characteristic or input impedance of a simple flat dipole

Hello,

I am trying to learn HFSS 12. I used the simple dipole design to start with. I am not getting the characteristic or input impedance of this simple flat dipole. Any help will be widely appreciated.

Dipole dimension:

Length = 63.05 mm because I want it to resonate at 2.26 Ghz, so taking length of the dipole to be 0.475 times the wavelength in free space.

So each arm of the dipole is 31.525 mm long. A 0.125mm gap between the arms for the lump port set to 50 ohm.

width of the dipole = 2.522 mm because the book says to approximate a circular dipole (wire) to a flat dipole use the equation:

length/diameter = 50 and diameter = 0.25*width

Dipole is in vacuum and a lump port of 50 ohm between the dipole arms.

Assuming all these, theoretically the input impedance should be 73 ohm.

While simulating in HFSS 12, I am not getting any input or characteristic impedance of the dipole antenna.

My idea of getting the input impedance is to draw the Z parameter and see where the imaginary curve crosses the zero line. But in simulation the imaginary curve never crosses zero line and it remains negative which means it is capacitive (while theoretically the imaginary part should be inductive or positive).

I have attached the simulation file.

Same problem when I put dipole on a dielectric of 1.17 mm thick TLY-5 substrate of relative permittivity 2.2. I know that for the dielectric I have to consider the effective wavelength which comes out to be 62.3 mm but my problem with the characteristic impedance remains the same - the imaginary curve of the Z parameter NEVER crosses zero line.

Please provide help. Any other method to find characteristic impedance? Any help will be highly appreciated.

Hello,

If you do a simulation (for example) between 2.0 and 2.6 GHz, it must cross the point where Im(Zin) = zero. I don't have HFSS, so I cannot read your file. Did you enter the correct frequency range? what happens when you increase the gap for the lumped feed to about 1mm?

Can you see in how many segments you divided the structure (you need to look into "mesh"). Your mesh should divide each arm of your dipole in at least 4 segments to get some useful results.

When you put it on a substrate with a large groundplane, the impedance at resonance drops, depending on (height above ground plane)/lambda. The smaller this number, the lower the impedance at resonance (can be below 5 Ohms).

Regarding permittivity. Your wavelength will not drop that much. The effective relative epsilon will be less the two, so your wavelength will not be below 94mm.

Re: characteristic or input impedance of a simple flat dipol

Hello WimRFP,

I should really thank for your reply. it helped me tremendously.

I used your advice and made the gap to be 1 mm between the dipole arms. I was doing the frequency sweep from 1 Ghz through 10 Ghz and in that range the Im(Z) crossed the zero line twice, one at 2.56 Ghz and one at 3.3 Ghz.

I kept the solution frequency to be 2.26 Ghz as theoretically the resonant frequency supposed to be at that. Could you please explain why there is other resonant frequency at 3.3 ghz and the resonant frequency is not exactly at 2.26 ghz.

Theoretically the input resistance should be around 73 ohms whereas I am getting the input resistance to be 202.8 ohm at 2.56 Ghz and 353.65 ohm at 3.3 Ghz

Please provide help with that. I really appreciate your help. I have attached the file again.

Re: characteristic or input impedance of a simple flat dipol

Also could you please tell me the "mesh" thing. How do I configure and find about it. I have looked into the HFSS manuals available online they talk more about the slave and master and not about the "mesh".

Any help will be highly appreciated.

Hello,

When setting the frequency range from 1 to 10 GHz and the number of frequencies that are used for simulation is too small, the interpolator algorithm for plotting the impedance on the smith chart may go wrong. Even intelligent adaptive algorithms may fail.

You need sufficient frequency simulation points around your resonant frequency. Check the actual frequencies that are used for the simulation.

It is correct that you get several resonances. The lowest resonant frequency is where the electrical length is 0.5 lambda (that means that the physical length is somewhat less depending on the thickness of your dipole). Impedance will be between around 50…70 Ohms. Thicker dipoles (that have size remarkably below 0.5 lambda are closer to 50 Ohms).

The second one appears around 1 lambda, that impedance is high end depends strongly on thickness/wavelength ratio. Very thin full wave dipoles will be around kOHms, where very thick ones (for example thickness/length = 0.2), will have several hundreds Ohms.

