Shielding -- Is it effective when floating? - Page 1
Forum section: electronics / projects

author: tree
author_rank: Regular Contributor
timestamp: April 03, 2014, 07:04:40 pm
content: I am reading this app note by analog devices:http://www.analog.com/static/imported-files/application_notes/41727248AN_347.pdfAt the bottom of the first page it says: "An electrostatic shield, to be effective..." "But grounding the shield is useless if the signal is not grounded"Why is grounding the shield useless if the signal is not grounded?

author: tsmith35
author_rank: Frequent Contributor
timestamp: April 03, 2014, 11:11:43 pm
content: Perhaps this may help:Practical Data Acquisition for Instrumentation and Control Systems

author: tree
author_rank: Regular Contributor
timestamp: April 03, 2014, 11:18:52 pm
content: Let me ask this question differently...If I had an electronic circuit (powered by batteries) sitting in sealed metal case and that metal case was not electrically connected to the circuit's ground, would the shield not provide shielding from EM radiation/noise to the electronic circuit?

author: DavidDLC
author_rank: Frequent Contributor
timestamp: April 03, 2014, 11:32:15 pm
content: Quote from: tree on April 03, 2014, 11:18:52 pmThe shielding should protect you from outside noise, but also from internal noise from different circuits to couple.David,

author: tree
author_rank: Regular Contributor
timestamp: April 03, 2014, 11:35:13 pm
content: So then why can't I used a cable that has a floating shield to interconnect two of these enclosed circuits, assuming that the shield is connected to the cases of the two circuits? Also assume that it uses differential signaling and that the cases and shields are floating with respect to the circuit ground.

author: georges80
author_rank: Frequent Contributor
timestamp: April 03, 2014, 11:37:51 pm
content: Quote from: tree on April 03, 2014, 11:18:52 pmGround does not mean the AC ground stake outside the building in this case. Ground is the 0V potential in the design. So, your 0V plane in your battery powered unit needs to be tied to the shield.If the shield is 'floating' then you can have a large charge build on it versus the 0V plane and then radiate in/out. You don't want the shield to become an antenna...cheers,george.

author: tree
author_rank: Regular Contributor
timestamp: April 03, 2014, 11:40:29 pm
content: Quote from: georges80 on April 03, 2014, 11:37:51 pmSo how can cell phone signal be attenuated when you're in an elevator? The cell phone's signal ground is floating...

author: georges80
author_rank: Frequent Contributor
timestamp: April 04, 2014, 12:14:29 am
content: Cell phone signal is also attenuated if you're surrounded by wood too... not just a steel elevatorRead your own AD URL that you linked - it even shows examples of having items charged and the coupling that occurs to circuitry. Seems pretty clear why you want to connect the shield to your circuit's 0V potential plane.Look at shields on sensitive circuitry - small radio modules, GPS units etc - you'll see the shield is soldered down and to the 0V plane. Shields can be prevent RFI in and or out and also coupling from one section of circuitry to another within the same design.cheers,george.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 12:18:01 am
content: How do Faraday cages work then? Do they need to be grounded (at the 0V reference) as well?Suppose I wanted to design a product that was well shielded, safe (by connecting PE to the case), whose output was floating. The only way I see to do that would be to either have two metal cases (one inside the other & electrically isolated from one another) or to use a 10Meg resistor between circuit ground and PE. By having two cases, the PE would protect the user by connecting to the outer case, and the inner case would be connected to the circuit's 0V reference.

author: tsmith35
author_rank: Frequent Contributor
timestamp: April 04, 2014, 03:16:39 am
content: Everything is relative. A shield connected to earth ground doesn't mean that the circuit reference that same ground. Now that you supposedly have a potential between the shield and the signal, the voltage differential could contribute to noise. Of course, I'm not an authority on this. so there's a good possibility that I'm wrong.

