Hi Todd, Thanks for your comments and your interest. Steve’s approaches, such as what you’ve described, sound like good ones to me. Regarding your questions: Wider spacing of the two conductors will increase inductance and decrease capacitance. The reason that inductance decreases when the conductors are twisted together, or at least brought closer together, is that since the currents in the two conductors are moving in opposite directions (one toward the load and one toward the source, at any given time, with the directions alternating between each half cycle of the signal in the case of AC), the magnetic fields created by those currents are in opposite directions and tend to cancel each other if the conductors are close together. To the extent that those fields do not cancel, the impedance presented to the signal will increase as the rate of change of the signal (i.e., its frequency) increases. For further explanation the Wikipedia writeup on Lenz’s Law may be helpful. Regarding capacitance, a capacitor consists basically of two conductive plates separated by a non-conductive dielectric, with each of the two terminals of the capacitor connected to one of the plates. As explained in the Wikipedia writeup on capacitance the closer those plates are to each other the greater the amount of capacitance, everything else being equal. Similarly, cable capacitance presents itself as a shunt (aka parallel) phenomenon between the two conductors, and therefore increases as the conductors become closer together. While inductance and resistance present themselves as series phenomena. In general, inductance is most likely to be significant in the case of speaker cables, to an increasing degree if the impedance of the speaker is low at high frequencies. (See my first post in this thread). Resistance may also be a significant factor in a speaker cable, of course, especially if the speaker has low impedance at many or most frequencies. Capacitance will usually be unimportant in the case of a speaker cable, unless it is very high, in which case it can cause problems for the amplifier. Especially if the amplifier is solid state and therefore most likely has low output impedance, and uses significant amounts of feedback, in which case even destructive oscillations can occur if the capacitance is extremely high. In general, resistance and inductance will be unimportant in the case of a line-level interconnect, aside from the possibility that the resistance and inductance of the ground/signal return conductor can make a difference with respect to ground loop-related low frequency hum or high frequency buzz or noise, if the components involved are susceptible to ground loop issues. Capacitance can be important in the case of a line-level interconnect, especially if the output impedance of the component providing the signal is high. The interaction of that output impedance and cable capacitance will form a low pass filter, whose rolloff will usually begin in the ultrasonic or RF region, and therefore be inconsequential, but if those parameters are too high the filter can start rolling off and/or introducing phase shifts at frequencies that are low enough to have audible consequences. For a given cable type, all of those parameters are of course proportional to length. All of this, btw, pertains to analog cables. Completely different considerations and effects come into play in the case of digital cables. And yes, crossing cables at right angles, or at least minimizing how much of their lengths are parallel and closely spaced, is good practice and will minimize or eliminate any effects the corresponding signals might otherwise have on each other. Best regards, -- Al |
Hi Jim,
As often occurs when this kind of subject comes up, ambiguity and/or imprecise use of terminology muddles the issue. If you replace his use of the word "charge" with the words "charge carrier," I think what he says then becomes pretty much correct.
As explained by Kijanki with the balls in a tube analogy, and as alluded to in my long post in this thread dated 8-23-2017 at 7:08 p.m. EDT (although what I said in that post was stated in terms signal energy rather than charge), charge propagates at near light speed, while charge carriers (electrons, in the case of a metallic conductor) move very slowly. And current, defined in terms of amperes, is proportional to the average number of charge carriers traversing a given cross-section of the conductor per unit time.
Best regards,
-- Al
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Thanks, Kijanki. Yes, that explains it pretty well. Although I can see how that statement could be misinterpreted. The reference to "in one direction" should say something like "in one direction for a given direction of the electric field," the direction of the electric field of course alternating every half-cycle in the case of AC. Also, the reference to "average velocity" is a bit misleading, because it could be interpreted as meaning that the much faster Fermi velocity of 1570 kilometers/second or so is numerically averaged in, even though (as I mentioned earlier) it cancels out of the average (the "net flow") since it is in random directions.
Best regards,
-- Al
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Geoffkait 9-3-2017
No one has answered my question, why there is a net velocity for electrons? Assuming electrons move back and forth with alternating current which I’m not convinced they actually do. I believe I did answer that, Geoff: Almarg 9-3-2017
...there is no **overall** net movement of the electrons, assuming that the DC component of the applied voltage is zero. However, within each half-cycle of the applied voltage there is net electron movement and net velocity in one direction or the other, the direction corresponding to the +/- polarity of the applied voltage at that instant. I had said that in one of my early posts in this thread. So yes, the electrons moving at drift velocity do move back and forth with alternating current, never moving very far from their original location, assuming no DC is present. The Wikipedia writeup you quoted even alluded to that: "... electrons flowing across the contact point in a switch will never actually leave the switch." Also, Geoffkait 9-3-2017
To say there is a Fermi velocity when current is present makes no sense since electrons travel axially.
