# What Every Electronics Engineer Needs to Know About: Shielding

You recall from your previous studies that a key parameter in shielding is what is called shielding effectiveness (SE) and that SE(dB) = 20 log E1/E2, where E1 = incident wave, and E2 = attenuated wave and that a higher SE(dB) number equates to better shielding material. You find out the product is constructed with an aluminum chassis and this material along with other conductive materials such as steel or copper should provide more than enough shielding for most applications. You remember that most high-frequency shielding issues are the result of gaps or slots in the shielding material, not usually the material itself.

You also recall what the two primary shielding mechanisms are: Reflection (R) and Absorption (A). Reflection occurs when energy from the electromagnetic wave encounters the shield and is reflected back. Absorption is what happens when non-reflected energy from the wave is absorbed in the shield with only the remaining energy emerging from the other side. These two effects (or losses) combine to form the total shielding effectiveness, often expressed as SE(dB)=R(dB) + A(dB). There’s a third term sometimes added to this equation called re-reflection (B) term. The B term could decrease the shielding expected however it only occurs in extremely thin shields. You decide to just ignore it in this case since the product is constructed of shielding material strong enough to support its own weight and the shield can obviously be considered thick.

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There are a couple of other key points regarding reflection and absorption losses that you take note of.

Reflection Loss

First, you recall that higher conductivity materials result in higher reflection losses and that the distance of the EMI source from the shield also affects reflection losses due to what the impedance of the electromagnetic wave is at certain points in space (Note that distance from the source is not an issue with absorption loss – discussed later). A higher wave impedance, combined with lower shield impedance (in Ohms per Square) results in a higher reflection loss. Because the primary reflection occurs at the first surface of the shield in the case of electric fields, even very thin materials provide good reflection losses.

For a plane wave (an electromagnetic field located in the far field at a distance of approximately l/2π, where l = wavelength) that is entering a shield at normal incidence, the following equation is used to determine R:

R = 20 log Zw/4Zs (dB)

Zw = impedance of wave prior to entering shield

Zs = impedance of shield

As the angle of incidence of the impinging electromagnetic wave increases so does the reflection loss. You note that a lower shield impedance also results in a greater reflection loss and vice versa.

In the near field of the electromagnetic wave (distance < l/2π), you recall that it’s imperative to know the characteristics of the source (is it an electric field or magnetic field?), the highest frequency of interest and distance from source and shield. The reflection loss of an electric field decreases with frequency until the separation distance l/2π, beyond which the reflection loss is the same as for a plane wave. In contrast, the reflection loss of a magnetic field increases with frequency until the separation becomes l/2π.

Because an understanding of the above information is important in the determination of reflection loss, you take note that you need to determine the characteristics of the source (electric – wave impedance > 377Ω, magnetic – wave impedance < 377Ω, or plane – wave impedance of the medium or free space = 377Ω) during your troubleshooting activity in the lab.

Absorption Loss

Second, you recall that the material’s permeability (µ), conductivity (r), thickness (t) and frequency of the electromagnetic wave all contribute to skin depth (SD) and therefore to absorption loss because A(dB) = 8.68 (t/SD). You note that skin depth is where electric current tends to flow on the outer surface of a conductor at higher frequencies and it is the depth where the field/surface current is attenuated by 37% or approximately 9 dB. You remember that another way to think about it is that the absorption loss in a shield one skin-depth thick is approximately 9 dB and that doubling the thickness of the shield doubles the loss in decibels. Also, absorption loss (dB) is proportional to the square root of the product of the permeability times the conductivity of the shield material and this loss increases with frequency and as well as with shield thickness. You recall that steel offers more absorption loss than copper of the same thickness.

The Effect of Apertures, Slots, Penetrations

Finally, you refresh your understanding of the guidelines associated with adding apertures and slots and other penetrations to shields. You know that at the high frequencies involved the inherent shielding effectiveness of the shield material is of less concern than what the leakage is through the apertures used to carry signals in and out of the product. You remember from your earlier studies that aperture control is the key to high-frequency shielding. You recall that the amount of leakage from an aperture is highly dependent on the maximum linear dimension and not the area of the aperture, the impedance of the electromagnetic wave and also its frequency.

Before you head to the lab to start troubleshooting, you jot down some of the prevailing rules of thumb to control leakage from shielded enclosures. They are: 1) Keep slots shorter than one-twentieth of a wavelength at the highest frequency of interest (1.5 cm or 2/3 in at 1000 MHz) to ensure at least 20 dB of attenuation; 2) Shield or filter all cables that enter or leave the shielded enclosure and tie shields to the enclosure; 3) Maintain electric contact across the seams of shielded enclosures in order to minimize leakage; 4) If possible, limit the number of apertures as shielding effectiveness decreases proportional to the square root of their number; 5) A large number of small holes produces less leakage than a large hole of the small total area; 6) Shields do not have to be grounded in order to be effective.

Summary

Now that you have refreshed your understanding of EMI/RFI shielding you’re ready to head down to the lab to see for yourself what is actually going on. Most likely the shielding material itself is adequate but there are excessively long openings or slots, unshielded/unfiltered cables entering or exiting the shield, or a combination of both that compromise the shielding effectiveness and contribute to the product failing radiated emissions testing. Since the product is only failing by a few dB it should be easy for you to diagnose and correct in time to ship the product on time. Good luck!References

1. Gerke, D., Kimmel B., The Designer’s Guide to Electromagnetic Compatibility, EDN, 2005.
2. Ott, H., Electromagnetic Compatibility Engineering, John Wiley& Sons, 2009.
3. Montrose, M., EMC Made Simple – Printed Circuit Board and System Design, Montrose Compliance Services, Inc., 2014.

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