Trouble is, we being kooks at this level of chemistry, is to hold any meaningful personal opinion concerning ideas about conduction mechanisms in metal-BaO solids.
However, such deep chemistry seems relevant to grid emission, because the nature of whatever compound, alloy or such which might form the grid surface is said to significantly affect thermionic emission - as does the external field at the surface for that matter, in principle.
Anyways, it's a good excuse to have a read and explore a backwater and learn a bit on the way. Wouldn't surprise me if assumptions like the detection of only Ba and Au on grid surface might have previously lead to conclusion that Ba was deposited, but whether such explanation is satisfying and true without what appears like obvious loose ends........?
We shall soon see !
I certainly would not argue against that. I don't actually see much problem in bending the discussion in threads away from the original topic posed by the OP - so long as the OP's original question is not forgoten.
Perhaps.
In the meantime, since authors who were deeply involved in tube engineering and manufacture at RCA etc in the 1960's (see the reference I provided, as well as Rod Colemans's posted page from Beck) wrote that gold is good because Ba atoms difuse into it, I'll go with that.
It is supported by the fact that all manner of authors have shown that the oxide layer consists of a BaO/SrO porous matrix infiltrated by Ba (& Sr) atoms (and Ni atoms from the sleeve), together with the facts that:-
a) The vapour pressure of BaO & SrO is minute compared to the vapour presure of Ba & Sr.
b) The continued vaporisation of Ba (and Sr) at the surface means it must be replaced by dissassociating BaO (and SrO), leaving the oxide layer increasingly porous and loosing mass over time (which is actually what happens), and freeing up Ba & Sr to diffuse up to the surface.
These facts mean that what becomes tube gas is Ba (which can contaminate the grid) and O, which ends up in the gettering. Not BaO, which pretty much dissassociates in situ.
You can check the relative vapour pressures with the references I gave in my earlier post. Or if you like Hermann/Wagener, you can compare the Ba & Sr vapour presure graphed in Fig 82 (Vol 2 page 154) with BaO and SrO in Fig 83 (Vol 2 page 155) - the vapour presures of the metals is about nine orders of magnitude greater than their oxides at 1050 K.
Ba having a vapour presure nine orders of magnitude greater than BaO leaves plenty of margin to compensate for the fact that the amount of Ba present in an oxide layer is a trace amount.
So, those interesting reactions involving BaO and Au are not particularly relavent.
We can forget about those formulae repeated ad nauseum by Popilin - he can copy from books but has yet to show he can do a real calculation, properly justified and applicable to a real tube with real tube geometry. The reality is that Schottky effect is minor.
Last edited:
That's an easy trap to fall into, because the concentration of Ba versus BaO in the original oxide is also minute. Posted previously, also directly quoted from H&W is the directly measured rate of evaporation for Ba and BaO based on the material concentrations at hand, which shows the opposite result, BaO more prevalent, as measured by mass per hour...........and it is this result that seems far more relevant to me.
We'll see.
BTW, based on what I've read, the process of adsorption of Ba at a gold surface at 600-880K seems to be one of ionic bonding, resulting in Barium Auride - not a process of solid diffusion. Amongst metals to form anions such as Au- is rare, and just happens to also happen with Pt- and Ag-, the other coating metals known to supress grid emission.........check it out.
However, if anyone amongst the proper chemists out there might chip in with what might happen to Ba or BaO vapour in interaction with a gold surface at 600-800K and why, that might be very interesting.
And whether or not the Shottky effect is a factor here depends on the external field strength, popilin's point I think.
Err.. I think you based that on H&W's description on what happens during pump-out. This is when the porous oxide matrix is being set up out of the carbonates. During this process there is pretty much no Ba or Sr present, and quite a bit of BaO and SrO dissappears down the sealex machine or gets left in the tube as loose material.
After pump-out, the tube is sealed off, and post-seal activation converts some BaO & SrO into Ba, Sr, and O, the O being already a gas is consumed by the gettering. Tube operation continues throughout the life of the tube to convert more BaO and SrO into Ba, Sr, and O to replace what gets evaporated.
For equal amounts of BaO (&SrO) in the tube gas during tube operation, the amount of Ba would have to be about 1 part in 10^9 of the oxide layer. I haven't actually seen a reference that gives or estimates the concentration, but it is quite unlikely to be that low.
Perhaps.
