1ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityLecture 23Multilayer StructuresIn this lecture you will learn:• Multilayer structures• Dielectric anti-reflection (AR) coatings• Dielectric high-reflection (HR) coatings• Photonic Band-Gap StructuresECE 303 – Fall 2005 – Farhan Rana – Cornell University1oZ2oZ1+V1−V2+V0=z()zkjzkjzeVeVzV11110+−−+<+=()zkjzeVzV220−+>=11121211+−==Γ+−ooooZZZZVVTransmission Line Junctions and Discontinuities - IBoundary conditions:211+−+=+ VVV221111oooZVZVZV+−+=−1) Continuity of voltage at z=0:1) Continuity of current at z=0:12121212+==++ooooZZZZVVTTransmission line discontinuities generate reflections2ECE 303 – Fall 2005 – Farhan Rana – Cornell University1oZ2oZ1+V1−V2+V0=z()zkjzkjzeVeVzV11110+−−+<+=Transmission Line Junctions and Discontinuities - IICan also replace the infinite transmission line on the right by a lumped impedance1oZ1+V1−V2oZ11121211+−==Γ+−ooooZZZZVVWhich gives:0=z()zkjzeVzV220−+>=+-2+V()()121101212121112+=Γ+==⇒Γ+=+===+++−++ooooZZZZVVTVVVzVVECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityUnmatched Transmission Lines - I1oZ3oZ2oZ0=zl−=z1+V1−V3+V2+V2−VQuestion: How does one solve a problem like this?()()()lll++−+−+−<+=zkjzkjzeVeVzV1111()zkjzeVzV330−+>=()zkjzkjzeVeVzV22220+−−+<<−+=l• In each segment (except the right most one), the wave is written such that the phase is zero at the right end of the segment• In each segment, the phase has the wavevector corresponding to that segment3ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityUnmatched Transmission Lines - II1oZ3oZ2oZ0=zl−=z1+V1−V3+V2+V2−VQuestion: How does one solve a problem like this?1oZ2oZ0=zl−=z1+V1−V2+V2−V3oZNow calculate the impedance Z(z=-ℓ ):11232322+−==Γ+−ooooZZZZVV()lll2222211kjkjoeeZzZ−−Γ−Γ+=−=STEP 2: Replace the middle line with the impedance Z(z=-ℓ )1oZ1+V1−V()l−=zZ()()111111+−=−−===Γ+−ooZzZZzZVVllSTEP 1: Replace the last line with a lumped equivalent impedance (corresponding to an infinite line)ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityQuestion: Is it possible to use a transmission line to perfectly match two dissimilar transmission lines so that there is no reflection?What is the appropriate impedance Zo2? What is the appropriate length ℓ ?Matching Transmission Lines - I0=zl−=zUse a Quarter-Wave Transformer:Suppose the length ℓ of the intermediate transmission line is quarter-wavelength 2422πλ=⇒= ll k()()()()01402232==⎟⎠⎞⎜⎝⎛−=⇒==⇒=zZzZZZzZZzZzZnnoononλA quarter-wavelength long transmission line inverts the normalized impedance1oZ3oZ2oZ1+V3+V2+V2−V4ECE 303 – Fall 2005 – Farhan Rana – Cornell University()()322230140oonnoonZZzZzZZZzZ===⎟⎠⎞⎜⎝⎛−===λThe actual impedance at z = -λ2/4 is then:32222244oonoZZzZZzZ =⎟⎠⎞⎜⎝⎛−==⎟⎠⎞⎜⎝⎛−=λλTo have no reflection we need:124oZzZ =⎟⎠⎞⎜⎝⎛−=λ31231221322oooooooooZZZZZZZZZ=⇒=⇒=⇒The impedance of the quarter-wavelength long transmission line must be the geometric mean of the impedances of the two transmission linesMatching Transmission Lines - II0=z42λ−=z1oZ3oZ2oZ1+V3+V2+V2−VECE 303 – Fall 2005 – Farhan Rana – Cornell University1oZ2oZ1+V1−V2+V0=z()zkjzkjzeVeVzV11110+−−+<+=()zkjzeVzV220−+>=11121211+−==Γ+−ooooZZZZVVWaves at Interfaces and Transmission Lineszkkiˆ−=rErHrzkkiˆ=rEiHizkktˆ=rEtHtioiηµεtotηµε()zkjrzkjiziieExeExzE+−<+=ˆˆ0rBoundary conditions:()2111+−+=+⇒ VVV()2211112oooZVZVZV+−+=−⇒()zkjtzteExzE−>=ˆ0rBoundary conditions:()triEEE=+⇒1()ttiriiEEEηηη=−⇒211+−==ΓititirEEηηηη0=z5ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityTri-layer Structure - IQuestion: How do we calculate the reflection coefficient for the above structure?1oZ3oZ2oZ0=zl−=z1+V1−V3+V2+V2−VE+1E+31η3ηE-10=zl−=zE+2E-22ηAnswer: Use the method employed earlier in the equivalent transmission line problem ?11==Γ+−EEECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityTri-layer Structure - IIE+1E+31η3ηE-10=zl−=zE+2E-22η()()()lll++−+−+−<+=zkjzkjzxeEeEzE1111()zkjzxeEzE330−+>=()zkjzkjzxeEeEzE22220+−−+<<−+=l• In each segment (except the right most one), the wave is written such that the phase is zero at the right end of the segment• In each segment, the phase has the wavevector corresponding to that segment6ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityTri-layer Structure - IIIE+1E+31η3ηE-10=zl−=zE+2E-22ηSTEP 1: Calculate the reflection at z = 0:11232322+−==Γ+−ηηηηEEAnd calculate the effective impedance at z=-ℓ :()lll2222211kjkjeez−−Γ−Γ+=−=ηηSTEP 2: Now the problem becomes:E+11ηE-1()l−=zηl−=zCalculate the reflection coefficient as:()()111111+−=−−===Γ+−ηηηηhlzzEEECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityDielectric Anti-Reflection (AR) Coatings - IQuestion: Is it possible to somehow make the reflection coefficient zero?Answer: Use the quarter-wave transformer concept:E+1E+31η3ηE-10=zE+1E+31η3η0=z42λ−=−= lzE+2E-22η7ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityDielectric Anti-Reflection (AR) Coatings - IIE+1E+31η3η0=z42λ−=−= lzE+2E-22ηSTEP 1: Calculate the normalized impedance at z = 0 and at z = -λ2 / 4STEP 2: Now the problem becomes:E+11η()l−=zηTo have zero reflection ⇒Γ= 0()3123222214ηηηηηληηηη=⇒=⎟⎠⎞⎜⎝⎛−==−==⇒ zznl()()()()3222320140ηηηληηηηηηη===⎟⎠⎞⎜⎝⎛−=⇒==⇒=zzzzznnnnA quarter-wavelength long segment inverts the normalized impedance42λ−=−= lzECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityDielectric Anti-Reflection (AR) Coatings - IIIQuestion: How do quarter-wavelength long matching layers work?0=z42λ−=z1η3η312ηηη=Wave reflected at the 1stinterfaceWave reflected at the 2ndinterfaceWaves reflected at the two interfaces cancel each other out in the backward direction8ECE 303 – Fall 2005 – Farhan Rana – Cornell UniversityFrequency Dependence of Dielectric AR Coatings: ExampleE+oE+3oη5.33oηη=0=zE+2E-25.332ooηηηη==l−=zSiliconConsider the AR coating on a Silicon photodetector that is designed for detecting green light which has a frequency ωof 3.625x1015rad/sec and a wavelength of 0.52 micro-meter in free-spaceQuestion: How good will the AR coating work for frequencies