Motional EMF from an Electron Flow Perspective

Status
Not open for further replies.

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Admittedly, 10 years old seems too young to get into too much detail, so some basic understanding of voltage combined with the concept of current flow may be useful and sufficient. Even when water flow is considered, the idea of pressure is needed to really get the gist of what is happening, and voltage is analogous to pressure.
Here's where the whole problem began for me (there's a question at the bottom). I'm reading one of the NEETS training manuals where it describes DeForest's experiments with triodes. This is the first of two graphics they use, this one showing how changes in grid voltage result in more or less electron flow across the void in the tube:

neets 1.jpg

Their description goes like this:

"It’s the old principle of 'opposites attract.' When the control grid is made positive [fewer electrons on it], electrons surrounding the cathode (negative charges) are attracted to the grid. But remember, the grid is a metal mesh. Most of the electrons, instead of striking the grid wires, are propelled through the holes in the mesh. Once they pass the grid, they are attracted to the positive charge [fewer electrons] on the plate."

And I was getting it. Loving it, in fact. What could be simpler? All we need to worry about is electrons and the fact that they don't like to be near each other. So I turned the page, expecting that I would find out what happened to those electrons after they hit the plate and exited the tube. I found this diagram with plate resistors added:

neets 2.jpg

Which was fine, because I had seen a similar configuration of parts in many guitar amp circuits. What was not fine was the description because it unexpectedly (and in my mind, unnecessarily) abandoned the electron-flow paradigm:

"Instead of amplification, De Forest had obtained 'conversion,' or in other words, converted a signal voltage to a current variation. This wasn’t exactly what he had in mind. As it stood, the circuit wasn’t very useful. Obviously, something was needed. After examining the circuit, De Forest discovered the answer—Ohm's law. Remember E = I × R? De Forest wanted a voltage change, not a current change. The answer was simple: In other words, run the plate current variation (caused by the voltage on the grid) through a resistor, and cause a varying voltage drop across the resistor."

And I thought to myself, Why didn't they simply continue describing the electron flow? Why talk about a voltage drop across the plate resistor when it's what's happening at the bottom of the resistor -- ie, the output of that stage of the amp -- that's of interest to us? In short, I thought they should have said something like this:

"If we put a resistance in the way of all those exiting electrons, it will obviously create 'congestion' or a 'bottleneck' at the bottom of that resistor. We will thus have a greater [negative] voltage (ie, a higher concentration of agitated electrons, all trying to repel one another) at that point -- the pressure increasing while the flow out of the tube continues, then decreasing as the electrons are able to eventually drain away through the resistor."

I think this latter description better for three reasons: first, because it doesn't change paradigm horses in mid-stream; secondly because it directly relates not only current but voltage to our little electron buddies; and thirdly because we can now predict what will happen, in a general sense, intuitively and without mathematics, if we change the value of that resistor (specifically, too much resistance will result in too much congestion and thus distortion of the signal; too little resistance will result in the electrons rushing by the "output" point and too little volume will result).

Question: Is that essentially what happens with those exiting electrons? If not, exactly what does happen to them?

 
Last edited:

Lool

Joined May 8, 2013
116
The problem is that your description is not physically correct. You are bothered by the so called "paradigm horse change", but one horse does not pull the carriage when you think in terms of circuit theory, which is what is relevant here.

The presence of the resistor basically is providing room for your signal to "swing", as they say. You need to bias the tube so that the voltage output (or current flow) at the junction of the anode and resistor has room to move up and down, so that your input voltage signal can create an amplified version of itself on the output. Imagine if there is no resistor, - then the potential on the output would always equal the anode voltage, and even if current changes, how will you use that current change to drive another amplifier stage?

Generally we can't have a bottleneck at the bottom of the resistor. When current flows, there can be certain charge distributions created on conductors, but any point of the circuit does not build up an overall excess or deficit of electrons. The charge distribution is secondary and one never thinks of it. All it does is allow charges to flow around bends, but again nobody ever thinks of it that way. Basically, the current flows and the current is the same in any series path. There is no bottleneck, other then the fact that more resistance correlates with less current flow (Ohm's law), or in this case, the current drive generates more voltage drop on the reisistor (again Ohm's law).

Without a good understanding of the rules of circuit theory, it is very hard to understand circuits like this.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
My question had two parts. First: "Is that essentially what happens with those exiting electrons?" To which you replied:

The problem is that your description is not physically correct.
Which I'll take as a "No." Now the second part of the questions was, "If not, exactly what does happen to those electrons exiting the tube?" And your reply was:

You are bothered by the so called "paradigm horse change", but one horse does not pull the carriage when you think in terms of circuit theory, which is what is relevant here. The presence of the resistor basically is providing room for your signal to "swing", as they say. You need to bias the tube so that the voltage output (or current flow) at the junction of the anode and resistor has room to move up and down, so that your input voltage signal can create an amplified version of itself on the output. Imagine if there is no resistor, - then the potential on the output would always equal the anode voltage, and even if current changes, how will you use that current change to drive another amplifier stage? Generally we can't have a bottleneck at the bottom of the resistor...
No mention of electrons yet...

...When current flows, there can be certain charge distributions created on conductors, but any point of the circuit does not build up an overall excess or deficit of electrons.
...and when you finally do mention electrons, what you say directly contradicts the usual and ubiquitous description of vacuum tubes: Doesn't the cathode of a tube sometimes have an "excess" of electrons on and about it? Doesn't the plate of a tube typically have a deficit of electrons at those same times? Isn't that what we mean when we say the plate is "positively charged"? And what about batteries? Batteries are frequently described as having an excess of electrons on one side, and a deficit on the other. Or how about charged capacitors? Or the metal rod moving through a uniform magnetic field described in the very first post of this thread. Everyone describes that in a fashion similar to this:

"The magnetic force acting on a free electron in the rod will be directed upwards. As a result, electrons will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, and the top of the rod will have an excess of electrons (negative charge) while the bottom of the rod will have a deficit of electrons (positive charge). This will result in a potential difference between the ends of the rod equal to LvB."

Note the portion in italics. Those are not my words; they can be found on many websites and in most basic electronic books.

