1
\$\begingroup\$

I understand how/why graded absorbers work - you achieve a match with free space by gradually varying the electromagnetic properties from "almost air" to strong absorption (either geometrically like pyramidal absorbers or through density like graded foam sheets).

How does this work for those (very thin compared to e.g. pyramidal absorbers) absorber sheets? How come they actually absorb energy and don't just reflect most of the waves due to this very sharp step?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

These sheets are made of a loss material – it has a \$\mu_r=\mu'+j\mu''\$ whose imaginary part is particularly high.

Reflections happen when the boundary conditions enforce that – for example, a metal surface reflects well, because there can't be (much) E-Field inside the metal, leading to equation describing an incident wave having a solution where the surface-normal component of the pointing vector is inverted, but the parallel part remains untouched.

Perfect conductivity is not the case here.

The other option to enforce reflection would be attenuate the magnetic field (set it to 0 to get total reflection) inside the material – and that's what these materials do. However, they are mainly lossy, typically the real part \$\mu'\$ of the permeability is lower in magnitude than the (loss-indicating) imaginary part \$\mu''\$ for the frequencies you're trying to shield. Take this figure from Würth as an example:

Figure showing reflection and absorption

As you can see, there is a lot of reflection – but seeing that you do this to shield a box, typically that reflected power in the end "just" gets reflected around within the device and gets absorbed eventually (if you're not in extremely bad luck).

\$\endgroup\$
4
  • \$\begingroup\$ Okay, I get that - so some sheets just care about a reduction of transmission. But then there are examples like this one (masttechnologies.com/mr1-tuned-frequency), which actually achieves a strong reflection loss rather than just lowering transmission. What about those? \$\endgroup\$
    – jfeis
    Commented May 23, 2021 at 18:59
  • \$\begingroup\$ well, you know the answer to that: graded material in such a way to minimized both transmission and reflection. For fixed frequencies rather than wideband absorption, things get a fair bit easier, as "graded" often becomes "two or three different layers". \$\endgroup\$ Commented May 23, 2021 at 19:10
  • \$\begingroup\$ Hm, I think they are too thin for any (meaningfully effective) gradient to happen, aren't they? As an example, there is a sheet optimised for 20.6GHz that is 1.9mm thick (masttechnologies.com/wp-content/uploads/2013/03/…). That would be one very sharp gradient at those frequencies? \$\endgroup\$
    – jfeis
    Commented May 23, 2021 at 19:13
  • \$\begingroup\$ Sounds pretty reasonable! Say, we have a layer of refractive index n=8, then the wavelength inside that is 1.875 mm. Now, play the funny thin-film tricks one plays with oil on water or lens coatings for optical light: total reflection on the inner-to-outer material interfaces, thus "ping-ponging" inside that layer, and a bit of absorbtion suddenly is applied to the wave for a large "virtual" length. \$\endgroup\$ Commented May 23, 2021 at 19:17

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.