Copper tape shielding and capacitance

[Rabbitz]

Hi All,

Just mooching around the shed and decided to experiment with something that has been bugging me.

I overlaid two pieces of copper tape:



As is done in most guitar construction efforts.

Now, a capacitor is essentially two pieces of metal overlapping each other separated by a dielectric material.

So two pieces of tape separated by the glue backing, does this equal a capacitor? In other word is the overlapping of tape inserting capacitance into our guitar circuitry.

A little surprisingly, the answer is Yes.

I simply measured the capacitance between the two pieces of copper.

In the randomly selected but probably indicative sizes used in my experiment it resulted in almost 0.4 nF (Nano-Farads).

A small amount, for sure, but in an AC circuit the reactance is also of concern. Reactance is the resistance presented by a capacitive circuit - as frequency decreases reactance increases, so if our signal is travelling across the shield, and as the ground component of the AC it can, will this affect our bottom end response?

I don't really know enough about the response side of things to say, and it would be difficult to objectively test.

So I'll throw it over to some of the more experienced wire fiddlers to shoot down my theorem - as the scientific method would ask me to do...

Oh, btw, the way the simple way to overcome this reactance issue would be to make a solder bridge across the joins, thus the path is no longer capacitive.

[dingobass]

Oh, btw, the way the simple way to overcome this reactance issue would be to make a solder bridge across the joins, thus the path is no longer capacitive.

And this is why I always solder the joins and only use good quality shielding tape... Not snail tape!

[Rabbitz]

So after a little more experimenting albeit at radio frequencies I came up with some surprising numbers.

At 1.6MHz through 0.4 nF showed a reactance of around 450Ω. (this is the lowest frequency my test gear will measure)

I then started digging through my radio theory books to find the equation, then I had a brainwave, maybe the interwebs has the equation? Better still it has a calculator:
0.4 nF at 300Hz = 1.3 MΩ
0.4 nF at 3kHz = 1.32 kΩ

The amount of reactance/resistance surprised me.

Bottom line, follow DingoBass' advice = solder the joins!!!!!!!!!!

[stan]

dude, you have too much time on your hands...

very interesting info though

[jarrod]

As in solder where the tape overlaps Rabbitz ?

[dingobass]

Spot on Jrod

You only need a dab of solder on each join, but if you are anally retentive like I am about these things do the whole join..

[Rabbitz]

dude, you have too much time on your hands...

very interesting info though

Well, today I've made rillettes, mowed the lawn, finished sanding the ST-1, potted some plants, put a clear coat on the ST-1, fixed a leak in the roof, worked out why the water tank doesn't seem to be filling - is it any wonder I was pottering in the shed (out of sight) waiting for the clear coat to dry?



Hey Jarrod, yes a blob will do, essentially we want a low resistance path between the pieces of copper, kind of shorting out the capacitance/reactance.

Now I hope we don't have to worry about "skin effect" surface flow between the layers in the unbalanced circuits...

[Alm_63]

Rabbits, I about to reply show my ignorance.
The tape is connected (or should be ) directly to earth., and the frequencies we would be draining out the earth would be either 60 Hz or 120 from some flouro's .?
So if we have a direct path for the pickups, and a filtered path for EMF, would it effect the signal from the pickups?

[Rabbitz]

Hi Bruce,

The ground is part of the circuit, it is the "other half" of the AC cycle if you like.

A lot of the EMI (noise, hum, etc) is lessened due to the handling of the ground outside of the guitar. There is still part of the signal also travelling in the earth.

I gather the cavity shielding forms a type of Faraday Shield, which aims to prevent the electrical noise from getting into the signal and becoming part of the amplified signal.

I am just trying to really understand this myself (a bit of a new field for me and is an idle curiosity), but for me the "drain analogy" seems a little flawed when you really start to dig in. However, it is a useful simplification.

The issue here, if it is in fact an issue, is not the ability of the EMI to be removed it is the fact that the dielectric effect of the overlaps could make it difficult for the earthing to happen due to the high resistance (reactance) presented at the join. 0.4 nF @ 60Hz will be seen as 6.6 MΩ. There are a million other factors, such as alternate paths in the wiring, which will probably mitigate the effects, but I think we really want to get rid of the reactance if we want the Faraday Shield to work as it should.

The capacitance/reactance thing is really a bit left field and probably is of little real world consequence, but as it will take about 10 minutes to solder some joins. After all the fiddly mucking around with the tape, I think it is worth the effort.

[dingobass]

Yep, it's a Faraday shield