Question 6.1: Q OF HALF-WAVE COAXIAL LINE RESONATORS A λ/2 resonator is ma...
Q OF HALF-WAVE COAXIAL LINE RESONATORS
A λ/2 resonator is made from a piece of copper coaxial line having an inner conductor radius of 1 mm and an outer conductor radius of 4 mm. If the resonant frequency is 5 GHz, compare the unloaded Q of an air-filled coaxial line resonator to that of a Teflon-filled coaxial line resonator.
We first compute the attenuation of the coaxial line, using the results of Examples 2.6 or 2.7. From Appendix F, the conductivity of copper is σ= 5.813 × 107 S/m. The surface resistivity at 5 GHz is
\ R_{s}=\sqrt{\frac{\omega \mu _{0}}{2\sigma } } =1.84\times 10^{-2}\Omega .The attenuation due to conductor loss for the air-filled line is
\ \alpha _{c}=\frac{R_{s}}{2\eta \ln b/a} \left(\frac{1}{a}+\frac{1}{b} \right)
\ =\frac{1.84\times 10^{-2}}{2\left(377\right)\ln \left(0.004/0.001\right) } \left(\frac{1}{0.001} +\frac{1}{0.004} \right) =0.022Np/m.
For Teflon,\ \epsilon _{r}=2.08 and\ \tan\delta =0.0004, so the attenuation due to conductor loss for the Teflon-filled line is
The dielectric loss of the air-filled line is zero, but the dielectric loss of the Teflonfilled line is
\ \alpha _{d}=k_{0}\frac{\sqrt{\epsilon _{r}} }{2}\tan \delta
\ =\frac{\left(104.7\right)\sqrt{2.08} \left(0.0004\right) }{2} =0.030Np/m.
Finally, from (6.27)\ Q_{0}=\frac{\omega _{0}L}{R} =\frac{\pi }{2\alpha \ell} =\frac{\beta }{2\alpha } , the unloaded Qs can be computed as
\ Q_{air}=\frac{\beta }{2\alpha } =\frac{104.7}{2\left(0.022\right) } =2380,\ Q_{Teflon}=\frac{\beta }{2\alpha } =\frac{104.7\sqrt{2.08} }{2\left(0.032+0.030\right) } =1218.
Thus it is seen that the Q of the air-filled line is almost twice that of the Teflonfilled line. The Q can be further increased by using silver-plated conductors.