author: Tinkerer
author_rank: Frequent Contributor
timestamp: April 04, 2014, 03:24:31 am
content: Quote from: tree on April 04, 2014, 12:18:01 amYou do not need to be connected to ground to shield something and this is a huge misconception among many people. Enclosing something in metal will stop all EM radiation from getting though, not counting things like infrared of course. Even sheilding that isnt grounded that only partially covers an object can have a huge effect in attenuating a signal.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 03:26:11 am
content: Quote from: Tinkerer on April 04, 2014, 03:24:31 amI want someone to confirm this because EVERYONE I asked has differing opinions

author: IanB
author_rank: Super Contributor
timestamp: April 04, 2014, 03:36:06 am
content: Faraday cages only work if they completely enclose the thing being shielded. Since the cable comes out at each end the shield is not a full enclosure and is not a Faraday cage.Here is my poor, non-technical attempt to explain what is going on.Suppose your cable has a floating shield. That means that it will act as a receiving antenna for any radiation intersecting it. As an antenna it will therefore experience induced voltage changes. The wires inside the shield are in close proximity to it, so that they form a capacitor with the shield. So if the shield experiences voltage changes, the wires inside the shield will experience electrostatically induced voltage changes in sympathy. The shield is letting things through.By tying the shield to an appropriate part of the circuit this effect is reduced. Now the whole system changes in voltage when the shield acts as an antenna and so differential voltages between the shield and the circuit are minimized.What the shield is basically doing is acting as an equi-potential bonding system. It shorts out any induced differences in voltage along the length of the cable, and it shorts out any differences in voltage between the outside and the inside.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 03:49:04 am
content: Quote from: IanB on April 04, 2014, 03:36:06 amDid you read reply #4, in which I proposed a thought experiment?

author: pa2ees
author_rank: Contributor
timestamp: April 04, 2014, 05:38:06 am
content: In answer to your question, yes, shielding is effective when floating. It might be more effective when grounded to your circuit ground, but it still is effective even if floating.First, I work with a handheld wireless device for my job, and if I bring a piece of metal close to the antenna, it definitely changes the signal coming off the antenna.Second, we have a shield room / anachoic chamber, and if you go in there, and close the door, your cell phone loses all communication with the outside, ie no 3g, no wifi, no bluetooth unless it's within the chamber. It is not grounded to your phone, but it certainly works well.Third, when we were getting one of our products tested for FCC certification, it failed because a microcontroller (ATTINY44 to be precise) in a QFN package was radiating too much noise (RF). So I took a piece of copper tape, cut it to a square just smaller than the package, and stuck it to the top of the microcontroller. It was not grounded, in fact it was not connected to anything, just stuck to the plastic case, and it reduced the emissions enough to bring the product into compliance.With regards to your other question, why is grounding the shield useless if the signal is not grounded. It's been answered, but I'll try to state it a different way. Connecting the shield to any potential that is not referenced to the signal of interest in some way (ie the ground of your circuit of interest), is like not connecting it to anything. You could connect the shield to earth ground, or to a computer's ground, or to a cow. There's stil no connection to your circuit, and thus it doesn't change how the shield behaves with regards to your circuit.Hope that clears things up.- Erik

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 04, 2014, 05:58:26 am
content: Quote from: pa2ees on April 04, 2014, 05:38:06 amCow ground:

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 02:16:29 pm
content: pa2ees,What I gather is that the way to connect shielding depends on the application, correct?

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 04, 2014, 04:08:18 pm
content: This is a good brain-teaser, tree. I've been thinking about it for the past few hours.Here's the question I've gotten stuck on. If the shield is not grounded -- like the coaxial one pictured in the app note -- then from where does the shield draw charge to counteract an external e-field?Obviously any sheet of metal will reject far-field EM waves. But I think this app note is talking about low-frequency (e.g. DC'ish) influences. The physics of the two effects are completely different.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 04:22:44 pm
content: Quote from: CaptnYellowShirt on April 04, 2014, 04:08:18 pmWell if the shield not connected to anything, i.e. floating, then an external E field will just cause charge to separate in the shield. Apparently this is how you get capacitive coupling, but then how do Faraday cages work, if not just like that, where charges separate to produce no internal E field.You're absolutely correct that this paper is mainly concerned with inductive coupling. I went back and reread the statement, "But grounding the shield is useless if the signal is not grounded"I think the trouble I had with that statement was a matter of semantics. I thought that statement was saying that the shield becomes useless, but all it was saying is that grounding the shield has no effect on attenuation.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 04, 2014, 04:31:53 pm
content: Quote from: tree on April 04, 2014, 04:22:44 pmSomeone else pointed out the notion that the Faraday cage is a complete enclosure. I think that's the ticket.If you have a coax and you bring a charged object near the middle, charges on the shield will rearrange themselves. The shield will 'borrow' enough charge to chancel out the local field from parts of the coax that are more remote. But what happens at that remote end? If there's no place to add charge you'll have a discontinuity? So the wire inside is going to experience the effects of an e-field at this point.So I think in your 'battery powered, completely enclosed' thought experiment, the shielding would work.That's the best I can do. If its not right, I'd love to be corrected.

