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Lecture 26 Dielectric Slab Waveguides

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Lecture 26 Dielectric Slab Waveguides 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 E()rr = yˆ E sin()k x e− j kz z x>0 o x x Ei Ei r r Ey k E r k i r kr i H Hi i z ε µo Hr r r ki = −kx xˆ + kzzˆ kr = kx xˆ + kzzˆ 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 m π k = where : m = 1, 2, 3, x d KK ECE 303 – Fall 2005 – Farhan Rana – Cornell University 1 Dielectric Waveguides - I Consider TE-wave undergoing total internal reflection: E i x ε1 µo r k E r i r kr H θ θ i i i z r H r ki = −kx xˆ + kzzˆ r kr = kx xˆ + kzzˆ Evanescent wave ε2 µo ε1 > ε2 r E()rr = yˆ E e− j ()−kx x +kz z + yˆ ΓE e− j (k x x +kz z) 2 2 2 x>0 i i kz + kx = ω µo ε1 Γ = 1 when θi > θc When θ i > θ c : kx = − jα x r E()rr = yˆ T E e− j kz z e−α x x 2 2 2 x<0 i kz − α x = ω µo ε2 ECE 303 – Fall 2005 – Farhan Rana – Cornell University Dielectric Waveguides - II x ε2 µo Evanescent wave cladding Ei E ε1 µo r i r k E r k i r kr i core Hi θi θi Hi ε1 > ε2 z Hr 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 y core cladding z ECE 303 – Fall 2005 – Farhan Rana – Cornell University Dielectric Vs Metal Waveguides m = 2 ε µo m = 1 E z y Ey Metal Waveguides (modes are tightly confined) ε2 µo cladding Ey core z Ey ε1 µo cladding ε2 µo Dielectric Slab Waveguides (modes are loosely confined) ECE 303 – Fall 2005 – Farhan Rana – Cornell University 3 Dielectric Slab Waveguides – TE Modes: Formal Solution x ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding symmetric ε 2 µo antisymmetric The TE solutions are of the form: r r ⎧cos(kx x )⎫ − j k z E()r = yˆ E ⎨ ⎬ e z x <d o sin()k x ⎩ x ⎭ The “sine” and “cosine” represent r E()rr = yˆ E e−α x ()x −d e− j kz z the symmetric and antisymmetric x >d 1 solutions w.r.t. the z-axis + r r ⎧ ⎫ +α x ()x +d − j kz z E()r = ⎨ ⎬yˆ E1 e e x <−d ⎩−⎭ Where: k2 + k 2 = ω2 µ ε z x o 1 Given a frequency ω, the values of kz , kx , and α are still not known 2 2 2 x kz − α x = ω µo ε2 ECE 303 – Fall 2005 – Farhan Rana – Cornell University TE Modes: Boundary Conditions x ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding ε 2 µo 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()kx d ⎫ ⇒ Eo ⎨ ⎬ = E1 (1) ⎩ sin()kxd ⎭ (2) At x =± d the component of H-field parallel to the interface (i.e. the z-component) is continuous for all z ⎧− kx sin()kx d ⎫ ⇒ Eo ⎨ ⎬ = −αxE1 (2) ⎩ kx cos()kxd ⎭ ECE 303 – Fall 2005 – Farhan Rana – Cornell University 4 TE Modes: Transcendental Equation x ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding ε 2 µo Dividing (2) by (1) on the previous slide gives: ⎧ tan()kx d ⎫ α ⎨ ⎬ = x ⎩− cot()kxd ⎭ kx But: 2 2 2 kz + kx = ω µo ε1 2 2 2 α x + kx = ω µo (ε1 − ε2 ) 2 2 2 kz − α x = ω µo ε2 So we finally get: 2 Transcendental equation that can ⎧ tan()kx d ⎫ ω µo ()ε1 − ε2 ⎨ ⎬ = − 1 be used to solve for k in terms of − cot()k d 2 x ⎩ x ⎭ kx the frequency ω ECE 303 – Fall 2005 – Farhan Rana – Cornell University TE Modes: Graphical Solution x ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding ε 2 µo Graphic solution of the transcendental equation Different red curves for Increasing ω values 2 2 ⎧ tan()kx d ⎫ ω µo ()ε1 − ε2 d ⎨ ⎬ = 2 − 1 ⎩− cot()kxd ⎭ ()kxd LHS RHS For the m-th TE mode (TE mode) m 0 π π 3π 2π 5π the value of kx is in the range kxd (depending on the frequency ω): 2 2 2 π π ()m − 1 ≤ k d ≤ m 2 x 2 m = 1,2,3,KK ECE 303 – Fall 2005 – Farhan Rana – Cornell University 5 TE Modes: Cut-off Frequencies x ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding ε 2 µo 2 2 Different red curves for Increasing ω values ⎧ tan()kx d ⎫ ω µo ()ε1 − ε2 d ⎨ ⎬ = 2 − 1 ⎩− cot()kxd ⎭ ()kxd 