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How to divide the output of PLL with RF to 5 GHz into two output channels?


     I wonder how to divide the output of PLL with RF to 5 GHz into two output channels effectively. As the picture shows, the recommanded circuit of ADF4113 uses three 18 Ohm resistors to do that. I wonder why 18 Ohm, why not 50 Ohm or other.

Thank you!

  • Hi,

    It would be possible to omit the resistors and instead use a simple "T" splitter. The issue with this is poor isolation and poor impedance matching. 

    The network you have highlighted is a three way resistive power divider, also known as Wye splitter. For a three port splitter, the resistor value is calculated by the characteristic impedance (50 Ohms in this case) divided by 3 = 16.6667, 18Ohm is a reasonable approximation. The two output ports will have ~6dB of loss, 3dB from splitting the signal in two and roughly 3dB loss from the resistances. In this case the additional 3dB attenuation is a benefit as it generally ensures the VCO signal is within sensitivity limits of the RFIN port of the PLL.

    Wye splitters are widely used, you can research online to find the derivation of the resistor value calculation, especially for slitter with more than 2 output ports - There is a good article on on resistive power splitters.