You should not see a resonance at 3.3 GHz. I think you should use more frequency points that are actually used for simulation and/or check the meshing.

Regarding meshing, there must be something that will show you how the simulator divided your structure before simulation. In IE3D it is the "show meshing" command. Maximum mesh size of 1/20 or 1/10 lambda is mostly used as default. Correct meshing is vital to get best results with minimum simulation time. Where you expect strong current amplitude change versus position, you need a finer mesh. You can be lucky, with your dipole, there are no strong current changes versus position, so a mesh size of 1/10 to1/20 lambda will give you practically useful results.

I hope someone else will comment on how to find/set meshing parameters (and other things) as I don't know HFSS.

Re: characteristic or input impedance of a simple flat dipol

Hello WimRFP,

As you said, that the second resonance at 3.3 Ghz appears to be at 1 lambda, so why shouldn't I suppose to see that. I mean, if resonance occur at half lambda, then it should also occur at 1 lambda, 1.5 lambda and so on.

I did the meshing for lambda/10, inside selection in HFSS and getting the second resonance at 3.5 Ghz and the first one at 2.16Ghz.

Please explain. You reply is highly appreciated and most awaited.

Hello,

The high impedance resonance at 3.5 GHz (instead of 4.32 GHz) isn't strange for your antenna.

At the full wave resonant frequency, you have high impedance between the arms. As at the feed point the arms are close together, there is capacitance also. This capacitance causes a significant frequency shift for the full wave case where Im(Z) = 0. Antennas with relative large thickness/length ratio show more frequency shift.

You will also find a 1.5 lambda resonance (low impedance, that will be more close to the theoretical 1.5lambda frequency).

Just for seeing the effect. Try to do the simulation for a dipole thickness of about 0.1 mm and you will see that the frequency for resonance will be closer to the physical half wave frequency and physical full wave frequency.

You will notice a significant increase of resonance impedance around the full wavelength and 2 wavelength resonance in you dipole. The impedance at the 0.5 and 1.5 wavelength will not change significantly. You will also notice that d(Im(Z))/df will be larger for the thin dipole case, hence the useful bandwidth will be less.

Re: characteristic or input impedance of a simple flat dipol

Hello WimRFP,

I did the simulation for increased thickness to 0.1mm

For the frequency sweep from 1Ghz to 10Ghz, the first resonance occur at 2.27 Ghz instead of 2.26 Ghz (0.01mm thick) and the impedance is down from 384 ohm (for 0.01mm thick) to 241 ohms. The impedance went down with increase of the thickness. This was unexpected.

i would appreciate if you let me understand a little bit of theory thing. Its simple.

I designed the half-wave antenna to make it resonant. So if the length is 62.3mm, it means the half-wave length is 62.3mm and the full wave length is 62.3*2 = 124.6mm

So the half-wave or the first resonant frequency should be speed of light/half-wave frequency which is (3*10^10 mm/s)/62.3mm which gives 4.8 Ghz. However in the simulation gives 2.26 Ghz. Please help in understanding this discrepancy - it is very frustrating.

Hello,

I assume you are simulating in air, no dielectric, no conducting ground.

The half wave impedance at resonance (so Im(Z) = 0) should be around 70 Ohms for very large length/thickness ratio. You have around 250..400 Ohms, so there must be something wrong in the simulation settings. You may check the conductivity and thickness of the layers also.

You did calculate the full wave frequency (62.3mm is a full wave at 4.8 GHz). So you have to halve your frequency to make 62.3mm a half wave.

You will get a resonance somewhat below 4.8 GHz (depending on thickness of conductors), but it will be high impedance. You can see it as a source that is in between two half wave transmission line sections that are open. As you know an open halve wave transmission line section has a high impedance. Due to field fringe, the actual frequency of high impedance will be lower then based on the physical length.

How much frequency point have you used? Try to experiment with that also.

Re: characteristic or input impedance of a simple flat dipol

mequitnever

You probably have found the problem by now.

Yoiur air box is too small, it should be at least 1/4 lambda.

Whne I made the box 200 x 200 x 200 mm, impedance dropped down to 83.5 + j*12.5.

This is still far from theortically 73 + j*42.

I made a circular dipole and it was the same problem.

Any suggestion?