Does AC current pass through perfectly shielded box?

[goodphy]

Hello.

I've recently have a lot of question of building Faraday cage for our high voltage gas discharge lab. I think this question is a kind of fundamental question for such a job.

Let's say I would make perfect Faraday cage with perfect shielding or have metal plate of infinity transverse size. Thickness of the plate is much larger than skin depth of the AC current. Then AC current path (wire) are connected to both surface of the metal plate in a way described in attached image to try to communicate between two sides with shielding radiation EMI.

I believe this is actually not conducting but just good insulation since thickness of the metal plate is much larger than current skin depth. As a result, if the current can flow, the current only can flows along the surface then...never succeed to fine a way to get inside of the shield as shield is assumed as perfect.

If this is true, when we make metal enclosure of electronics and the enclosure is to be used as a current return path, it is always better to drill some hole for AC current to easily pass.

Could you please confirm this is right logic?

 

[Jeff Rosenbury]

I think you are mistaken. What you say would be true for a perfect electric conductor.

But the difference between a conductor and an insulator is in the amount of conductance. They aren't really different materials at heart. (Though there are many orders of magnitude in the difference.)

So a magnetic field impinging on on the plate will set up eddy currents. These currents will cause resistive losses. This will allow some of the magnetic field to pass through the plate. The amount of signal passing through is characterized by the skin depth. The skin depth is defined as the depth where the signal is attenuated by 1 Neeper. (8ish dB). So unless the skin depth is infinite, some signal will get through. How much depends on how many skin depths thick the plate is. (Which is frequency dependent.) Each skin depth drops the signal by 8ish dB. To stop everything, you would need an infinitely thick plate as well.

There is an additional loss effect though. The metal plate would act as a wave guide to steal power from the signal. How much it takes is complicated. It would depend on the frequency vis a vis the geometry of your current path. I'll leave that explanation to some wave guide wonk. I is too complicated for me.

If you do want current to pass into your real world Faraday cage, you will need to make some holes. These should be thought of as wave guides. (Not really a problem at 60 Hz, but for higher frequencies...) If this is a problem, put metal around your current path as well.

 

[the_emi_guy]

goodphy,
You instincts are correct, the current will not penetrate the faraday shield. However, as the previous poster mentioned, it is not about "insulation".

You are overcomplicating the analysis. This is a simple circuit problem, no need to invoke surface currents of skin effect for first order analysis.

Something I learned a long time ago that has served me well over the years when dealing with noise issues:
1 - Draw *complete* circuits.
2 - Follow the current

You are not drawing a complete circuit. You are drawing a wire going into the box and imagining that there is a return path but you have failed to include it in your drawing.

Imagine a light bulb connected by 2 wires to a battery. We have a complete circuit including return path. The bulb lights.

Now imagine placing the light bulb inside of the faraday shield and bonding *both* of the wires to the shield on their way in. Will the bulb light? Why or why not?

What happens if we bond one of the wires to the shield?

 

[goodphy]

Thanks for replying.

Apparently making closed loop is enough to get the circuit work in first order analysis as you said. But I'm dealing with high voltage experiment in which discharge pulse occurs for ~400 ns in single cycle, I think I need make system quickly response as much as possible.

Thus you have a point that closed loop eventually work to flow current But in this case, without hole, do you think current has to take a longer time for finding a way to get inside (or from inside to outside) compared with case that there is hole near contact points?

My most concern is how to make system faster to deal with transient voltage.

 

[the_emi_guy]

goodphy,
You have not really described what you are trying to accomplish. Are you trying to transfer a pulse into the faraday cage?

 

[goodphy]

We have metal box (Faraday cage) in which discharge circuits are installed. There are thyratron as fast electrical switch and capacitor as a storing energy to discharge. We have grounded this metal box for discharge current to safely escape to ground for safety in a way told in the attached image. Yes these two red wires are actually ground wires.

I've always curious this is really good connection geometry as transient current can overpass the shield instead of direct flowing between two wires..

 

[the_emi_guy]

goodphy,
So you are trying to transfer this pulse out of the cage?

 

[goodphy]

Hello.

Basically, I think yes.

goodphy,
So you are trying to transfer this pulse out of the cage?

 

[the_emi_guy]

goodphy,
You are not sure?

How high is the voltage?

 

[goodphy]

goodphy,
You are not sure?

How high is the voltage?

Voltage is about 30 kV. What I'm was sure is the role of the ground of the discharge metal box. I'd said that this would be possible escaping route for discharge current, which means this ground is carrying current. But I also had another thought that current would choose to go toward capacitor instead of Earth when I replied to you. In this case ground only gives us reference potential for discharge and prevent the box from floating in voltage.

Just right now, I found old data measuring current through the ground wire using current transformer and yes, certainly current is flowing to the ground. Then the current must quick flow to the Earth as much as possible for safety.

 

[the_emi_guy]

goodphy,
If you are discharging a 30kV capacitor, current is leaving one terminal of the capacitor, going through the thyratron switch, then eventually returning to the other end of the capacitor. This is your circuit. You need to diagram this complete circuit including where you what this current to go before it returns back to the capacitor.

Draw *complete* circuits.
Follow the current