How are signals reflected in a wire? - Page 1
Forum section: electronics / general

author: TimNJ
author_rank: Super Contributor
timestamp: April 21, 2013, 11:36:23 pm
content: If electrons move all at once when a potential is applied, how do you explain signal reflections in a circuit?

author: IanB
author_rank: Super Contributor
timestamp: April 22, 2013, 12:34:01 am
content: "If I make a false assumption, how come I reach a false conclusion?"Change your initial assumption and the problem will resolve itself.

author: BravoV
author_rank: Super Contributor
timestamp: April 22, 2013, 12:42:59 am
content: Quote from: IanB on April 22, 2013, 12:34:01 am+1Here, an old video to rescue ..

author: lemmegraphdat
author_rank: Frequent Contributor
timestamp: April 22, 2013, 12:55:12 am
content: False assumptions in this case are probably not assumptions. We get lied to a little in the beginning.

author: lemmegraphdat
author_rank: Frequent Contributor
timestamp: April 22, 2013, 01:25:29 am
content: Does that taper section remind you of anything?

author: TimNJ
author_rank: Super Contributor
timestamp: April 22, 2013, 01:35:06 am
content: Thanks. About half way through it. Love old videos.

author: IanB
author_rank: Super Contributor
timestamp: April 22, 2013, 02:01:59 am
content: Quote from: lemmegraphdat on April 22, 2013, 12:55:12 amMaybe so. But analysis and deductive reasoning are powerful tools to help you learn when you have been lied to.For instance, reflections are a property of waves, and waves have a propagation velocity; ergo, if we get signal reflections in a circuit there ought to be waves in the circuit, which implies the electrons do not all move at once when a potential difference is applied. Voltage changes around a circuit must have a finite propagation velocity.We ought to deduce this up front, since "everyone knows" nothing can travel faster than the speed of light. If we could make all the electrons move at once we could transmit information down a wire faster than the speed of light, and we would have violated one of the fundamental theorems of physics. Therefore this can't be true.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: April 22, 2013, 03:28:50 am
content: Thinking of fundamental atomic particles as "knowing things" is a definite leap for some folks at first. I find another marvelous example of that principle is what happens when you let go of a stretched slinky:When I was first asked what would happen, I was close, but not quite right. Here's the answer, in glorious slow mo:

author: amyk
author_rank: Super Contributor
timestamp: April 22, 2013, 10:50:22 am
content: The fact that there is an absolute "speed limit" in the universe is actually why many things work the way they do, and even when modeling digital logic you need to keep in mind that events do not all occur "immediately"; without propagation delays, things like ring oscillators and shift registers could not work at all.

author: jpb
author_rank: Super Contributor
timestamp: April 22, 2013, 11:54:46 am
content: To confuse matters further, electrons are moving all the time between collisions and in random directions. Applying an electric field causes a drift in the direction of the field (hence the drift diffusion model for semiconductor junctions).And then you might need to worry about the wave nature of the electrons, some of them might tunnel out the end of the wire or in fact might be at the other end of the universe if their wave function extends that far.To try and answer the original question, if you have a wire which is at uniform potential and open circuit at one end, if a voltage pulse is applied to one end there will be a local field which will cause a net current flow in that direction which in turn will cause a field further down. Until the pulse reaches them, the electrons further down the wire will be moving in random directions unaware of what is about to hit them:%-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B %-B %-B %-B %-B%-B %-B %-B %-B %-B %-B %-B %-B %-B %-B %-B %-B

author: AlfBaz
author_rank: Super Contributor
timestamp: April 22, 2013, 12:34:19 pm
content: Now there's a use for emoticons you don't see every day

author: G7PSK
author_rank: Super Contributor
timestamp: April 22, 2013, 03:37:39 pm
content: Just to really confuse things electrons are both a solid matter and a wave just depends on the conditions and as far as I can understand it they swap from one state to the other at will.

author: BravoV
author_rank: Super Contributor
timestamp: April 22, 2013, 03:41:26 pm
content: This thread is a fine example that a pure innocent thread asking for a basic things gets fubared because of overcomplicated junk replies.

author: mswhin63
author_rank: Frequent Contributor
timestamp: April 23, 2013, 01:08:01 am
content: Quote from: TimNJ on April 22, 2013, 01:35:06 amI have to agree, the explanations seem to be far better than schools now-a-days.