Again, I believe that is incorrect. Under the influence of an applied voltage/electric field, I believe that ALL electrons do NOT travel axially, in a direction corresponding to the polarity of the applied voltage, just SOME of them do. I believe that some of them continue to move in a random manner, at the Fermi velocity. If that were not true, how much voltage would have to be applied for ALL of the electrons to suddenly cease moving in a random manner at Fermi velocity, and obediently start moving in an axial manner at drift velocity? 1000 volts? 120 volts? 1 volt? 1 millivolt? 1 microvolt? 1 nanovolt? 1 picovolt? How much current is necessary to be able to say that "current is present"? 10 amperes? 1 ampere? 0.0000000000001 amperes? I hope you see my point. In any event, barring further questions from others I’m done with this discussion. Regards, -- Al |
Kijanki, I of course agree with your post. I suspect, though, that Geoff intended his "Exhibit A" post to be a rebuttal (mainly in its second paragraph) of the second paragraph of my post which immediately preceded it. To recapitulate the relevant paragraphs: Almarg 9-3-2017
Also, I believe that your [Geoff’s] statement that "Fermi velocity (random) applies only to materials when no current is applied" is incorrect, and that there is always random movement of some electrons, at Fermi velocity and in random directions. That is why the word "net" comes into play. Since the movements at Fermi velocity are in random directions, that velocity does not factor into (or average into) the drift velocity. Geoffkait 9-3-2017 (quoting Wikipedia)
By comparison, the Fermi flow velocity of these electrons (which, at room temperature, can be thought of as their approximate velocity in the absence of electric current) is around 1570 km/s. My response to that, assuming it was intended as a rebuttal of my statement: The quoted Wikipedia paragraph, referring to Fermi "velocity in the absence of electric current," says nothing about whether or not random movement of electrons at Fermi velocity occurs when a current is present. And I believe that such movement does in fact occur when a current is present, which is why drift velocity corresponds to **net** electron movement, past any given point. Regards, -- Al |
Geoff, re your pop quiz, as you appear to realize there is no **overall** net movement of the electrons, assuming that the DC component of the applied voltage is zero. However, within each half-cycle of the applied voltage there is net electron movement and net velocity in one direction or the other, the direction corresponding to the +/- polarity of the applied voltage at that instant. I had said that in one of my early posts in this thread.
Also, I believe that your statement that "Fermi velocity (random) applies only to materials when no current is applied" is incorrect, and that there is always random movement of some electrons, at Fermi velocity and in random directions. That is why the word "net" comes into play. Since the movements at Fermi velocity are in random directions, that velocity does not factor into (or average into) the drift velocity.
Regards,
-- Al
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Geoffkait 9-2-2-2017
Uh, I’ve already stated that it’s a tie. As indicated by the
mathematical paper from the Journal of Physics on the dodgy subject of
whether the energy of the signal is located outside or inside the
conductor the energy is actually partly outside and partly inside. And
the mathematics for that conclusion is provided in the first couple of
paragraphs. Don’t tell me you didn’t read it. GASP
Yes, I had read the paper you are referring to. There is nothing in it that is inconsistent with what I have said. If you'll notice, it deals with a hypothetical situation in which the wire ***is*** the load. In other words, a single piece of "long" wire is connected directly across the terminals of a voltage source. (The numerous references in the paper to the wire being "long" presumably imply that its resistance is high enough to limit the resulting, um, current, to an amount that can be provided by the voltage source, and that would not cause the wire to melt). In that situation the Poynting Vector would point inward to the conductor, at all points along its length, as shown in Figure 1 of the paper. The energy carrying photons would therefore enter the conductor, causing the conductor would heat up. Note the references to energy flowing **into** "the cylinder," resulting in "Joule heating." "The cylinder" referring to the geometry of the wire. As the paper says:
The picture that emerges from these considerations is that the electromagnetic field around a current carrying wire is such that the energy dissipated in the wire is brought into it by the corresponding Poynting vector through each point of its surface.
