Design, Fabrication and Characterization of a Microwave Resonator ...

2.1 The Coplanar Transmission LinewithOne can calculate the effective permittivityε e f f = 1 + q 1 (ε r1 − 1) + q 2 (ε r2 − ε r1 )q 1 = 1 [K(k 1diel ) K(k)2 K(k1diel ′ ) K(k ′ ) + K(k ] −11)K(k1 ′ ) ,q 2 = 1 [K(k 2diel ) K(k)2 K(k2diel ′ ) K(k ′ ) + K(k ] −11)K(k1 ′ ) .The impedance here is equal to the one of a single layer substrate given by (2.10).It can be shown that the second dielectric layer of SiO 2 with ε r2 = 4.7 influences theimpedance for values of H 2 larger than 1.7 µm [12].2.1.5 DispersionWith the help of the spectral domain method an empirical formula to compute ε e f f of a CPWis obtained in reference [18].√ε e f f ( f ) = √ ε e f f + (√ ε r − √ ε e f f )1 + aF −1.8 (2.14)whereF = f / f T E is the normalized frequencywith f T E = c 0 /(4H √ ε r − 1), the cutoff frequency for the lowest-order TE mode,ε e f f is the effective permittivity at the quasi-static limit.The parameteris calculated with( ) Slog(a) ≈ ulog + vWu ≈ 0.54 − 0.64q + 0.015q 2 ,v ≈ 0.43 − 0.86q + 0.540q 2 ,q = log S H .With an accuracy of 5 %, the formula is valid into the THz regime in the ranges given inTable 2.1.Calculations show that for our designs, explained in detail in chapter 3, all ranges, exceptfor the ratio of the inner strip to height of the substrate, are in the application range. Sinceno better formula for the designs with parameters which are out of range could be found,(2.14) was applied to these designs, too. The cutoff frequency f T E for Sapphire as substratematerial is 45.4 GHz, for Silicon as substrate material we evaluate 54.4 GHz.Figure 2.7 shows the effective dielectric constant for Silicon and Sapphire as substratematerials. We see that dispersion is negligible in the measurement range.9