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Hello I am confused about how EM shielding works, especially after checking out my VGA cable.

The end of the VGA cable has a metal chassis around the 15 pins, covered by a plastic coat, and 2 screws on the 2 sides. The 15 pins connect to the 15 wires inside the cable. The core wires are insulated by a layer of plastic coat and twisted together, and then shielded by a layer of aluminium foil, and then by a layer of braided aluminium, and then covered by the outside rubber coat. This is how it's made.

Now my assumption is that the shield (the foil+ braided aluminium) has to be insulated from the core of the signal wire otherwise it short circuit's it, or perhaps cause interference and act as a receiver antenna that would absorb the noise to destroy the signal inside the cable.

But that is not true, there is electric connection between the 15 pins, the end-chassis, and even the screws, therefore I assume that the shield itself is also connected to the signal wire. How is this even possible?

Can somebody explain how the electromagnetic shielding works for a VGA cable, because it sounds illogical to me how the inside signal is connected to the shield, I thought they have to be insulated from one another.

Especially how does this shield the inside of the cable from outside noise, doesn't this make the cable an ever bigger antenna if the metal surface is bigger?

And note that this is not a defect cable, I have checked it on 2 different cables, they are all the same and they all work perfectly fine with my monitor.

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  • \$\begingroup\$ How are you measuring this connection? The screws aren't normally connected to anything, so I suspect you may be doing something wrong. \$\endgroup\$ – JRE Aug 19 '16 at 12:54
  • \$\begingroup\$ I connected a little LED light to a small 5V battery and the 2 wires: 1 to the end-side metal chassis of the VGA cable, and the other to the pins (carefully to not touch the chassis). The LED blinked, so there is electric connection. \$\endgroup\$ – Kevin77 Aug 19 '16 at 12:58
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    \$\begingroup\$ Was the other end of the VGA cable plugged in to something? \$\endgroup\$ – HandyHowie Aug 19 '16 at 13:01
  • \$\begingroup\$ No. The cable is on my table, unplugged. I took a small LED light connected to a small 5V battery. The 2 wires that come out of the LED, 1 is connected to the 5V battery, the other is to the end metal chassis of the VGA cable, then another extended wire coming from the 5V battery is carefully connected to the pins (to not touch the chassis). Therefore there is connection between the chassis and the pins, because the LED was lighted up. \$\endgroup\$ – Kevin77 Aug 19 '16 at 13:05
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When you say signal is tied to the shield, which pins? The ground pins are tied to the shield. See pinout below

I am surprised to see that the grounds of the vga cable are tied to the shield, on the cables I have some of the grounds are tied to the shield. This is a more noisy solution, but with the cable making industry its about what works not what is best. They may be cutting corners by eliminating some conductors in the cable and running the ground through the shield.

A shielding conductor is used to prevent electrical fields to capacitively couple to the conductor they are protecting.

As far as turning the cable into an antenna:

From the cable shielding wiki:

The best method to wire shielded cables for screening is to ground the shield at ONE end of the cable

This is to avoid ground loops, ground loops are a big problem because current flowing through the cable causes current through the shield conductor to couple to the signals inside the cable magnetically. So to prevent this current flowing on the outside of the cable a choke can be placed on the outside of the shield to increase the high frequency impedance of the cable and cause high frequency signals to find a different "pathway" (as high frequency currents will "prefer" the path of lowest impedance/inductance).

The cable is shield is probably not tied to ground on both ends (computer and monitor sides), and one would hope that it is isolated somehow. Otherwise DC and AC currents will flow through the cable and cause problems.

So the shield shouldn't be tied to ground, it is probably introducing noise but it is and it still works, as far as we can tell. It would be interesting to find the VGA specs and see if this is a standard practice.

enter image description here

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VGA uses 75 Ohm terminations on R,G,B signals to both signal ground and earth ground.

This creates common mode EMI from all the shield currents.

The solution is to raise the shield impedance on ALL VGA cables with a large CM choke using a molded clamshell ferrite core (split C cores to make a torroid). This reduces the CM current radiated to acceptable emission levels.

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    \$\begingroup\$ Solution to what? The question was "why are the pins connected to ground when the cable is disconnected at both ends?" \$\endgroup\$ – JRE Aug 19 '16 at 14:40
  • \$\begingroup\$ Even if disconnected, the antenna impedance is raised with the CM choke on the shield to cut emanation current. Sometimes it is connected. \$\endgroup\$ – Sunnyskyguy EE75 Aug 19 '16 at 14:44
  • \$\begingroup\$ capacitive coupling to ground is a much lower impedance above 1MHz. Sometimes it is connected in cable design. But the common solution is the clamshell choke. \$\endgroup\$ – Sunnyskyguy EE75 Aug 19 '16 at 14:52
  • \$\begingroup\$ But how does shield the cable from outside EMI or shield other devices from the cable's EMI. If the surface is bigger, doesnt that make the cable an antenna to create noise and disrupt other devices? \$\endgroup\$ – Kevin77 Aug 19 '16 at 16:36
  • \$\begingroup\$ No. The 75 ohm load shunts the stray ingress due to the high series impedance of the choke. It works both ways for egress too. \$\endgroup\$ – Sunnyskyguy EE75 Aug 19 '16 at 17:04

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