1
\$\begingroup\$

I have a near-field loop antenna that I'm applying a microwave field to at 10 GHz. I'd like to have it interact with a small magnet that is much smaller than the loop. Here's a picture with the little magnet shown as a small circle in the figure:

enter image description here

To do some calculations, I need to find the magnitude of the small magnet some distance away from the microwave loop coil.

Is there a simple approximation for the B-field strength of a loop antenna when driven at 10 GHz?

Some extra details:

  1. The loop is made from a 10GHz microwave SMA cable, with one end cut off and tied in a loop, and is about 1 cm.
  2. The drive is via a VNA, and can go up to +27dbm
  3. The small magnet can be placed directly inside the loop, or on the side.

Any ideas how I can get started here?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

loop is made from a 10GHz microwave SMA cable, with one end cut off and tied in a loop, and is about a cm.

That's not a small loop antenna – a small loop antenna has a circumference that is relatively small, compared to the wavelength. Wavelength at 10 GHz is 30 mm, so the circumference \$l\$ needs to be \$l\ll 30\,\text{mm}\$, and 1cm is not \$\ll 30\,\text{mm}\$.

This is more of a medium-sized loop: not small enough for the loop not to matter to the far-field, not large enough for the feed to not matter to the far-field.

Also, just tying an end in a loop will not yield a good loop antenna - it will first and foremost leave you with a massive impedance mismath, and field mostly radiating from the cable's sleeve, not your loop.

This isn't some LF band loop antenna, where the cable, distance to the transmit amplifier and so on are inherently tiny compared to wavelength. You'll probably want to design a PCB that has a proper, impedance-matched connector for your coax cable, and impedance-controlled lines to a microwave distributed element Balun, which directly integrates into your loop, and also applies relatively massive impedance transformations.

So is there a simple approximate for B-field strength of a loop antenna when driven at 10 GHz?

Even when properly designed, this will depend on influences such as the feed line, which physically can't be so small that it's negligible to the size of the loop. So, no, there's no simple formula, you'll need to sit down and learn how to do an electromagnetic simulation of the whole system.

Also note that basically all "simple" formulas you find for antennas describe the far field, more than a wavelength away from the antenna. Now look at where your magnet sits: it's right in the reactive near field! Depending on just how small the magnet is, it will directly interact with the antenna, changing its properties.

So, there's really no way around a proper simulation, probably of a performance RF-material PCB with the loop, balun, impedance matching, and feed line printed on it, and probably also the magnet material in the right place.

Doing this all on a PCB also has the very distinct experimental advantage that things actually become repeatable – again, your wavelengths are in the millimeters, so small changes in mechanical bending will not allow you to produce the same antenna twice. Even a barely OK-quality board would improve that situation and save you a lot of time, and consequently money.

\$\endgroup\$
4
  • \$\begingroup\$ I've fixed the language to make it clear that the loop is big relative to the ROI. The loop and the magnet are both physically small, being 1cm and <1mm respectively. \$\endgroup\$ Commented Jul 18 at 12:58
  • \$\begingroup\$ yes, OK, but that basically only changes the wording. Instead of "antenna" think "coil" in my answer. The rest still 100% applies - your coil will receive little of the energy, a significant amount of magnetic field will be caused by mantle current. You need to build a balun, ideally match the impedance of all this better, and do a simulation. \$\endgroup\$ Commented Jul 18 at 13:01
  • \$\begingroup\$ No way I can play with the loop size to make things easier, right? \$\endgroup\$ Commented Jul 18 at 13:06
  • \$\begingroup\$ Nope, not that I see any. At some point it would just start to look like an inductive termination of the unbalanced line, not what you really want \$\endgroup\$ Commented Jul 18 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.