Its a good reference published in English in 1951 but containing mostly stuff dating back to the 1930's, and somewhat out of date given improvements in both theoretical understanding and manufacturing technique that came in the 1950's and 1960's. For instance it doesn't have the understanding of how the oxide emission layer works (they thought it was some kind of semiconductor back then. In the 1950's it was realised it was not - conduction occurs by a combination of electron tunneling and the spaces between grains acting as microscopic thermionic diodes. H/W's data on radial conductivity is hopelessly out wrt modern oxide cathodes).
It is by no means the only reference on vacuum tube theory. You need to read more widely.
This is something that can happen, but not under the relativley low voltager conditions of domestic tubes, as has been explained to you before.
As and I have pointed out before, it does not refute your claim that there is no electron cloud (space charge) in tubes when operating at anode currents less than the cathode capability.
So, why are they not used in noise diodes then? These are operated saturated but at quite low anode voltages, circa 200 V, far less stress than many oxide cathode audio power tubes. The fact is, as any text on noise figure testing or tube application notes will tell you, the much less eficient pure metal cathodes have to be used, because an oxide cathode woukld be rapidly destroyed by ion bombardment.
That is correct.
Not likely. Not relavent under warmup conditions in normal tube applications. Under these conditions anode current is lower than for normal operation. While the oxide layer radial resistance is higher at lower temperatures, the increase is not great enough to matter. Emission is neglible until cathode temperature, normally around 1050K, has reached around 800K. At 800K the radial resistance is about 30x normal, but emission is <2% of normal. Dissipation is proportional to I^2 x R, so a drop in curent overcomes a much larger change in resistance. The highest joule loss in the oxide layer during warmup typically occurs very close to operating temperature.
If you are going to draw high currents in a warmed-up tube, that's a different story.
It certainly doesn't look like it.
As I and others have pointed out in this thread, you consistently misread and misunderstand the reference you claim to love, Herrmann/Wagener 1951.
Post #150: Popilin pointed out you misuderstood the context of electrostatic stress
Several posters have pointed out your understanding of the term 'space charge' is incorrect and makes no sense.
Post #126: I showed that you misunderstood implications of grid control
Several post by myself and DF96 pointed out you do not understand the difference between electron emission quantity and emission launch velocity wrt position of space charge
Post #108: I showed that your claim that heater temperature is not self regulated is rediculous. If you are so good on physics of vacuum tubes, you should have no trouble with Worthing's Equation and the Stefan-Bolzman Law.
In several posts you showed you do not understand shot noise and the implications for it of temperaturte limitted and space charge limiiited operation.
You don't seemn to understand the concept of work function. You talked in several posts about a "barrier potential". It isn't a potential, it's an energy. It is measured and quoted in joules or electron-volts - both units of energy. (I note that high-school students exposed to science classes sometimes think 'electron-volt' is a measure of potential.) A person who understands physics would not make that mistake. Certainly not several times.
And I could add more of your slipups.
And you are unable to post a calculation you claim to have done.
Youy have been unable to explain various aspects of tube behaviour that follows from the existence onf an electron cloud.
I'm here to help. But your claim to 'comprehensively understand' the physics of vacuum tubes is absurd!
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Luckthedog said:
OK Keit, so you disagree with the definitive reference on Oxide Cathodes concerning an unambiguous explanation of reason for advice on delaying B+ until after cathode-warm up. That's your prerogative. But makes good sense to me and comes from the best source one might hope for, and I totally buy it. It's also your prerogative to believe in ion bombardment as a cathode hazard, mitigated by a dense free electron cloud, but I see no basis in serious technical literature, plus clear contradictions and that's not good enough for me.
Quite possibly. I agree in practice there's no apparent harm from not delaying B+ in most domestic applications. But understanding what's going on at least leads to being able to weigh up the risks in an application, for me that's very useful.
This is because you've consistently misunderstood, and still misunderstand the very concept of space charge, and what I have posted on the matter - as I have pointed out before. I genuinely deeply understand all concepts, physics, maths, involved in papers and discussions here. Too deeply perhaps. Of course there are electron charge carriers in operational valves........!😉 But it's moot now, in any event.Keit said:
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You have made it quite clear in numerous posts now your claim that there is no electron cloud. And in response to DF96 in his post #113:-
in your post #122 you said:-
Dow 1937 Fig 4 happens to be a set of J-curves for triodes under various conditions.
