wilkinson divider circle length in HFSS
时间：03-29
整理：3721RD
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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
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
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.
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.
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.