I'm trying to work out why a 0.7pF capacitor was chosen. The circuit operates at 12GHz.

At 12GHz, the capacitor has an impedance of about \$18\Omega\$. Also at 12GHz, TL3 is \$0.259\lambda\$ (which is roughly a quarter of a wavelength. I worked out the characterstic impedance of TL3 to be \$74\Omega\$.

I thought that C1 was meant to be a DC blocking capacitor and prevent the 12GHz signal from going into the DC supply; however, I don't understand why an \$18\Omega\$ capacitor was chosen, as a quarter wavelength piece at \$74\Omega\$ does not transform \$50\Omega\$ to \$18\Omega\$...


Could someone point me in the right direction?



The whole point of the DC bias network (TL3 and C1) is to present a very high impedance at the junction of the three TLs, in order to perturb the RF signal flowing through TL1 and TL4 as little as possible.

At these frequencies, the parasitic effects of every component must be considered. For example, even a tiny SMT capacitor has a certain amount of parasitic inductance in series with it that creates a self-resonant frequency at which its impedance is lowest. Above this frequency, the inductive reactance dominates and the impedance starts rising again.

My guess would be that the capacitor chosen has a self-resonant frequency at or above 12 GHz, presenting the minimum possible impedance at the node between C1 and TL3. This gets transformed by TL3 into the highest possible impedance at the junction of the three TLs.

  • \$\begingroup\$ Curiously, could TL3 have been chosen so that the very low impedance of the capacitor at resonance be presented at the TL1-TL2-TL3 junction? (So, say, the length be made to be either \$\lambda/2\$ or \$\lambda\$?) Would this have the same affect? (Because the RF signal would reflect off the low impedance like it does with the high impedance?) \$\endgroup\$ – user968243 Apr 8 '14 at 6:54
  • \$\begingroup\$ @user968243: No. There are three connections at that junction, and two of them are 50 ohms. We want the third one to be very high impedance precisely so that it does NOT affect the RF. As far as the RF is concerned, that third connection doesn't exist. \$\endgroup\$ – Dave Tweed Apr 8 '14 at 12:22
  • \$\begingroup\$ Thanks for the reply. I thought that a short circuit would not affect the RF signal because I'm thinking that the reflection coefficient at that point for an open circuit has the same magnitude as that of a short circuit. \$\endgroup\$ – user968243 Apr 8 '14 at 13:03

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