Dielectric Slab Waveguide

Transcription

1 Chapter Dielectric Slab Waveguide We will start off examining the waveguide properties of a slab of dielectric shown in Fig... d n n x z n Figure.: Cross-sectional view of a slab waveguide. { n, x < d/ n(x) = n, else (.). Propagating Ray We will initial look at the light traveling in the slab as a propagating ray. Even though this is not technically accurate, it provides some intuitive feel for what is going on. Figure. shows that if the propagation angle is greater than the critical angle then the ray will bounce off of the surface and will be conned to the core region. Therefore, the propagation is conned to be θ > θ c = sin ( n n ). (.) In order to maintain that the propagation angle is greater than the critical angle, the entrance angle into the optical ber must be less than θ a. sin θ a = n sin (90 θ ) (.3) = n cos (θ ) (.4) ECEn 56 January 7, 007

2 cladding n θ <θ c θ =θ c θ >θ c core n cladding n Figure.: Cross-sectional view of a slab waveguide. Since θ > θ c sin θ a < n cos θ c (.5) sin θ a < < n sin θ c (.6) ( ) n < n (.7) n < n n n n (.8) n n NA (.9) n=.0 n θa 90 θ c θ =θ c n n Figure.3: Numerical aperture of an slab waveguide. In addition to requiring the propagation angle to be greater than the critical angle, there are also only a descrete set of propagaton angles that remain in phase as illustrated in Fig..4. These allowable propagation angles are called the modes of the waveguide. In this ray optics analysis the The particular modes of a waveguide can be characterized by their propagation angle. The mode can be thought of as a plane wave that is either traveling upwards or downwards in the waveguide. The resulting plane waves are given by E(x, z) = E o e jk on(± cos θ x+sin θ z). (.0) The mode is essentially a standing wave pattern in the x-direction and a traveling wave in the z-direction as given by E(x, z) = E m (x) exp (j (ωt βz)), (.) ECEn 56 January 7, 007

3 n λ θ θ n d n Figure.4: The rays must remain in phase after multiple reections. β is called the propagation constant and is given by β = k o n sin θ. (.) Since the propagation angle is in the range given by the propagation angle is in the range given by θ c < θ < 90, (.3) k o n sin θ c < β < k o n sin (90 ) (.4) n k o n < β < k o n n (.5) k o n < β < k o n (.6) If you divide the propagation angle by the free-space wavevector you get the effective index of the mode as given by n eff β k o. (.7) n < n eff < n (.8). Wave Equation Now that we have a qualitative understanding of waveguide modes, we want to calculate the exact values of the supported mode, which we will characterize by the propagation constant β m and the transverse mode eld E m (x). We start with Maxwell's equations in the sinusoidal steady state. E = jωb = jωµh D = ɛe = ρ v (.9) H = jωd + J = jωɛe + J B = µh = 0 (.0) ECEn 56 3 January 7, 007

4 First, we rewrite Ampere's Law for the case of no sources resulting in Likewise, if we have no free charges ρ v = 0 and thus D = 0 If we take the curl of Faraday's law: There is a vector identity so that H = jωɛe (.) E = jωµ H (.) = jωµ(jωɛe) = ω µɛe (.3) E = ( E) E (.4) ( E) E = ω µɛe (.5) From Gauss' law we get D = 0 since ρ v = 0. Since D = ɛe we get (ɛe ) = 0. If ɛ is independent of position then we can pull it outside of the spatial derivatives resulting in ɛ ( E ) = 0 and thus E = 0. (.6) Plugging Eq..6 into Eq..5 and rearranging results in the Homogeneous Wave Equation given by E + ω µɛe = 0 (.7).3 Dielectric Slab Waveguide Since the waveguide is homogeneous along the z axis, solutions to the wave equation can be taken as E(x, t) = E m (x) exp (j (ωt βz)) (.8) H(x, t) = H m (x) exp (j (ωt βz)). (.9) In time harmonic form the eld equations become E(x, t) = E m (x) exp ( jβz) (.30) H(x, t) = H m (x) exp ( jβz). (.3) Plugging the general eld solutions into the wave equation (Eq..7) results in x E + z E + k on i E = 0 (.3) x E + ( jβ) + kon i E = 0 (.33) x E + ( kon i β ) E = 0 (.34) n i is either n or n depending on which region we are dening the eld in. ECEn 56 4 January 7, 007