Yes, I know that is his point. But it was my point that he hasn't shown how to correctly calculate the external (to the oxide) field strength (taking into account tube geometry), and it is the field inside the oxide layer that is the relavent field anyway, not the field in the inter-electrode space. And there's more: that field is depressed by the space charge.
Last edited:
Err..no. It's based on published measured evaporation rates for BaO and Ba in real cathode material versus temperature, surely that is the crux. In units of mass per hour per cm^2 in vacuum.
From what I read, most BaO evaporation happens during activation, when cathode temperature is intentionally elevated. Grid current is then said to stabilise within 25s or so of first use, ie there is an initial peak followed by long period of normality.
Then, there is no basis to have an opinion based on relative vapour pressures. Don't just sit there man, go find out !😉keit said:
lucky, have you seen the paper by Naumovets on desorption of Ba and BaO from W driven by the Schottky effect?
Page no? Section?
It's not an unreasonable opinion, and it aligns with textbooks that say the grid gets contaminated with Ba, not BaO.
Oh, I have been looking. I am stuck at home ill, though, so I'm limitted to the half of my existing library that is at home, plus what I can find on the web.
Don't forget that there is plenty of metallic barium present in the valve: the getter flash on the glass.
I'm not surprised that the rate of evaporation of BaO is higher than that of Ba.
(vapour) pressure is the net result of both evaporation and condensation. As long as the equilibrium vapour pressure has not been reached, there will be a measurable nett evaporation. At the equilibrium evaporation and condensation are equal: no net evaporation to be measured.
Metallic barium reached its equilibrium vapour pressure (by definition) in the process of getter flashing.
This of course is in a closed system (like a valve).
How were the rates of evaporation measured?
No, the vapour pressure -and therefore the number of particles per unit of volume- in the gas phase depends only on temperature and total pressure, not on the number of particles in the solid phase.
Strictly speaking that statement of mine was wrong. I was trying to inject some appreciation of the scale of things.
In strict equilibrium, you are correct, and you are assuming each component has plenty of excess solid phase not captured (eg diffused in something else).
Thus in such equilibrium, the gas concentration of BaO (and thus reaching the grid) would have to be 1 part in 10^9 parts of Ba, as the vapour presure of Ba is 10^9 times greater.
Vapour pressure is not the whole story. Apart from capturing in other substances (eg in tubes, Ba cannot evaporate until it has dissociated from BaO and diffused to the surface) Each substance has its own rate of evaporation, a function of the substance and the temperature.
Tubes are not always in equilibrium in any case. The effectiveness of the gettering and other metal parts depends on their temperature, with the gettering being the slowest to warm up.
Each time equipment is switched on, tube gas typically increases in the first few minutes than decreases over the following 15 to 30 minutes.
(Old octal tubes not using dumet seals excepted - they can slowly gain oxygen from the air while not in use).
It appears from his other posts that Lucky was relying on Hermann/Wagener page 101, which gives a graph (Fig 54, page 101 in Vol 1) of rate of evaporation of BaO, SrO, CaO, taken from Claassen A & Veenemans C F, Physik 80 (1933), 342.
Unfortunately I cannot fluently read German. It's a struggle.
It appears they measured the evaporation rate in a vacuum chamber (i.e., not an electron tube) by heating a sample to temperatures much higher than normal in a vacuum tube (so the experiment could be done in days, not years). Present in the vacuum chamber was a neighbouring surface that evaporate from the sample condensed on. After a suitable time the condensing surface was assayed to determine the amount of condensate (presumably way way too small to determine by simple weighing).
Just what the condensate was (BaO?, Ba? etc) was not reported. In other words we know that for a BaO sample, BaO was lost from the heated sample, but whether is was lost by direct evaporation, or it first dissociated into Ba & O, is unknown.
Importantly, no emission current was drawn off from the sample. Thermionic emission current is known to be important in the diffusion of Ba within the oxide matrix and also important in loss of Ba at the emissive surface.
Hermann/Wagener point out that at typical tube temperatures, the rate of evaporation would be minute, by extrapolation of Claassen & Veenemans' results, about 0.001 microgram per hour for a typical small power tube. This should correspond to a typical cathode design life of 20,000 hours.
Hermann/Wagener on the following page goes on to say that evaporating BaO is bad because it will condense on the the other electrodes (eg grid), forming emissive surfaces activated by bombarding electrons (not applicable to control grids except in large transmitting tubes) or by evaporated Ba. So Ba is necessary to get a problem with grid current??
