Most people of my generation, and most certainly me, were exposed to precision beam balances (sometimes known as pendulum balances) in high school or college. If not they can start with https://en.wikipedia.org/wiki/Analytical_balance
which gives the precision and accuracy as up to 0.1 microgram.
Even the local weights and measures store near me sells calibrated masses to 100 PPM accuracy ex stock.
Students would be doing well to get 0.01 eV accuray out of PES (in part because the process involves subtracting a big number from another similar number to get a small number answer. In any case 0.01 eV accuracy means only 3 significant figures - nothing like the precision to which we can measure mass, length, frequency, and all manner of things.
My students had no problem in that regard. To be fair, they had the advantage of actually operating the instruments.
False.
Seems to me that you have put the wrong numbers, a very rough calculation showed that the emission for various metals that could be used for grids can vary over several orders of magnitude.
False, seems to me that you use your valves without any applied +B.
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.
e φ = E - EF
Where
e φ is the work function of the metal
E is the total energy of the electron
EF is the Fermi energy
Since electrons are fermions, using the Fermi-Dirac distribution to calculate the number of electrons that reach the surface of the metal with enough energy and proper motion direction to escape, we can obtain the thermoelectric current density leaving metal surface as a function of its temperature T.
The result is known as Richardson-Dushman equation and has the form
Jo = Ao T² exp ( - e φ / k T ) ….. (*)
Where
Jo is the emission current density
Ao is the Richardson constant
k is the Boltzmann constant
T is the absolute temperature of the metal
Ao = 4 π m e k²/ h³
Where
e is the charge of electron
m is the mass of electron
h is the Planck constant
Applying an accelerating potential Va, to a sufficiently high value thereof, the current density reaches the saturation value of Js, then it can be shown that
Js= Jo exp ( e Va / k T ) ….. (**)
Between (*) and (**) there is a lot of grid current that was escaped from your calculations. 😀
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Hello Rod,
Yours is an interesting post about the practical origin and use of gold grids.
The way I read it the gold grid avoids primary grid emissions by absorbing the barium oxide that evaporates from the cathode. The absorption continues and grid emissions are prevented until the gold grid coating is saturated with barium oxide, this sound similar to the function of the getter.
I looked up the Bell System reference.
https://archive.org/stream/bellsystemtechni26amerrich#page/834/mode/1up
The gold grid coating was developed empirically to avoid grid emissions. The Bell System Technical Journal reports that for the war effort many combinations of grid materials and coating were tested. Gold tested best for preventing grid emissions.
The short version is that there was a mountain of empirical test data in search of a supporting theory.
DT
I barely could read Rod's attachment, my new glasses will be ready next tuesday, seriously! 😀
The supporting theory should include both three, work function, electron affinity and contamination. IMHO.
The key is to know the influence of each one.
Edit: If contamination is the main reason, a new valve would not exhibit grid current, and this is not correlated with reality.
Here a very informative book, start at page 26
https://books.google.com.ar/books?i...nepage&q=grid emission in vacuum tubes&f=true
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Instead of copying formulae out of a textbook (You didn't need to - I already gave you the formula), which proves you can ciopy from a book and nothing more, why not insert actual values into the Richardson-Dushman equation and see what the answer is. Turns out that at 1000 K, emission is not even nanoamps.
I note that, whereas I did supply numbers, you have not. And you have not identified which if any or my numbers are wrong.
Unfortunately this "formulae" was not discovered by me, but Richardson, it is not a secret. 😀
Between Richardson-Dushman equation and derived equation for current density saturation, there is a lot of current density, more than your alleged "nanoamps".
BTW, derived equation for saturation was experimentally verified by myself, measuring...a work function. 😉
As I said before in another thread, it is very difficult a serious technical discussion if you begin to act as a cheater lawyer. 🙄
Seems like Sy can't stand being wrong.
He posted very early: in post#2 he asserted that control grids get gold plated because of "work function".
The thread went off topic foe a while, then in post#24 DualTriode brought it back, repeating the original question "why glod plated (control) grids?" DualTriode notes that Sy said it was work function. DualTRiode also noteed that teh work function of gold is the same as nickel.
In post #25 I posted that work function has nothing to do with it, becaasue the thermionic emission of any metal at oxide cathode temperatures is nanoamps ie negligible.
In post #27 Sy posted again that it is work function, but gave no explanation, other than a request to plug the numbers into the Richardson-Dushman equation - something clearly he hadn't tried.
In subsequent posts, references provide by various folk clearly established that grid emission occurs because of migration of cathode material. Without that, grid emission is insignificant.
In post #51, DualTriode asked if work function has a standard distribution, not doubt noticing that quoted values for work function vary somewhat from reference to reference, and some references quote it as a range of values.
In post #53 I answered in detail, stating that it is not that work function has a standard deviation, its that measuring it with any degree of precision is very difficult. As it seemed likley that someone would say "why not make up a special test diode, with the filament made of the metal to be tested, heat it, and measure the resulting current" I explained why that is in practice very inaccurate, and not even possible for metals with melting points below the temperature at which easily measurable emission occurs.
I did mention in post #53 that work function crops up in a number of applications and veruous alternative ways of measuring work function have been devised - all of them subject to significant error.
In post #55, although of no particular relavence to teh thread, Sy decided to rubbish my answer, saying that PES is "the technique of choice".
In post #56, I pointed out that PES is dependent on knowing the work function of metal used as a calibration, thus depended on measurements made by other (inaccurate) methods.
