7.7.7 Losses of the microstrip line

Conductor loss

At microwave frequencies the current flows in a skin of depth ${\delta }_s=1/\sqrt{\pi f{\mu }_0{\sigma }_c}$ on the conductors, which have the surface resistance $R_s=1/({\sigma }_c{\delta }_s)=\sqrt{\pi f{\mu }_0/{\sigma }_c}$ (Section 6.2). The current crowds on the underside of the strip and on the ground directly beneath it. As a first estimate let the current $I$ spread evenly over the width $w$ on both, with the surface current density $I/w$. The power lost per unit length is then $2\times \frac{1}{2}{R}_s|I/w{|}^{2}w=R_s|I{|}^2/w$, while the line carries $P=\frac{1}{2}{Z}_0|I{|}^2$. Since the power decays as ${\mathrm{e}}^{-2\alpha z}$, $\alpha =P_{\mathrm{loss}}/(2P)$:

Dielectric loss

Only the field in the substrate is lossy. For a line completely filled with a dielectric of loss tangent $\tan \delta$ the attenuation is ${\alpha }_d=\frac{1}{2}\beta \tan \delta$. In the microstrip the same holds with only the part $q$ of the electric energy that is stored in the substrate:

Radiation

Bends, open ends and steps in width radiate, and the radiated power grows roughly as $(h/{\lambda }_0{)}^2$: thin substrates radiate less.

Figure 7.7.6 shows both losses for the $50\Omega$ FR-4 line. At $2.4\mathrm{GHz}$, ${\alpha }_c=0.72dB/m$ and ${\alpha }_d=7.2dB/m$. A low-loss laminate (RO4003C, $\tan \delta =0.0027$, $h=0.508\mathrm{mm}$, $w=1.14\mathrm{mm}$) loses $1.9+0.9=2.8dB/m$: less dielectric loss, but more copper loss on the narrower strip. Rough copper raises ${\alpha }_c$ further in practice.