So your post #122 makes it very clear you think that the dip for positive grid conditions only. Well, there can be a dip under positive conditions, depending on tube grid spacing and distance from cathode, but none of the graphs in Fig 4 are staed as being for +ve grid. There is one (Fig 4a) for "large plate current" clearly showing for zero grid bias that there is a dip due to space charge. I can only assume you think Fig 4 shows no evidence of a space charge / electron cloud because the curves (a) and (b) do not go below the cathode potential - an incorrect misinterpretation.
In Fig 4 (d) the space charge dip is displaced further negative by the superimposed electric field from the grid. Unfortunately teh artist did not show a notional space charge free line for comparison. It's actually much the same size dip due to space charge as in Figs (a) and (b), but it has been shifted downward by the grid field.
In my post #159 by simple graphical construction I showed that the presence of an electron cloud near the cathode results in a J-curve just like those published in textbooks. I showed that with positive anode voltage, the anode electric field displaces the dip caused by the electron cloud, so it may still go below cathode potential, but with sufficient anode voltage, it need not. The displacement results in a curve just like in the textbooks in basic shape.
In short, it is obvious that, far from having a "genuine deep understanding of all concepts, physics, maths, involved", you have no idea what you are talking about, and you cannot even read J-curve graphs - probably THE most important tool for understanding tube internal operation.
It doesn't matter how many times you go on about charge carriers, me misunderstanding, or that the authors of books have gone astray, the fact is: there is an electron cloud under normal operation, it is very dense compared to the anode stream, it is by convention called a space charge, and that is the long and the short of it.
I suggest you abandon your nonsensical claims, start again, and study the material properly.
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Case 1: Vak = 0, Hot Cathode
The condition Vak = 0 was taken for simplicity and to avoid misunderstandings, as a rule of dumb I never left a valve element floating freely; maybe it is a bit restrictive condition, but it is valid and serves to illustrate the point.
Into this condition, you do not need a potential difference between cathode and anode, Maxwell’s equation (AKA Gauss Law)
∇ . D = 4 π ρ … (1)
Assures you that, if you have a charge distribution, then you have an electric field, e.g. as on post#151
E(x) = ∑ qi (x - xi) / ∣x - xi∣³ … (2)
Case 2: Vak = B+, Cathode warms from cold
In this case, I did supposed only one electron emitted; its own electric field is also given by solving (1), but for simplicity it was neglected, then for the simplest case
E = - V / d … (3)
Then you seem to refer to this case
Case 3: Vak = B+, Hot Cathode
Here you can solve (1) with adequate boundary conditions, or use the linear superposition principle and take the sum, e.g. (2) and (3)
BTW, this case was treated before, here
http://www.diyaudio.com/forums/tubes-valves/264781-getter-heater-b-sequencing-12.html#post4128561
That’s why you must use the Fermi-Dirac distribution.
As a consequence, using the Richardson-Dushman equation
J = A T² exp ( - e φ / k T ) … (4)
You can measure temperature and work function of cathode and anode, contact potential and Stefan-Boltzmann constant IIRC, I made the experiment a long ago with a 5R4GY, and the most sensitive magnitude to measure was low current.
Paradoxically, the electron cloud can make your life miserable in order to make accurate measurements.
I use delayed B+ and regulated soft-started PSUs, even for heaters, never had a problem.
Energy barrier and potential barrier are the same thing, and in the worst case they are equivalent concepts, because the later can be referred to potential energy barrier, even when you treat with work functions, i.e. W = e φ, some people take only φ as the work function (in Volts), particle physicists speak about mass in eV, and it is understood.
Authors of serious technical theoretical texts had it sorted years ago, there's no question otherwise and no need to look further so long as one properly understands the physics, maths and concepts there. It's no surprise that such texts, and indeed my posts, might get misinterpreted or taken out of context. It's entirely your prerogative for you to hold your own interpretations and believe what you will, Keit, whatever floats the boat. It doesn't matter in the scheme of things - the texts worth reading remain true as references, as does physics and maths behind them and that is what is ultimately convincing and useful for anyone who understands them properly. Those texts clearly define charge carrier distribution in operational valves - and I choose to use those rather than either your interpretation or what you've misunderstood me to say, thanks.
@Keit example to illustrate:
But Popilin correctly gets it straight away:
Misunderstanding concepts is more than half the problem here, IMO. HTH!