Hermann/Wagener's comment here should not be relied on. It has the air of a speculation or extraploation made with the level of knowlege available in the 1930's. It's not like other topics in H/W where they explain in detail. Later workers say its evaporating Ba (without BaO) that is the problem and that is not excluded by what is in Claasen & Veenemans results described in H&W.
Last edited:
Sorry to know that, I sincerely hope your speedy recovery. 🙂
Does not makes any sense to talk about where an electric field starts or terminates, because it depends on where you put the coordinate system.
This is also the third time that we discuss the surface issue, and we only agree on that the electric field inside the conductor
E = - ∇ φ = - ∇(constant) = 0
It results more than evident that the electric field must be evaluated at the surface, outside the conductor, regardless if it starts or terminates there.
So far, I did not solve Poisson's equation in any coordinate system in this thread, nor I will, because it does not matter, let me explain and I hope you can understand correctly Schottky effect.
Let’s suppose an electron (charge –e) at a distance x from the conductor surface (i.e. outside the conductor), then the image charge (charge e) is at a distance –x inside the conductor, from Coulomb’s law, the force on charge e due to external charge -e is
F = e Ec = e { -e / 4 π ε0 [ x – (-x) ]²} = e ( -e / 16 π ε0 x² )
Then
Ec = -e / 16 π ε0 x²
Integrating, this corresponds to an electric potential
φ = -e / 16 π ε0 x
Let’s suppose now that we add an external electric field E=constant.
The total electric potential will be
φ = φ(barrier) - e / 16 π ε0 x - E x
Taking the derivative
dφ/dx = e / 16 π ε0 x² - E = 0
Then φ reaches a maximum at
x(max) = √ (e / 16 π ε0 E )
Then
φ = φ(barrier) - √ (e E / 4 π ε0 )
As a result the potential barrier is lowered, and Richardson-Dushman equation
Jo = Ao T² exp ( - e φ / k T )
Takes the form
Js = Jo exp{e √[e E / (4 π ε0)] /(k T)}
Contrary to your belief, "free" electrons inside the conductor (E=0) can "feel" the presence of an external electric field outside the conductor.
That’s why Schottky effect is also shown in conductors.
Even though Va definition was repeated all over the thread, and you are the one who did not understood, I will clarify one more time.
Richardson-Dushman equation is valid with no applied external electric field; note that ex profeso I write the work function as e φ, where φ is a potential, in this case corresponding to the potential barrier.
If we apply an accelerating potential, Va, for a high value enough, current density reaches its saturation value, Js
Js = Jo exp [(α √ Va ) / (k T)]
Obviously, the accelerating electric potential, Va must be referred to the emitter surface, now if the emitter surface is a valve cathode surface; Va becomes the electric potential of the anode with respect to the cathode, aka anode voltage, as you clearly understood a long ago
Seeing this in perspective, your little gopher was right. 😀
Tarzan-English! I meant multimeter readings, sorry. 😛😛
BTW, your writing is not precisely Shakespeare… 😛😀
As seems that thermodynamics is your strong point, I thought that it would be interesting that you prove your assertion, resistive heating is the easy part.
I cannot think that you are not able to understand a datasheet.
Please stop to act as a teenager, this is not a contest for who has the right answer, it is just a technical discussion. 😉
Last edited:
Thanks. Unfortunately my part of the world is in the grip of an unusually bad influenza bug. And it my age, getting rid of it is a slow process.
Quick! Inform the electric power industry! For 120 years they have been calculating skin effect and proximity effect in cables based on magnetic fields! They have been ignoring the electric fields! My goodness!
I am familiar with the concept of images in electrostatics. I am in the electric power generation industry - we use it in solving what is known as power coordination problems. (If there is a pole route down one side of the street carrying 66kV at height x above ground 50Hz eletric power distribution, and on the same street a telephone company cable at distance y away for the powerline and z metres above ground, and the ground is substantailly conductive, what is voltage of 50Hz coupled into the telephone cable? This is a routine question is it may be enough to make the phones hum, and it may even be enough to electrocute the phone company technicians, requiring more spacing or the addition of shield wires.) We use the method of images to calculate the electrostatically coupled voltage. Magnetic coupling also occurs.
