plates and propagating in the z-direction !
The x-component of the wavevector can have only discrete values – its quantized
kx =
mπ
d
where : m = 1, 2, 3, KK
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
1
Dielectric Waveguides - I
Consider TE-wave undergoing total internal reflection:
Ei
x
r
ki
H
θi θi
r i
k i = − k x xˆ + k z zˆ
ε1
r
kr
Er
z
r
k r = k x xˆ + k z zˆ
Hr
Evanescent wave
r r
E (r )
x >0
= yˆ E i e − j (− k x
x + kz z )
µo
+ yˆ ΓE i e − j (k x
ε 2 µo
ε1 > ε 2
x +kz z )
k z2 + k x2 = ω 2 µo ε 1
Γ = 1 when θ i > θc
When
r r
E (r )
θ i > θc :
x <0
= yˆ T E i e
− j kz z
e
k x = − jα x
−α x x
k z2 − α x2 = ω 2 µo ε 2
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Dielectric Waveguides - II
x
ε2
cladding
Evanescent wave
Ei
Hi
r
ki
θi θi
r
kr
Er
µo
Ei
Hi
Hr
r
ki
ε1
µo
core
ε1 > ε 2
z
cladding
Evanescent wave
ε2
µo
One can have a guided wave that is bouncing between two dielectric interfaces
due to total internal reflection and moving in the z-direction
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
2
Dielectric Slab Waveguides
W
2d
Assumption: W >> d
x
cladding
core
y
cladding
z
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Dielectric Vs Metal Waveguides
Ey
m=1
ε
m=2
µo
z
Ey
Metal Waveguides
(modes are tightly confined)
ε2
µo
ε1
µo
ε2
µo
cladding
Ey
Ey
core
z
cladding
Dielectric Slab Waveguides
(modes are loosely confined)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
3
Dielectric Slab Waveguides – TE Modes: Formal Solution
x
ε2
µo
ε1
µo
ε2
µo
cladding
Ey
2d
core
Ey
z
cladding
symmetric
antisymmetric
The TE solutions are of the form:
r r
E (r )
r r
E (r )
r r
E (r )
x <d
x >d
⎧cos(k x x )⎫ − j k z z
= yˆ Eo ⎨
⎬e
⎩ sin(k x x ) ⎭
= yˆ E1 e
x < −d
−α x ( x − d )
e
The “sine” and “cosine” represent
the symmetric and antisymmetric
solutions w.r.t. the z-axis
− j kz z
⎧+ ⎫
= ⎨ ⎬ yˆ E1 e +α x ( x + d ) e − j k z z
⎩− ⎭
Where:
k z2 + k x2 = ω 2 µo ε 1
k z2
− α x2
Given a frequency ω, the values of kz , kx ,
and αx are still not known
2
= ω µo ε 2
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Modes: Boundary Conditions
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Ey
Ey
core
z
cladding
Boundary conditions:
(1) At x =± d the component of E-field parallel to the interface (i.e. the y-component)
is continuous for all z
⇒
⎧cos(k x d )⎫
Eo ⎨
⎬ = E1
⎩ sin(k x d ) ⎭
(1)
(2) At x =± d the component of H-field parallel to the interface (i.e. the z-component)
is continuous for all z
⇒
⎧ − k x sin(k x d )⎫
Eo ⎨
⎬ = −α x E1
⎩ k x cos(k x d ) ⎭
(2)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
4
TE Modes: Transcendental Equation