author: Marco
author_rank: Super Contributor
timestamp: April 04, 2014, 04:33:43 pm
content: Quote from: tree on April 03, 2014, 11:18:52 pmWouldn't the case act like an antenna? Receiving and retransmitting the EM. A lossy antenna at best, so it should still attenuate the noise some ... but not as well as if it was at virtual ground and a proper shield.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 04:39:29 pm
content: Quote from: CaptnYellowShirt on April 04, 2014, 04:31:53 pmLike I said earlier, I think it now depends on the application. There were people who said that it will NEVER work and others said it will ALWAYS work, but the truth is, it depends.

author: Marco
author_rank: Super Contributor
timestamp: April 04, 2014, 04:40:16 pm
content: Quote from: Tinkerer on April 04, 2014, 03:24:31 amI'm too lazy to do an experiment and too stupid to have an intuitive grasp of the underlying physics ... but while it might be a poor substitute for either I can google stuff.Hereungrounded shielded UTP provided 20 dB attenuation at 100 MHz compared to unshielded, but properly grounded it provided 40 dB. So it seems that an ungrounded shield certainly helps, but not as much as a grounded one, would you agree with that observation?

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 04, 2014, 04:50:54 pm
content: Yeah, but its the *why* that's bothering me.

author: tree
author_rank: Regular Contributor
timestamp: April 04, 2014, 04:51:12 pm
content: Quote from: Marco on April 04, 2014, 04:40:16 pmI would agree with that because the test results for a floating shield still showed attenuation as I suspected. Had this been a test setup that provided a Faraday cage as in my thought experiment then I'm sure an ungrounded and grounded shield would show very similar results.

author: jerry507
author_rank: Regular Contributor
timestamp: April 04, 2014, 05:27:28 pm
content: This doesn't depend on the application, it just depends on how good you need it.A Faraday cage works because, as others have said, itcompletelyencloses an area or volume from an electric field. Because charge is free to move around the enclosure, it can move to cancel the internal field. If you have penetrations for cables and such, it will degrade the effect with the magnitude of degradation depending on many factors such as frequency and the geometric layout.In this respect, a fully enclosed shield such as a braid or foil wrap will form a very effective Faraday cage around the signal. What I think you're missing is that while the volume inside the shield may be at zero potential, the shield itself is not restricted. In fact, the shield will be at the same potential as the electric field it's within. The Analog Devices article tells you to ground the shield because you don't want the shield to be at a potential. Ground it so it's potential is as close as possible.Why? Well, your shield isn't infinite. Let's reference figure 5 from the AD article. Let's say that this three segment system is a RS-485 link inside a building (something I deal with every day) and then let's break it in different ways and see what the effect would be.If we first remove the ground reference at the left hand side, we have three connected shields. The conductors inside the shield will be effectively shielded from a noise source that is located significantly far away from either end but now that noise can travel up and down the connected portion and couple into the conductors at the far left and far right.Now let's instead break the connections between the shields, leaving the ground reference at the far left. Now noise can be injected at the middle of the middle shield and will radiate down the center shield. It is now free to couple into the conductors at the breaks between the shields.Now the shield does have loss and so noise signals can be attenuated simply by having to travel along the shield. This may be enough to eliminate problems if the noise is small in magnitude or the shield is very long. Also, the shield is not a perfect radiator so the coupling to the internal conductors may not be that strong. But grounding it is the best solution to attenuating that noise signal.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 04, 2014, 05:46:45 pm
content: I think jerry nailed it.My hang up was the difference between voltage and electric field:The un-grounded shield may reject an internal electric field, but it does nothing to reject an overall potential with respect to some other (remote) part of the circuit.

