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I know that a transmission line is essentially two wires with some series resistance and inductance and some parallel capacitance - when there is a change in the voltage between them at a point the charge in the capacitance moves and thus a current is flowing in opposite directions in both wires.

In the case that a transmission line is terminated with an open circuit - the voltage and current waveform are reflected back when they reach the termination point.

But what happens if one wire is longer than the other? - I assume that the voltage and current reflect at the point at which the short wire ends, and the voltage and current in the extra section of long line never sees any change - is that right?

EDIT:

To add some information, these are long wires I'm talking about - roughly 1KM in total. I'm looking at the effect of having open circuits at different points and how this changes the voltage and current on both wires and on both sides of the break.

The signal propagating down the wire is of the order 10s of MHz

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Considerations:

In my opinion The other answers I saw in this post accurately define the practical results of your question, I very much liked @Norm's answer since he likened it to a cutting the internal wire of a co-axial cable. I will attempt to explain a perhaps more theory based solution by starting with a transient discussion of waves to which steady state could then be interpreted from.

Consider the simple schematic at the end of this post as a reference for my answer, the "Mutual C" represents the break point.

Assumptions:

Please pardon the fact that I didn't model the transmission line using all circuit parameters, it is enough just to place that parasitic capacitance to get my point across.

Background:

In traveling wave theory, you'd expect a wave to propagate to the end of the conductor (just like water would in a pipe), but to really represent this moving voltage and current wave we have to use inductors and capacitors to show that there is a "time delay" in the voltage and current signals. This is a transfer of energy.

Answer:

I know that a transmission line is essentially two wires with some series resistance and inductance and some parallel capacitance

You nailed the point right here, and if you use an entirely ideal model the theory will fall apart here. Notice that by simply adding in some parasitic and non-ideal devices we can get a better idea as to what will happen.

the voltage and current in the extra section of long line never sees any change - is that right?

Not quite, consider my follow up question; if there was no voltage or current change at that longer line, then what do you think the voltage would be at the end of that longer line at steady state?

Let's look into this:

A voltage impulse will travel from the source to the break point and charge up the line. There will be reflection at the break point as if it was an open circuit (slightly different but safe to assume).

The beauty of transients is that the circuit does not "know" that that wire "goes nowhere" until the energy reflects and cancels. So there will be an energy wave in the form of a voltage impulse that then travels to the end of the longer line. From here you would find a partial reflection and partial dissipation by an "antenna-like" effect and by parasitic impedances. The current in this path will be negligible from the other sources but it's existence is what validates our schematic. This will continue until there is a standing voltage across all points.

Now remember that question I asked up there? Well now you can say that at steady state the voltage at the end of both wires will approximately be equal in magnitude to the source voltage (given some losses). This agrees with our open-circuit theory of voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Ok, I like this, "So there will be an energy wave in the form of a voltage impulse that then travels to the end of the longer line. From here you would find a partial reflection and partial dissipation by an "antenna-like" effect " That's what I was looking for. So if I didn't have the extra section 100% of the current/voltage gets reflected at the break, but with an extra section on one side, a small portion of voltage/current doesn't get reflected and travels up the longer section as an antenna? \$\endgroup\$ – Tim Mottram Feb 27 '18 at 9:59
  • \$\begingroup\$ Yes! The point is that that extra section has to get energized so that it can get to the same voltage as the rest of the line voltage (with a little leakage) or else you would have a constant main current flow, which makes no sense for an open circuit at steady state. To justify that there is some initial "flow" to create the voltage, you need to add these non-idealities like antennas and parasitic impedances. Otherwise you have a source potential at the "break even" point and then who knows what at the end! \$\endgroup\$ – Commanderson Feb 27 '18 at 12:40
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Yes - its the same as having a coax cable with a broken center conductor. Past the break there is no longer a "transmission line".

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  • \$\begingroup\$ I don't understand why this answer has been down voted. Have one back +1. \$\endgroup\$ – Andy aka Feb 26 '18 at 14:05
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Consider what having one wire 'longer' means. It means it has moved away from the other wire. This means the line impedance has increased severely, and you will get a large reflection.

A transmission line still only has two 'ports', four terminals with two at each port, with the signals being seen differentially across each pair of terminals. If you start trying to think what's happening on one conductor, and on the other conductor, then you no longer have a transmission line, you have two coupled transmission lines, which is much more complicated.

Having one conductor projecting beyond the other means that this port has widely spaced terminals, and so is only suitable for a high impedance connection.

We sometimes get this situation by accident when a high speed digital track crosses a ground plane. We might need to squeeze one last connection into the board, and so route it through the ground plane, running perpendicularly to the track above. This breaks the ground plane, increasing the path length the return current has to take under the track. The fact that the current has been moved away from where it wanted to flow which was directly under the track means there is a large increase in line impedance at that point, resulting in reflections and corrupted signals on that track.

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  • \$\begingroup\$ "Having one conductor projecting beyond the other means that this port has widely spaced terminals, and so is only suitable for a high impedance connection." - Maybe I explained it badly, Lets say a have a very long loop of wire connected to a voltage source - if I cut the wire in the middle id have an open circuit transmission line where the two conductors were of equal length. If I cut the wire right next to the positive terminal of the source I have an open circuit transmission line with unequal wire lengths. - but the port terminals are still close (only a knife width apart ;)) \$\endgroup\$ – Tim Mottram Feb 26 '18 at 14:14
  • \$\begingroup\$ Or did I misunderstand your answer? \$\endgroup\$ – Tim Mottram Feb 26 '18 at 14:34
  • \$\begingroup\$ what do you expect if you don't draw a picture! My mental model was two parallel straight wires, one longer than the other, aligned at this end, staggered at that end. However, the principle still applies. A transmission line only has two ports, and what goes on between them can be expressed differentially in terms of impedance versus distance. \$\endgroup\$ – Neil_UK Feb 26 '18 at 14:45
  • \$\begingroup\$ Thanks Neil, and sorry for not providing a picture, I actually did the same, the reality is a loop which is cut towards the positive source end, but I was considering the problem as two parallel conductors, I didn't realise the distance between the end ports was important. \$\endgroup\$ – Tim Mottram Feb 26 '18 at 14:54
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The voltage and current doesn't suddenly disappear, it will travel to the end normally. You will get inequal signals in the receiver as there's a phase difference between the wire pulses. Depending on your line speed and the difference in lengths this may or may not be significant.

To counteract this you simply add more "wiggles" to the shorter trace to equalize the line length. This will hurt the coupling and can cause reflections but these are often less of a problem than phase shift. Depending on the severity of course.

As a trivia, you can have differential bus with completely isolated wires. This is how high speed transmission lines e.g. V-by-one are implemented, they use individual coaxial wires for each bus line. You don't obviously get field cancellation advantage that way but since it's a coaxial, it doesn't really matter.

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  • \$\begingroup\$ I understand why this has been down voted but it wasn't me! \$\endgroup\$ – Andy aka Feb 26 '18 at 14:06
  • \$\begingroup\$ @Andyaka Well, please do enlighten me? \$\endgroup\$ – Barleyman Feb 26 '18 at 14:52
  • \$\begingroup\$ Read the question again and ask yourself if he needs an answer that focusses on correcting for inequal length traces. \$\endgroup\$ – Andy aka Feb 26 '18 at 14:54
  • \$\begingroup\$ @Andyaka Hmm.. I think I misinterpreted what he said - "and the voltage and current in the extra section of long line never sees any change", probably means "charge keeps traveling down the wire as before", not "there is no change in the line state after that point". \$\endgroup\$ – Barleyman Feb 26 '18 at 14:59

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