What happens if shielded cable has a short to the shield?

[solvejskovlund]

While I was rolling out a shielded cable, a though came to my mind - what happens to the current flow in the cable if there came a short between the wire and the shield in both ends of the cable?

For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the shield has the same cross section area. There are insulating material between them, and in both ends there is a short between them.

My first thought, the total resistance of the cable would be reduced. Seeing the copperwire as an resistor, adding the shield is like connecting two resistors in parallel. Despite having the same cross section area, the aluminum has higher resistance, so the total resistance would go down, but not very much.

My next thought was about the skin effect. Remembering that electrons like to travel along the outer edge of a wire. Now that they have the ability to chose a path with a even more outer edge would that make them prefer the aluminum path of the inner copper? If so, would they actually chose the path of HIGHER resistance? Could there be some conditions where the total resistance of the cable actually increases when the shield is included?

Maybe the behavior is dependent of if the current is DC or high frequency AC?

Could someone shed some light on this?

 

[.Scott]

It's worse than that.
A single shielded cable normally expects the return current to pass through the shield. The shield will often be connect to the circuit ground or perhaps even the chassis ground.
So at the signal source end, the signal will be grounded - possibly resulting in an overcurrent. At the destination end, any residual signal or stray noise will be similarly grounded. So, no signal will pass.

There are cases where you could get a distorted signal, but in the simplest cases you will get nothing.

 

[Baluncore]

For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the shield has the same cross section area. There are insulating material between them, and in both ends there is a short between them.

I believe you are talking about a coaxial transmission line.

A signal current on the outside of the inner conductor, will be equal and opposite to the signal current on the inside of the outer conductor. The volume of the internal insulation, between the two conductor surfaces, is an orderly, but isolated transmission line cavity, a universe, connected to the outside, only at the two ends, but you have shorted those end ports, preventing internal currents.

The signal on the outside of the outer conductor, is not related to the balanced signal on the inside of the transmission line. In effect, signals do not flow through the holes in the shield. The current on the outside of the shield is equal and opposite to the current flowing on the surface of the conductive environment that surrounds the cable. That includes all the local sources of RF, that will be reflected from the outside of the outer conductive shield, screening the inner world. That outside universe makes an ugly and mismatched transmission line, separate from the universe inside.

 

[solvejskovlund]

So at the signal source end, the signal will be grounded to nothing - possibly resulting in an overcurrent.

Well, if the shield was connected to something else, then surely that something would be affected as well. But I was thinking of what happens to just the cable, for the simplification used as a 1-wire only. This implies that there must be another path/cable for the return current.

 

[solvejskovlund]

I believe you are talking about a coaxial transmission line.

It's quite similar to a coax, yes. Just that the shield in a coax tend to be woven copper, while the cable I was handling, and got me into this thinking, was a aluminum sheet wrapped around.

A signal current on the outside of the inner conductor, will be equal and opposite to the signal current on the inside of the outer conductor. The volume of the internal insulation, between the two conductor surfaces, is an orderly, but isolated transmission line cavity, a universe, connected to the outside, only at the two ends, but you have shorted those end ports, preventing internal currents.

The signal on the outside of the outer conductor, is not related to the balanced signal on the inside of the transmission line. In effect, signals do not flow through the holes in the shield. The current on the outside of the shield is equal and opposite to the current flowing on the surface of the conductive environment that surrounds the cable. That includes all the local sources of RF, that will be reflected from the outside of the outer conductive shield, screening the inner world. That outside universe makes an ugly and mismatched transmission line, separate from the universe inside.

Either I don't understand what you are explaining, or we are thinking two very different scenarios. What are the "holes in the shield"?

 

[Baluncore]

What are the "holes in the shield"?

They are the gaps between the woven wire of a coaxial outer braid, or the very narrow slots between the edges of a metal foil tape, wound or wrapped around the outside of the insulation, to provide an internal return circuit, and to screen the cable from external influences.

 

[.Scott]

Well, if the shield was connected to something else, then surely that something would be affected as well. But I was thinking of what happens to just the cable, for the simplification used as a 1-wire only. This implies that there must be another path/cable for the return current.

The "to nothing" part was basically a typo. The signal source will be shorted to ground, the same ground as the shielding. So, that will be the end of your signal.

@Baluncore described how this kind of cable is used. The purpose of having the shielding fully encircle the core is to protect the voltage difference between the shield and the core from being affected by outside electromagnetic fields. If you short the core and shield together, that voltage difference goes to zero.

There is another shielding configuration called "shielded twisted pairs". And since the name describes the design of the cable, you can guess what they look like. For that kind of cable, the outside shielding is not involved in carrying current - in fact, it is often grounded at one end and left open on the other. It is especially useful is keep magnetic pulses in the environment or crosstalk from other wire pairs it is running with from inducing any current in the twisted pair. And since the pair of signal-carrying wires are twisted together, any external influence will likely affect both wires equally - keeping their voltage difference the same.

 

[Rive]

For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield.

The issue is, that different cables/shieldings can have different purpose and so simplification might make the question miss the mark entirely.
This:

or this:

or this:

...are meant to be very different.

 

[solvejskovlund]

The signal source will be shorted to ground, the same ground as the shielding. So, that will be the end of your signal.

That assumes the shield was connected to ground, which I though was clearly stated in the question - and the follow up - that the shield was not connected to anything other than the wire (cable core) in both ends.

 

[solvejskovlund]

The issue is, that different cables/shieldings can have different purpose and so simplification might make the question miss the mark entirely.
This: View attachment 366264 or this: View attachment 366265or this: View attachment 366266
...are meant to be very different.

I think all of those cables can be used to send a current in one end, and you'll get it out at the other end. Sure they have different "advanced" properties - like wire cross section area, internal interference, how much current they handle, how much voltage you can apply before you may experience internal short, how easy they are to bend, and so on - but the basic properties are quite similar - they are all copper wrapped in none-conductive material. You can apply a voltage at one end, and measure the voltage at the other end. And if you send current through, the current you measure at one end will be identical to the current you measure at the other end, they all have a wire resistance. The value of that wire resistance is different, but they all have a resistance of some value.

And for the multi-wire cables - if you connect two of the wires together at both ends, you will find that the resistance from one end to the other end has become 50% of what you found with only one wire in use. The two wires connected together at both ends, will mostly behave like one wire with the double cross section area.
Also the coax - if you connect the inner wire with the outer wire, the cable will behave mostly like a single wire with a cross section equal to the sum of the two wires cross section. Although, electrons may prefer to travel in the outer wire because of the skin effect.

All the above has been said about copper wire. For my question, however, think of it as any of the cables in the pictures. Assume you only intend to use one wire in the cable. Assume you did not intend to connect the shield to anything. You just needed one wire, and used the cable you had available. While you put the cable in place the shield made of aluminum (not copper) happened to get connected to the one copper wire you intended to use at both ends. The shield did not get connected to anything else (even though thats what you would have done if you were using the cable for its intended usage).
How would the aluminum wrapping affect the electrical properties of the cable, compared with the properties of the cable with the shield not connected to anything at all?

 

[Baluncore]

Assume you only intend to use one wire in the cable. Assume you did not intend to connect the shield to anything.

Then there is no return circuit. Without the return circuit, you have an end-fed antenna.

There is a good reason why all conductors in a circuit must pass through the same shield, and the same hole in conductive or magnetic bulkheads. The equal and opposite circuit currents will cancel, and so not radiate energy or induce eddy currents in conductive bulkheads.

 

[.Scott]

That assumes the shield was connected to ground, which I though was clearly stated in the question - and the follow up - that the shield was not connected to anything other than the wire (cable core) in both ends.

It wasn't clear enough for me.

So, what you are describing is a cable that is not connected to any circuit at all. It is just a length of cable sitting on a bench with shielding applied. Or a similar length of shielded cable sitting on a store shelf.

We will also assume that this shielding has been applied in such a way that it forms an electrically seamless metal cylinder.

So, when such a cable is used as if it was a single wire, it would have more available surface to support skin effect. So, for those very high frequencies, there would be an advantage - and you would expect that most of the very high frequency components of the current would be bourn by the shielding.

In all cases, the addition of the shorted-out shielding would decrease the total resistance.