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Ey
core
Ey
z
cladding
Dividing (2) by (1) on the previous slide gives:
⎧ tan(k x d ) ⎫ α x
⎨
⎬=
⎩ − cot (k x d )⎭ k x
But:
k z2 + k x2 = ω 2 µo ε 1
k z2
− α x2
α x2 + k x2 = ω 2 µo (ε 1 − ε 2 )
2
= ω µo ε 2
So we finally get:
⎧ tan(k x d ) ⎫
ω 2 µo (ε 1 − ε 2 )
−1
⎨
⎬=
k x2
⎩ − cot (k x d )⎭
Transcendental equation that can
be used to solve for kx in terms of
the frequency ω
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Modes: Graphical Solution
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Ey
core
Ey
z
cladding
Graphic solution of the transcendental
equation
Different red curves for Increasing ω values
⎧ tan(k x d ) ⎫
ω 2 µo (ε 1 − ε 2 ) d 2
−1
⎨
⎬=
(k x d )2
⎩ − cot (k x d )⎭
LHS
RHS
For the m-th TE mode (TEm mode)
the value of kx is in the range
(depending on the frequency ω):
π
π
2
2
(m − 1) ≤ k x d ≤ m
0
π
π
2
3π
2
2π
5π
2
k xd
m = 1, 2, 3,KK
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
5
TE Modes: Cut-off Frequencies
x
ε2
µo
ε1
µo
ε2
µo
cladding
Ey
2d
core
Ey
z
cladding
Different red curves for Increasing ω values
⎧ tan(k x d ) ⎫
ω 2 µo (ε 1 − ε 2 ) d 2
−1
⎨
⎬=
(k x d )2
⎩ − cot (k x d )⎭
LHS
RHS
Cut-off frequency of the m-th TE
mode can be obtained by setting the
RHS equal to zero for kx d = (m-1)π/2
ωm =
π
0
3π
2
π
2
(m − 1)π
5π
2
2π
k xd
1
µo (ε 1 − ε 2 )
2d
m = 1, 2, 3,KK
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Dielectric
Waveguides – What is Cut-off ?
x
ε2
µo
cladding
Ei
2d
r
ki
Hi
θi
θi
r
kr
Er
Ei
Hi
r
ki
ε1
kz
µo
core
z
Hr
cladding
θi k x
ω µo ε 1
ε1 > ε2
ε2
µo
So what does “cut-off” really mean? It means that the wave is no longer being
guided through total internal reflection since θi < θc
A wave will not be guided if:
θ i < θc
⇒
⇒
⇒
⇒
⇒
kz
ω µo ε 1
sin(θ i ) < sin(θc ) = ε 2 ε 1
< ε 2 ε1
For the m-th TE mode (TEm mode)
the smallest value of kx is:
(m − 1)
So TEm mode will not be guided if:
k z < ω µo ε 2
ω<
ω 2 µo ε 1 − k x2 < ω µo ε 2
ω<
kx
µo (ε 1 − ε 2 )
π
2d
⇒
ωm =
kx
µo (ε 1 − ε 2 ) k
(m − 1)π
2d
x=
(m −1)π
2d
1
µo (ε 1 − ε 2 )
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
6
TE Modes: Near Cut-off Behavior
ε2
µo
ε1
µo
ε2
µo
cladding
Ey
core
Different red curves for
increasing ω values
z
cladding
ω >> ω2 (TE2 mode well confined in the core)
ε2
π
0
µo
cladding
Ey
core
3π
2
π
2
ε1
µo
ε2
µo
k xd
z
cladding
ω >≈ ω2 (TE2 mode near cut-off – mode not well confined in the core)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Modes: Dispersion Curves
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Ey
core
Ey
z
cladding
How does one obtain dispersion curves?