The charge distribution is secondary and one never thinks of it.
Unless they're describing the operation of a vacuum tubes, or explaining batteries and capacitors to beginners. But it's obvious, I agree, that they put it all out of their minds after that. Otherwise, it wouldn't be so hard to get anyone to tell me what happens to those exiting electrons. And just because nobody thinks of things that way, doesn't mean we shouldn't: maybe thinking that way will give us a simpler way to teach kids, or insights we otherwise would have missed. I know it helped me decide whether to put a bigger or smaller plate resistor on my guitar amp when I thought it was plenty clear but not loud enough.

Basically, the current flows and the current is the same in any series path. There is no bottleneck, other then the fact that more resistance correlates with less current flow (Ohm's law), or in this case, the current drive generates more voltage drop on the reisistor (again Ohm's law). Without a good understanding of the rules of circuit theory, it is very hard to understand circuits like this.
Which is the equivalent of saying, "Don't think of it like that (even though that's how things are invariably described inside the tube)." Which, unfortunately, is not an answer to the question, "Exactly what does happen to those exiting electrons?"

I'm sorry if I sound frustrated, but I am. I'm consistently told I'm wrong about those exiting electrons, and yet none of those correcting me seem to be able to tell me when and where those electrons actually go. It seems to me they must temporarily "pile up" at the base of the plate resistor. After all, they have to go somewhere, right? And it's harder for them (and takes them longer) to get through a resistor than a plain wire, right? Surely you're not saying that they move at exactly the same pace whether the plate resistor is 1 ohm or 1 megaohm... I don't see how they can help but accumulate when that resistance is large.
 

Lool

Joined May 8, 2013
116
OK, you are making some good points here, and I know my wording is not perfectly describing everything exactly correctly. Keep in mind that I'm not a teacher by profession, so I'm not used to trying to present things both correctly for the advanced student and simply for the new student. Many of your explanations show an advanced level, but a few holes in your knowledge would be more typical of a new student.

So as far as charge distributions accumulating, this can and does happen in the real world. The simplest example is a capacitor which accumulated positive charge on one plate and negative charge on the other plate, but the total charge on the capacitor remains zero. The vacuum tube is an even more advanced device that requires field theory to understand and describe precisely. However, when we use circuit theory, the assumption is that each device (resistor, capacitor, tube etc) does not accumulate charge as a whole, and the conductors that connect the devices are assumed ideal zero ohm resistance.

As far as where electrons go, they typically do not pile up, or if they do, as in the case of a capacitor, they do it in a balanced way. The electrons simple move in a circuit and that's why a series connected closed circuit has the same current at every point in the circuit, whether conductor or device is considered. Charge is conserved in a circuit, and this is expressed as Kirchoff's current law that says all the currents entering a node sum to zero. Then there is Kirchoff's voltage law which says that the voltage drops in any circuit loop sum to equal the sum of EMFs in the loop.

What you have to realize is that there is a formal subject of "circuit theory" that simplifies the more accurate field theory. Much of your confusion probably comes from the fact that people usually don't explicitly state the rules of circuit theory. They just know them from experience, and do not necessarily express them in a consistent and complete formalism. However, these descriptions do exist. You can find various lectures on youtube and many text books discuss the subject formally. I have left out many details, but Kirchoff's laws, Ohms law, no accumulation of charge for a total device, lumped device elements, ideal conductors for connections, and no magnetic coupling between devices and no electromagnetic radiation are the major assumptions of circuit theory.

I'll stress again, that vacuum tubes really require field theory to understand correctly, and solid state physics is needed to understand transistors correctly. Still, both types of devices can be approximately described by circuit theory.
 
Last edited:

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
As far as where electrons go, they typically do not pile up, or if they do, as in the case of a capacitor, they do it in a balanced way.
I understand the need for "balance". The electrons that gather on the cathode of the tube must come from somewhere, so if there's an excess of electrons on the cathode, there must be a corresponding deficit of electrons somewhere else in the circuit -- though not necessarily all in one spot.

The electrons simple move in a circuit and that's why a series connected closed circuit has the same current at every point in the circuit, whether conductor or device is considered.
Sure. When an amp is idling, for example, there's a steady flow of electrons from cathode to plate, through the plate resistor, through a diode, through the power transformer secondary, and back around (though, if they're not needed to replenish the cathode, they may end up temporarily waiting on the bottom of the reservoir capacitors).

But that's not what makes the music play. Here's how I see it. The incoming signal is "splurges" of electrons, moving first one way, and then the other, changing direction more often with higher-frequency signals, less often with low-frequency signals; bigger splurges for louder sounds. These electrons accumulate on the grid when the splurges are incoming, and are sucked back off the grid when the splurges are going in the opposite direction. The grid thus has, at some points in time, an excess of electrons (which causes the flow from cathode to plate to decrease below idle-level), and at other times has a deficit of electrons (which causes the flow from cathode to plate to exceed the idle level). When the cathode-to-plate electrons leave the tube in a bunch they and accumulate both at the base of the plate resistor and on the input side of the coupling capacitor to the next stage (giving us a more negative voltage reading on a meter or scope); when there is less than the normal idle flow, the deficit of electrons at the base of the plate resistor and on the input side of the coupling capacitor gives us a more positive voltage reading on the meter or scope).

Charge is conserved in a circuit, and this is expressed as Kirchoff's current law that says all the currents entering a node sum to zero. Then there is Kirchoff's voltage law which says that the voltage drops in any circuit loop sum to equal the sum of EMFs in the loop.
Sure, at any point in time. But the actual collection of numbers that sum to zero are significantly different in different spots depending on whether we're at idle, or riding a peak of the signal wave, or at a trough of the signal.

