Exercise 7.7.3: The Q of a half-wave microstrip resonator

$k_0=2\pi f/c=121.5rad/m$, ${\lambda }_0=51.7\mathrm{mm}$, $R_s=\sqrt{\pi f{\mu }_0/{\sigma }_c}=19.87\mathrm{m\Omega }$.

RO4003C: ${\epsilon }_{\mathrm{eff}}=2.780$, $Z_0=50.25\Omega$, $w=1.136\mathrm{mm}$, $\beta =121.5\times 1.667=202.6rad/m$, ${\lambda }_g=31.02\mathrm{mm}$ and $\ell ={\lambda }_g/2=15.51\mathrm{mm}$.

$q=\frac{1-1/2.780}{1-1/3.55}=0.8915$ and ${\alpha }_d=q\beta \tan \delta /2=0.8915\times 202.6\times 0.0027/2=0.244Np/m=2.12dB/m$.

a $3\mathrm{dB}$ bandwidth of $f/Q_u=34\mathrm{MHz}$. Since ${\alpha }_d=q\beta \tan \delta /2$, $Q_d=\beta /(2{\alpha }_d)=1/(q\tan \delta )=1/(0.8915\times 0.0027)=415$: it depends neither on $f$ nor on the length.

FR-4 ($w=3.06\mathrm{mm}$, ${\epsilon }_{\mathrm{eff}}=3.330$): $\beta =221.7rad/m$, $\ell =14.17\mathrm{mm}$, ${\alpha }_c=0.129Np/m$ ($1.12dB/m$), $q=0.9055$, ${\alpha }_d=2.007Np/m$ ($17.4dB/m$), so

On RO4003C the copper loss dominates (thin substrate, narrow strip), on FR-4 the dielectric loss dominates by far.

$Q_d=1/(q\tan \delta )$ for FR-4: $1/(0.9055\times 0.02)=55.2$. The thicker FR-4 board has the wider strip ($3.06$ against $1.14\mathrm{mm}$), hence the smaller current density and the higher $Q_c$.