Would a small hole in a Faraday cage drastically reduce its effectiveness at blocking interference?
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Would a small hole in a Faraday cage drastically reduce its effectiveness at blocking interference?
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$begingroup$
I'm shielding the pickup cavity of a guitar to minimize interference. I'm doing this with copper tape that has conductive adhesive, and I'm connecting this tape to ground.
Without getting into too much detail, let's say there's a small spot I miss or purposely leave uncovered because it's difficult to reach, so that the Faraday cage is not completely "sealed" all the way around. Would this drastically reduce the effectiveness of the cage, or would it simply reduce it proportionally to the size of the hole?
I ask because I'm fine if it's just slightly less effective, but if it ruins the whole thing, then I'll put in the extra effort.
faraday-cage
edited May 3 at 16:48
Glorfindel
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
$endgroup$
|
show 4 more comments
$begingroup$
I'm shielding the pickup cavity of a guitar to minimize interference. I'm doing this with copper tape that has conductive adhesive, and I'm connecting this tape to ground.
Without getting into too much detail, let's say there's a small spot I miss or purposely leave uncovered because it's difficult to reach, so that the Faraday cage is not completely "sealed" all the way around. Would this drastically reduce the effectiveness of the cage, or would it simply reduce it proportionally to the size of the hole?
I ask because I'm fine if it's just slightly less effective, but if it ruins the whole thing, then I'll put in the extra effort.
faraday-cage
edited May 3 at 16:48
Glorfindel
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
$endgroup$
1
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
3
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
3
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
2
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58
|
show 4 more comments
$begingroup$
I'm shielding the pickup cavity of a guitar to minimize interference. I'm doing this with copper tape that has conductive adhesive, and I'm connecting this tape to ground.
Without getting into too much detail, let's say there's a small spot I miss or purposely leave uncovered because it's difficult to reach, so that the Faraday cage is not completely "sealed" all the way around. Would this drastically reduce the effectiveness of the cage, or would it simply reduce it proportionally to the size of the hole?
I ask because I'm fine if it's just slightly less effective, but if it ruins the whole thing, then I'll put in the extra effort.
faraday-cage
edited May 3 at 16:48
Glorfindel
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
$endgroup$
I'm shielding the pickup cavity of a guitar to minimize interference. I'm doing this with copper tape that has conductive adhesive, and I'm connecting this tape to ground.
Without getting into too much detail, let's say there's a small spot I miss or purposely leave uncovered because it's difficult to reach, so that the Faraday cage is not completely "sealed" all the way around. Would this drastically reduce the effectiveness of the cage, or would it simply reduce it proportionally to the size of the hole?
I ask because I'm fine if it's just slightly less effective, but if it ruins the whole thing, then I'll put in the extra effort.
faraday-cage
faraday-cage
edited May 3 at 16:48
Glorfindel
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
edited May 3 at 16:48
Glorfindel
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
edited May 3 at 16:48
Glorfindel
4071510
edited May 3 at 16:48
Glorfindel
4071510
edited May 3 at 16:48
Glorfindel
4071510
4071510
asked May 3 at 2:49
AnthonyAnthony
29726
asked May 3 at 2:49
AnthonyAnthony
29726
asked May 3 at 2:49
AnthonyAnthony
29726
29726
1
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
3
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
3
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
2
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58
|
show 4 more comments
1
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
3
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
3
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
2
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58
1
1
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
3
3
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
3
3
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
2
2
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58
|
show 4 more comments
5 Answers
5
active
oldest
votes
$begingroup$
It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. But the holes in the mesh must be much smaller than the wavelength you're trying to screen.
In particular, it's the largest dimension of the hole, not its area, that matters. A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide.
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
$endgroup$
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
|
show 6 more comments
$begingroup$
Regarding the actual use case in question:
Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. (How's your soldering?)
If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring (unbeknownst to you!).
If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads.
answered May 3 at 20:04
schadjoschadjo
748212
$endgroup$
add a comment |
$begingroup$
The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. Reality check: a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less.
If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) thick is essentially transparent to such waves.
At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. A wavelength at that frequency is still around 300m, so small holes will not matter. Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear).
At 100 MHz the copper foil is really effective (with skin depth of only 6μm). The wavelengh is around 3m, so holes of reasonable size will not be of your concern.
Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic.
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
$endgroup$
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
add a comment |
$begingroup$
As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
$endgroup$
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
add a comment |
$begingroup$
Faraday cages aren't entirely closed boxes. Like cages, they have gaps. The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount.
For example: a microwave oven always has a mesh in front of the window. The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through.
Another example: your car can be considered a Faraday cage in the event of lightning. It will protect you from the strikes, because the wavelength is way too large. However... because of the huge gaps in the cage (glass windows) we can still receive cell phone signals through it.