What you have to realize is that there is a formal subject of "circuit theory" that simplifies the more accurate field theory. Much of your confusion probably comes from the fact that people usually don't explicitly state the rules of circuit theory. They just know them from experience, and do not necessarily express them in a consistent and complete formalism. However, these descriptions do exist. You can find various lectures on youtube and many text books discuss the subject formally. I have left out many details, but Kirchoff's laws, Ohms law, no accumulation of charge for a total device, lumped device elements, ideal conductors for connections, and no magnetic coupling between devices and no electromagnetic radiation are the major assumptions of circuit theory. I'll stress again, that vacuum tubes really require field theory to understand correctly, and solid state physics is needed to understand transistors correctly. Still, both types of devices can be approximately described by circuit theory.
I understand that there are different ways of understanding and analyzing circuits, each with various strengths and weaknesses of it's own. But I'm not really interested "circuit theory" or "field theory" except as they help me further understand and develop "electron flow theory" -- which, at least to me, appears (a) to be the way vacuum tubes were first understood and explained (when the circuits in question were first designed and built and productively used by millions), and which appears (b) to be the most accessible way of thinking about electronics for the absolute beginner (since "electron flow theory" is used to describe not only vacuum tubes, but batteries, and capacitors, and generators, etc, in almost all beginning texts).
 

Lool

Joined May 8, 2013
116
I understand the need for "balance". The electrons that gather on the cathode of the tube must come from somewhere, so if there's an excess of electrons on the cathode, there must be a corresponding deficit of electrons somewhere else in the circuit -- though not necessarily all in one spot.
Somewhere else yes, and yes not all in one spot, but more particularly on the other plate of the capacitor. If charges accumulate, then you are describing capacitance. There is most definitely capacitance in a vacuum tube because there are large plates.

However, and this is a key point, the capacitance is not the essential thing that causes the operation of the vacuum tube. This is a secondary effect and very often you would prefer to reduce the capacitance, if you could. The AC variation that results in frequencies you mentioned is determined by the frequencies present on the grid, which is the input drive. If capacitance were too high, then the tube response might not even be able to follow the higher frequencies. For audio this is not likely to be a problem, but for video frequencies it might be.

So, you are focusing all your thought on a secondary effect, as if it were the primary effect. It is not charge build up of charge on capacitor plates that matters here fundamentally. Even with capacitance, the current on each capacitor terminal is identical because the charges are equal and opposite in sign on each plate.


Sure. When an amp is idling, for example, there's a steady flow of electrons from cathode to plate, through the plate resistor, through a diode, through the power transformer secondary, and back around (though, if they're not needed to replenish the cathode, they may end up temporarily waiting on the bottom of the reservoir capacitors).
I'm not talking about steady flow, but I am talking about flow being equal at all points in a series circuit. Steady flow refers to constant DC, but equal current applies to all AC frequencies. Capacitance will matter more when frequency goes up and it could be detrimental sometimes.

But that's not what makes the music play. Here's how I see it. The incoming signal is "splurges" of electrons, moving first one way, and then the other, changing direction more often with higher-frequency signals, less often with low-frequency signals; bigger splurges for louder sounds. These electrons accumulate on the grid when the splurges are incoming, and are sucked back off the grid when the splurges are going in the opposite direction. The grid thus has, at some points in time, an excess of electrons (which causes the flow from cathode to plate to decrease below idle-level), and at other times has a deficit of electrons (which causes the flow from cathode to plate to exceed the idle level). When the cathode-to-plate electrons leave the tube in a bunch they and accumulate both at the base of the plate resistor and on the input side of the coupling capacitor to the next stage (giving us a more negative voltage reading on a meter or scope); when there is less than the normal idle flow, the deficit of electrons at the base of the plate resistor and on the input side of the coupling capacitor gives us a more positive voltage reading on the meter or scope).
The movement one way and the other is not at all dependent on charge accumulation. You may have capacitors that do accumulate charges on the plate, but if capacitance is low, then it is insignificant, yet you would still hear audio frequencies through your amp without charge accumulation. Your concept is just wrong.

I understand that there are different ways of understanding and analyzing circuits, each with various strengths and weaknesses of it's own. But I'm not really interested "circuit theory" or "field theory" except as they help me further understand and develop "electron flow theory" -- which, at least to me, appears (a) to be the way vacuum tubes were first understood and explained (when the circuits in question were first designed and built and productively used by millions), and which appears (b) to be the most accessible way of thinking about electronics for the absolute beginner (since "electron flow theory" is used to describe not only vacuum tubes, but batteries, and capacitors, and generators, etc, in almost all beginning texts).
I disagree that this is the best for beginners. Most beginners start with circuit theory. Batteries , capacitor and generators can not be well understood in terms of electron flow. Potential, EMF and fields are too important for their understanding. Beginners often don't try to understand these at a deep level and often use circuit theory models for these devices. I can't argue with you about vacuum tubes, but nobody uses those any more, and as fascinating as those devices are, basing a whole approach around the fortunate accident that vacuums tubes can be crudely understood in terms of electron flow seems wrong, particularly when we rarely even discuss those devices any more.

Still much will be missed in vacuum tube theory if one does not consider electric fields and voltage. Often, for example your tube amplifier example, the voltage is used mearly as an information signal. Since the current is not necessarily conveying the information from stage to stage, then current flow alone misses a key aspect of the circuit operation. Further, consider what happens when you get to the speaker. Their you need real power to drive the speaker and current alone does not give power. It is voltage times current that gives the power. So, why all the effort to push voltage out of the description?

And what of EMF from batteries and generators. How can you understand that with current flow alone. These things provide power and there must be an energy source to generate both the current and the voltage that gives the power. This implies chemical energy for the battery and mechanical energy (which might come from chemical energy such as oil, coal, natural gas, food, or some other source) for the generator.
 
Last edited:

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
And what of EMF from batteries and generators. How can you understand that with current flow alone.
Again, I'm not talking current flow alone. I'm talking both current and voltage:

1. One amp of current being 6.2x10^18 electrons moving past a given point per second.

2. One volt being some yet-to-be-determined deficit of free electrons per proton. My theory here is numerically, but not conceptually, incomplete.

Everyone agrees on (1). Regarding (2) the literature is unfortunately sketchy. But let's refer back to the first post in this thread where it is said of the moving rod in a uniform magnetic field:

"The magnetic force acting on a free electron in the rod will be directed upwards. As a result, electrons will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, and the top of the rod will have an excess of electrons (negative charge) while the bottom of the rod will have a deficit of electrons (positive charge). This will result in a potential difference between the ends of the rod..."

Clearly, this "potential difference" between the ends of the rod, which is measured in volts, is caused by there being more free electrons per proton at the top of the rod than the bottom. It thus seems reasonable to conclude that number of free electrons per proton and voltage are directly related. Yes?

