# Why grounding the shield helps reducing noise?

#### Marcelo Besch

Joined Mar 25, 2015
10
Hello!

This is a matter that I have never understood. Inside a conductor (the shield in this case) there is not supposed to be any electrical field. If there was, the charges on the conductor would move in order to balance it, making it zero on the inside. So, it is perfectly understandable that using a shield will reduce the noise. However, why is it important to ground it??

Joined Jul 18, 2013
23,381
The shield is supposed to be protection from external EMI, if it were ungrounded and of any significant length any external EMI field could present a potential between one end of the shield and the other and impose this on the internal conductors so essentially ineffective..

Max.

#### #12

Joined Nov 30, 2010
18,222
Max has it, but here's another way to say it. If it isn't grounded, it isn't a shield, it's an antenna, and an antenna radiates from every surface.

#### Marcelo Besch

Joined Mar 25, 2015
10
I didn't understand. The external emi generates a potential between one side of the shield and the other in a way that the internal field generated by the shield would oppose the external field. That way, there would be no internal field, which is exactly what we want.

#### studiot

Joined Nov 9, 2007
4,998
This is a matter that I have never understood. Inside a conductor (the shield in this case) there is not supposed to be any electrical field
Think again.

An external field reduces to zero across a conductor, yes.

But apply a voltage along the conductor, what causes the current if not an electrical field?

Now suppose you have a metal body interposed between a signal carrying conductor and an electric field.

If that body is grounded it is at fixed potential (ground).

If that body is ungrounded than the field can (and does) induce a voltage in the body.

If that field now varies (randomly?) the induced voltage varies.

This induced voltage is equivalent to an unwanted signal (noise) in series with the signal fed to some device.

Joined Jul 18, 2013
23,381
I didn't understand. The external emi generates a potential between one side of the shield and the other in a way that the internal field generated by the shield would oppose the external field.
The external field would be imposed on whatever signal information was being carried by the internal conductors.
Max.

#### #12

Joined Nov 30, 2010
18,222
the internal field generated by the shield would oppose the external field. That way, there would be no internal field
Do you see the error in that logic? The internal field opposes, therefore there is no internal field.

I seem to be slow today. See post #5
A pdf is worth a thousand words.

#### Marcelo Besch

Joined Mar 25, 2015
10
@#12 I got a little confusing with the names I used to reference the fields

It would be better to have named them external field, shield generated field and internal field. The shield generated field is opposite to the external field, so the internal field becomes zero. I made a little explanation of my logic using paint, its attached to this reply.

@studiot yes, there is a field on a conductor when you apply a voltage, which generates the current. However, if you don't provide a way for the charges not to accumulate, the conductor would soon gather enough to create an electrical field opposite to the one you're applying, interrupting the current. It's what happens when you charge a capacitor, for example. But I see what you meant, I was very informal when I presented my reasoning.

@nsaspook thank you for the material, I will study it.

@MaxHeadRoom I believe the image I am attaching to this reply will make it clear as to why I still didn't understand why an ungrounded shield would allow external fields to interfere on the internal cables. Grounded or not, the shield should be able to move its charges to oppose the external field. I will read the pdf that nsaspook has sent to see if it presents some other phenomenon that I am not accounting for.

#### Attachments

• 17.8 KB Views: 30

#### Marcelo Besch

Joined Mar 25, 2015
10
By the way, that terrible drawing I attached is a cross section of a shield

I am not good at drawing, specially using ms paint.

#### Marcelo Besch

Joined Mar 25, 2015
10
@MaxHeadRoom I am aware of the current preferred methods. I have seen in experiments and in some situations at my job that not grounding the shield causes noise problems. I just don't understand why, from the point of view of physics.

#### studiot

Joined Nov 9, 2007
4,998
Marcelo, you only seem to have read the first part of my post.

I haven't time to add diagrams now, maybe later.

#### studiot

Joined Nov 9, 2007
4,998
Interesting document, Max, but it does not explain why, although it explains how rather fully.
Earth loop interference is proportional to the area enclosed by the loop.
So grounding at both ends is good for small earth loops, but less good for large ones.

Here is an interesting exercise.

Take a long shielded cable (say 30 metres or more in length), carrying a 10 volt signal.
Subject a one metre section to an interference of 1 volt and consider where would the ideal grounding be located for different interference points along the cable run?

#### nsaspook

Joined Aug 27, 2009
8,531
I have seen in experiments and in some situations at my job that not grounding the shield causes noise problems. I just don't understand why, from the point of view of physics.
Maybe you are talking about two things here. A Faraday cage with perfect shielding provides protection from electric fields to objects isolated inside but shielding (electromagnetic) as we normally use the term with shielded conductors and circuits not completely enclosed by shielding requires a shield reference ground connection for the noise currents to loop and be redirected into instead of the signal path loop. You can't eliminate the noise energy but you can shunt it away with a shield ground and/or neutralize it by canceling the fields.

Ungrounded Faraday cage with electrostatic field cancellation.

#### studiot

Joined Nov 9, 2007
4,998
Pretty, spook

Now connect the reference ground for my 10v signal to one side of your shield, run a signal through cable in the box to a receiver grounded at the other side.

#### Marcelo Besch

Joined Mar 25, 2015
10
@nsaspook nice GIF! That's exactly what I thought a shield did. I thought a shield behave as a Faraday cage.

I mean, the cables are not 100% enclosed in the shield, but their length inside the shield is so much greater than what is left outside that I thought it could be approximated.

I still don't understand why isn't shielding just like using a Faraday cage on the wires. And, as the gif showed, there is no need to ground the cage, it works without this connection.

#### studiot

Joined Nov 9, 2007
4,998
I still don't understand why isn't shielding just like using a Faraday cage on the wires. And, as the gif showed, there is no need to ground the cage, it works without this connection.
And you still have not read?? my explanations or responded to them.

#### Marcelo Besch

Joined Mar 25, 2015
10
@studiot I'm sorry! About what you said: I simply don't understand what difference would exist if the metal body was grounded or not. Look at the GIF sent by @nsaspook: what difference would there be if that faraday cage was grounded?

#### nsaspook

Joined Aug 27, 2009
8,531
@studiot I'm sorry! About what you said: I simply don't understand what difference would exist if the metal body was grounded or not. Look at the GIF sent by @nsaspook: what difference would there be if that faraday cage was grounded?
The difference would be that (most of) the electrostatic field energy would be shunted into ground instead of charging the enclosure and the inside would still be neutral.
Look at a typical RS232 cable.

Pin1 is the drain/electrostatic shield connection. The signals (including the signal ground) are inside the shield. If an external noise field were to charge the ungrounded shield it would act like a coaxial capacitor outside plate with the wires as the inside plate coupling energy into the signal lines to develop a noise current and voltage across the signal receiver resistance.