Keit said:
But Popilin correctly gets it straight away:
popilin said:
Misunderstanding concepts is more than half the problem here, IMO. HTH!
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Maths is a tool to help bring a quantitative and deterministic approach to understand what is going on but there is the danger that it becomes an end until itself and you forget the underlying physics.
Go back to the physics that underlies the situation at hand.
Without a potential difference between cathode and anode there can be no current flow (there is no quantum state here akin to superconductivity to allow otherwise).
First you need an applied B+ at the anode. Then you can have current flow. This current flow consists of space charge across the vacuum which results in a non-uniform electric field - the shape of which can be estimated fairly accurately from Maxwell's equations. Without the current flow there is no reason for the existence of any space charge to enter your into your formula.
The risk I alluded to is where you have a valve with a heater at it's normal operating temperature but no B+ and as a result of the circuit topology the necessary operating voltages are not ready - perhaps the grid bias is not yet applied and the grid is not sufficiently negative wrt the cathode - then application of B+ would result in a surge of current through the tube (until operating conditions have settled to the correct values) and this surge exceeds the maximum ratings for the valve or other elements in series with the valve. It's a risk that depends on the details of the circuit and is therefore, avoidable with proper design.
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I'm not sure what you are talking about here. My comment was in reference to your statement in post 155 about the Fermi level.
You said: "In metals, at normal temperature, the conduction band is essentially filled of electrons only up to the Fermi energy EF, to extract an electron from the metal is therefore necessary to give a starting energy eφ, but at high temperatures the occupation of electronic states extends above EF.
If the temperature is high enough, some electrons reach energies greater than EF + eφ, and escape from the metal."
I was trying to say that your statement was wrong. Let me explain my thoughts. The Fermi level in a conductor is defined as the energy level at which there is a 50% chance of it being occupied by an electron. It is a thermal distribution of electron energies. Therefore, by definition of 'Fermi level' there are electrons with energies above it. This means it is not correct to say that the conduction bad is filled up only to the Fermi energy. I do understand why you talk about the Fermi level, it is a measure of the energy required to lift an electron from the conductor - and in this case, into the vacuum. Hence, it is directly and simply related to the Work function of the metal.
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If you read the narrative associated with fig 4 of Dow 1937 page 3 et seq, you'll find that what you interpret as a potential minimum is actually an artefact of the grid wire - there are two solid lines drawn, one for mid-way between grid wires and one for on the grid wire axis. Obviously on the grid wire axis at the grid location is grid potential, and that is what you have misinterpreted as a potential 'dip' in that diagram. In 4(a) and (b) the dotted line shows potential correction for real operation ie conduction with charge carriers (space charge). In 4(c) and 4(d) there is no dotted line because there are no charge carriers, the valve is at cut-off in (c) and beyond cut-off in (d), - it is not that the artist didn't show it, it has no meaning because there are no charge carriers for those conditions.
Dow 1937 is excellent IMO, a careful read with proper interpretation is well worth it if one is interested in bottoming out some of these loose ends.
True. But you don't have this understanding.
In your post #103, you wrote "Nope, cathode temperature is not regulated.." in reponse to my earlier statement that the temperature is self regulating. With statements like that you cannot claim "I genuinely deeply understand all concepts, physics, maths, involved in papers". This is akin to a doctor not knowing that exercise influences muscle power.
Yet again you demostrate that you have no understanding of what the graphs are telling you.
Sure the there is a pronounced negative region right on the grid wires, having a lessened influence midway between the grid wires. As you would expect. But I wasn't talking about that. I was talking about the potential depression on the cathode side of the grid (roughly half way betwen cathode and grid in Dow's Fig 4). This is the region around the green reference line on my diagrams in my post#159. In that post I showed that the space charge potential dip is displaced upwards by the anode electric field and then doesn't necessarily go below the cathode voltage. Go back and study my post #159 carefully. Have you done that?
In Dow's Fig 4 (a) and (b), the artist used a dotted line to indicate the notional change btween space charge and no space charge, so he didn't need to draw as many digrams as I did.
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I guess you can call it whatever you like, but what has to be supplied from somewhere to get electrons to leave a substance is an energy, akin to the enegy needed to get molecules to evaporate from a liquid.
When molecules evaporate from a liquid, they leave the liquid cooler. When electrons leave a substance (whether by thermionic emision, field emission, photo-emission, secondary emission) they leave the substance cooler. This phenomena is well documented althgouigh it has not much influence on tube operating conditions.
