Of shielding and apertures and seams

[cyclone24]

Of shielding and apertures and seams...

I am working on a project that deals with the 400 MHz RF source that is interfering with a diagnostic device close to it which is a complete DC circuit.

Shielding with thin copper is suggested. Read the theory behind the influence of apertures and seams at high frequencies.

If I have to use the numbers, at 400 MHz, the wavelength is 75 cm. If I consider 1/20th of the wavelength, it would be 3.75 cm. Now the holes that I have are less than 3 mm. Considering the shield is perfect except for the 5 holes that are evenly spaced, I see the EMI.

Now is there any rational reasoning to assume that the RF can leak into those tiny holes at that frequency?

Any pointers would be helpful...

 

[marcusl]

You'll need to provide more information. Here are some questions:
1) Is the shield a complete closed surface ("Faraday cage")? Sticking pieces of foil here or there isn't much help.
2) How did you treat the seams? A rule of thumb is
RF tight = water tight
No drips allowed.
3) How are leads entering / exiting the cage? You need to use EMI filters on each and every lead.

 

[cyclone24]

Thanks for the reply:

1. It is a faraday cage except for the holes that are 3 mm in diameter.
2. I liked your analogy. Given the geometry, I tried putting copper tabs to cover the seam. But it is more or less open. Any other suggestions here? I think that is the main hurdle for me now. Again what is the max. open seam length/area beyond which the RF can leak?
3. The leads are entering through a hole that has 1 cm linear length, which is less than 1/20th of the wavelength at that frequency.

Value of EMI filters is what I am exploring at the moment. I posted my queries in the thread "Decoupling filters"

 

[vk6kro]

You might need to be using feed through capacitors:

A suitable hole is drilled in a shield and the capacitor is soldered into position.
It provides a bypassing and feedthrough function as the wire is continuous from one side to the other.

 

[cyclone24]

RF tight = water tight
No drips allowed.

Can anyone please elaborate the above statement? Can RF still penetrate into seams and holes even at wavelengths less than 1/4 or 1/100 etc.

Thanks

 

[marcusl]

RF tight = water tight
No drips allowed.

Can anyone please elaborate the above statement? Can RF still penetrate into seams and holes even at wavelengths less than 1/4 or 1/100 etc.

Thanks

Lambda/4 is too big, a seam or slit that long still acts enough like an antenna. Slots and holes should be no bigger than lambda/10 for good shielding.

Thanks for the reply:
3. The leads are entering through a hole that has 1 cm linear length, which is less than 1/20th of the wavelength at that frequency.

Sticking a wire though a hole or tube blows everything--the wire conducts EMI straight through. (It's called a transverse electromagnetic or TEM waveguide.) You need a proper EMI filter as I mentioned in bullet #3. The feedthrough capacitor suggested by vk6kro is perfect for a DC current. An AC signal would require a different feedthrough.

 

[cyclone24]

Well the leads are well shielded as they go in or come out of the faraday cage.

I have circular holes that are 3 mm in diameter on the faraday cage and I have 400 MHz source near it. Now are you saying RF can leak through those holes? I am still stuck at this fundamental question.

"Is it the components inside the faraday cage that are still radiating at 1/200 wavelength or is the harmonics of the fundamental frequency that is influencing the components to radiate?

 

[marcusl]

3 mm holes should be fine. Two questions: 1) Did you put the high frequency circuit in the shield? (It won't do much good to try to shield the low frequency instrument.) 2) What do you mean by "the leads are well shielded as they go in or come out of the faraday cage"? Apparently you are not using EMI feedthrough filters. Can you describe the penetrations in detail?

 

[cyclone24]

Sure...

1. I have a high freq. (DC) PCB inside a faraday shield covered with copper. An RF source generating 400 MHz freq. is placed near the shield as close as 3 mm. Now the faraday cage has multiple screw holes where plastic screws go in. The diameter of the screws is 3 mm.

In this scenario, I see the RF being picked up by the PCB that is inside the shield. So is it correct to say that RF is leaking through these holes that are 1/200 wavelength (assuming its a reasonable faraday cage)?

My point is can RF really leak into those small holes. the literature says lambda/10 is good etc., but looks like even that is not enough. Is there anything else that I am missing?

2. There was a rectangular opening that is 1 cm long through which the PCB trace comes out. The whole trace is encapsulated by the copper shield without any openings etc. So its an extended faraday cage.

Note: Its an example of near-field condition

 

[f95toli]

But again: Are you using EMI filters for ALL feedthroughs?
Even a perfect shield will be ruined by a cable entering it, unless the feedthrough is properly filtered.
You can't just drill a small hole (regardless of how small it is) and have a cable going straight through, it will inevitably act as a path for the RF interference.

