Rectangular to circular waveguide converter loaded with dielectric material

[Soni Singh]

Hello to all,
Do you know how to decide transition length from rectangular to circular waveguide converter in dielectric loaded case in HFSS.Here in my case it is loaded with (titanium dioxide) and works in s band(2 -3 GHz). Any idea from the group.

 

[jiripolivka]

All depends upon device bandwidth. For a full-band of the rectangular waveguide, usual design calls for 3-4 wavelengths of the transition section. With a dielectric filling, it is possible to shorten it.
The problem is that then you need two more transitions from the air-filled waveguide to the dielectric-filled one, and back. Maybe someone invented a refined design.

 

[Soni Singh]

Thanks for your response...
I have to work at 2.45 GHz but I am unable to determine the length of transition section and diameter of circular waveguide for the case which i have written above.Could you please provide me any information regarding my problem

 

[vinoth8051]

Hi
Diameter of circular waveguide for 2.45 GHz would be 93.3mm(kindly refer Microwave Engineering by D. M. Pozar)
Initially the Transition length can be choosen such that, length is greater than twice the guided wavelength. Then transition length can be optimized to get better return VSWR.

 

[jiripolivka]

Thanks for your response...
I have to work at 2.45 GHz but I am unable to determine the length of transition section and diameter of circular waveguide for the case which i have written above.Could you please provide me any information regarding my problem

If you use air-filled coaxial line, you can adjust its length to exactly one-half or one-quarter wavelength. For dielectric-filled coaxial cables, you need to know the propagation velocity in it, or, the permittivity of that dielectric. Check manufacturer's data for it. There are significant differences for e.g. Teflon (k ~ 2.5), polyethylene (k ~ 2.2), polystyrene (k ~ 3.0), etc.
If you have no data, with some meters of cable to spare you can find the "correct" quarter wave as follows:

1. Short one cable end, sheath to center conductor, by soldering the metal through. With the other end open, cut approximately one-quarter wavelengh from the short.

2. At the open end, separate sheath from the center conductor , approx. 1-2 cm from each other.

3. Have a RF signal source ready, with ~ 1..10 mW (at e.g.; your 123 MHz) and connect a diode detector to its coaxial output. Detected voltage, ~ 30-300 mW, should be indicated by an analog voltmeter or mA meter.

4. If you touch the point where the detector is connected to the generator output, by your finger, the detector output should drop a bit. NOw, connect the open end of your shorted coax cable to the same point. The output will swing higher or lower than without anything touching the point.

5. Then cut an inch of the open end of the cable and separate the sheath from the center conductor. Touch the point again. Continue cutting and touching the sensitive point till there is no change at all in the detector output indication. You now have an exactly quarter-wavelength cable section!

No change means that your device - shorted section of the cable- has an extremely high impedance compared to cable's 50 or 75 Ohms. This is the resonant effect a quarter-wave pad has. Ideally it should have an infinite impedance.

A half-wave shorted section of a cable should have a "zero" impedance (the short repeats every half-wavelength). Using the same sensitive point, you can adjust cable half-wave length to achieve a no output of the detector, like a short by a screwdriver.