The unit in the SI system for this is the Joule - a unit of energy.
You cannot change the horse at the middle of the river, on post#156 you said
In a nutshell, the electron cloud (space charge) cannot create an electric field, but it can produce its non-uniformity.
As I showed you before, this is wrong.
Now you say a totally different thing.
Mathematics is THE language of physics, and you seem cannot see the physics behind the maths, no offence.
This is nonsense, and it has nothing to do with quantum state akin to superconductivity.
As I showed you before, if you have a charge distribution, you have an electric field and this electric field exert a force over the charges, as the charges are free into the vacuum, they move, moving charges produce a current.
Think about a pool table, with a lot of white balls toward the other balls, or better, read a book about physics, or an old book about valves, no offence.
Again, this is a different matter.
Speaking of Maxwell’s equations
Again, taking the divergence on (4) and combining with (1) into vacuum, i.e. D=E, B=H
Without current flow means J=0, then
Hence, the original charge distribution, ρ, is static, and I would call it an electron cloud.
Then, you must think about a properly designed PSU.
Hint: Delayed B+, not delayed for grid bias and heaters.
Typical confusion, Fermi energy and Fermi level are very confusing terms and often are used interchangeably to refer each other, let me explain
Fermi energy: It is the highest occupied level at absolute zero, does not depend on temperature.
Fermi level: It is the energy level with 50% chance of being occupied at finite temperature T, as you said correctly.
Electrons in the metal experience a constant confining potential of depth U.
Electrons fill all the available states up to the Fermi energy, EF.
The work function, W, is defined as the minimum energy needed to remove an electron from the metal
Good to know that energy has units of energy.
I meant that, when you read a physics’ book, “potential barrier” is equivalent to “energy barrier” 😉
In a nutshell, the electron cloud (space charge) cannot create an electric field, but it can produce its non-uniformity.
As I showed you before, this is wrong.
Now you say a totally different thing.
Mathematics is THE language of physics, and you seem cannot see the physics behind the maths, no offence.
This is nonsense, and it has nothing to do with quantum state akin to superconductivity.
As I showed you before, if you have a charge distribution, you have an electric field and this electric field exert a force over the charges, as the charges are free into the vacuum, they move, moving charges produce a current.
Think about a pool table, with a lot of white balls toward the other balls, or better, read a book about physics, or an old book about valves, no offence.
Again, this is a different matter.
Speaking of Maxwell’s equations
∇ . D = 4π ρ … (1)
∇ . B = 0 … (2)
∇ x E + (1/c) ∂B/∂t = 0 … (3)
∇ x H - (1/c) ∂D/∂t = (4π / c) J … (4)
Again, taking the divergence on (4) and combining with (1) into vacuum, i.e. D=E, B=H
∂ρ/∂t + ∇ . J = 0 … (5)
Without current flow means J=0, then
∂ρ/∂t = 0 … (6)
Hence, the original charge distribution, ρ, is static, and I would call it an electron cloud.
Then, you must think about a properly designed PSU.
Hint: Delayed B+, not delayed for grid bias and heaters.
Typical confusion, Fermi energy and Fermi level are very confusing terms and often are used interchangeably to refer each other, let me explain
Fermi energy: It is the highest occupied level at absolute zero, does not depend on temperature.
Fermi level: It is the energy level with 50% chance of being occupied at finite temperature T, as you said correctly.
Electrons in the metal experience a constant confining potential of depth U.
Electrons fill all the available states up to the Fermi energy, EF.
The work function, W, is defined as the minimum energy needed to remove an electron from the metal
W = U - EF = e φ
Good to know that energy has units of energy.
I meant that, when you read a physics’ book, “potential barrier” is equivalent to “energy barrier” 😉
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Take this just as a mental picture, a rough analogy, because…
This is not valid for common valves, let me explain
At low energies, electron-atom (ion) collisions and ion-atom collisions (except perhaps for a proton) cannot be treated classically.
Even if you insist with the classical treatment, in an electrodynamic system of charged particles into an electric field (AKA a valve) neither kinetic momentum (p=mv), nor kinetic energy are conserved, so forget elastic collisions.