author: Rick Law
author_rank: Super Contributor
timestamp: April 23, 2013, 04:25:02 am
content: Quote from: TimNJ on April 21, 2013, 11:36:23 pmRemember from Einstein, light is a wave and light is a particle (He was awarded his only Nobel prize on that). Electrons are particles and they are also waves. If electrons are not waves, you won’t have electron microscopes. (Well, actually, you and I won’t exist.)When there is a transition of medium, there is a change of speed of propagation. That means a change in energy as well. Any of these will create an opportunity for the wave to reflect.Try this analogy: Imagine during a marathon, if the path involves a bridge, some may not cross it and choose that point to terminate his/her race. Or when say one leg of the race is crossing a city boundary, some may choose that as their stopping point. This is human psychology involving probabilities in human decision making. Electrons work with probably as well. Similar to a door, not everyone in the crowd will choose to walk through - whereas, if the path is unchanged, it is less likely someone will just stop and turn around. Electrons may “choose” to stop or turn back when it has to make a decision to transit or not.When there is any change in propagational properties, whatever is propagating will encounter energy change. Any such change will result in reflection, refraction, or simply lost.To visualize the point, tie a thick heavy rope to a thinner lighter rope. Tie the heavy rope to the door (or some stationary thing). Take the lighter end and pull until the combo-rope is almost tight but with some slack. Now start creating a wave with the combo-rope and you will see a reflection from where your lighter rope meets the thicker rope. Or try this: next time you take a bath, try creating a small barrier in the tub of water with a large opening and push the water through, you will see some waves reflecting.When there is a joint (say a soldered joint), your electrons have to leave the copper (more on this shortly), go into the tin-lead solder, then exit the solder and enter the other copper. Plenty of energy change, plenty of collision, plenty of reflections, plenty of “decision making” for an electron.Back to the copper joint - some what related but some what merely side notes:(1)Since electron is negative and negative repel negative. Thus, electrons repel each other. The repulsive force pushes the electrons to the outer layer of your solid strand of copper. Any change in thickness of the circular wire (say a good squeeze with a pair of pliers), however small, reduces the circumference of the outer layer thus changes the propagational-property thus increasing the possibility of reflection.(2)Actually, the situation with wires is worst most would think. When (for example) you have a pair (+ and -) of copper wires connecting an AA battery to a light. You cut the (say the red +) copper wire in the middle, strip the ends and now twist the stripped copper ends together, so you have your copper wires making a “complete circuit” again this time through the twisted-together joint. Your light come on, but should it?When you stripped the wire and expose the copper, however fast you do it and however well you twist the ends together, the copper will oxidize. Copper oxide is not a conductor. Copper oxide is a semi-conductor. How then does it complete the circuit? Copper light switch should not even work since the copper plate will be covered with copper oxide – not a good conductor at all.In fact, a significant effect in what actually completes the circuit is quantum tunneling. Your electron actually tunnels through the non-conductive layer. The thinner the non-conductive layer of copper oxide, the easier it is to tunnel. Without tunneling effect, assuming you can exist at all in this altered universe, you will find light switches work much better in true vacuum, and practically useless in our oxygen-filled environment. (High voltage will burn through the oxide layer, lower voltage will not.)(3)Just because you push 20 electron “through” it doesn’t mean 20 electrons will come out. It is all probability. Each electron being a wave, there will be interference between waves. A certain position, there should be a point where the combined wave crosses the zero - no electron at the point at all. It doesn’t cross that point, it just magically appear beyond that point.

author: AlfBaz
author_rank: Super Contributor
timestamp: April 23, 2013, 04:36:18 am
content: I'll give it a go...Undoubtedly analogies are good but it's also good to understand what happens electrically, in fact, in this instance I don't know why it's not explained like this more often. So I’ll attempt to give a very simplistic descriptionHopefully you have dispensed with the assumption that electricity is instantaneous, in fact it doesn't even travel at the speed of light. In a normal bit of copper it can travel anywhere between 90-95% the speed of light and in coax (if I remember w2eaw's video correctly) it’s about 66%The speed of light is 300 000 000 m/s so let’s just say about 280 000 000m/s for electricity, so it takes about 3.5ns to travel a meterLet’s take a simple circuit that has a voltage generator that produces a step voltage from 0 to 1V feeding a resistor via a wire. We are going to raise the voltage from 0 to 1V quicker than the time it takes to reach the resistor. At this rate of rise, parasitic capacitance and inductance in the wire that are normally ignored come into play causing it to now have impedance (a transmission line). It has ohmic resistance as well but we won’t worry about that here.When the voltage transitions from 0V to 1V at the generator all that voltage see’s is the impedance of the wire because it hasn’t reached the resistor yet so the current into that bit of wire is equal to the voltage divided by impedance (ohms law). This current travels down the wire and reaches the resistor. The resistor has a higher value than the impedance of the wire (impedance mismatch), but the amount current coming through the wire was determined by the generator’s voltage and the wire’s impedance, so what happens now?This current flows through the resistor and due to ohms law (E=IR) we now have a higher R but the same current so therefore E is now higher at the entry to the resistor.The wires voltage is still at the lower voltage. This difference in voltage (higher at the resistor, lower on the wire) causes current to flow from high to low (ie the other way). This current may bounce back and forth but will eventually settle to a value determined by the resistor.So there you have it, reflection and the added bonus of overshoot

author: TimNJ
author_rank: Super Contributor
timestamp: April 23, 2013, 09:56:01 pm
content: Thank you all. Alfbaz, when you say "it" moves at 90-95% the speed oflight. It is "electricity" but what what does that actually mean? Theelectric field? The electrons?I just don't understand *what* is delayed. So the electrons arepushing on eachother? Does it make sense to say that one electron isenergized and then that electron passes its energy on to an adjacentelectron and so forth? So is it like a domino effect kind of? Thevoltage source can deliver energy into a wire even if the wire doesnot make a complete circuit?Thanks everyone.