I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). So as long as the gap isn't too big, I don't think it'll be too much of an issue.
answered May 4 at 10:05
OpifexOpifex
1955
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. But the holes in the mesh must be much smaller than the wavelength you're trying to screen.
In particular, it's the largest dimension of the hole, not its area, that matters. A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide.
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
$endgroup$
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
|
show 6 more comments
$begingroup$
It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. But the holes in the mesh must be much smaller than the wavelength you're trying to screen.
In particular, it's the largest dimension of the hole, not its area, that matters. A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide.
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
$endgroup$
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
|
show 6 more comments
$begingroup$
It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. But the holes in the mesh must be much smaller than the wavelength you're trying to screen.
In particular, it's the largest dimension of the hole, not its area, that matters. A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide.
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
$endgroup$
It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. But the holes in the mesh must be much smaller than the wavelength you're trying to screen.
In particular, it's the largest dimension of the hole, not its area, that matters. A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide.
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
answered May 3 at 3:01
The PhotonThe Photon
89k3104210
89k3104210
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
|
show 6 more comments
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
3
3
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
$begingroup$
@MSalters, you may also be worried about higher frequency RF signals carrying audio modulation, or combinations of RF signals with beat frequencies in the audio range.
$endgroup$
– The Photon
May 3 at 13:43
3
3
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
$begingroup$
@MSalters In what universe does a 60Hz signal have a wavelength of 15 meters?
$endgroup$
– Dmitry Grigoryev
May 3 at 14:22
4
4
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
$begingroup$
@DmitryGrigoryev: at least 15 meters, but I can indeed tighten that limit to at least 15 kilometers (for 20 khz waves). Yeah, that's basically the realm where we're better of treating it as electro-static. Ground that cage!
$endgroup$
– MSalters
May 3 at 14:26
4
4
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
$begingroup$
@MSalters, also if you're trying to shield something from frequencies below about 50 kHz, remember the shield thickness needs to be more than a couple of skin depths, which might not be practical for such low frequencies.
$endgroup$
– The Photon
May 3 at 15:47
3
3
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
$begingroup$
@pipe, remember that microstrip lines don't radiate much (at appropriate frequencies), and they're only shielded on one side. And that radiation is reciprocal (an antenna works equally well receiving as transmitting). So yes, a "shield" that only surrounds one side of a structure could (within limits) reduce the radiation from that structure.
$endgroup$
– The Photon
May 3 at 15:51
|
show 6 more comments
$begingroup$
Regarding the actual use case in question:
Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. (How's your soldering?)
If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring (unbeknownst to you!).
If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads.
answered May 3 at 20:04
schadjoschadjo
748212
$endgroup$
add a comment |
$begingroup$
Regarding the actual use case in question:
Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. (How's your soldering?)
If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring (unbeknownst to you!).
If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads.
answered May 3 at 20:04
schadjoschadjo
748212
$endgroup$
add a comment |
$begingroup$
Regarding the actual use case in question:
Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. (How's your soldering?)
If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring (unbeknownst to you!).
If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads.
answered May 3 at 20:04
schadjoschadjo
748212
$endgroup$
Regarding the actual use case in question:
Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. (How's your soldering?)
If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring (unbeknownst to you!).
If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads.
answered May 3 at 20:04
schadjoschadjo
748212
edited yesterday
answered May 3 at 20:04
schadjoschadjo
748212
answered May 3 at 20:04
schadjoschadjo
748212
answered May 3 at 20:04
schadjoschadjo
748212
748212
add a comment |
add a comment |
$begingroup$
The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. Reality check: a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less.
If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) thick is essentially transparent to such waves.
At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. A wavelength at that frequency is still around 300m, so small holes will not matter. Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear).
At 100 MHz the copper foil is really effective (with skin depth of only 6μm). The wavelengh is around 3m, so holes of reasonable size will not be of your concern.
Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic.
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
$endgroup$
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
add a comment |
$begingroup$
The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. Reality check: a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less.
If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) thick is essentially transparent to such waves.
At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. A wavelength at that frequency is still around 300m, so small holes will not matter. Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear).
At 100 MHz the copper foil is really effective (with skin depth of only 6μm). The wavelengh is around 3m, so holes of reasonable size will not be of your concern.
Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic.
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
$endgroup$
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
add a comment |
$begingroup$
The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. Reality check: a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less.
If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) thick is essentially transparent to such waves.
At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. A wavelength at that frequency is still around 300m, so small holes will not matter. Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear).
At 100 MHz the copper foil is really effective (with skin depth of only 6μm). The wavelengh is around 3m, so holes of reasonable size will not be of your concern.
Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic.
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
$endgroup$
The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. Reality check: a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less.
If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) thick is essentially transparent to such waves.
At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. A wavelength at that frequency is still around 300m, so small holes will not matter. Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear).
At 100 MHz the copper foil is really effective (with skin depth of only 6μm). The wavelengh is around 3m, so holes of reasonable size will not be of your concern.
Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic.