5 The portion in parenthesis is a constant in terms of x. The differential equation is a constant coefcient equation. For the elds in the core region ( x < d/) n i = n and the solution is given by or E m = Ae jhx + Be jhx, (.35) E m = A sin(hx) + Bcos(hx), (.36) h = k on β (.37) For the elds in the cladding region ( x > d/) n i = n and the solution is given by E m = Ae jgx + Be jgx, (.38) g = k on β. (.39) However, since β > k o n the argument of the square root is actually negative resulting in E m = Ae qx + Be qx, (.40) q = β k on. (.4) The total electric eld of the mode is given by A sin hx + B cos hx x < d E m (x) = C exp( qx) x > d D exp(qx) x < d (.4) The unknowns are A, B, C, D, q, and h. The solution of the unknows requires applying the boundary conditions. Since the boundary conditions depend on the vector quantities, we will break up the mode into two orthogonal polarization cases. The directions of both the electric and magnetic elds need to be perpendicular to the rays shown in Fig..4. One possible solution is to have the electric eld in the ŷ-direction. In this case the electric eld is perpedicular to the direction of power ow (z-direction). This case is called Transverse Electric (TE). For TE-polarization the magnetic eld has both x and z components. The other case is when the magnetic eld is in the ŷ-direction. In this case the magnetic eld is perpedicular to the direction of power ow (z-direction). This case is called Transverse Magnetic (TM). For TM-polarization the magnetic eld has both x and z ECEn 56 5 January 7, 007

6 .3. TE Modes The electric eld for TE polarization is in the y-direction as given by (A sin hx + B cos hx) e jβz x < d E y (x) = C exp( qx jβz) x > d D exp(qx jβz) x < d. (.43) The magnetic eld is H = E jωµ (.44) resulting in H z (x) = j E y ωµ x. (.45) The boundary conditions are that the tangential components of both E and H are equal across a boundary. The tangential component of the electric eld at x = d/ is given by ( ) ( ) A sin hd + B cos hd = C exp ( ) and at x = d/ it is given by A sin ( ) ( ) hd + B cos hd = D exp ( ) (.46) (.47) The continuity of the tangential components of the magnetic magnetic eld essentially becomes continuity of the derivative of the electric eld across the boundary resulting in ( ) ( ) ha cos hd hb sin hd = qc exp ( ) (.48) at x = d/ and at x = d/. ( ) ( ) ha cos hd + hb sin hd = qd exp ( ) (.49) These four equations can be combined to produce ( ) A sin hd = (C D) exp ( ) ( ) ha cos hd = q (C D) exp ( ) ( ) B cos hd = (C + D) exp ( ) ( ) hb sin hd = q (C + D) exp ( ) (.50) (.5) (.5) (.53) ECEn 56 6 January 7, 007

7 The solutions of the TE modes may be divided into two classes: (a) Symmetric (A = 0 and C = D): ( ) h tan hd = q (.54) (b) Antisymmetric (B = 0 and C = D): h cot ( ) hd = q (.55) There are now four unknowns (A or B, C, h, and q). The rst term (A or B) can be thought of as the amplitude of the mode. Let call this term E o. The last two terms (h and q) are both related to β so they are actually only one unknown. Let's combine these two together as given by h + q = ( k on β ) + ( β k on β ) (.56) = k on k on (.57) and C is just the continutity of the electric eld at the boundary. Putting all of this together we get E y = E e { qx jβz } x > d sin hx E 0 e cos hx jβz x d { + } E e +qx jβz x < d (.58) ( E exp ) { sin hd = E o cos hd E = E o exp ( } ) { sin hd cos hd (.59) }. (.60) So now the only unknown is β. We determine β by solving these two equations h + q = ko ( n n ) (.6) ( ) ( ) hd hd h tan = q OR h cot = q (.6) We can solve these nonlinear transcendental equations using a nonlinear solver on a computer or calculator. However, they can also be solved graphically to calculate the number of modes and estimate the approximate solutions. Since the argument of the tan and cot is in terms of hd/ we will plot the term / along the x-axis and hd/ along the y-axis. The rst equations becomes ( ) hd ( ) + = ( (ko n d) (k o n d) ) (.63) = ( π ) (n λ d n ) V (.64) ECEn 56 7 January 7, 007

8 This is the equation of a circle with a radius of V as given by x + y = V. The boundary condition equation for the symmetric modes is ( ) hd h tan = q (.65) ( ) hd hd tan = (.66) which becomes x tan (x) = y. (.67) and for the antisymmetric modes it is ( ) hd h cot = q (.68) hd ( ) hd cot = (.69) which becomes x cot (x) = y. (.70) In summary the equations are h + q = ko ( n n ) x + y = V (.7) ( ) hd h tan = q x tan (x) = y (.7) ( ) hd h cot = q x cot (x) = y (.73) The zero crossing of the tan are 0, π,...mπ and the zeros of the cot are π, π, 3π,... π ( + m)..3. TM Modes We can repeat the whole process for TM modes. In this case, we have H y (x, z, t) = h m (x) exp (j (ωt βz)) (.74) E x (x, z, t) = j ωµ z H y (.75) E z (x, z, t) = j ωµ x H y (.76) and H m (x) = A sin hx + B cos hx x < d C exp( qx) x > d D exp(qx) x < d (.77) ECEn 56 8 January 7, 007