That is all perfectly consistent with what I have said on the subject previously, assuming a more real world scenario involving low resistance wires conducting energy to a resistive load. In that situation the electromagnetic wave, and the photons comprising it, travel outside the conductors, aside from (as I said in my previous post) "the very small fraction of the photons corresponding to the very
small amount of energy that is absorbed by the resistance of the
conductor and converted to heat." Regards, -- Al |
Hi Jim, I always have great respect for Herman’s opinions and insights, and as usual I don’t disagree with any of his comments that you quoted. And as you’ve no doubt gathered over the years, like him I happen to be someone who generally tries to be as precise as possible with words and terminology, which he is certainly trying to be in the quoted passages. However, it seems to me that there are circumstances which can make some latitude in the use of terminology appropriate. Including this one, given the extremely widespread use (or arguably misuse) of the term "current." As well as the fact that for nearly all practical purposes, other than perhaps providing fodder for Internet debates about wire directionality (which Herman has not expressed a strong opinion about either way), the more widely used concept of "current" works fine. So as I said earlier: Almarg 9-1-2017
It is energy absorbed **from** the electromagnetic wave by the non-zero resistance of the conductor in the fuse, which as I said causes the Poynting vector to tilt slightly toward the conductor, that causes it to blow....
... Since the amount of energy that is absorbed from the electromagnetic wave by the conductor in the fuse and converted into heat (causing it to blow if excessive) is proportional to both the energy that is being conveyed by that wave and to "the current," it is reasonable (and of course far more practical) to analyze the situation in terms of amperes and ohms, rather than in terms of joules (a unit of energy) and Poynting Vectors.
And correspondingly, since in the case of electrical signals (or power) being conducted via wires the slow moving "current" and the very fast moving electromagnetic wave go hand-in-hand (as I’ve explained), IMO it would be meaningless to think of one but not the other as being the cause of the fuse blowing. Now, regarding: Geoffkait 9-2-2017
One thing I will sign up to is that if anything is traveling down the conductor it’s photons, not electrons. Free free to concur with comment, concur without comment or non concur. As I’ve stated on previous occasions, I agree fully that the energy of an electrical signal (or power) being conducted via wires is conducted at near light speed in the form of an electromagnetic wave that is comprised of photons. We’ll have to agree to disagree, however, as to whether those photons propagate within or outside of the conductor, aside from the very small fraction of the photons corresponding to the very small amount of energy that is absorbed by the resistance of the conductor and converted to heat. Regards, -- Al |
Gentlemen, I believe that at this point we are all on the same page regarding what occurs when an electrical signal propagates. To the extent that there is disagreement I believe it just revolves around terminology, and its interpretation. Regarding "how could Drift Velocity be the average electron velocity?" I think that if the word "average" is changed to the word "net" we could all agree. At least I hope so. The word "net" in this context implying that random electron movements at Fermi velocity would cancel out of the drift velocity calculation, with electron movement caused by the applied voltage remaining in the calculation. Also, regarding the mention in the article that Jim quoted to the effect that drift velocity is a function of wire thickness, that is correct, and related specific calculations can be seen in the Wikipedia article on drift velocity I linked to earlier. Also, Kijanki, thanks for the excellent and very informative perspective you provided a few posts back on the Poynting Vector, E and H fields, etc. Jim, re your Fluke 87 multimeter, I have an 87V I purchased a couple of years ago, which I assume is a similar but more recent model. Great meter, although certainly not cheap (I think I paid around $375 for it). When the tips of the leads on mine are held together it reads either 0.1 ohms or 0.2 ohms, depending on exactly how the tips are held against each other. I don’t know what the gauge of the leads is, but given that the total length of the two leads is about 8 feet I suspect the lead resistance is a significant contributor to the 0.1 ohms, together with round-off due to the limited resolution. I previously had a small Triplett model 310 analog multimeter, which was ridiculously inaccurate (e.g. it indicated my AC as being around 95 volts; the Fluke indicates about 118 or 119 depending on time of day, etc). Which was surprising because I had read that many electricians use that particular Triplett model. Guess I just had a bad example of it. Regarding fuse resistance, you might find the information on page 2 of this Littelfuse datasheet to be of interest. For the 4 amp 250 volt slow blow 6.3 x 32 mm glass fuse which is among the many listed, the "cold" resistance (meaning the resistance with negligible current being conducted) is indicated as 0.0311 ohms. So for a design which puts say half the rated max current through it the voltage drop would be a bit more than 0.06 volts. Best regards, -- Al |
Thanks, Jim. Regarding... Am I correct in assuming watts is a measurement of electrical energy? Watts is a unit of power, as you of course realize. Power is a quantity that is defined at a specific instant of time, although its average value over some interval of time can of course be calculated. Energy is defined as the product (multiplication) of power and time, and can be expressed as some number of joules, as well as in various other units. The electrical energy will be greater at 120V than at 24V for a circuit using the same 2 amp fuse for overcurrent protection.