Thus power (current times voltage) depends not just on how fast electrons are flowing, but on whether there is a excess or a deficit of electrons (relative to the number of protons) in the areas under consideration, as well. But note that both current and voltage are intimately related -- in this model, in simple circuits comprised of things like copper wire and resistors and capacitors and vacuum tubes and transformers -- with the moment-by moment movement and accumulation of free electrons. Yes?
 

Lool

Joined May 8, 2013
116
I would not say current and voltage are intimately related, as you did. I would say that charge distribution and voltage are intimately related.

Voltage can be thought of as energy per unit charge. Imagine there is only one electron in the universe and it is located at the center of the universe. It took zero energy to put it there because there was no other charges to resist the movements. Strictly, the electron has mass, so to move it requires energy to accelerate it and then decelerate it to place it at the center of the universe, but we are talking about electrical energy, not mechanical energy, so energy is zero to place the electron there.

Now imagine that you try to take another electron from an infinite distance away and place it near the first one. That first electron is going to repel the next electron. And the electric field of the first electron is defined to be the force per unit charge for the next electron that tries to sit near it. The potential (or voltage) caused by the first electron is the energy per unit charge that the next electron has at whatever position it is at.

It is not really the deficit of electrons relative to protons, but the total net charge distribution that results in the voltage. Two electrons sitting near one proton is not all that different than one electron by itself. Voltage is basically indicating potential energy. A charge at a high potential is like a still rock sitting on top of a hill. That rock has the potential energy and if you start it moving, when it hits the ground is has a kinetic energy proportional to it's mass times the height of the hill. Similarly, the kinetic energy of an electron released in a potential will develop a kinetic energy proportional to its charge times the potential, assuming it is in vacuum and nothing impedes it.

For this reason, if you charge a capacitor you are storing energy in the capacitor and you can use that energy if current is allowed to flow in a circuit. EMF is a bit different because it is the equivalent voltage that can do work. It may not be fully represented as an accumulated charge, the way the capacitor does. So a battery is not really like a capacitor, even though it is similar. The chemical reactions in a battery allow ions inside the battery to flow against the electric field internally. This is like a rock rolling up a hill. Somehow the charges are pushed up the potential hill as needed and the the EMF can be used to do electrical work in your circuit. If you think about electron flow and potential alone, you will not understand why ions are moving rather than electrons and you will not understand why they are rolling up the hill internally. It is a chemical reaction that does this. So, it's not electron flow in the battery and EMF is not quite the same at potential of a capacitor, although it is functionally similar.

I disagree with your last statement about accumulation of electrons. In a circuit, where are the charges accumulated? Only on plates of a capacitor, if there is a capacitor, and if you think of the entire capacitor (both plates) there is no accumulation of charge in any device in a circuit.
 
Last edited:

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
I would not say current and voltage are intimately related, as you did.
I don't recall saying that.

I would say that charge distribution and voltage are intimately related.
But that I did say: "It thus seems reasonable to conclude that number of free electrons per proton and voltage are directly related."

Voltage can be thought of as energy per unit charge. Imagine there is only one electron in the universe and it is located at the center of the universe. It took zero energy to put it there because there was no other charges to resist the movements. Strictly, the electron has mass, so to move it requires energy to accelerate it and then decelerate it to place it at the center of the universe, but we are talking about electrical energy, not mechanical energy, so energy is zero to place the electron there.
Okay.

Now imagine that you try to take another electron from an infinite distance away and place it near the first one. That first electron is going to repel the next electron. And the electric field of the first electron is defined to be the force per unit charge for the next electron that tries to sit near it. The potential (or voltage) caused by the first electron is the energy per unit charge that the next electron has at whatever position it is at.
Not sure I understand that last sentence. I would have said something like, "The potential (or voltage) increases as the two electrons get closer to one another because like charges repel (no one knows why) and the closer they are to one another, the stronger the repulsion." I would also guess that the curve of voltage vs distance is not linear but has something to do with the square of the distance (or something like that) between the electrons. In any case, it takes more and more force or energy to push the electrons closer and closer together, which means -- looking at it the other way around -- the closer they get, the more they want to get away from each other (ie, more "potential energy" is stored in the configuration as the electrons are forced closer together).

It is not really the deficit of electrons relative to protons, but the total net charge distribution that results in the voltage.
I'm thinking those are essentially the same thing, since (as I understand it) a free electron has to be in (or missing from) the outer shell of some atom. So to get, say, twice as much "charge" at one end of a rod vs the other end, we would need twice as many free electrons around the atoms at one end vs the other end. I would call that relative voltage, (ie, the potential at one spot versus another spot). Absolute voltage is simply a measure of the potential of the electrons around a given collection of atoms in one spot (and is relative to whatever the norm would be for that kind of conductor).

For example, here is a very tiny copper rod with just three atoms, moving through a uniform magnetic field at a uniform velocity (which, let's say, tends to push free electrons to the right):

copper rod atoms 1.jpg
Note that the left atom has no free electrons in its outer shell; the center has one; and the right atom has two. I would define the relative EMF or voltage in this rod, from end to end, to be "2 electron's worth." From the center to either end would be "1 electron's worth." Etc. The absolute EMF or voltage in the center would be zero (because this is the "normal" or "neutral" state of a copper atom -- 29 electrons, 1 per proton). The absolute EMF or voltage would be "-1 electron's worth" for the atom on the left and "+1 electron's worth" for the atom on the right (you can switch the signs if you prefer the conventional current perspective).

Two electrons sitting near one proton is not all that different than one electron by itself.
I should have been clearer. I'm talking, as above, only about free electrons -- the outermost electrons in a conductor that can easily jump from one atom to another; the electrons that an atom can have too few or too many of and still be the the same kind of substance.

Voltage is basically indicating potential energy. A charge at a high potential is like a still rock sitting on top of a hill. That rock has the potential energy and if you start it moving, when it hits the ground is has a kinetic energy proportional to it's mass times the height of the hill. Similarly, the kinetic energy of an electron released in a potential will develop a kinetic energy proportional to its charge times the potential, assuming it is in vacuum and nothing impedes it.
Agreed. The atom on the right above has too many electrons and will want to get rid of one (repulsive potential); the atom on the left has too few and will want to get one back from somewhere (attractive potential). We can get work out of either one.

