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This is another question related to magnetic fields. I'm trying to determine the thickness of copper foil that would be required to shield common electronic items from H fields of about 133 A/m over a range of frequencies from 1-1,000 MHz. I'm shooting for 50 dB of attenuation, and I'm trying to include cost and other practicalities among the considerations.

Based on the fact that 35 microns of copper foil attenuates an H field by 8.6 dB at 4 MHz, I calculated that a 407 micron thickness would be required to provide 50 dB of attenuation at 1 MHz. This is twice as much shielding as what's required at 4 MHz, four times as much as what's needed at 16 MHz, and much more than what's required at 1 GHz. In short, it's overkill for most frequencies.

So, my question is: what kinds of electronic items are most likely to couple damaging currents in a very strong H field at 1-36* MHz? Any item that contains a coil antenna that's sensitive on any of these frequencies would make the list. One item that comes to mind immediately is an AM radio (which might lower the lowest frequency of interest to around 500 kHz or lower, although this would require a thickness of copper that I think would be impractical). I can't think of any other kinds of items that would contain loops or coils long enough to couple an H field at these relatively low frequencies. Any suggestions? (Again, please consider a range of 1-36* MHz.)

* I chose 36 MHz as an upper limit, because 68 microns of copper would be required at that frequency, and I think this might be a reasonable thickness. It might not be the best thickness based on how many types of items would be susceptible to damage at lower frequencies, though, so I'm open to suggestions.

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  • \$\begingroup\$ What formulas (or web page calculators) did you use to calculate the attenuation of the H field versus frequency? \$\endgroup\$ – Andy aka Feb 23 '18 at 9:34
  • \$\begingroup\$ I calculated thicknesses of copper based on the ratios that analogsystemsrf provided in his answer to a previous question. If I use the formula at learnemc.com/shielding-theory and 5.96e+7 S/m for the conductivity of copper, I get a skin depth of 65.19 microns at 1 MHz. At 8.7 dB per skin depth, it would take 374 microns of copper foil to provide 50 dB of attenuation. \$\endgroup\$ – dcorsello Feb 24 '18 at 6:17
  • \$\begingroup\$ @Andy aka, can you comment on Marcus Müller's answer? Also, is there a different way to calculate thickness in a high strength H field? Or, does the calculated thickness represent only the skin depth, meaning that an additional thickness is required to make the skin effect possible? \$\endgroup\$ – dcorsello Feb 26 '18 at 4:49
  • \$\begingroup\$ it's been years since I did that math! \$\endgroup\$ – Andy aka Feb 26 '18 at 8:24
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An H-field changing at 36 MHz will induce a 36 MHz current in your copper foil.

Maxwell's equations plus Stokes give you

$$ \oint_{\partial \Sigma} \mathbf{E} \cdot d\mathbf{l} = -\mu\int_\Sigma \frac{\partial \mathbf{H}}{\partial t} \cdot d\mathbf{A} $$

I.e. the electric field integrated along a contour \$\Sigma\$ (which gives you a potential) is the integral of the time-derivative of your H-field times your medium's \$\mu\$. At 130 A/m, and remembering that derivative of \$\sin(ft)=f\cos(ft)\$, that's bad news for your copper, and for the things it protects. The induced currents will be high enough to fill the whole couple of microns of thickness, and will couple into the things you wanted to protect.

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  • \$\begingroup\$ Thanks. If t is very small, what is the effect? It looks like it makes for a very big number, but my calculus skills are nil. \$\endgroup\$ – dcorsello Feb 23 '18 at 13:30
  • \$\begingroup\$ I'm working on getting up to speed on the equation you provided. In the meantime, I've learned that the relative permeability of a material tends towards 1 at magnetic saturation in a high H field strength. But the relative permeability of copper is roughly 1, so it doesn't seem to me that copper is affected much in this way. Are there other properties that change at 133 A/m that diminish the benefits of the skin effect? Or, do I just need a thicker barrier than what I can calculate using the formulas I find? \$\endgroup\$ – dcorsello Feb 26 '18 at 4:32

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