Note that work functions are given in reference books only to 2 or 3 figure accuracy (usually 2 figures).
So Sy then makes it personal....
Sy, as a moderator, don't you think you need to show a higher standard, instead of clogging up threads and following me about with pointless dissagreement?
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A common sign that someone who is wrong, and doesn't like it, is that as soon as they are challenged to do the calculation, instead of just doing the calculation, they begin to turn it into a personal attack.
You know you are wrong, Popilin, without doing the calc, because people have posted details of reference books that state that grid emission is negilible until migration of cathode oxide material has occured. Go back and look.
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No, you are wrong, Richardson-Dushman equation is valid for thermionic emission without an applied external potential, with an external potential, it takes the general form
J= Jo exp ( e V / k T )
With V>0 for an accelerated potential, V<0 for a retarded potential, Jo is still given by Richardson-Dushman equation.
No need any calculation to prove that you are wrong, ignoring elementary physics.
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A more accurate expression for Richardson-Dushman equation with an external electric field E can be obtained replacing the work potential φ with the effective work potential
φeff = φ – φSch = φ – √[e E / (4 π ε0)]
Then
J = Jo exp{e √[e E / (4 π ε0)] / k T}
The applied electric field increases thermionic emission lowering the potential barrier, this is the Schottky effect.
Can you hear me now?
φeff = φ – φSch = φ – √[e E / (4 π ε0)]
Then
J = Jo exp{e √[e E / (4 π ε0)] / k T}
The applied electric field increases thermionic emission lowering the potential barrier, this is the Schottky effect.
Can you hear me now?
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Schottky effect requires very high amounts of external electric field. It isn't noticeable in pure metal cathodes (as an uncontaminated emitting grid would be) for very simple reason: You cannot have an electric field of any magnitude inside a good electric conductor.
The externally sourced electric field terminates on the surface of the conductor. In an oxide cathode, the electric field penetrates the surface and exists down to the surface of the cathode cylinder (usually nickel). It thus contributes to accelerating electrons towards the surface, so they may have a launch energy by the time they get there.
Note that while schottky accelleration is noticeable in oxide cathodes if tested for, it is insignificant in the practical application of oxide cathode vacuum tubes, because in order to eliminate loss of emission due to depletion of of metal atoms faster than diffusion can replace them, oxide cathodes are operated well below the Richardson-Dushman limit (typically 40% or less for power tubes. For small signal tubes, much less).
Oxide emission or pure metal emission, schottky effect is far too small to have any noticeable effect on the emission of a grid.
The externally sourced electric field terminates on the surface of the conductor. In an oxide cathode, the electric field penetrates the surface and exists down to the surface of the cathode cylinder (usually nickel). It thus contributes to accelerating electrons towards the surface, so they may have a launch energy by the time they get there.
Note that while schottky accelleration is noticeable in oxide cathodes if tested for, it is insignificant in the practical application of oxide cathode vacuum tubes, because in order to eliminate loss of emission due to depletion of of metal atoms faster than diffusion can replace them, oxide cathodes are operated well below the Richardson-Dushman limit (typically 40% or less for power tubes. For small signal tubes, much less).
Oxide emission or pure metal emission, schottky effect is far too small to have any noticeable effect on the emission of a grid.
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Yes and no, in that order. 😀
-Yes, Schottky effect requires a very high external electric field, and just to avoid put any arbitrary number, I found my own measurements, it shows for the Schottky term, at most, about an order of magnitude, I expected at least two or three. 🙁
From elementary Electrodynamics, inside a conductor the potential is constant, then
E = - ∇ φ = 0
-No, Schottky effect has nothing to do with the electric field inside the metal, but on its surface.
Before conclude that uncontaminated grid emission is negligible, I want to know about a new valve, just below the contact potential, where it supposedly grid emission predominates, why the hell there is still grid current? 😕
I prefer to gather realistic numbers to plug into calculations before doing them. I don't have all the ones I need yet, been a bit busy with tornadoes and their aftermath. However, I have shown that the ratio of emission between gold and molybdenum is several orders of magnitude.
B.O Baker "Gold as a grid emission inhibitor in the presence of an oxide coated cathode British Journal of Applied Physics, Vol 4, No 10, 1953"
Abstract :
Emission measurements have been made on gold-plated molybdenum and gold-plated manganesenickel grids in the presence of an oxide-coated cathode. For grids which cannot be designed to operate below 350°C, a minimum thickness of 1 μ of gold will suppress grid emission. Silver is not so reliable, but is effective in some cases.
I haven't downloaded the article, but the abstract seems relevant and informative.
Abstract :
Emission measurements have been made on gold-plated molybdenum and gold-plated manganesenickel grids in the presence of an oxide-coated cathode. For grids which cannot be designed to operate below 350°C, a minimum thickness of 1 μ of gold will suppress grid emission. Silver is not so reliable, but is effective in some cases.
I haven't downloaded the article, but the abstract seems relevant and informative.
I don't have much experience with new valve grid current measurement. The ones where I do have experimental data are ECC83. What I'm missing, and if someone could kindly measure it or provide a good reference, is the length and diameter of the grid wire used in that tube type.
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We're OK, but there was a lot of property damage in our area. I could only access the internet through my phone up until the past couple of hours. No-one seems to have been injured, but the local school lost its roof, many trees came down, and most of the houses on our street have large sections of roof missing, siding ripped off, and windows smashed. We've been helping the clean-up effort today. Appreciate your concern.
My attitude, yes, for sure. The interruption did upset my son, who was about three moves away from checkmating me.😀
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