(1) For a given frequency ω find kx using:
kz
kz = ω
TE3 mode
dispersion relation
µo ε 1
⎧ tan(k x d ) ⎫
ω 2 µo (ε 1 − ε 2 ) d 2
−1
⎨
⎬=
(k x d )2
⎩ − cot (k x d )⎭
TE2 mode
dispersion relation
kz = ω
(2) Then find kz using:
k z = ω 2 µo ε 1 − k x2
TE1 mode
dispersion relation
ω1
ω2
µo ε 2
ω
ω3
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
7
TM Modes: Formal Solution
x
ε2
µo
ε1
µo
ε2
µo
cladding
Hy
2d
core
Hy
z
cladding
The TM solutions are of the form:
r r
H (r )
r r
H (r )
r r
H (r )
x <d
x >d
The “sine” and “cosine” represent
the symmetric and antisymmetric
solutions w.r.t. the z-axis
= yˆ H1 e −α x ( x − d ) e − j k z z
x < −d
Where:
⎧cos(k x x )⎫ − j k z z
= yˆ Ho ⎨
⎬e
⎩ sin(k x x ) ⎭
⎧+ ⎫
= ⎨ ⎬ yˆ H1 e +α x ( x + d ) e − j k z z
⎩− ⎭
k z2 + k x2 = ω 2 µo ε 1
Given a frequency ω, the values of kz , kx ,
and αx are still not known
k z2 − α x2 = ω 2 µo ε 2
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Modes: Boundary Conditions
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Hy
Hy
core
z
cladding
Boundary conditions:
(1) At x =± d the component of H-field parallel to the interface (i.e. the y-component)
is continuous for all z
⇒
⎧cos(k x d )⎫
Ho ⎨
⎬ = H1
⎩ sin(k x d ) ⎭
(1)
(2) At x =± d the component of E-field parallel to the interface (i.e. the z-component)
is continuous for all z
⇒
⎧ kx
⎫
⎪⎪ − ε sin(k x d )⎪⎪
αx
H
Ho ⎨ 1
⎬=−
kx
ε2 1
⎪
cos(k x d ) ⎪
⎩⎪ ε 1
⎭⎪
(2)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
8
TM Modes: Transcendental Equation
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Hy
core
Hy
z
cladding
Dividing (2) by (1) on the previous slide gives:
⎧ tan(k x d ) ⎫ α x ε 1
⎨
⎬=
⎩ − cot (k x d )⎭ k x ε 2
But:
k z2 + k x2 = ω 2 µo ε 1
k z2
− α x2
α x2 + k x2 = ω 2 µo (ε 1 − ε 2 )
2
= ω µo ε 2
So we finally get:
⎧ tan(k x d ) ⎫ ε 1 ω 2 µo (ε 1 − ε 2 )d 2
−1
⎨
⎬=
(k x d )2
⎩ − cot (k x d )⎭ ε 2
Transcendental equation that can
be used to solve for kx in terms of
the frequency ω
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Modes: Graphical Solution
x
ε2
µo
ε1
µo
ε2
µo
cladding
2d
Hy
core
Hy
z
cladding
Graphic solution of the transcendental
equation
Different red curves for Increasing ω values
⎧ tan(k x d ) ⎫ ε 1 ω 2 µo (ε 1 − ε 2 ) d 2
−1
⎨
⎬=
(k x d )2
⎩ − cot (k x d )⎭ ε 2
LHS
RHS
For the m-th TM mode (TMm mode)
the value of kx is in the range
(depending on the frequency ω):
π
π
2
2
(m − 1) ≤ k x d ≤ m
0
π
π
2
3π
2
2π
5π
2
k xd
m = 1, 2, 3,KK
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
9
Fiber Optical Communications: Optical Fibers
cladding
9.0 µm
core
z
125 µm
cladding
cladding
core
z
Silica (SiO2) core and cladding materials
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Integrated Optics
2 um
An optical micro-ring filter (separates out
light of a particular color) – SEM
An optical micro-splitter (splits light two
ways) – SEM
Slab Waveguide
Si
Si
SiO2
SiO2
Si
Si
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
10
Integrated Optics: Semiconductor Quantum Well Lasers
top metal
InP/InGaAsP
Waveguide
1 .0 µm thick
polyimide layer
3 .0 µm
The dielectric waveguide for a
semiconductor quantum well laser
with metal on top (for electrical
connection) is shown
A microchip containing several
semiconductor laser stripes running in
parallel is shown.
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
11