wilkinson divider circle length in HFSS

[Sumer19]

Hello everybody
I am designing a wilkinson divider in HFSS.

In HFSS how can we make /edit incomplete circle of width 70.7ohm (say 1.846mm for 5.2 GHZ)
and incomplete circumference of half lambda (λ/2) for inner and outer circumference.
Can it be drawn by using equation by curve.
Please help if you can.

Thank you

 

[PlanarMetamaterials]

I would draw a circle, subtract it from a smaller one for form a ring. You could then remove a small portion of this to allow space for the resistor.

 

[Sumer19]

sir
how to remove a small portion in HFSS and what is measurement of this small portion.
Will circumference of your ring be half lambda.
Resistor will be as lumped element in HFSS.

 

[PlanarMetamaterials]

You can remove portions of elements using the subtract operator.

Yes, for any Wilkinson, the total path length of the arms should be about half a wavelength.