120V x 2A = 240 watts
24V x 2A = 48 watts
240V x 2A = 480 watts The power and the energy being conveyed to the load will of course be much greater in the 120 and 240 volt cases than in the 24 volt case. But as I’m sure you realize but others may not, the only voltage that the fuse "knows about" is the one that appears between its two terminals, which when it is not blown corresponds to the amount of current it is conducting times its resistance. In the case of audio equipment operating normally that voltage will typically be a small fraction of a volt. If and when the fuse were to blow, however, the full 120 volts would then appear across its terminals. Although no current would be conducted then since the resistance of the blown fuse would be essentially infinite. Relevant to all of this, it’s worth noting that in the detailed specifications that are provided by the major fuse manufacturers, such as Littelfuse and Eaton/Cooper Bussmann, the "melting point" (i.e., the point at which the fuse is nominally rated to blow) is specified as i^2 x t (e.g., amperes squared x seconds). As you of course realize, power into a resistive load = i^2 x R, and i^2 x t is therefore proportional to energy. Best regards, -- Al |
Hi Jim, As always you ask good questions. Regarding the first one, though...
Could you please explain in more detail the relationship of the
electromagnetic wave, that travels in the space outside of the
conductor, (At near the speed of light), and the "current" that travels
very slowly slightly vibrating back and forth at 60Hz in the conductor.
... I'm not sure what I can add to what I said in my long post above dated 8-23-2017 at 7:08 p.m. EDT. Regarding your other questions:
From what I understand the movement of the current in the conductor is quite slow.... Correct?
Correct, assuming "current" is defined as the movement of charge carriers (i.e., electrons in a metallic conductor). An example described in the Wikipedia writeup on Drift Velocity indicates that for a current of 1 ampere in a copper conductor of 2 mm diameter the velocity calculates to 23 um/second ("um" = millionths of a meter). As noted in the writeup, btw, random movement of electrons even in the absence of "current" occurs at a far greater velocity (the Fermi velocity) than the "drift velocity" of current, although the Fermi velocity is still vastly slower than the speed of electromagnetic wave propagation.
Am I correct in saying you can’t have the electromagnetic wave without having current?
Yes, in the case of electrical energy that is being conveyed via wires. Electromagnetic waves can of course propagate in free space, as in the cases of radio waves and light waves.
The bigger the load, the more current in the conductor. The more current
in the conductor the larger the electromagnet wave.... Correct?
Yes, assuming "larger" is interpreted in the sense of having "more energy."
IF the wire is too small to handle the amount of current in the wire is
it the current that causes the wire to overheat or is it the energy of
the electromagnetic wave? Please explain in detail.
The Poynting Vector, which describes the direction in which energy is being propagated, would be perfectly parallel to the conductor if the conductor's resistance were zero. Since that resistance is non-zero, the Vector will tilt slightly toward the conductor, resulting in a small amount of energy being transferred into conductor, absorbed by its resistance, and converted to heat. In effect, the resistance of the conductor causes it to become part of the load.
... if only a voltage, (potential), is present, an electromagnet field will
exist outside of the conductor/s without there being current... Correct?
I'm not 100% certain, but I believe in that situation an electric field would be present, but not a magnetic field.
I know it is the energy, from the electromagnetic wave, that makes a
heating element heat up and gives off its’ heat into the surrounding air
around it. It is not the "current" directly causing the resistance
element to heat up.... Correct?
As I've said, in the case of electrical signals (or power) being conducted via wires the electromagnetic wave and "the current" go hand-in-hand, and one would not exist without the other. So the question is essentially just an academic/philosophical one IMO, not unlike the classical question of whether the chicken or the egg came first. Your succeeding statements involving E, I, P, etc. are of course correct.
herman said it is the energy of the electromagnetic wave passing on the
outside of the fuse element link that causes it to melt and blow open
when the fuse is overloaded.