For this reason, if you charge a capacitor you are storing energy in the capacitor and you can use that energy if current is allowed to flow in a circuit.
Agreed. The capacitor has too many electrons on one side (which repel each other and want to separate) and not enough electrons on the other side (which is attractive to any free electrons within range).

EMF is a bit different because it is the equivalent voltage that can do work. It may not be fully represented as an accumulated charge, the way the capacitor does.
Only if we don't distinguish between "potential electrical energy" and "total potential energy of all kinds." See next point.

So a battery is not really like a capacitor, even though it is similar. The chemical reactions in a battery allow ions inside the battery to flow against the electric field internally. This is like a rock rolling up a hill. Somehow the charges are pushed up the potential hill as needed and the the EMF can be used to do electrical work in your circuit. If you think about electron flow and potential alone, you will not understand why ions are moving rather than electrons and you will not understand why they are rolling up the hill internally. It is a chemical reaction that does this. So, it's not electron flow in the battery and EMF is not quite the same at potential of a capacitor, although it is functionally similar.
I see a battery has having two kinds of potential energy: electrical and chemical. The electrical potential of the battery at any point in time is the the same as it would be for a capacitor -- it's the result of too many electrons per proton on one side, and not enough on the other. In other words, the electrical potential of a battery only takes into account the arrangement of free electrons that has already been created by the chemical reaction within the battery. The rest of the potential in the battery is chemical potential and does not become electrical potential until the chemical reaction succeeds in further modifying the arrangement of the free electrons in the battery.

I disagree with your last statement about accumulation of electrons. In a circuit, where are the charges accumulated?
All over the place. In the wires that go from the guitar to the amp: the electrons first "bunch up" in one direction, then in the other, as the strings are plucked. On the cathodes of the tubes. On the grids of the tubes. At the base of the plate resistor. Etc. You can see the accumulation on a scope:

electron crowd 2.jpg
The lower wave (in red) shows the gathering/dispersal of electrons on the tube's grid; the upper wave (also red) shows the same at the base of the plate resistor. The electrons in the wire from the guitar pickup are moving first one way, then the other, at a frequency determined by the note plucked. Electrons are thus alternately sucked off of the tube's grid and pushed onto the grid as shown on the scope. Since the valve opens when the grid has fewer electrons and closes when the valve has more electrons, the gathering/dispersal of electrons at the base of the plate resistor is not only bigger, but inverted.

Now let's say the note was played on a bass guitar and had a frequency of about 60 cycles/second. That would mean that one complete cycle would take 1/60 of a second, and that each gathering/dispersal would take 1/120 of a second (each reaching its utmost peak or trough after 1/240 of a second). So the excesses and deficits I'm talking about are admittedly temporary and of relative short duration. Yet it's these gatherings and dispersals that generate the voltages that make the thing work.

If you think of the entire capacitor (both plates) there is no accumulation of charge in any device in a circuit.
I really don't know what that means. A charged capacitor, by definition, has an excess of electrons on one side and a deficit of electrons on the other. And sure, the excess minus the deficit may be zero -- but that doesn't mean that electrons haven't accumulated on the negative side of the thing. Touch a big charged-up cap sometime and see for yourself as the accumulated electrons try to get to the other side through your body.
 
Last edited:

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
I've found a page that describes voltage in terms of "displaced electrons" (http://www.tpub.com/neets/book1/chapter1/1k.htm). Here are the main points (I'm quoting now):

"The charge of one electron might be used as a unit of electrical charge, since charges are created by displacement of electrons; but the charge of one electron is so small that it is impractical to use. The practical unit adopted for measuring charges is the COULOMB, named after the scientist Charles Coulomb. One coulomb is equal to the charge of 6,280,000,000,000,000,000 (six quintillion two hundred and eighty quadrillion) or (6.28 x 1018) electrons."

"When a charge of one coulomb exists between two bodies, one unit of electrical potential energy exists, which is called the difference of potential between the two bodies. This is referred to as ELECTROMOTIVE FORCE, or VOLTAGE, and the unit of measure is the VOLT."

"Electrical charges are created by the displacement of electrons, so that there exists an excess of electrons at one point, and a deficiency at another point. Consequently, a charge must always have either a negative or positive polarity. A body with an excess of electrons is considered to be negative, whereas a body with a deficiency of electrons is positive."

"A difference of potential can exist between two points, or bodies, only if they have different charges. In other words, there is no difference in potential between two bodies if both have a deficiency of electrons to the same degree. If, however, one body is deficient of 6 coulombs (representing 6 volts), and the other is deficient by 12 coulombs (representing 12 volts), there is a difference of potential of 6 volts. The body with the greater deficiency is positive with respect to the other."

I believe this to be consistent with what I've said above. Further, I can now fill in the missing number in my previous statement regarding voltage in post #47, specifically:

1. One amp of current is 6.28x10^18 electrons moving past a given point per second.

2. One volt is an excess or deficit of 6.28x10^18 free electrons (where excess means more electrons than protons per atom, and deficit means less electrons than protons per atom).

So if we had a copper rod with 6.28x10^18 atoms, half like the one pictured on the left at one end (one missing electron per atom), and half like the one pictured on the right at the other end...
copper rod atoms 2.jpg

...we'd have an overall voltage difference, end to end, of 1 volt.

 
Last edited:

Lool

Joined May 8, 2013
116
OK, you have quite a bit above correct. In your previous post you did say "current and voltage are intimately related", but I see now you must have misspoke, so no problem there.

Note that electric field (or force) falls off as radius squared for a point charge and potential falls off as radius for a point charge. One thing I mentioned previously, but which may not have gotten through is that electric field is the negative gradient of the potential, or voltage is the distance (line) integral of electric field. This is just another way to say the well known fact that energy (or work) is integration of force over distance.

Now lets deal with these comments.

The lower wave (in red) shows the gathering/dispersal of electrons on the tube's grid; the upper wave (also red) shows the same at the base of the plate resistor. The electrons in the wire from the guitar pickup are moving first one way, then the other, at a frequency determined by the note plucked. Electrons are thus alternately sucked off of the tube's grid and pushed onto the grid as shown on the scope. Since the valve opens when the grid has fewer electrons and closes when the valve has more electrons, the gathering/dispersal of electrons at the base of the plate resistor is not only bigger, but inverted.