Context is Dow 1937 Fig 4(d) where potential line drawn shows a depression between cathode and grid in the shape of a J. Fig 4(d) illustrates a triode beyond cut-off, ie heavily negative grid. There are no charge carriers (hence no dotted line) and the potential map is simply determined by combination of two electrostatic fields from anode and grid. Anode field has a positive gradient, grid field has a negative gradient. As drawn, the potential line mid way between the grid wires thus has a contour which resembles a J, being a combination of electrostatic fields from grid (negative) and anode (positive). Whereas the potential line for the 'on grid wire' location does not show this since the effect of the grid wire's potential screens the anode field. Hence the potential 'dip' in this case has nothing to do with charge carriers or space charge - the valve is in cut-off, not in normal operation such as figure 4(b) 😉
Re luckythedog post #174:-
The context is all four of the graphs in Dow Fig 4. You kept claiming that there is no electron cloud under normal operation, and have claimed Fig 4 supports this. Normal operation is depicted in Dow Fig 4 (a) large anode current and (b) small anode current as well as (c) just cutoff and in some cases (d) beyond cutoff.
Again you have not understood what happens inside vacuum tubes. Cutoff means the grid is not allowing any electrons to pass through to the anode. It does NOT mean that there is no space charge electron cloud beween it and the cathode - though, as I said before, if the grid (or anode for that matter) is sufficiently negative the space charge electron cloud will be pushed so far back almost all of it retreats inside the cathode.
Whether or not the space charge electron cloud is a) substantially pushed back into the cathode coincident with grid cut-off, or b) cut-off occurs before substantial space charge push-back into the cathode (a la low-mu power triodes), depends on the physical design of the tube.
The situation may be easier to understand if you assume the grid to be a plane structure without the rubber sheet (Telegen) effect, like the accelerating anode/grid in some CRT's.
I note that you have not identified any error in my Post #159 "J-Curves for Dummies", which start with an assumption of a space charge electron cloud, and end up with a curve just like those publised in textbooks. You have found error neither in the text nor the graphs.
The context is all four of the graphs in Dow Fig 4. You kept claiming that there is no electron cloud under normal operation, and have claimed Fig 4 supports this. Normal operation is depicted in Dow Fig 4 (a) large anode current and (b) small anode current as well as (c) just cutoff and in some cases (d) beyond cutoff.
Again you have not understood what happens inside vacuum tubes. Cutoff means the grid is not allowing any electrons to pass through to the anode. It does NOT mean that there is no space charge electron cloud beween it and the cathode - though, as I said before, if the grid (or anode for that matter) is sufficiently negative the space charge electron cloud will be pushed so far back almost all of it retreats inside the cathode.
Whether or not the space charge electron cloud is a) substantially pushed back into the cathode coincident with grid cut-off, or b) cut-off occurs before substantial space charge push-back into the cathode (a la low-mu power triodes), depends on the physical design of the tube.
The situation may be easier to understand if you assume the grid to be a plane structure without the rubber sheet (Telegen) effect, like the accelerating anode/grid in some CRT's.
I note that you have not identified any error in my Post #159 "J-Curves for Dummies", which start with an assumption of a space charge electron cloud, and end up with a curve just like those publised in textbooks. You have found error neither in the text nor the graphs.
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Just to play devil's advocate, are you sure you can have space charge in a triode at cut-off? Is not the point at which the space charge 'retreats into the cathode' the very definition of cut-off? Because, if a some space charge did still exist, that very fact means electrons are still attaining enough energy to leave the cathode and therefore, statistically, some of them will also attain enough escape energy to overcome the potential well in the grid plane and so reach the anode, hence the valve is not fully cut off (?)
If cut-off means strictly zero anode current then you may be right. This would require a very negative grid bias to ensure that even the most thermally energetic electrons are kept at bay. I suspect that cut-off really means 'the point at which anode current becomes very small' (with perhaps an exponential reduction for further negative grid bias?) in which case a space charge could still exist.
Merlimb and DF96 both have very good points.
What exactly is "cut-off" depends a bit on what book or datasheet you are reading, because in practice it is impossible to cut off a tube completely. Not only does the retarding field current reduce as the logarithm of the negative grid or anode voltage(which is just another way of saying there is really no limit to how energetic a tiny minority of electrons can be), real tubes have "leakage" - that is some electrons don't go to the anode via the grid at all, they go around it thru gaps between the micas etc.
As DF96 suspects, engineers regard a tube as cut off, when the grid bias is such that the anode current that flows is not significant or useful in the application to which the tube has been put.
