VIEWS: 7 PAGES: 11 POSTED ON: 11/25/2011 Public Domain
Chapter 1 Dielectric Slab Waveguide We will start off examining the waveguide properties of a slab of dielectric shown in Fig. 1.1. x n1 z d n2 n1 Figure 1.1: Cross-sectional view of a slab waveguide. n2 , |x| < d/2 n(x) = (1.1) n1 , else 1.1 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 1.2 shows that if the propagation angle is greater than the critical angle then the ray will bounce off of the surface and will be con ned to the core region. Therefore, the propagation is con ned to be n2 θ1 > θc = sin−1 . (1.2) n1 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 θa . sin θa = n2 sin (90 − θ1 ) (1.3) = n2 cos (θ1 ) (1.4) ECEn 562 1 January 17, 2007 cladding n1 θ1>θc θ1<θc θ1=θc core n2 cladding n1 Figure 1.2: Cross-sectional view of a slab waveguide. Since θ1 > θc sin θa < n2 cos θc (1.5) < n2 1 − sin2 θc (1.6) 2 n1 < n2 1− (1.7) n2 n2 − n2 2 1 < n2 (1.8) n2 2 sin θa < n2 − n2 ≡ N A 2 1 (1.9) n=1.0 n1 θ1=θc θa 90−θc n2 n1 Figure 1.3: 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. 1.4. 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−jko n(± cos θ1 x+sin θ1 z) . (1.10) 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)) , (1.11) ECEn 562 2 January 17, 2007 n1 λ n2 d θθ n1 Figure 1.4: The rays must remain in phase after multiple re ections. where β is called the propagation constant and is given by β = ko n2 sin θ1 . (1.12) Since the propagation angle is in the range given by θc < θ1 < 90◦ , (1.13) the propagation angle is in the range given by ko n2 sin θc < β < ko n2 sin (90◦ ) (1.14) n1 ko n2 < β < ko n2 (1.15) n2 ko n1 < β < ko n2 (1.16) If you divide the propagation angle by the free-space wavevector you get the effective index of the mode as given by β nef f ≡ . (1.17) ko n1 < nef f < n2 (1.18) 1.2 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 βm and the transverse mode eld Em (x). We start with Maxwell's equations in the sinusoidal steady state. × E = −jωB = −jωµH ·D = · E = ρv (1.19) × H = jωD + J = jω E + J ·B = · µH = 0 (1.20) ECEn 562 3 January 17, 2007 First, we rewrite Ampere's Law for the case of no sources resulting in × H = jω E (1.21) Likewise, if we have no free charges ρv = 0 and thus ·D =0 If we take the curl of Faraday's law: × × E = −jωµ ×H (1.22) 2 = −jωµ(jω E) = ω µ E (1.23) There is a vector identity 2 × ×E = ( · E) − E (1.24) so that 2 ( · E) − E = ω2µ E (1.25) From Gauss' law we get · D = 0 since ρv = 0. Since D= E we get · E = 0. If is independent of position then we can pull it outside of the spatial derivatives resulting in · E = 0 and thus · E = 0. (1.26) Plugging Eq. 1.26 into Eq. 1.25 and rearranging results in the Homogeneous Wave Equation given by 2 E + ω2µ E = 0 (1.27) 1.3 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)) (1.28) H(x, t) = H m (x) exp (j (ωt − βz)) . (1.29) In time harmonic form the eld equations become E(x, t) = E m (x) exp (−jβz) (1.30) H(x, t) = H m (x) exp (−jβz) . (1.31) Plugging the general eld solutions into the wave equation (Eq. 1.27) results in ∂2 ∂2 2 E + 2 E + ko n2 E = 0 i (1.32) ∂x2 ∂z ∂2 E + (−jβ)2 + ko n2 E = 0 2 i (1.33) ∂x2 ∂2 2 E + ko n2 − β 2 E = 0 i (1.34) ∂x2 where ni is either n1 or n2 depending on which region we are de ning the eld in. ECEn 562 4 January 17, 2007 The portion in parenthesis is a constant in terms of x. The differential equation is a constant coef cient equation. For the elds in the core region (|x| < d/2) ni = n2 and the solution is given by Em = Aejhx + Be−jhx , (1.35) or Em = A sin(hx) + Bcos(hx), (1.36) where h= ko n2 − β 2 2 2 (1.37) For the elds in the cladding region (|x| > d/2) ni = n1 and the solution is given by Em = Aejgx + Be−jgx , (1.38) where g= ko n2 − β 2 . 