We've already discussed the fact that your description works reasonably well for a vacuum tube, but one point that I've tried to make, which i don't think is getting through is that when you say the grid charges and discharges, you are talking about the fact that the grid forms capacitors between itself and the other conductors in the tube, particularly the anode and cathode. So, when you charge the grid, you have taken that charge from the cathode (for example) and the grid and cathode form a capacitor. If you consider the entire tube as one device, it has a total charge of zero always. Yes, the grid charges and discharges in response to AC, and this is just AC current. The grid charge creates electric field and potential that changes the current flow between the cathode and anode.

Now let's say the note was played on a bass guitar and had a frequency of about 60 cycles/second. That would mean that one complete cycle would take 1/60 of a second, and that each gathering/dispersal would take 1/120 of a second (each reaching its utmost peak or trough after 1/240 of a second). So the excesses and deficits I'm talking about are admittedly temporary and of relative short duration. Yet it's these gatherings and dispersals that generate the voltages that make the thing work.

Yes.

Lool said: "If you think of the entire capacitor (both plates) there is no accumulation of charge in any device in a circuit."
I really don't know what that means. A charged capacitor, by definition, has an excess of electrons on one side and a deficit of electrons on the other. And sure, the excess minus the deficit may be zero -- but that doesn't mean that electrons haven't accumulated on the negative side of the thing. Touch a big charged-up cap sometime and see for yourself as the accumulated electrons try to get to the other side through your body.


Yes. You don't seem impressed by the fact that the net charge on a tube or capacitor is zero. Okay, that's fine as long as you understand it. But this is a crucial assumption in circuit theory. It means that the current entering one terminal equals the sum of the currents leaving all other terminals. For a two terminal device, like a capacitor, the current going in one side equals the current going out the other side. As long as you understand that, we are in good shape.

So, what now remains? I'm now again confused on what the problem is. If you can think in terms of currents and voltage, you are in good shape to use circuit theory to understand how any circuit works. One of your previous misconceptions was that a resistor has charge buildup just like a capacitor. This was incorrect. Do you see that now? Once you understand that, you will understand resistors and capacitors and tubes, and then you can consider coils and transformers, which also have no charge accumulation.

Are we making progress? It seems to me we are, if your misconception about resistors is resolved.
 
Last edited:

Lool

Joined May 8, 2013
116
With more time to read what you wrote, I can add the following comments.

All over the place. In the wires that go from the guitar to the amp: the electrons first "bunch up" in one direction, then in the other, as the strings are plucked. On the cathodes of the tubes. On the grids of the tubes. At the base of the plate resistor. Etc. You can see the accumulation on a scope:
This is a confusing and misleading way to describe the situation and nobody talks about it this way. To the extent that charges are "all over the place", that is basically a statement that paracytic capacitance is all over the place, which is true but hopefully small enough to be ignored and not an essential feature of the operation. It is the electrons that flow in one direction and then the other, not the electrons that bunch up in one direction and then the other. The first wording implies current, while the second wording implies charge accumulation. "AC means alternating current, not alternating congestion" might be a simple way to say it.

The lower wave (in red) shows the gathering/dispersal of electrons on the tube's grid; the upper wave (also red) shows the same at the base of the plate resistor. The electrons in the wire from the guitar pickup are moving first one way, then the other, at a frequency determined by the note plucked. Electrons are thus alternately sucked off of the tube's grid and pushed onto the grid as shown on the scope. Since the valve opens when the grid has fewer electrons and closes when the valve has more electrons, the gathering/dispersal of electrons at the base of the plate resistor is not only bigger, but inverted.
Since the valve opens when the grid has fewer electrons, and closes when the value has more electrons, the flow of electrons from cathode to anode is NOT inverted. Current is higher when voltage is higher. The inversion is cause by the fact that voltage at the anode goes down when current goes up.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Are we making progress? It seems to me we are...
I also think we are making progress, and I greatly appreciate your time and patience.

It is the electrons that flow in one direction and then the other, not the electrons that bunch up in one direction and then the other. The first wording implies current, while the second wording implies charge accumulation. "AC means alternating current, not alternating congestion" might be a simple way to say it.
This appears to be the point where we still don't see eye-to-eye. It seems to me that sometimes the movement of electrons results primarily in alternating current, and sometimes it results primarily in alternating voltage! (But usually both, of course.) Please bear with me. Here we have a typical guitar-to-amp input circuit:

pickup circuit 1.jpg
The input impedance (resistance will do for this exercise) is very large. And I've replaced the rest of the amp with a scope showing one cycle of an input wave. The pickup is high-performance, able to generate a 2-volt swing.

Now I'm pretty sure we agree that when a string is plucked, electrons move around the loop, first in one direction and then in the other. I propose -- using the term "gazillion" to mean 6.28x10^18 -- that when the string is plucked:

1. About a gazillion electrons attempt to move counterclockwise around the loop, but are stymied by the huge resistor and that we thus end up (at time t1) with a deficit of about a gazillion electrons in the top of the circuit (and a surplus of about a gazillion electrons in the bottom half). The deficit shows up on the scope (which is wired up conventionally) as +1 volt (relative to zero).

2. Then the electrons reverse and attempt to move in a similar fashion clockwise around the loop, but are again stymied by the huge resistance and we end up (at time t2) with a surplus of a gazillion electrons on the top of the loop (and a deficit of a gazillion on the bottom). This surplus shows up on the scope as -1 volt (relative to zero).

Now why do I think this?

First of all, because the scope is showing us voltage and voltage is a difference of charge, and the only way we're going to get a difference in charge between the top and the bottom of this circuit is if there are more free electrons on one side than the other.

Secondly, because if we replace the resistor with a wire, we get a straight line on the scope (no voltage differential) even though we know the electrons are still trying to move in the same ways. Only now they can succeed and make it all the way around the loop. So why no voltage? because the electrons don't accumulate -- they just keep moving.