It is energy absorbed **from** the electromagnetic wave by the non-zero resistance of the conductor in the fuse, which as I said causes the Poynting vector to tilt slightly toward the conductor, that causes it to blow.
Is not P the energy of the electromagnetic wave?
They are proportional, but strictly speaking energy corresponds to power x time.
Here is where I get hung up. As you know a 2 amp 250V fuse can be used
for any voltage 250V or less. It could be used where the voltage is 24V.
The ampere rating of the fuse is still 2 amps. So to me the current has
to be some component that causes the fuse to blow when the current that
passes through the fuse link and exceeds 2 amps in the time curve set
by the fuse manufacture. NOTE I did not say current flow.
In my earlier long post I defined "the current" as follows:
What can be referred to as "the current," as opposed to "the signal,"
can be considered as corresponding to the number of electrons traversing
a given cross-section of a conductor in a given amount of time. One
ampere of current, for example, corresponds to one coulomb per second,
where one coulomb corresponds to the amount of charge possessed by about
6.2 x 10^18 electrons.
Since the amount of energy that is absorbed from the electromagnetic wave by the conductor in the fuse and converted into heat (causing it to blow if excessive) is proportional to both the energy that is being conveyed by that wave and to "the current," it is reasonable (and of course far more practical) to analyze the situation in terms of amperes and ohms, rather than in terms of joules (a unit of energy) and Poynting Vectors. And correspondingly, since in the case of electrical signals (or power) being conducted via wires the slow moving "current" and the very fast moving electromagnetic wave go hand-in-hand (as I've explained), IMO it would be meaningless to think of one but not the other as being the cause of the fuse blowing. Best, -- Al |
Chalmersiv, the answer to your question is of course not predictable with any kind of certainty, in part because cable effects are dependent to a significant degree on the technical characteristics of what is being connected. And in the case of a speaker cable, among other things on how the impedance of the speaker varies as a function of frequency, as well as the nominal impedance of the speaker, and perhaps in some cases also on the amplifier's output impedance and on how much feedback its design incorporates.
But as I mentioned earlier the inductance of a speaker cable can be drastically degraded (i.e., increased) if the + and - conductors are not in very close proximity. Although whatever significance that may have will depend on the impedance of the speaker at high frequencies, and also on the length of the cable. And I note that the description of the T-14 cites low inductance as one of its key features, while the description of the Q-10 does not. That despite the fact that the materials used for their conductors and dielectrics are apparently the same.
So for that reason, together with what has been said earlier about the + and - conductors having essentially equal importance, I would recommend against using any of the cheaper wires you listed for the negative conductors.
What I would suggest that you try, if you already haven't, is using BOTH the T-14 and the Q-10 in parallel. And in each case with the conductors that are enclosed within a given cable jacket being used for BOTH + and - (which if I understand your last post correctly may not have been what you were doing), rather than for just one polarity. Also, I'm not sure if I understand whether your Q-10 is configured with a sufficient number of connections for biwiring, but if not use it in parallel with the T-14 for the bass connections, and use the additional T-14s alone for the mid/hi connections.
On another note: Jim & Kijanki, thanks again for the nice words. Kijanki, in response to your question, no, I have never been a teacher, and I have not ever had any particular desire to be one. But in my career working in a corporate environment I have always found it advantageous to be able to communicate in as clear and precise a manner as possible.
Also, speaking of being knowledgeable, I'll mention that Jim's (Jea48's) knowledge of all things electrician-related continually amazes me. And I've certainly learned more than a few things from his posts over the years. As well as having had the pleasure in various threads here of the two of us successfully resolving more than a few problems people have had with their systems.
Best regards,
-- Al
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Thanks once more, Kijanki. Your post just above is excellent IMO, as usual.
As far as the audibility of dielectric absorption in audio cables is concerned, my reading of various technical and anecdotal references to that effect I've seen over the years suggests to me that it stands a good chance of being audibly significant in many applications. However I have never seen either an analysis or measured data that would provide a quantitative perspective on it, in the context of audio cables.
So FWIW my own "expectation bias" is in the direction of that effect being great enough in degree to be an audibly significant contributor to cable differences, in many systems. But as far as I am aware information doesn’t seem to be available that would provide insight that is more specific.