 

[marcusl]

Is there anything else that I am missing?

Yes, you are not paying attention to the comments and advice you are getting. Suggest you reread the prior responses carefully. Then let's try one more time:

1. There is little radiation through holes that are smaller than lambda/10. Holes that are lambda/200 in diameter are not your problem. Move on.

2. You should shield the source of noise, that is, your 400 MHz system. The reason is that it will radiate noise into your DC circuit, and into the wires going to/from the DC circuit, and into everything that those wires connect to (power supplies, measurement instruments, processors, readouts, AC mains, etc.) Shielding your DC circuit alone, as you are discovering, is inadequate.

3. How are you treating the wires going into/out of your shield? They need to pass through feedthrough capacitors (DC) or EMI filters (AC) that are properly screwed or soldered into the shield.

Your last post brings up another question.
4. You say "In this scenario, I see the RF being picked up by the PCB that is inside the shield." How do you see this pick up?

 

[cyclone24]

I appreciate your responses. lambda/20 is a standard. I had the question of whether this standard is true for PCBs too compared to wires.

1. Thanks for making it clear. Holes or slits of that dimension is not the problem then. Got it!

2. You are suggesting that shielding my DC circuit components is NOT the same as shielding the 400 MHz source. I imagined them to be the same thing because of the proximity of the shield to the RF source. My RF source is NOT meters or centimeters away from the shield that is protecting the PCB, but it is 2-3 mm.

3. Imagine my shield is a doughnut size. I just covered the inlet/outlet coming from the doughnut with copper tape with no compromise on shield integrity. Btw, there is only one inlet. Tried EMI filters, made it worse.

4. RF source is generating RF bursts at recurrent times (not continuous). So I see the pickup in my data whenever the source is ON. The magnitude of the pickup was minimized by adding another layer of shield covering the holes (again this made me think that holes are the culprit).

Now the pickup minimized because I added the shield (thickness?) or because I made it 'RF tight'?

 

[marcusl]

1. We need to revise this comment in light of #2 and your new information in #4.

2. Having the source just 2-3 mm from the holes could compromise their shielding. It is significant if the receiving circuit is 2-3 mm from the hole, for the following reason. Even when the hole is small compared to the wavelength, a little electromagnetic field (called an "evanescent wave") protrudes approximately 1 hole diameter into the shield before attenuating away. If there is something sitting in this field, it could have power coupled onto it.

2. My point is that an unshielded RF source can couple noise onto other things that your DC circuit might be connected to, like displays, power supplies, etc. I don't know if you have such a setup, or whether this is a problem for you, but as general advice it is better to eliminate the noise from everywhere if you can than from just one part of a system.

4. A pulsed source can have considerably higher frequency content than a continuous one, so this is important new information. If your RF pulses have fast rise/fall times, frequency content could extend well into the microwave regime. For example, if you had components at 6 GHz, then 3 mm holes are just lambda/17. With source and receiver so very close together, the holes act as significant leaks.

 

[cyclone24]

marcusl

Thank you for your inputs for all the queries. The mechanics of my setup doesn't allow me to shield the RF source. However, I agree with your explanation. RF transition times and edges do matter.

In summary, I wanted to understand how a second layer worked. Is it the thickness or the layers. There is lot of theory about multilayer shielding etc. But basically I am making my system RF tight even though 3mm hole is 1/200*wavelength.

On a closing note, I know that 1/20*wavelength is a rule of thumb, a standard measure that is followed everywhere. Can you point me where did this '1/20' come from? Is there a documented literature or paper that you could share? why is it 1/20 and not 1/50 or 1/34 etc. Thanks.

 

[cyclone24]

Just a thought,

Can I do an analogy that RF tight = water tight = light tight?

I have a diode inside the RF shield. Can I say vaguely that if the Faraday cage shield is so good that even light cannot penetrate through the shield? But if there is a light leak in the shield that means RF may pass through the shield?

Or is it irrelevant...? thanks

 

[marcusl]

On a closing note, I know that 1/20*wavelength is a rule of thumb, a standard measure that is followed everywhere. Can you point me where did this '1/20' come from? Is there a documented literature or paper that you could share? why is it 1/20 and not 1/50 or 1/34 etc. Thanks.

Ott, in Noise Reduction Techniques in Electronic Systems, cites J. Quine, "Theoretical Formulas for Calculating Shielding Effectiveness...," Proc. 3rd Conf on Radio Interference Reduction, p. 315-329, Feb. 1957. I haven't seen it but the title sounds good. Also look at books on EM compatibility (EMC).

But if there is a light leak in the shield that means RF may pass through the shield?

Or is it irrelevant...? thanks

Solid metal is a perfect shield, while as you've seen, penetrations can cause leakage.