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 0 π π 3π 2π 5π kxd 2 2 2 ()m − 1 π 1 ωm = 2d µo ()ε1 − ε2 m = 1,2,3,KK ECE 303 – Fall 2005 – Farhan Rana – Cornell University Dielectricx Waveguides – What is Cut-off ? ε 2 µ o cladding kz Ei E ε 1 µ o r i r k E r k 2d i r k r i core H θ θ H θ k i i i i z i x Hr ω µo ε1 cladding ε 1 > ε 2 ε 2 µ o 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: For the m-th TE mode (TEm mode) the smallest value of kx is: θi < θc ⇒ sin()θi < sin(θc ) = ε2 ε1 π ()m − 1 kz 2d ⇒ < ε2 ε1 So TE mode will not be guided if: ω µo ε1 m kx ⇒ kz < ω µo ε2 ω < µo ()ε1 − ε2 ()m−1 π 2 2 kx = ⇒ ω µo ε1 − kx < ω µo ε2 2d m − 1 π 1 kx ( ) ⇒ ω < ⇒ ωm = 2d µ ε − ε µo ()ε1 − ε2 o ()1 2 ECE 303 – Fall 2005 – Farhan Rana – Cornell University 6 TE Modes: Near Cut-off Behavior ε2 µo cladding Ey Different red curves for core z ε1 µo increasing ω values cladding ε2 µo ω >> ω2 (TE2 mode well confined in the core) ε2 µo 0 π π 3π kxd cladding 2 2 Ey core z ε1 µo cladding ε2 µo 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 ε 2 µo cladding Ey 2d core z Ey ε 1 µo cladding ε 2 µo How does one obtain dispersion curves? kz TE3 mode dispersion relation k = ω µ ε (1) For a given frequency ω find kx using: z o 1 2 2 TE2 mode ⎧ tan()kx d ⎫ ω µo ()ε1 − ε2 d dispersion relation ⎨ ⎬ = 2 − 1 ⎩− cot()kxd ⎭ ()kxd kz = ω µo ε2 (2) Then find kz using: 2 2 TE mode kz = ω µo ε1 − kx 1 dispersion relation ω ω 1 ω 2 ω 3 ECE 303 – Fall 2005 – Farhan Rana – Cornell University 7 TM Modes: Formal Solution x ε 2 µo cladding Hy 2d core z Hy ε 1 µo cladding ε 2 µo The TM solutions are of the form: r ⎧cos()k x ⎫ H()rr = yˆ H x e− j kz z x <d o ⎨ ⎬ ⎩ sin()kx x ⎭ The “sine” and “cosine” represent r r −α x ()x −d − j kz z the symmetric and antisymmetric H()r = yˆ H1e e x>d solutions w.r.t. the z-axis + r r ⎧ ⎫ +α x ()x +d − j kz z H()r = ⎨ ⎬yˆ H1e e x<−d ⎩−⎭ Where: k2 + k 2 = ω2 µ ε z x o 1 Given a frequency ω, the values of kz , kx , and α are still not known 2 2 2 x kz − α x = ω µo ε2 ECE 303 – Fall 2005 – Farhan Rana – Cornell University TM Modes: Boundary Conditions x ε 2 µo cladding Hy 2d core z Hy ε 1 µo cladding ε 2 µo 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()kx d ⎫ ⇒ Ho ⎨ ⎬ = H1 (1) ⎩ sin()kxd ⎭ (2) At x =± d the component of E-field parallel to the interface (i.e. the z-component) is continuous for all z ⎧ kx ⎫ ⎪− sin()kx d ⎪ ⎪ ε1 ⎪ α x ⇒ Ho ⎨ ⎬ = − H1 (2) kx ε2 ⎪ cos()kxd ⎪ ⎩⎪ ε1 ⎭⎪ ECE 303 – Fall 2005 – Farhan Rana – Cornell University 8 TM Modes: Transcendental Equation x ε 2 µo cladding Hy 2d core z Hy ε 1 µo cladding ε 2 µo Dividing (2) by (1) on the previous slide gives: ⎧ tan()kx d ⎫ α ε ⎨ ⎬ = x 1 ⎩− cot()kxd ⎭ kx ε2 But: 2 2 2 kz + kx = ω µo ε1 2 2 2 α x + kx = ω µo (ε1 − ε2 ) 2 2 2 kz − α x = ω µo ε2 So we finally get: 2 2 Transcendental equation that can ⎧ tan()kx d ⎫ ε1 ω µo ()ε1 − ε2 d ⎨ ⎬ = − 1 be used to solve for k in terms of − cot()k d ε 2 x ⎩ x ⎭ 2 ()kxd the frequency ω ECE 303 – Fall 2005 – Farhan Rana – Cornell University TM Modes: Graphical Solution x ε 2 µo cladding Hy 2d core z Hy ε 1 µo cladding ε 2 µo Graphic solution of the transcendental equation Different red curves for Increasing ω values 2 2 ⎧ tan()kx d ⎫ ε1 ω µo ()ε1 − ε2 d ⎨ ⎬ = 2 − 1 ⎩− cot()kxd ⎭ ε2 ()kxd LHS RHS For the m-th TM mode (TM mode) m 0 π π 3π 2π 5π the value of kx is in the range kxd (depending on the frequency ω): 2 2 2 π π ()m − 1 ≤ k d ≤ m 2 x 2 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 An optical micro-splitter (splits light two light of a particular color) – SEM 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 3 .0 µm 1 .0 µm thick polyimide layer A microchip containing several semiconductor laser stripes running in parallel is shown.
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