EMI Shielding Effectiveness with Various Conductor Thicknesses

[Ntip]

TL;DR Summary

I would like to know how a conductors thickness effects EMI shielding performance

Let's assume that I have have one conductor carrying current to the load in one direction and the return line with current in the opposite direction from the load. If the current has high di/dt due to switching, may have EMI issues with nearby circuits. I know that I can put a "ground" plane on the outside of the current carrying conductors which would ideally have no current to shield the outside electrical circuits. I'm wondering how the thickness of the conductor effects the shielding effectiveness.

 

[berkeman]

Summary:: I would like to know how a conductors thickness effects EMI shielding performance

Let's assume that I have have one conductor carrying current to the load in one direction and the return line with current in the opposite direction from the load. If the current has high di/dt due to switching, may have EMI issues with nearby circuits. I know that I can put a "ground" plane on the outside of the current carrying conductors which would ideally have no current to shield the outside electrical circuits. I'm wondering how the thickness of the conductor effects the shielding effectiveness.

You will have B-field interference issues unless you do things like twisting the two conductors. That is why there is twisted pair wire -- to null out the B-field pickup from- and interference induced in- other nearby circuits. It's hard to get twisted pair after you get above 16AWG, though. How much current are you talking about? What is the application?

 

[Ntip]

I don't really have an application as of now. I was talking about the theory behind it with a friend and he had a similar answer to yours. I still don't understand the reason why though. I thought its common to find ground planes in PCBs to reduce the noise on signal layers. What's the difference in this?

 

[Baluncore]

I still don't understand the reason why though.

It is all done with mirrors.

When an EM wave hits a conductive sheet it is reflected and reversed, which is exactly what happens when you look in a mirror, and see your reversed image behind the conductive silver coating.

A ground plane is a conductive sheet that reflects the EM field from conductors in other parallel planes. The effect is to construct a reversed virtual image behind the mirror. For wavelengths that are long compared with the separation, the sum of the field generated by the conductor, and the field generated by the image will be zero, because one is reversed they cancel.

Shielding, or the braid on a coaxial cable is also a mirror. The screen prevents fields passing through the mirror from either side, which reduces interference to both sides.

 

[berkeman]

I don't really have an application as of now. I was talking about the theory behind it with a friend and he had a similar answer to yours. I still don't understand the reason why though. I thought its common to find ground planes in PCBs to reduce the noise on signal layers. What's the difference in this?

You can use ground shield layers between PCB trace layers to lower E-field crosstalk, but if you are using high-speed transmission lines (TLs), you have to be careful about maintaining consistent Zo all along the TL (and not lowering it with shield layers that are not accounted for in the TL layout).

Information about B-field shielding with twisted pair cables can be found with a Google search, resulting in pages like this one:

https://en.wikipedia.org/wiki/Twisted_pair

 

[Ntip]

A constant current would have a constant B-field but wouldn't switching current have a changing magnetic field which would induce an electric field? In that way, it seems like you could say that the ground shield layer also minimized noise from di/dt.

 

[berkeman]

A constant current would have a constant B-field but wouldn't switching current have a changing magnetic field which would induce an electric field? In that way, it seems like you could say that the ground shield layer also minimized noise from di/dt.

Are you familiar with the concept of skin depth with frequency? If not, it would be good to do a Google search to learn more about the effectiveness of a conducting layer at shielding B-fields and EM fields with frequency.

 

[Ntip]

Are you familiar with the concept of skin depth with frequency? If not, it would be good to do a Google search to learn more about the effectiveness of a conducting layer at shielding B-fields and EM fields with frequency.

Yes I am familiar with skin depth. I'm actually kind of expecting this to have something to do with skin depth. I looked at http://www.magnetic-shield.com/pdf/magnetic_fields_shields_overview.pdf and they also discuss the need to use conductive layers for shielding >100 kHz frequencies. This website https://www.strouse.com/blog/importance-of-emi-shielding mentions that the material thickness should be thick enough to prevent the frequency from penetrating. Is there a rule of thumb for like that at least 2 skin depths?