The substantial retreat of the space charge electron cloud is NOT part of the definition of "cut off". However, depending on the tube design, the two phenomena may or may not happen together. Yes, depending on the tube design, you can have a space charge electron cloud present when the grid has cut off. It is a question of electrode spacing and grid pitch.
A good comparison is a triode intended for audio preamp service, compared to an RF amplifier tube. Both have anode currents of similar order under design centre conditions - circa 1 to 1.5 mA. The audio triode, having a sharp cut-off characteristic, would be regarded as cut-off at the point where distortion is significant - this might be at an anode current of 200 to 300 uA. You'd still get a large fraction of maximum gain out of an RF pentode at 200 uA. For RF pentodes, which have variable pitch grids to give a remote cutoff effect (cut off starts in the fine pitch part of the grid, and with increasing negative grid voltage, gradually spreads along the grid to the coarse pitch part(s) ), cut-off would be regarded as 10 uA or even less.
Datasheets don't typically quote a "cut off bias voltage" stated as such. What they do is just not mention cut-off at all, because for an audio tube you can estimate it from the design centre anode current (at design centre voltage) and gm, or they quote the grid bias to get a specific low value anode current (such as 10 uA), or they quote the grid bias required for a 100:1 reduction in gm.
What exactly is "cut-off" depends a bit on what book or datasheet you are reading, because in practice it is impossible to cut off a tube completely. Not only does the retarding field current reduce as the logarithm of the negative grid or anode voltage(which is just another way of saying there is really no limit to how energetic a tiny minority of electrons can be), real tubes have "leakage" - that is some electrons don't go to the anode via the grid at all, they go around it thru gaps between the micas etc.
As DF96 suspects, engineers regard a tube as cut off, when the grid bias is such that the anode current that flows is not significant or useful in the application to which the tube has been put.
The substantial retreat of the space charge electron cloud is NOT part of the definition of "cut off". However, depending on the tube design, the two phenomena may or may not happen together. Yes, depending on the tube design, you can have a space charge electron cloud present when the grid has cut off. It is a question of electrode spacing and grid pitch.
A good comparison is a triode intended for audio preamp service, compared to an RF amplifier tube. Both have anode currents of similar order under design centre conditions - circa 1 to 1.5 mA. The audio triode, having a sharp cut-off characteristic, would be regarded as cut-off at the point where distortion is significant - this might be at an anode current of 200 to 300 uA. You'd still get a large fraction of maximum gain out of an RF pentode at 200 uA. For RF pentodes, which have variable pitch grids to give a remote cutoff effect (cut off starts in the fine pitch part of the grid, and with increasing negative grid voltage, gradually spreads along the grid to the coarse pitch part(s) ), cut-off would be regarded as 10 uA or even less.
Datasheets don't typically quote a "cut off bias voltage" stated as such. What they do is just not mention cut-off at all, because for an audio tube you can estimate it from the design centre anode current (at design centre voltage) and gm, or they quote the grid bias to get a specific low value anode current (such as 10 uA), or they quote the grid bias required for a 100:1 reduction in gm.
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Let me repeat my statement in a different way because I think if I word it differently I will find that I agree with you and you may find that you can agree with me.
I am saying that the application of B+ causes a current to flow in the external circuit until there are charges on the anode and cathode which creates an electric field in the vacuum. When the cathode emits electrons this electric field causes a current to flow. This current is made of electrons, charge carriers, which constitutes a space charge. The space charge causes the electric field in the vacuum to be non-uniform.
I should clarify that I agree that a space charge will generate an electric field since electric fields are created by charge. However, the prime cause can be traced back to the application of B+. I say that the application of B+ is a necessary condition for the creation of the electric field in the vacuum that exists when the tube is in operation.
Is this a description we both agree with ?
You might be right, it's been awhile since I used very much math. It's not that I don't like math - I did plenty of it in the past (Child-Langmuir, Maxwell etc. etc.). But on a DIY audio forum I think we should do our best to explain things that many can understand without them needing such background skills. I think it's an achievable goal.
Thanks for that tid bit - something learned.
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If the collision is inelastic then we have kinetic energy being transferred to excitation or ionization. But I have not seen any references to show that this is the majority of collisions between ions and electrons. An electron can have an elastic collision with an ion.
Why are low energies not able to be treated as classical ?
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