author: SArepairman
author_rank: Frequent Contributor
timestamp: April 04, 2014, 06:31:52 pm
content: So, how about lossy shields? Like those RF gaskets Dave did a blog about, some of them had a few hundred ohm resistance, or more. The foamy conductive things.When the shield is lossy, is it better if its grounded?http://www.eevblog.com/2013/06/04/eevblog-477-mailbag/**********this is a great episode if yoou wanna learn about shielding technology btw.

author: skipjackrc4
author_rank: Regular Contributor
timestamp: April 04, 2014, 06:32:54 pm
content: Jerry is absolutely right.The shielding requirements for low-frequency magnetic and electric fields and TEM waves (RF) are somewhat different. The following is an extension of what Jerry said.There are two primary mechanisms when shielding against RF--reflection of the incident wave and internal absorption due to what Jerry was talking about.I'm not sure if you are familiar with the concept of electromagnetic transmission line reflection, but if not, here is the basic idea. If a wave is traveling through cable of one impedance (say 50 ohm, which is the most common for lab use) and it encounters another impedance (such as an imperfect connector of maybe 48 ohm), part of the wave will continue to travel through the connector, but part of it will reflect back and travel down the cable in the opposite direction. The amount of reflection depends on the differences between the two impedances. This is not usually noticeable at low frequencies, but every time you hook a DMM up to a circuit, you are getting reflections. The DMM is just too slow to see them.An EM wave has an impedance that depends on the material that it is traveling through. This impedance is simply the ratio between the electric and magnetic fields, just like resistance is the ratio of voltage to current. In free space (which is basically air), the wave impedance is 377. In a conductor, this impedance is lower. The better the conductor, the lower the impedance. This means that when a wave hits the surface of a conducting shield, part of it will just reflect off.The portion of the wave that does not reflect is then attenuated as it travels through the material. This attenuation is influenced by material thickness and conductivity, as well as the frequency of the wave. An increase in any one of these three parameters will improve shielding effectiveness. However, at very high frequencies, shielding effectiveness starts to break down due to the fact that it takes the free charges in the conductor a finite amount of time to respond to the high frequency incident wave. If the incident wave is changing faster than the charge carriers can reorient themselves to oppose it, then the attenuation will not be as effective.If you are interested in learning more, Dr. Clayton Paul has an excellent book calledIntroduction to Electromagnetic Compatibilitythat describes this process in more detail. I believe it is at least partially available on Google Books.

author: jerry507
author_rank: Regular Contributor
timestamp: April 04, 2014, 06:44:22 pm
content: I read the article a bit more while munching on my lunch and realized that the article does a better explanation of the coupling at the ends. You can effectively think about the shield and the conductor as the two plates of a huge parallel plate conductor. If you're dealing with cable that actually specifies this, such as Belden 3105A (https://edeskv2.belden.com/Products/index.cfm?event=printPrev&pnum=3105A&ut=english), then this effect becomes pretty obvious.As an aside, I have a friend who worked at a company who did a lot of cellular networking stuff. They had a Faraday cage with a mini cell tower inside that they would use to test their devices in isolation from the regular "live" networks. They had a lot of problems with the < 2 sq in slit that they used to run serial cables and coax through. Given that the models have a typical sensitivity of -110dbm, they would sometimes lock onto the outside networks instead of the intended one. It took a significant amount of fiddly work to seal that gap around the cables. Even when they did use metal fingers to extend the shield, the signals would couple in on the serial cables. Ferrites were the key to that problem.

author: SArepairman
author_rank: Frequent Contributor
timestamp: April 04, 2014, 10:37:42 pm
content: Quote from: jerry507 on April 04, 2014, 06:44:22 pmDid they stuff the opening with ferrite foam?Would a box made of a lossy material like ferrite be a better equipment housing then a box made of something like aluminum? could it be worse in certain cases?