Let me put it this way: Let us suppose an electric field created by a voltage carrying conductor is above a horizontal perfectly conductive ground plane. Let us suppose there is another separate conductor, also above the ground plane. A voltage will exist on the second conductor as it is within the field. The closer it is to the source conductor the higher the voltage; the closer to the ground plane the lower the voltage. You can plot contours of equipotentail in the sapce aove the ground plane. To work out the voltage, we need to take into account the image of the source conductor, which appears mathematically as though it is underneath the ground plane, and the ground plane isn't there.
Now, let us suppose there is a third conductor, in the space underneath the ground plane. What is the voltage on this conductor? Answer: Nothing, nada, zip, naught.
It is nothing becasue there is no electric field below the ground plane - the field terminated on the ground plane. The image that appears to be below the ground plane is just a mathematical fiction that gives the right result in calculating what goes on above the ground plane.
An analogy: Lets say there is a lamp illuminating an object. Nearby is a very large mirror. As far as the object is concerned there is a second lamp adding to the illumination. The second lamp appears as though it is behind the mirror. That does not mean that there is light behind the mirror. No light reaches there.
Give your head another bang, Popilin - the electrons within a conductor are not affected by the electric field outside it. Your bricks evidently are not hard enough - try a piece of Krupp cemented armour.
Schottly effect applies to oxide coated cathodes becasue the oxide layer is an insulator masquerading as a resistance due to internal thermionic emission in the space between grains of BaO and SrO, plus some tunnelling assistence by diffused metal atoms. Thus there is an electric field within the oxide layer.
You never did concisely state just what you meant by Va.
Ah, at last. Your English is a little imprecise - do you mean Va is the voltage at the cathode surface?
No, that's incorrect. The anode voltage is not the voltage at the emission surface.
The voltage of the emission surface may be millivolts or less above the cathode nickel sleeve. The anode voltage may be several hundred volts.
The electric field within the oxide layer may be quite high, as a typical thickness is a few tens of micrometers. But the field just inside the emission surface is not the field just outside the surface.
Nah, he's your gopher. And another poster noticed he died -couln't draw a breath.
Thank christ for that. Shakespeare wrote in boring rhyme in a 16th century style. Something we had to endure in high school - definitely NOT by choice!
I freely admit to being a poor typist though.
Seems you are the guilty one there. A glance at the pure tungsten type datasheets show the Ia vs Va line is practically horizontal once saturatioj is past, varing only a few percent attributable to end effects & such.
You haven't proved wrong my assertion that a Va/Ia plot for an oxide cathode is nothing like the plots for the pure tungsten types - the rate of increase is far greater.
You are the teenager here. You made a couple of claims. I showed why they are wrong. You keep on posting, but you almost never directly address the things I point out.
All that nonsense about datasheets and scaling - just a childish attempt to distract - it doesn't matter what the scale is, the shape is the same, the fraction change is the same.
Last edited:
By measuring mass displacement versus time, in vacuum, for the range of temperatures presented ie evaporation and condensation. Don't know where Keit got the idea of BaO vapour dissociation, mass was conserved, nor where any idea comes from that the measurement was extrapolated for acceleration............seems straightforward and directly to the point to me.
Last edited:
What alternative universe are you in?
It certainly appeared from your various posts that you took your view from Hermann/Wagener.
And, as I explained, if you look at the measured data, it is for temperatures (1300 to 1750 K) way above what would be encountered in tubes (1050 K). Hermann/Wagener simply extrapolates the results down to tube temperatures. Look at H/W Fig 54 and see those dotted lines. Nothing to do with acceleration - a word I didn't use.
Last edited:
So, mass displacement vs time was measured by measuring mass displacement vs time! Gee, that's informative. Of course it was in vacuum. Coudn't be done otherwise.
My post #270 on how Claassen & Veenemans did it is perfectly clear, though if anyone is fluent in German, they may get a detail I missed.
Last edited:
I suggest you go back and read more carefully. Kindly don't say I said things I didn't say.
Except......during activation, when it is said that most of the BaO evaporation takes place.
And yes, it seems data below 1300 deg is extrapolated. Still seems reasonable to me.
But the topic in this thread is about why grids in some tubes are gold plated. And it is established in the thread discussion that the gold plating is about preventing the grid from becoming an electron emitter.
That is about what happens in the tube during its normal use over its' service lifetime. Temporary conditions existing for a few minutes during factory activation have nothing to do with it. Most tubes do not have gold grids. Never the less, they still have negligible grid emission - when reasonably new in service.
I agree that the extrapolation down to 1050 K is reasonable - never said it wasn't. The reverse in fact.
Last edited:
- Status
- Not open for further replies.