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
answered May 3 at 14:20
Dmitry GrigoryevDmitry Grigoryev
18.9k22878
18.9k22878
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
add a comment |
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
2
2
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
$begingroup$
Hm, sounds like I now need to ask a question about all this skin depth of copper and how that affects blocking..
$endgroup$
– pipe
May 3 at 15:19
add a comment |
$begingroup$
As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
$endgroup$
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
add a comment |
$begingroup$
As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
$endgroup$
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
add a comment |
$begingroup$
As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
$endgroup$
As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
edited May 3 at 3:23
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
answered May 3 at 3:16
analogsystemsrfanalogsystemsrf
16.8k2823
16.8k2823
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
add a comment |
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
2
2
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
$begingroup$
Just less than 1 wavelength? A half-wavelength slot antenna radiates pretty well, doesn't it?
$endgroup$
– The Photon
May 3 at 3:29
2
2
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
$begingroup$
The rule of thumb more commonly used is that the largest dimension of the hole must be less than 1/10 the wavelength of the radiation you're concerned about, and depending on application even that can be insufficient
$endgroup$
– llama
May 3 at 23:54
1
1
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
$begingroup$
Notice I included "And your circuit is at least a wavelength inside the shield". This is from Feynman Lectures series.
$endgroup$
– analogsystemsrf
May 4 at 10:02
add a comment |
$begingroup$
Faraday cages aren't entirely closed boxes. Like cages, they have gaps. The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount.
For example: a microwave oven always has a mesh in front of the window. The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through.
Another example: your car can be considered a Faraday cage in the event of lightning. It will protect you from the strikes, because the wavelength is way too large. However... because of the huge gaps in the cage (glass windows) we can still receive cell phone signals through it.
I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). So as long as the gap isn't too big, I don't think it'll be too much of an issue.
answered May 4 at 10:05
OpifexOpifex
1955
$endgroup$
add a comment |
$begingroup$
Faraday cages aren't entirely closed boxes. Like cages, they have gaps. The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount.
For example: a microwave oven always has a mesh in front of the window. The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through.
Another example: your car can be considered a Faraday cage in the event of lightning. It will protect you from the strikes, because the wavelength is way too large. However... because of the huge gaps in the cage (glass windows) we can still receive cell phone signals through it.
I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). So as long as the gap isn't too big, I don't think it'll be too much of an issue.
answered May 4 at 10:05
OpifexOpifex
1955
$endgroup$
add a comment |
$begingroup$
Faraday cages aren't entirely closed boxes. Like cages, they have gaps. The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount.
For example: a microwave oven always has a mesh in front of the window. The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through.
Another example: your car can be considered a Faraday cage in the event of lightning. It will protect you from the strikes, because the wavelength is way too large. However... because of the huge gaps in the cage (glass windows) we can still receive cell phone signals through it.
I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). So as long as the gap isn't too big, I don't think it'll be too much of an issue.
answered May 4 at 10:05
OpifexOpifex
1955
$endgroup$
Faraday cages aren't entirely closed boxes. Like cages, they have gaps. The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount.
For example: a microwave oven always has a mesh in front of the window. The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through.
Another example: your car can be considered a Faraday cage in the event of lightning. It will protect you from the strikes, because the wavelength is way too large. However... because of the huge gaps in the cage (glass windows) we can still receive cell phone signals through it.
I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). So as long as the gap isn't too big, I don't think it'll be too much of an issue.
answered May 4 at 10:05
OpifexOpifex
1955
edited May 4 at 18:44
answered May 4 at 10:05
OpifexOpifex
1955
answered May 4 at 10:05
OpifexOpifex
1955
answered May 4 at 10:05
OpifexOpifex
1955
1955
add a comment |
add a comment |
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1
$begingroup$
I have to wonder, knowing little of music, just how sensitive is this pickup to interference anyway? Or is it the other way around, that it produces interference?
$endgroup$
– Hearth
May 3 at 3:00
3
$begingroup$
@Hearth My understanding is that it depends on the type of pickup. Some pickups (single coils) do nothing to reduce noise. I've heard stories of people picking up radio stations with their guitar.
$endgroup$
– Anthony
May 3 at 3:03
3
$begingroup$
@Toor If you have an answer you need to write it in the Answers section. Your current remark is just a random opinion and can't be vetted by the community or accepted as correct, but will still be the first that everyone will see when the come here.
$endgroup$
– pipe
May 3 at 13:26
$begingroup$
Look at your microwave. It's a Faraday cage (so you can stand in front of it while it's running without getting cooked), but you can see through the door...
$endgroup$
– Sean
May 4 at 21:03
2
$begingroup$
@Sean: Absolutely, but while "there are holes in the shielding" is literally true, it's misleading because it implies the shielding is compromised. It would be more accurate to say that the holes are part of the shielding.
$endgroup$
– Ben Voigt
May 5 at 2:58