9 The eigen equations become ( h tan ( h cot hd ) hd ) = n n q (.78) = n n q (.79).3.3 Parameter Meanings What are the physical meanings of h, q, and β? If we look back at the ray optics treatment, then β is the z-component of the wave, h is the x-component, and q species the rate at which the eld decays with distance away from the core. β k z (.80) h k x (.8) q α (.8) Dielectric Waveguide Example How many modes exist in a dielectric waveguide that has the following parameters? index of refraction of the core n =.6, index of refraction of the cladding n =.5, wavelength λ =.0µm, waveguide core thickness d = 0µm. The equations are Using k y d = x and αd = y these equations become αd = k y d tan (k y d) (.83) αd = k y d cot (k y d) (.84) (k y d) + (αd) = (k o d) ( n n ) (.85) y = x tan x (.86) y = x cot x (.87) x + y = (k o d) ( n n ) (.88) For this example the radius of the circle is given by r = π 0 n.0 (.89) r = 7.5µm (.90) The equation x tan x is equal to zero when x = 0π, π, 3π,... mπ and is equal to when x = π, 3π, 5π,... π + mπ. The equation x cot x is equal to zero when x = π, 3π, 5π,... π + mπ and is equal to when x = π, π, 3π,... mπ. And when x = 0 x cot x =. ECEn 56 9 January 7, 007

10 The radius of the circle for this problem is r = 7.5 = 5.56π. There are 6 even modes (0, π, π, 3π, 4π, 5π ) and 6 odd modes (0.5π,.5π,.5π, 3.5π, 4.5π, 5.5π). What is the waveguide thickness for single mode operation? We need r < 0.5π (.9) π.0 d.6.5 < π (.9) d < (.93).4 Asymmtric Slab Waveguides In practice most slab waveguides are asymmetric. An asymmetric slab waveguide is given by n, x < 0 n(x) = n, t < x < 0 n 3, x < t (.94) Sometimes rather than using n, n, and n 3 these indices are labeled as cover index n c, waveguide index n w, and substrate index n s. If we assume that n < n 3 < n then the range for β is given by k o n 3 < β < k o n. The process used to calculate the mode eld prole is similar to the process describe above except that the boundary conditions will be different at the top and bottom boundary. For a TE mode the electric eld is given by E y (x, z, t) = E m (x)e j(ωt βz), (.95) the mode prole is given by C exp qx x > 0 E m (x) = C ( cos(hx) q h sin(hx)) t < x < 0 C ( cos(ht) + q h sin(ht)) exp[p(x + t)] x < t, (.96) h = k β (.97) q = β k (.98) p = β k3. (.99) The mode condition equation is given by h sin(ht) q cos(ht) = p (cos(ht) + q ) h sin(ht) (.00) For a TM mode the elds are given by E x (x, z, t) = H y (x, z, t) = H m (x)e j(ωt βz) (.0) i H y ωµ z = β ωµ h m(x)e j(ωt βz) (.0) E z (x, z, t) = j ω ɛ H y x (.03) ECEn 56 0 January 7, 007

11 the mode prole is given by C h q exp( qx) x > 0 ) H m (x) = C ( h q cos(hx) + sin(hx) t < x < 0 ( ) C h q cos(ht) + sin(ht) exp[p(x + t)] x < t, (.04) q n n q (.05) p n n p (.06) 3 The mode condition equation is given by tan(ht) = h( p + q) h p q (.07).5 Effective Index Theory A slab waveguide only connes light in one dimension. In practive it is necessary to conne light in both directions. Exact analytic treatment of rectangular dielectric waveguides is not possible for arbitrary structures. These type of waveguides can be analyzed using numerical techniques. There are also several approximate analytical approaches. One of the simplest approaches is the effective index theory. Figure.5 shows a ridge waveguide. The three regions of the ridge waveguide (I, II, I) are treated as slab waveguides resulting in three different effective indices (n eff,i, n eff,ii, and n eff,i ). Referring to Fig..5 n eff,i is calculated by solving for the mode of a slab waveguide with a thickness of d and for n eff,ii the waveguide thickness is t. The ridge waveguide effective index is then calculated by treating the effective indices as the cover, waveguide, and substrate indices with the waveguide thickness being the ridge width a. n n t d n 3 a I II I y=-a/ y=a/ Figure.5: Rectangular waveguide. Example: Consider a ridge waveguide made of GaAs (n = 3.5) waveguiding layer on an AlGaAs (n = 3.) substrate. The thicknesses are t = 0.4λ, d = 0.5λ, and a = 0.5λ. ECEn 56 January 7, 007