Why is this HFSS model now showing similar results to VNA?

[drkirkby]

I tried to create a loop antenna, consisting of a single closed loop for 144.3 MHz. I wanted to set the diameter so the real part of the input impedance was 50 Ohms, and then tune out the inductive part with a capacitor. So I set up this HFSS model, and found a radius for the loop where the input Z is 50 + j X Ω. So I built the loop according to those dimensions. When I measure this on a VNA, I am seeing 119 + j 138. Ω

I've tried converging on both the usual delta S of 0.02, as well as setting up expressions for R and X, and converging on them. Whatever I do, this is sufficiently far from measurements to alert me something is wrong.

For the measurements an N connector was put directly across the loop. After calibrating with an N calibration kit, I applied a small port extension to move the reference plane to the loop. I did this by putting a short across the loop, and adjusting for a phase of 180 degrees. At this frequency (wavelength around 2 m), the N connector is pretty insignificant.

The feed cable was taken out horizontally. I don't know if the fact the computer model does not include the feed is the problem. I could simulate with a waveport and include some coax in it. Also there is no balun. These might be the cause of the problem, or it might just be the model is very wrong.

Dave

 

[drkirkby]

Re: Why is this HFSS model NOT showing similar results to VNA?

Somehow I stuck the title wrong. The model is NOT showing results like the VNA is measuring

 

[SLK001]

The loop is balanced, while your VNA input is unbalanced. That is probably most of your problem. Also, I thought that the intrinsic impedance of a single loop antenna was ~300 ohms.

 

[drkirkby]

I did wonder if the balance/unbalance issue. I'm not convinced that is the case though.

This loop is totally closed, rather than have a gap in the middle. You are thinking of something quite different when you say 150 Ω.

I want to make an omnidirectional pattern. I think this means the loop circumference needs to be << λ. From what I have read, then the radiation resistance comes very low, so loss resistance causes low efficiency. I may have to put up with that.

 

[SLK001]

There is no such thing as a realizable omnidirectional pattern. You would probably have better luck with a dipole. If your loop is totally closed, how are you energyzing it? Yes, I assumed that your loop had a gap where it was energyzed.

 

[drkirkby]

I want this horizontally polarized. A horizontal dipole has two nulls.

A small loop, with a circumference << λ has a virtually constant current, so the pattern is close to symmetrical. One book I have says < 0.016 λ, but I've no idea what that number comes from. That does lead to unacceptably low radiation resistance and so efficiency. Multiple turns increases radiation resistance as square of number of turns, although I don't know what it will do the radiation pattern.

I wish I had faith in the HFSS model, but I am a bit concerned there's some problem I have overlooked.

The problem with EM simulators is it is very easy to get nice looking but wrong results. At least with Word or Photoshop, you can see immediately if the results are wrong!

Dave

 

[drkirkby]

FWIW, the book Antenna Theory, 3rd edition, by Constantine A. Balanis, givens in equation 5.14-a for the radiation resistance Rr of a loop as

Rr = 20 π² (C/λ)⁴N² Ω

where λ is the wavelength, C is the circumference, N the number of turns. Sticking that in, for λ=2.08 m, diameter=225 mm, N=1 turn, one gets 2.52 Ω. So I have a nice set of conflicting results.

• Balanis 2.52 Ω
• HFSS 50 Ω
• VNA 119 Ω

The Balanis formula is being used outside the range of applicability, of C/λ < 0.1, where a constant current can be assumed But an example in the Balanis book gives the theoretical results for a loop of radius λ/25 with 1 and 8 turns, and gets 0.788 Ω for one turn and 50.43 Ω for 8 turns. In the book, λ/25 =0.04 λ. A loop radius of 0.04 λ means a circumference of 0.25 λ, so C/λ=0.25. My own antenna has C/λ=0.34.

I will make the loop smaller in HFSS, such that C/λ < 0.1 and see if the HFSS results agree with Balanis.

One thing Balanis does say is the loss resistance can be higher than the radiation resistance for a single turn, which is why multiple turns are preferable, as the radiation resistance goes up with the square of the number of turns, and the loss resistance roughly as the number of turns.

I assumed a perfect conductor in HFSS, but I'm using copper. I don't know if that could be the problem.

On thing I did notice on the VNA is that for low frequencies, below about 80 MHz, the resistance was very close to 0. That would be indicated by the Balanis formula, as at lower frequencies, C/λ would be less than 0.1.

Oh well, something to ponder over the weekend. Good job its only an amateur radio project, and not earning my living doing it !!

Dave