Now it is widely known that a guitar pickup in such a circuit generates only nano-amperes of current. Yet we must be moving at least a gazillion electrons to get the readings we do on the scope. What's going on with that? I conclude that the nano-ampere figure is describing the number of electrons that actually make it through the resistor and thus all the way around the loop; it does not take into account those that get "hung up" and "reversed" before making it through the resistor. But note that in this case -- where we're trying to generate alternating voltage (as opposed to current) -- it's the electrons that get "hung up" that make the difference. Literally: the electrons that get "hung up" are the ones that make up the potential difference that we see on the scope.

I wish I had the equipment to test this theory with a little experiment. Specifically, equipment to measure the maximum number of electrons the pickup is able to push (as pure current) when the string is plucked. The result should be very close to the number of electrons needed to account for the charge (voltage) difference shown on the scope at t1 when the resistor is in place.
 

Lool

Joined May 8, 2013
116
One way to see the flaw in your thinking is to study capacitance. C=Q/V which means that capacitance is the amount of charge per unit voltage. Now you describe a gazillion electrons which is approaching 1coulomb of charge , which is an enormous charge and at 1 volt you have nearly 1 farad of capacitance implied by your theory. This is a huge capacitance that would never be seen in the circuit you are describing. The actual paracytic capacitance in reality is quite small and at 1volt the actual charge is small and typically negligible. the effect of current dominates.
 

Lool

Joined May 8, 2013
116
Now based on the above you can note that I'm not saying that the principle you describe is wrong, but I say that you can neglect it and typically people do neglect it by making an idealized circuit model with no capacitance. If you add the capacitance to the analysis, then you describe something real but at the same time add more complication that does not help you understand the dominant effect of the current and voltage drop that results.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
One way to see the flaw in your thinking is to study capacitance. C=Q/V which means that capacitance is the amount of charge per unit voltage. Now you describe a gazillion electrons which is approaching 1 coulomb of charge , which is an enormous charge and at 1 volt you have nearly 1 farad of capacitance implied by your theory.
I'm no expert on these formulas, but it seems to me you're misapplying that one. I just took a 1 microfarad capacitor and charged it up with a 9 volt battery. Then I disconnected the battery and checked the voltage across the cap with a meter; it was about 8 volts. Which means there was a surplus of about 4 gazillion electrons on one plate of the cap, and a corresponding deficit of about 4 gazillion electrons on the other plate -- total potential difference, 8 volts. That's the plates. The charge stored in the dielectric material, the Q of the formula, would thus be:

Q = CV = 0.000001 farad * 8 volts = 0.000008 coulombs.

That capacitor is thus like a battery if we think of the chemical part of the battery as the stored charge Q. But in this case, a battery that's almost dead. It will put out 8 volts, but not for very long because there are only 0.000008 coulombs of stored/chemical energy left inside.

But I don't see how any of this applies to the circuit below. There's no dielectric in that circuit, and there are no plates big enough and close enough to each other to behave like the plates of a capacitor. But let's not spend a lot of time on this issue now -- what I really want to know is at the bottom of the next part of this post.

Now based on the above you can note that I'm not saying that the principle you describe is wrong, but I say that you can neglect it and typically people do neglect it by making an idealized circuit model with no capacitance. If you add the capacitance to the analysis, then you describe something real but at the same time add more complication that does not help you understand the dominant effect of the current and voltage drop that results.
As I said above, I don't see how talking about capacitance at this juncture helps. Please just answer one or two simple questions, given the same circuit as before:

pickup circuit 1.jpg

1. Does that +1 volt reading at time t1 on the scope indicate a deficit of 6.28x10^18 of electrons in the top half of the circuit (and a corresponding surplus of 6.28x10^18 electrons in the bottom half of the circuit) or not?

2. If not, to what physical property of the circuit does that +1 volt reading correspond?
 
Last edited:

Lool

Joined May 8, 2013
116
I'm no expert on these formulas, but it seems to me you're misapplying that one. I just took a 1 microfarad capacitor and charged it up with a 9 volt battery. Then I disconnected the battery and checked the voltage across the cap with a meter; it was about 8 volts. Which means there was a surplus of about 4 gazillion electrons on one plate of the cap, and a corresponding deficit of about 4 gazillion electrons on the other plate -- total potential difference, 8 volts. That's the plates. The charge stored in the dielectric material, the Q of the formula, would thus be:

Q = CV = 0.000001 farad * 8 volts = 0.000008 coulombs.

That capacitor is thus like a battery if we think of the chemical part of the battery as the stored charge Q. But in this case, a battery that's almost dead. It will put out 8 volts, but not for very long because there are only 0.000008 coulombs of stored/chemical energy left inside.
You are claiming that there are 4 gazillion electrons. One gazillion is 6.28x10^18 electrons according to you. The electron charge is 1.6 x 10^(-19) Coulombs. So, 4 gazillion electrons, is a charge of 4 Coulombs. Somehow you don't realize how ridiculously large that number is.

So you are stating (whether you realize it of not) that there is 4 Coulombs of charge on the capacitor plates, but you then calculated 8 microCoulombs using the correct formula. Thank you for making my point. This is what I was trying to tell you.

But I don't see how any of this applies to the circuit below. There's no dielectric in that circuit, and there are no plates big enough and close enough to each other to behave like the plates of a capacitor. But let's not spend a lot of time on this issue now -- what I really want to know is at the bottom of the next part of this post.
.
Exactly the point I was trying to make. You are describing capacitance when you talk about charge, but there is no significant capacitance in your example.

As I said above, I don't see how talking about capacitance at this juncture helps. Please just answer one or two simple questions, given the same circuit as before:

1. Does that +1 volt reading at time t1 on the scope indicate a deficit of 6.28x10^18 of electrons in the top half of the circuit (and a corresponding surplus of 6.28x10^18 electrons in the bottom half of the circuit) or not?

2. If not, to what physical property of the circuit does that +1 volt reading correspond?
No problem answering these.

1. No way does 1 V indicate a deficit of 1 C of charge. To find the charge, you would need to know the paracytic capacitances at those points. That's why I bring up capacitance as a way for you to see the problem. But, usually nobody worries about this. We usually don't worry about or think about the fact that paracytic capacitance (which always exists) must be charged up when any circuit is operating. We worry about the voltage and current and use those as either signals to convey information or as an energy source to do work.