Best regards,
-- Al
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Thank you, Kijanki. I was just about to post that numerous references can be found on the web indicating that the dielectric constant of Teflon is in the vicinity of 2.0, or even a bit more, not 1.0. Also, the 70-85% figure Geoff cited is of course at best an average or typical propagation velocity, and examples of audio cables having propagation velocities that are significantly slower and significantly faster are easily found.
Steve (Williewonka), the book you referenced looks like an excellent read! I note, btw, that the section your link goes to was authored by Bill Whitlock, of Jensen Transformers, who like Ralph Morrison is a noted authority on such matters. And I note that Mr. Morrison himself is referred to in Mr. Whitlock’s writeup.
Also, if I may be a bit presumptuous, let me extend kudos for your interest in gaining as thorough a technical understanding of such matters as possible, to complement what I know is your very extensive practical experience experimenting with various cable configurations.
Regards,
-- Al
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Kijanki, thanks for your characteristically excellent technical input. Williewonka 8-24-2017
Let’s take the IC connecting two components as an example...
- the signal conductor has an AC signal on it
- the neutral conductor is connected to the neutral sides of each component
On well designed components the neutral side of the circuit should always be at zero vaults - especially if grounded
If both components are well designed, then the neutral sides of the their respective circuits would be at zero volts,
Therefore, the neutral conductor of the IC should also be at zero volts - yes?
Hi Steve, First, be sure to keep in mind, as you no doubt realize, that a voltage must always be defined with respect to some reference. Given that, in the example you cite above the neutral conductor would indeed be at or very close to zero volts, **relative to the circuit grounds/signal grounds of the two components.** And probably in most (but not all) designs relative to AC safety ground and earth ground as well. But those facts do not have any inconsistency with what I said in my earlier posts. Consider the simple example of a 120 volt light bulb. When it is turned on via the switch on the wall, if you were to individually measure the current in the "hot wire" and the "neutral wire" that are connected to it you would measure exactly the same amount of current in both. Even though the neutral wire is at or very close to zero volts relative to earth ground and to AC safety ground. I’ll take a look at the book you referenced later today or tonight. Best regards, -- Al |
Jim (Jea48), Jc4659, and Kijanki, thanks very much for your kind words. Jim (Jea48), I’m not sure if your most recent post is suggesting that I try to explain why Geoff’s comment is incorrect, or that I refrain from doing so to avoid having this heretofore constructive thread go downhill the way the recent thread on wire directionality has. But I’ll assume the former, perhaps incorrectly. Geoffkait 8-24-2017
If the audio signal travels through the *dielectric* and not (rpt not) through the metal conductor I suppose we can throw out the whole skin effect idea, which says most audio frequencies travel *inside* the metal conductor at some depth, with only very high frequencies, perhaps above "audio frequencies," traveling near the surface, I.e., skin. How can audio frequencies travel inside the conductor when the audio signal - the electromagnetic wave - travels outside the conductor?
Geoff, to be precise, skin effect means that as frequency progressively increases above a certain frequency (which depends on the diameter of the conductor), " current density" (with "current" defined as in one of my earlier posts) decreases to a greater degree at progressively greater depths. That causes a progressive increase in the resistance of the conductor at progressively higher frequencies. "Audio frequencies travel inside the conductor," to use your words, in the sense that the movement of electrons, at the very slow drift velocity I referred to, is a very small back and forth oscillatory motion occurring at the same frequency or frequencies for which energy is being conveyed in the electromagnetic wave. As the +/- polarity of the applied voltage changes, at a given frequency, the direction of that slow movement of electrons changes correspondingly. (And actually, to be precise, I should say "net movement of electrons," because random movement of some electrons is always occurring to some degree). In the recent wire directionality thread which you participated in extensively, Jim (Jea48) quoted a statement by Ralph Morrison, a world renowned authority on such matters, and the author of several textbooks, which contradicts your assertion that the electromagnetic wave travels within metallic conductors. That assertion was also contradicted in another thread here by a noted designer of highly respected world class audio electronics, as well as by me and several other technically knowledgeable posters. During the course of my lengthy career and schooling in electrical engineering I have never seen such an assertion ever made by anyone other than yourself. If you can cite a seemingly credible reference supporting your contention I will attempt to explain why it is either incorrect or is being misinterpreted. Regards, -- Al |
Thanks, Jerry!
Steve, when I refer to currents in the two conductors that are equal except that they are moving in opposite directions (i.e., current in one conductor is moving toward the load when current in the other conductor is moving toward the source), another way to look at it, that amounts to saying the same thing but may make it more clear, is that the currents in both conductors are moving in the same direction but around a loop. The loop consisting of the two conductors plus the input circuit of the load plus the output circuit of the source. And between each half-cycle the direction the current is traveling around that loop reverses.