2 1 (1.39) However, since β > ko n1 the argument of the square root is actually negative resulting in Em = Aeqx + Be−qx , (1.40) where q= β 2 − ko n 2 . 2 1 (1.41) The total electric eld of the mode is given by d A sin hx + B cos hx |x| < 2 d Em (x) = C exp(−qx) x> 2 (1.42) D exp(qx) x < −d2 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. 1.4. One possible solution is to have the electric eld in the ˆ y -direction. In this case the electric eld is per- pedicular 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 ˆ y -direction. In this case the magnetic eld is per- pedicular 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 562 5 January 17, 2007 1.3.1 TE Modes The electric eld for TE polarization is in the y-direction as given by d (A sin hx + B cos hx) e−jβz |x| < 2 Ey (x) = C exp(−qx − jβz) x> d . 2 (1.43) D exp(qx − jβz) x < −d 2 The magnetic eld is ×E H= (1.44) −jωµ resulting in j ∂Ey Hz (x) = . (1.45) ωµ ∂x 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/2 is given by 1 1 1 A sin hd + B cos hd = C exp − qd (1.46) 2 2 2 and at x = −d/2 it is given by 1 1 1 −A sin hd + B cos hd = D exp − qd (1.47) 2 2 2 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 1 1 1 hA cos hd − hB sin hd = −qC exp − qd (1.48) 2 2 2 at x = d/2 and 1 1 1 hA cos hd + hB sin hd = qD exp − qd (1.49) 2 2 2 at x = −d/2. These four equations can be combined to produce 1 1 2A sin = (C − D) exp − qd hd (1.50) 2 2 1 1 2hA cos hd = −q (C − D) exp − qd (1.51) 2 2 1 1 2B cos hd = (C + D) exp − qd (1.52) 2 2 1 1 2hB sin hd = q (C + D) exp − qd (1.53) 2 2 ECEn 562 6 January 17, 2007 The solutions of the TE modes may be divided into two classes: (a) Symmetric (A = 0 and C = D): 1 h tan hd =q (1.54) 2 (b) Antisymmetric (B = 0 and C = −D): 1 h cot hd = −q (1.55) 2 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 Eo . 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 h2 + q 2 = 2 2 ko n2 − β 2 + β 2 − ko n2 − β 2 2 1 (1.56) = 2 ko n2 2 − ko n2 2 1 (1.57) and C is just the continutity of the electric eld at the boundary. Putting all of this together we get d E1 e−qx−jβz x> 2 sin hx E0 e−jβz |x| ≤ d Ey = cos hx (1.58) − E1 e+qx−jβz x < −d 2 + where qd sin hd 2 E1 exp − = Eo (1.59) 2 cos hd 2 qd sin hd 2 E1 = Eo exp . (1.60) 2 cos hd 2 So now the only unknown is β. We determine β by solving these two equations 2 h2 + q 2 = ko n2 − n2 2 1 (1.61) hd hd h tan = q OR − h cot =q (1.62) 2 2 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/2 we will plot the term qd/2 along the x-axis and hd/2 along the y -axis. The rst equations becomes 2 2 hd qd 1 + = (ko n2 d)2 − (ko n1 d)2 (1.63) 2 2 22 π 2 2 = d n2 − n2 ≡ V 2 1 (1.64) λ ECEn 562 7 January 17, 2007 This is the equation of a circle with a radius of V as given by x2 + y 2 = V 2 . The boundary condition equation for the symmetric modes is hd h tan =q (1.65) 2 hd hd qd tan = (1.66) 2 2 2 which becomes x tan (x) = y. (1.67) and for the antisymmetric modes it is hd −h cot =q (1.68) 2 hd hd qd − cot = (1.69) 2 2 2 which becomes −x cot (x) = y. (1.70) In summary the equations are 2 h2 + q 2 = ko n2 − n2 2 1 ⇒ x2 + y 2 = V 2 (1.71) hd h tan =q ⇒ x tan (x) = y (1.72) 2 hd −h cot =q ⇒ −x cot (x) = y (1.73) 2 The zero crossing of the tan are 0, π, ...mπ and the zeros of the cot are π , π , 3π , ... π (1 + 2m). 