In this case how exactly is the EMI being shielded? Here are my thoughts:
As the B-field crosses the conductor it induces a current. Energy from the B-field is then reduced since it was required to induce the current.

 

[berkeman]

DC E-field shielding is just done with a conductor, and the thickness doesn't matter much. DC B-field shielding is done with high-mu materials, typically a layering of netic and co-netic high-mu metals.

EM shielding is done with a varying thickness of a conductor/metal, with the thickness depending on the skin depth for the EM frequencies that you want to shield.

 

[alan123hk]

In this case how exactly is the EMI being shielded? Here are my thoughts:
As the B-field crosses the conductor it induces a current. Energy from the B-field is then reduced since it was required to induce the current

You can consider referring to the intuitive explanation of skin effects on https://en.wikipedia.org/wiki/Skin_effect

From wiki :-

Conductors, typically in the form of wires, may be used to transmit electrical energy or signals using an alternating current flowing through that conductor. The charge carriers constituting that current, usually electrons, are driven by an electric field due to the source of electrical energy. A current in a conductor produces a magnetic field in and around the conductor. When the intensity of current in a conductor changes, the magnetic field also changes. The change in the magnetic field, in turn, creates an electric field which opposes the change in current intensity. This opposing electric field is called “counter-electromotive force” (back EMF). The back EMF is strongest at the center of the conductor, and forces the conducting electrons to the outside of the conductor, as shown in the diagram on the right.

The skin depth of copper is about $\left( \frac {66} {\sqrt {f}} \right) mm$ , which means 0.066mm at 1MHz.
The skin effect is why the inside of the telescopic antenna can be hollow.

 

[Ntip]

DC E-field shielding is just done with a conductor, and the thickness doesn't matter much. DC B-field shielding is done with high-mu materials, typically a layering of netic and co-netic high-mu metals.

EM shielding is done with a varying thickness of a conductor/metal, with the thickness depending on the skin depth for the EM frequencies that you want to shield.

My thoughts were that if I have a power plane with switching currents on it, there could be EMI due to this plane. I thought a ground plane could be placed between the signal plane and the power plane to eliminate switching noise from coupling to the signal traces. I thought I understood why that helps, but I guess I don't understand the reasoning behind it. Doesn't the B-field from the switching current create eddy currents in the ground plane? If there are eddy currents in the ground plane, why does that not couple through the mutual inductance with the signals?

I know that it works, but I'm trying to understand why instead of just placing them like this and moving on without understanding.

 

[berkeman]

My thoughts were that if I have a power plane with switching currents on it, there could be EMI due to this plane. I thought a ground plane could be placed between the signal plane and the power plane to eliminate switching noise from coupling to the signal traces. I thought I understood why that helps, but I guess I don't understand the reasoning behind it. Doesn't the B-field from the switching current create eddy currents in the ground plane? If there are eddy currents in the ground plane, why does that not couple through the mutual inductance with the signals?

I know that it works, but I'm trying to understand why instead of just placing them like this and moving on without understanding.

The currents in PCB planes are very spread out, so they tend not to create noise coupling issues. If you can post your PCB layer stackup and your layout, we can answer more specific questions.

 

[Baluncore]

Doesn't the B-field from the switching current create eddy currents in the ground plane? If there are eddy currents in the ground plane, why does that not couple through the mutual inductance with the signals?

Because a good conductor only has induced high frequency currents flowing on the surface on the side with the B-field. Skin effect prevents the current reaching the other side of the copper plane. A ground plane is like two mirrors, back to back, that isolate the universe on the two sides of the plane.

Because the induced current in the plane is opposite to the current that caused the B-field, the B-filed of the reflected image in the conductive surface cancels with the current and B-field that caused it.