Best regards,
-- Al
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Hi Steve, You raise good questions, which get into some complexities that are not obvious. "The signal," and the energy that it conveys, is conducted through neither of the conductors. It is conducted in the form of an electromagnetic wave, which propagates at a substantial fraction of the speed of light in a vacuum, and propagates through the dielectric which surrounds the conductors. The exact propagation speed is dependent primarily on what is known as the "dielectric constant" of the particular insulation. Putting aside reflection effects that can occur mainly at RF frequencies as a result of impedance mismatches, and assuming that the load is essentially resistive, that energy propagates in just one direction, from the source of the signal to the load. However, that propagation of the signal and its energy is intimately related to movement of electrons within both of the conductors, which takes place in both directions (the direction alternating in each of the two conductors, assuming we’re not dealing with DC), and which takes place at an ***extremely*** slow velocity that is referred to as "drift velocity." In the case of electrical signals that are conducted via wires (as opposed, for example, to being radiated through the air or a vacuum), the extremely slow movement of electrons within the conductors and the near light speed movement of the signal and its energy are intimately related, as I said, and one would not occur without the other. A way to visualize it is that at the instant a signal voltage is applied to the source end of a cable, a **very** slow movement of electrons will occur into one of the two conductors at that end of the cable, and out of the other of the two conductors at that end of the cable, corresponding to the +/- polarity of the signal at that instant. At the other end of the cable, and at all points in between, there will be a similar slow movement of **different** electrons, with the response of those electrons being delayed from the response of the electrons at the source end of the cable by the amount of time it takes "the signal" to traverse the corresponding cable length (at near light speed). What can be referred to as "the current," as opposed to "the signal," can be considered as corresponding to the number of electrons traversing a given cross-section of a conductor in a given amount of time. One ampere of current, for example, corresponds to one coulomb per second, where one coulomb corresponds to the amount of charge possessed by about 6.2 x 10^18 electrons. So assuming that only two paths exist between the source and the load, namely the two conductors in a single cable, "the current" being conducted by both conductors in response to an applied signal is in fact identical, except that when it is moving in one direction in one conductor it is moving in the other direction in the other conductor. And in the case of audio signals, or any kind of signal other than DC, the directions in the two conductors alternate between each half-cycle of the waveform. So with the slight possible exception I mentioned earlier about RFI/EMI pickup, in the case of a speaker cable the two conductors are of equal importance. In the case of a line-level analog interconnect, on the other hand, IMO the "ground" or "return" conductor should if anything be considered to be **more** important than the "signal" or "hot" conductor. The reason being that the characteristics of the return conductor may affect susceptibility to ground loop-related high frequency noise or low frequency hum, depending on the internal grounding configuration and other aspects of the designs of the particular components that are being connected. So given the foregoing it hopefully becomes clear that your statement that... ... when I think about speaker cables, the "energy" in the signal conductor must be very different from the neutral side simply because by the time the signal gets through the speaker voice coil, most of it has been converted into the movement of the driver, so the neutral must be quite different - doesn’t it? ... is not a correct statement because the transfer of energy to the load goes hand-in-hand with current (movement of charge carriers, i.e., electrons) in **both** conductors. With that movement being equally important in the two conductors, and (putting aside the possible ground loop and RFI/EMI effects I’ve mentioned) being identical in the two conductors aside from being in opposite directions at any instant of time. Hopefully that clarifies more than it confuses :-) Best regards, -- Al |
Good comments by Steve (Williewonka). I would add that if the + and - conductors referred to in the OP are not in close proximity, and preferably twisted together in some manner, the inductance of the cable will be considerably increased. If the impedance of the speakers is low at high frequencies (as it is, for example, in the case of many electrostatics), and/or if the cable length is long (inductance is proportional to length, for a given cable type), that may result in perceptible rolloff of the upper treble, and dull or sluggish sounding transients.
Also, in terms of wire quality I would consider the negative conductor to be no less important than the positive conductor. After all, they are conducting the same current, just in opposite directions at any given instant. A conceivable exception to that, however, is that the amplifier might be more susceptible to RFI/EMI picked up by the cable and introduced into its feedback loop (if it has one) from the positive conductor than from the negative conductor.
Regards,
-- Al
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