2 2 2 2 1.3.2 TM Modes We can repeat the whole process for TM modes. In this case, we have Hy (x, z, t) = hm (x) exp (j (ωt − βz)) (1.74) j ∂ Ex (x, z, t) = Hy (1.75) ωµ ∂z j ∂ Ez (x, z, t) = − Hy (1.76) ωµ ∂x and d A sin hx + B cos hx |x| < 2 Hm (x) = C exp(−qx) x> d2 (1.77) D exp(qx) x < −d 2 ECEn 562 8 January 17, 2007 The eigen equations become 1 n2 2 h tan hd = q (1.78) 2 n2 1 1 n2 h cot hd = − 2q (1.79) 2 n2 1 1.3.3 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 speci es the rate at which the eld decays with distance away from the core. β ≡ kz (1.80) h ≡ kx (1.81) q≡α (1.82) Dielectric Waveguide Example How many modes exist in a dielectric waveguide that has the following parameters? index of refraction of the coren1 = 1.6, index of refraction of the cladding n2 = 1.5, wavelength λ = 1.0µm, waveguide core thickness 2d = 10µm. The equations are αd = ky d tan (ky d) (1.83) αd = −ky d cot (ky d) (1.84) 2 2 (ky d) + (αd) = (ko d) n2 1 − n2 2 (1.85) Using ky d = x and αd = y these equations become y = x tan x (1.86) y = −x cot x (1.87) 2 2 2 x + y = (ko d) n2 1 − n2 2 (1.88) For this example the radius of the circle is given by 2π 10 r = n2 − n2 1 2 (1.89) 1.0 2 r = 17.5µm (1.90) π 3π 5π The equation x tan x is equal to zero when x = 0π , 2π , 3π , ... mπ and is equal to ∞ when x = 2, 2 , 2 , π ... 2 + mπ . π 3π 5π π The equation −x cot x is equal to zero when x = 2 , 2 , 2 , ... 2 + mπ and is equal to ∞ when x = π , 2π , 3π , ... mπ . And when x = 0 −x cot x = −1. ECEn 562 9 January 17, 2007 The radius of the circle for this problem is r = 17.5 = 5.56π . There are 6 even modes (0, π , 2π , 3π , 4π , 5π ) and 6 odd modes (0.5π , 1.5π , 2.5π , 3.5π , 4.5π , 5.5π ). What is the waveguide thickness for single mode operation? We need r < 0.5π (1.91) 2π π d 1.62 − 1.52 < (1.92) 1.0 2 d < 0.449 (1.93) 1.4 Asymmtric Slab Waveguides In practice most slab waveguides are asymmetric. An asymmetric slab waveguide is given by n1 , x<0 n(x) = n2 , −t < x < 0 (1.94) n3 , x < −t Sometimes rather than using n1 , n2 , and n3 these indices are labeled as cover index nc , waveguide index nw , and substrate index ns . If we assume that n1 < n3 < n2 then the range for β is given by ko n3 < β < ko n2 . The process used to calculate the mode eld pro le 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 Ey (x, z, t) = Em (x)ej(ωt−βz) , (1.95) where the mode pro le is given by C exp −qx x>0 q Em (x) = C cos(hx) − h sin(hx) −t < x < 0 , (1.96) q C cos(ht) + h sin(ht) exp[p(x + t)] x < −t where h= 2 k2 − β 2 (1.97) q= 2 β 2 − k1 (1.98) p= 2 β 2 − k3 . (1.99) The mode condition equation is given by q h sin(ht) − q cos(ht) = p cos(ht) + sin(ht) (1.100) h For a TM mode the elds are given by Hy (x, z, t) = Hm (x)ej(ωt−βz) (1.101) i ∂Hy β Ex (x, z, t) = = hm (x)ej(ωt−βz) (1.102) ωµ ∂z ωµ j ∂Hy Ez (x, z, t) = − (1.103) ω ∂x ECEn 562 10 January 17, 2007 where the mode pro le is given by −C h exp(−qx) q ¯ x>0 h Hm (x) = C − q cos(hx) + sin(hx) ¯ −t < x < 0 , (1.104) −C h cos(ht) + sin(ht) exp[p(x + t)] x < −t ¯ q where n2 2 ¯ q≡ q (1.105) n2 1 n2 2 p ≡ 2p ¯ (1.106) n3 The mode condition equation is given by p ¯ h(¯ + q ) tan(ht) = 2 − pq (1.107) h ¯¯ 1.5 Effective Index Theory A slab waveguide only con nes light in one dimension. In practive it is necessary to con ne light in both di- rections. 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 1.5 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 (nef f,I , nef f,II , and nef f,I ). Referring to Fig. 1.5 nef f,I is calculated by solving for the mode of a slab waveguide with a thickness of d and for nef f,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. n1 t n2 d a n3 I II I y=-a/2 y=a/2 Figure 1.5: Rectangular waveguide. Example: Consider a ridge waveguide made of GaAs (n = 3.5) waveguiding layer on an AlGaAs (n = 3.2) substrate. The thicknesses are t = 0.4λ, d = 0.25λ, and a = 0.5λ. ECEn 562 11 January 17, 2007