Shielding -- Is it effective when floating?

[tree]

I am reading this app note by analog devices:
http://www.analog.com/static/imported-files/application_notes/41727248AN_347.pdf

At 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?

[tsmith35]

Perhaps this may help: Practical Data Acquisition for Instrumentation and Control Systems

[tree]

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?

[DavidDLC]

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?

The shielding should protect you from outside noise, but also from internal noise from different circuits to couple.

David,

[tree]

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.

[georges80]

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?

Ground 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.

[tree]

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?

Ground 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.

So how can  cell phone signal be attenuated when you're in an elevator? The cell phone's signal ground is floating...

[georges80]

Cell phone signal is also attenuated if you're surrounded by wood too... not just a steel elevator

Read 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.

[tree]

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.

[tsmith35]

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.

[Tinkerer]

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.

You 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.

[tree]

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.

You 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.

I want someone to confirm this because EVERYONE I asked has differing opinions

[IanB]

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.

[tree]

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.

Did you read reply #4, in which I proposed a thought experiment?

[pa2ees]

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

[CaptnYellowShirt]

You could connect the shield to earth ground, or to a computer's ground, or to a cow. 

Cow ground:

[tree]

pa2ees,

What I gather is that the way to connect shielding depends on the application, correct?

[CaptnYellowShirt]

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.

[tree]

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.

Well 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.

[CaptnYellowShirt]

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.

Someone 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.

[Marco]

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?

Wouldn'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.

[tree]

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.

Someone 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.

Like 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.

[Marco]

You 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.

I'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.

Here ungrounded 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?

[CaptnYellowShirt]

Yeah, but its the *why* that's bothering me.

[tree]

You 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.

I'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.

Here ungrounded 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?

I 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.