2. The voltage of 1 V is the same as 1 Joule/Coulomb. This means that any charge there has a potential energy of 1V times the charge. This means that an electron has an energy of 1.602 x 10^(-19) Joules when it is there.
 
Last edited:

Lool

Joined May 8, 2013
116
Let's look at the voltage source in your example. How are you getting that voltage? I know it is a guitar pickup (Faraday's law) but let's be more general.

You could get it with a battery, in which case a small charge would exist on the battery plates, but the chemical energy is there ready to drive your current if you need it too.

You could get it with a generator, in which case the EMF is due to Faraday's law. There may be charges at the wires of the generator that counter the Lorentz force, but if you need current to flow, the device will drive current provided that mechanical energy is there to keep driving the generator.

You could get it with a capacitor, in which case the EMF is due to actual stored charge. The extent to which you can drive current depends on how large the capacitance is and how much energy is actually stored.

In all these cases. that actual charge at the terminals is related to the capacitance. You correctly described how the capacitance may be very small (and hence the charge is small for a given voltage) if it is just wires and air dielectric with large separation.

Remember that capacitance is completely dependent on geometry of the conductors and the dielectric materials.
 
Last edited:

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
1. No way does 1 V indicate a deficit of 1 C of charge.
I got that figure from the NEETS manual:

"Electrical charges are created by the displacement of electrons, so that there exists an excess of electrons at one point, and a deficiency at another point... A difference of potential can exist between two points, or bodies, only if they have different charges. In other words, there is no difference in potential between two bodies if both have a deficiency of electrons to the same degree. If, however, one body is deficient of 6 coulombs (representing 6 volts), and the other is deficient by 12 coulombs (representing 12 volts), there is a difference of potential of 6 volts. The body with the greater deficiency is positive with respect to the other."

That sure sounds to me like 1 coulomb of displaced electrons (6.28*10^18) equals 1 volt.

2. The voltage of 1 V is the same as 1 Joule/Coulomb. This means that any charge there has a potential energy of 1V times the charge. This means that an electron has an energy of 1.602 x 10^(-19) Joules when it is there.
I don't see how that answers the question, "to what physical property of the circuit does that +1 volt reading correspond?"

The only thing moving in those wires is free electrons. So the only physical difference in the circuit at time t1 versus time t2 is the relative positions of those electrons. The voltage readings on the scope must therefore be explicable in terms of the relative positions of those electrons. In other words, we ought to be able to say something like, "When the electrons are arranged like this, we get +1 volt; when they're arranged like that, we get -1 volt." I think I've described the corresponding arrangements correctly; you disagree. So just tell me how those electrons actually are arranged at times t1 and t2, and how those arrangements correspond to the voltage readings on the scope, and it will all be cleared up.
 

Lool

Joined May 8, 2013
116
I got that figure from the NEETS manual:

"Electrical charges are created by the displacement of electrons, so that there exists an excess of electrons at one point, and a deficiency at another point... A difference of potential can exist between two points, or bodies, only if they have different charges. In other words, there is no difference in potential between two bodies if both have a deficiency of electrons to the same degree. If, however, one body is deficient of 6 coulombs (representing 6 volts), and the other is deficient by 12 coulombs (representing 12 volts), there is a difference of potential of 6 volts. The body with the greater deficiency is positive with respect to the other."

That sure sounds to me like 1 coulomb of displaced electrons (6.28*10^18) equals 1 volt.
Sure, it sounds like it is 1 Coulomb because it is one Coulomb, but only in that example.

You are misreading what they said. I can't say I blame you for that because their explanation is not very good and even misleading.

Look at what they said more carefully. " ... one body is deficient of 6 coulombs (representing 6 volts)" This means that in this example, one body has a capacitance of 1 Farad because C=Q/V=(6C)/(6V). The other body also has a capacitance of 1 F because it is 12/12.

But not all examples will have 1 F capacitors and 1 F is huge in electronics. Of course nowadays we have supercapcitors with those levels of capacitance, but these are very special devices. Generally, 1 F is considered a large capacitance.


So, I think we have now identified one of the major misconceptions that is misleading you. As I mentioned previously, I think there is a couple of misconceptions getting in the way. This is one of them.

I don't see how that answers the question, "to what physical property of the circuit does that +1 volt reading correspond?"
It is analogous to the height of a hill. What is the potential at the top of a hill? That is a property of the location at the top of the hill due to the gravitational field. The potential is there whether a rock is there or not. If the rock is there, it has potential energy. If the rock is not there it does not have potential energy, and the potential tells you the energy that a rock would have, if it were there. Electric field creates potential, just as gravitational field does. But there is another complication. Electric fields are conservative, just as gravitational fields are, but only for static electric fields. Once, you have Faraday's law and induction, the fields are not conservative, so our intuition is upset quite a bit by this.

The only thing moving in those wires is free electrons. So the only physical difference in the circuit at time t1 versus time t2 is the relative positions of those electrons. The voltage readings on the scope must therefore be explicable in terms of the relative positions of those electrons. In other words, we ought to be able to say something like, "When the electrons are arranged like this, we get +1 volt; when they're arranged like that, we get -1 volt." I think I've described the corresponding arrangements correctly; you disagree. So just tell me how those electrons actually are arranged at times t1 and t2, and how those arrangements correspond to the voltage readings on the scope, and it will all be cleared up.
I'm not fully clear on what you are asking here. You say the voltage readings depend on the relative positions of those electrons, but what about the current from those moving electrons? Don't they also affect the voltage reading? If current flows through a resistor, there is a voltage drop across that resistor. This voltage drop is not caused by accumulation of charges on each side of the resistor, which is true for a capacitor only. The voltage drop is cause by the loss of potential energy the electron experiences as it goes through the resistor. Where does this lost energy go? It goes into heating up the resistor.

How is the energy lost to heat? The electrons are accelerated by the electric field in the conductor/resistor. This means potential energy is converted to kinetic energy of the electron. But, the electron eventually bumps into a atom and loses that kinetic energy. This creates lattice vibrations which ultimately is converted to heat.
 
Last edited:
Status
Not open for further replies.
Top