author: IanB
author_rank: Super Contributor
timestamp: April 23, 2013, 10:54:48 pm
content: What moves at close to the speed of light is the "wave" in the electric field.Imagine all the electrons in the wire as trying to push themselves apart and space themselves out evenly. Since they have an electric charge, and like charges repel, they don't like being pushed closer together. Pushing them closer together creates a "pressure" that is measured as voltage.So if you raise the voltage at one end of a length of wire, you are pushing the electrons a little bit closer together at that end of the wire. The electrons then start pushing at the next electrons a bit further down, and this push travels down the wire like a wave until it reaches the other end.You can also measure the closeness of the electrons as an electric field strength. The more closely packed the electrons the stronger the field. If you pack the electrons too closely together the desire to escape becomes very strong and a spark may jump away from the conductor to a nearby object with lower potential.

author: lemmegraphdat
author_rank: Frequent Contributor
timestamp: April 24, 2013, 01:21:48 am
content: Charge carriers don't have to move far or fast. In wire I think is it something like a few centimeters per hour. There are a lot moving at the same time though. Want to see electric charges? Look at a shiny piece of metal. That shine is electric charges. Without a push they just move about randomly.

author: 4to20Milliamps
author_rank: Regular Contributor
timestamp: April 24, 2013, 02:08:17 am
content: Build one of these :http://electronicdesign.com/displays/build-your-own-cable-radar*NOTE*this is just a suggestion, you don't need to build equipment to actually study things, and I know everyone already knew this, so there's no reason to jump my ass about it, just take it with a grain of salt and move on.

author: IanB
author_rank: Super Contributor
timestamp: April 24, 2013, 03:15:11 am
content: Just for grins, here is a rough illustration of what happens if you suddenly increase the voltage at one end of a wire:This is not accurate or to scale or anything like that, but it gives an idea of how the voltage wave travels along the wire and gets reflected at the other end.

author: AndyC_772
author_rank: Super Contributor
timestamp: April 24, 2013, 06:48:40 am
content: It's not accurate at all, I'm afraid. The reflected wave isn't anything like the full 360 deg cycle you've shown, it's a doubling of amplitude and reversal of direction. In a low-loss transmission line, the signal at the receiving end doesn't decay to its final (dc) value until the wavefront has bounced many times up and down the entire length of the wire.It's well worthwhile downloading a copy of LTSpice if you haven't already, and modelling a wire by simulating a long chain of L-C elements. Try applying a fast rising edge to one end of a long chain of 1nH inductors and 1pF capacitors, then probe at various points to see what the voltage at that point in the wire would be doing. Start by simulating a wire without dc resistance just to isolate the transmission line effects.You should then be able to work out what the impedance of your transmission line is, by terminating it at one end in a resistor and then varying the value of that resistor until the size of the reflected wave is minimised.IMHO this is a really good way to see what the true effect of series and parallel termination resistors actually is. You can also see why it's a bad idea to use a single series terminator if you need a track to run from an output to multiple inputs, why the problem doesn't occur if you're using parallel termination, and why even a short stub is a really bad thing to have on a high speed signal trace.

author: jpb
author_rank: Super Contributor
timestamp: April 24, 2013, 08:27:28 am
content: The impedance should be sqrt(L/C) or 10.sqrt(10) or approximately 32 ohms in your example.

author: AlfBaz
author_rank: Super Contributor
timestamp: April 24, 2013, 10:15:05 am
content: Quote from: AndyC_772 on April 24, 2013, 06:48:40 amDoubling? Wouldn't the amplitude be dictated by the ratio of line to load impedance?

author: AndyC_772
author_rank: Super Contributor
timestamp: April 24, 2013, 10:49:39 am
content: Yes, it would - I'm considering the case of an unterminated transmission line, or one terminated in a very high impedance like a CMOS input.

author: IanB
author_rank: Super Contributor
timestamp: April 25, 2013, 04:27:07 am
content: Quote from: AndyC_772 on April 24, 2013, 06:48:40 amThanks for putting me right that my prediction was off the mark.Just for grins, I did a simple numerical simulation of an open ended wire where I could vary parameters and look at the spacial voltage profile. The example below (purely qualitative) shows the time evolution in a wire with significant resistance (so heavily damped).Here is what happens when the wire has less resistance: