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How do reflections work, physically to the charge carriers level?

Considering a square wave generator connected to a long, open cable, I've made a quick chronological diagram of the propagation of the generator voltage, but I can't explain how electrons and holes bounce off the open end. For light (so... coax cables?) and sound it sort of makes sense to reflect when there is a sudden change in impedance, but here...

Based on those diagrams, the generator should start producing a higher voltage when the line is "full" (can't push any more electrons in), therefore after a single delay; however it does that after a return delay right? Does it "push" electrons in until there is an open voltage across it? How should that diagram be modified so that it is accurate? enter image description here

The diagram is helping me to understand the "how" of this video.

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  • \$\begingroup\$ Reflections and signal propagation in coax (also waveguides and twisted pair to name a few others) is an electromagnetic wave. \$\endgroup\$
    – Andy aka
    Apr 16 '15 at 12:50
  • \$\begingroup\$ You mean like a fiberoptic? If so, I changed the question to normal cables, I am more interested in electrical signals. \$\endgroup\$
    – user42875
    Apr 16 '15 at 12:55
  • \$\begingroup\$ It's actually a lot like light reflection, but in a different part of the spectrum. \$\endgroup\$
    – pjc50
    Apr 16 '15 at 12:56
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    \$\begingroup\$ Any cables produce an EM wave that propagates and potentially reflects. \$\endgroup\$
    – Andy aka
    Apr 16 '15 at 13:08
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    \$\begingroup\$ You might find this article (actually a 3 part series) useful. sigcon.com/Pubs/news/14_03.htm \$\endgroup\$ Apr 16 '15 at 15:37
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First, it's not really right to say that positive charges are holes in regular (not semi-conductor) metals. However, that's not the point here.

You are OK thru the fourth diagram, but the 5th one is wrong. Think about it. Any affect at the far end of the line has to propagate to the near end before it matters there. The line being open matters first at the far end, then the affect of that eventually propagates back to the near end.

As the step is propagating down the line, the capacitance of the line is being charged up. That takes a steady current, since the same amount of capacitance will be charged per unit time as the step propagates. When the step reaches the open end, there is no place for the charges to continue flowing to, but due to the series inductance the charges keep coming in the short term. The net result is that for a ideal transmission line, the voltage at the end rises, absorbing some charge in the process because the inherent capacitance is charged up. This higher voltage now pushes charges back up the line to the source, which causes the current to decrease. The series inductance and parallel capacitance work such that the voltage builds up to double the original value, and the current is decreased just to the point where it becomes 0. This step now propagates back to the source.

When the step reaches the source, assuming ideal termination there, the line now looks like a open circuit. The line voltage and source voltage are equal, and the current is 0. The line will remain in this state as long as the same voltage is applied.

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    \$\begingroup\$ Wow. When will you not know something? Thanks a lot! \$\endgroup\$
    – user42875
    Apr 16 '15 at 13:12
  • \$\begingroup\$ Actually, the red line should travel roughly 200 million times faster than the blue line. Otherwise it's an OK diagram (current travels at close to speed of light but electrons are slow) \$\endgroup\$
    – slebetman
    Apr 16 '15 at 20:29
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This is another case where the travelling electron model of electricity will lead you astray.

The thing that propagates down a cable at a significant fraction of light speed is an electromagnetic wave. This has the effect of moving electrons back and forth, because they are charged particles, but it does not mean that if you somehow labelled an electron at the point of injection you would see it arrive at the other end along with the signal. It looks a lot more like a "Newton's Cradle". Electro-motive force applied at one end pushes through the mutual repulsive fields to apply a force at the other end.

You can send RF signals for some frequency ranges with low loss without using a center conductor at all: this is called a "waveguide". It relies on total internal reflection off the internal surfaces.

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    \$\begingroup\$ @user42875: The speed of electrons in a wire is around 1m/s. Which means that on average there should be a 5 second delay between switching on a light and the bulb producing light. But that's not what we experience in the real world. It's the RF/EM wave in the wire travelling at close to the speed of light that makes electric signals almost instantaneous. So electric signals != electron travel \$\endgroup\$
    – slebetman
    Apr 16 '15 at 20:25
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Think of a bunch of initially-motionless identical hanging weights connected by springs. If the top weight is momentarily jerked upward, that will pull up the weight below, which will some time later pull up the weight below that, etc. creating a downward-traveling tension wave. In an ideal conceptualized model (real springs don't behave quite perfectly) the amount of energy any spring other than the first or last spring receives from the weight above will precisely balance the energy passed along to the weight below, so each spring will be motionless after it passes along its energy.

If the last weight is fastened rigidly so it can't move, it will pull back against the previous spring twice as hard as it would if it could move smoothly. The net effect is that the previous weight will behave as though it had pulled on a succeeding spring (feeling a normal amount of reverse pull) and the succeeding spring had in turn pulled back with that much force again. This would cause that weight to then pull on the one above, which would pull on the one above, etc. causing a tension wave to propagate upward.

If the last spring is not fastened, it will try to pull on the next spring but will not receive the expected resistance (or any). The behavior will be equivalent to what would happen if it pulled against the expected resistance and was then pushed by that same force; consequently, it would push the weight above, which would push the weight above that, etc. causing a compression wave to propagate upward.

If one doesn't want any sort of wave to propagate upward, something must provide just the right amount of resistance to the last weight such that its kinetic and potential energy both get perfectly absorbed. Too much resistance and a tension wave will propagate upward; too little and a compression wave will do so.

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  • \$\begingroup\$ Excellent mechanical analogy. I'm not sure how accurate that is, but it's very well explained, thanks. \$\endgroup\$
    – user42875
    Apr 16 '15 at 18:10

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