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 (εr = 2.08, tan δ = 0.0004), with a thickness of 0.159 cm, and the conductors are copper. Compute the required length of the line for

Question

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 [latex]\ \left( \epsilon _{r}=2.08, an \delta =0.0004 ight)[/latex], 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.

Accepted Answer

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 [latex]\epsilon _e = 1.80[/latex]. The resonant length can then be calculated as

[latex]\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} ight\} ight] & for W/d\gt 2 \end{cases} [/latex] (3.197)

[latex]\ l=\frac{\lambda }{2}=\frac{v_{p} }{2f}=\frac{c}{2f\sqrt{\epsilon _{e}} }=\frac{3 imes 10^{8}}{2\left(5 imes 10^{9} ight)\sqrt{1.80} }=2.24cm.[/latex]

The propagation constant is

[latex]\ \beta =\frac{2\pi f}{v_{p}} =\frac{2\pi f\sqrt{\epsilon _{e}} }{c} =\frac{2\pi \left(5 imes 10^{9}\sqrt{1.80} ight) }{3 imes 10^{8}} =151.0rad/m.[/latex]

From (3.199)

[latex]\ \alpha _{c}=\frac{R_{s}}{Z_{0}W} Np/m[/latex],

the attenuation due to conductor loss is

[latex]\ \alpha _{c}=\frac{R_{s}}{Z_{0}W}=\frac{1.84 imes 10^{-2}}{50\left(0.00508 ight)}=0.0724 Np/m[/latex]

where we used Rs from Example 6.1. From (3.198), the attenuation due to dielectric loss is

[latex]\ \alpha _{d}=\frac{k_{0}\epsilon _{r}\left(\epsilon _{e}-1 ight) an \delta }{2\sqrt{\epsilon _{e}\left(\epsilon _{r}-1 ight) } } =\frac{\left(104.7 ight)\left(2.08 ight)\left(0.80 ight)\left(0.0004 ight) }{2\sqrt{1.80}\left(1.08 ight) }=0.024 Np/m.[/latex]

Then from (6.35) the unloaded Q is

[latex]\ Q_{0}=\frac{\beta }{2\alpha } =\frac{151.0}{2\left(0.0724+0.024 ight) } =783[/latex].

Question 6.2: A HALF-WAVE MICROSTRIP RESONATOR Consider a microstrip reson...

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