Question 6.2: A HALF-WAVE MICROSTRIP RESONATOR Consider a microstrip reson...
A HALF-WAVE MICROSTRIP RESONATOR
Consider a microstrip resonator constructed from a λ/2 length of 50 open-circuited microstrip line. The substrate is Teflon \ \left( \epsilon _{r}=2.08,\tan \delta =0.0004\right), with a thickness of 0.159 cm, and the conductors are copper. Compute the required length of the line for resonance at 5 GHz, and the unloaded Q of the resonator. Ignore fringing fields at the end of the line.
From (3.197), the width of a 50 microstrip line on this substrate is found to be W = 0.508 cm, and the effective permittivity is \epsilon _e = 1.80. The resonant length can then be calculated as
\frac{W}{d} = \begin{cases} \frac{8e^A}{e^{2A}-2} & for W/d\lt 2 \\ \frac{2}{\pi } \left[B-1-\ln(2B-1)+\frac{\epsilon _r-1}{2\epsilon _r}\left\{\ln (B-1)+0.39-\frac{0.61}{\epsilon _r} \right\} \right] & for W/d\gt 2 \end{cases} (3.197)
\ l=\frac{\lambda }{2}=\frac{v_{p} }{2f}=\frac{c}{2f\sqrt{\epsilon _{e}} }=\frac{3\times 10^{8}}{2\left(5\times 10^{9}\right)\sqrt{1.80} }=2.24cm.The propagation constant is
\ \beta =\frac{2\pi f}{v_{p}} =\frac{2\pi f\sqrt{\epsilon _{e}} }{c} =\frac{2\pi \left(5\times 10^{9}\sqrt{1.80} \right) }{3\times 10^{8}} =151.0rad/m.From (3.199)
\ \alpha _{c}=\frac{R_{s}}{Z_{0}W} Np/m,
the attenuation due to conductor loss is
\ \alpha _{c}=\frac{R_{s}}{Z_{0}W}=\frac{1.84\times 10^{-2}}{50\left(0.00508\right)}=0.0724 Np/mwhere we used Rs from Example 6.1. From (3.198), the attenuation due to dielectric loss is
\ \alpha _{d}=\frac{k_{0}\epsilon _{r}\left(\epsilon _{e}-1\right)\tan \delta }{2\sqrt{\epsilon _{e}\left(\epsilon _{r}-1\right) } } =\frac{\left(104.7\right)\left(2.08\right)\left(0.80\right)\left(0.0004\right) }{2\sqrt{1.80}\left(1.08\right) }=0.024 Np/m.Then from (6.35) the unloaded Q is
\ Q_{0}=\frac{\beta }{2\alpha } =\frac{151.0}{2\left(0.0724+0.024\right) } =783.