Question 3.5: STRIPLINE DESIGN Find the width for a 50 Ω copper stripline ...
STRIPLINE DESIGN
Find the width for a 50 Ω copper stripline conductor with b = 0.32 cm and \epsilon _r = 2.20. If the dielectric loss tangent is 0.001 and the operating frequency is 10 GHz, calculate the attenuation in dB/λ. Assume a conductor thickness of t = 0.01 mm.
Because \ \sqrt{\epsilon _{r}} Z_{0}=\sqrt{2.2} \left(50\right) =74.2 \lt 120 and x=30\pi /\left(\sqrt{\epsilon _{r}}Z_{0} \right) -\left(0.441\right) =0.830 , (3.180) gives the strip width as W = bx = (0.32)(0.830) = 0.266 cm.
\frac{W}{b} = \begin{cases}x & for \sqrt{\epsilon _{r}}Z_{0}\lt 120\Omega \\ 0.85-\sqrt{0.6-x} & for \sqrt{\epsilon _{r}}Z_{0} \gt 120\Omega \end{cases} (3.180a)
x=\frac{30\pi }{\sqrt{\epsilon _{r} }Z_0}-0.441. (3.180b)
At 10 GHz, the wave number is
\ k=\frac{2\pi f\sqrt{\epsilon _{r}} }{c} =310.6m^{-1}.
From (3.30) the dielectric attenuation is
\ \alpha _{d}=\frac{k\tan \delta }{2}=\frac{\left(310.6\right)\left(0.001\right) }{2}=0.155Np/m.
The surface resistance of copper at 10 GHz is\ R_{s}=0.026\Omega . Then from (3.181)
\alpha _{c}= \begin{cases} \frac{2.7\times 10^{-3}R_{s}\epsilon _{r}Z_{0}}{30\pi \left(b-t\right) }A & for \sqrt{\epsilon _{r}}Z_{0}\lt 120\Omega \\ \frac{0.16R_{s}}{Z_{0}b}B & for \sqrt{\epsilon _{r}}Z_{0} \gt 120\Omega \end{cases} Np/m (3.181)
the conductor attenuation is
\ \alpha _{c}= \frac{2.7\times 10^{-3}R_{s}\epsilon _{r}Z_{0}A}{30\pi \left(b-t\right)}=0.1222Np/msince A = 4.74. The total attenuation constant is
\ \alpha=\alpha_{d}+\alpha_{c}=0.277Np/m.
In dB,
At 10 GHz, the wavelength on the stripline is
\ \lambda =\frac{c}{\sqrt{\epsilon _{r}f} } =2.02cm,so in terms of wavelength the attenuation is
\ \alpha\left(dB\right)=\left(2.41\right)\left(0.0202\right)=0.049dB/\lambda