author: skipjackrc4
author_rank: Regular Contributor
timestamp: April 05, 2014, 01:19:09 am
content: Quote from: SArepairman on April 04, 2014, 10:37:42 pmThe best way to shield enclosure openings is to run the cables through a waveguide structure. Below a certain cutoff frequency, waveguides exhibit strong attenuation that is proportional to their length. If you are trying to shield everything below 10GHz for example, you would use a waveguide with a cutoff of 25GHz (just throwing a number out there--the actual cutoff frequency depends on the length and desired attenuation). It doesn't have to be an expensive commercial waveguide--copper pipe that is securely connected to the enclosure will work nicely. Take a look at your microwave oven--each of those holes in the door screen act as little waveguides operating below their cutoff frequency.As for the signal being picked up by the cables, ferrite beads are indeed the easiest way to handle it. The lab that I work in has several shielded anechoic chambers, and all the RF and control lines are heavily ferrited (new word!).Using ferrite for a shield material is effective against magnetic fields. In this case, aluminum or copper are completely useless (with DC fields) because they are non-ferromagnetic. Steel, ferrite, and mumetal/permalloy are all good for magnetic shielding, depending on the frequency of the field. Against RF waves, you are better off with aluminum or copper for pure shielding purposes.Ferrite tiles and foam do have their purposes against RF, though. Ferrite can be made so that it has a similar intrinsic impedance to free space (see my post above explaining this) which means that it absorbs RF energy without reflecting. Ferrite tiles and foam pyramids are used in EMI anechoic chambers to prevent reflections from the walls. A typical EMI chamber will be aluminum or copper on the outside for external shielding and have the inside walls lined with ferrite pyramids and possibly solid tiles for internal reflection damping.Here is some info from a ferrite absorber supplier if you are interested:http://www.djmelectronics.com/rf-absorber.html

author: SArepairman
author_rank: Frequent Contributor
timestamp: April 05, 2014, 06:43:17 am
content: I never thought about the impedance of free air before.

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 05, 2014, 03:36:41 pm
content: Quote from: SArepairman on April 05, 2014, 06:43:17 am

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 05, 2014, 04:09:28 pm
content: ^ lolBut yes, everything has impedance; and geometry shapes that. Free air has some impedance. A single wire in free space has a similar impedance, give or take radiation -- a 1/4 wave resonant wire has a lower impedance, but not by too much, because radiation damps the resonance. Put another wire near it and the impedance between them becomes a function of their separation; put a solid round shield on it and you get coax; etc.Or maybe you have a trace over a ground plane, which is a (microstrip) transmission line; suppose you wanted to short it out, so you drop a via to the ground plane. Maybe you're bypassing plane-to-plane with a capacitor and as little trace length as possible: but the traces, capacitor and vias all have impedance. The planes aren't even direct shorts: the impedance has an odd dependency with frequency, because the via doesn't connect to the entire plane at once, it connects first to the round zone it's touching. The inductance of a plane goes as ln(R/r), for an outer radius R and via radius r. You can imagine a very high frequency wave interacting with maybe the first 1/4 wavelength of the plane, while the rest acts like a short; as the wave expands over the plane, the apparent impedance drops and more and more gets reflected back at a progression of angles (the reflection is smeared out over time because it's dispersive).In EMC testing, there's a reason they test against 50 ohms: because sooner or later, the power line is going to look like a 'random wire' antenna and radiate stuff, at which point it's going to be in the vicinity of 50 ohms. Basically, the assumed load is one that hopefully radiates at all frequencies.You might think your switching power supply is isolated and happy and, how can I even filter this if it's all floaty and doesn't even have an impedance? (You can't just *make* a filter: a filter only does its job into a specified range of impedances.) Ah, but those transformer windings have impedance, within themselves and to everything nearby! The circuit board above a metal chassis (if you don't provide one, FCC Part 15 specifies a ground plane!) has an impedance. So it's not as bizarre as it sounds, and in fact is a quite tractable problem.Tim