Little Gopher Effect
Hello,
I am sorry to inform you that the little gopher has passed away due to the Vacuum Effect. However because of the work of the little guy the cathode surface is no longer a homogeneous emitter of electrons. Prior to his death the little guy configured grains or pores at the surface of the coated cathode. Depending on the presence and or proximity of the negative voltage of the grid some cathode pores emit fewer electrons because of the influence of the large and scary grid wires. The electrons do not have the courage or energy to escape from these localized pores. The little gopher called this the Island Effect.
Then there is the influence or potential of the even larger anode to call to his electrons, as he comes closer or calls with increase voltage in his voice more of the electrons gain the courage or energy to overcome their resistance to take the leap from their pore homes into the great vacuum beyond. The little gopher called this a corollary to the Schottky Effect.
This incredible story was retold to me by the great story teller Dr. Aldert Van Der Ziel in Chapter 2 ‘Low-Frequency Noise in Vacuum Tubes’ (Flicker Effect), in the book ‘Noise in Electron Devices’, The M.I.T. Press 1959.
Short version:
The number of electrons emitted by a given coated cathode at a given temperature is not homogeneous or constant. I call this the Little Gopher Effect.
Tongue in Cheek
DT
Hello,
I am sorry to inform you that the little gopher has passed away due to the Vacuum Effect. However because of the work of the little guy the cathode surface is no longer a homogeneous emitter of electrons. Prior to his death the little guy configured grains or pores at the surface of the coated cathode. Depending on the presence and or proximity of the negative voltage of the grid some cathode pores emit fewer electrons because of the influence of the large and scary grid wires. The electrons do not have the courage or energy to escape from these localized pores. The little gopher called this the Island Effect.
Then there is the influence or potential of the even larger anode to call to his electrons, as he comes closer or calls with increase voltage in his voice more of the electrons gain the courage or energy to overcome their resistance to take the leap from their pore homes into the great vacuum beyond. The little gopher called this a corollary to the Schottky Effect.
This incredible story was retold to me by the great story teller Dr. Aldert Van Der Ziel in Chapter 2 ‘Low-Frequency Noise in Vacuum Tubes’ (Flicker Effect), in the book ‘Noise in Electron Devices’, The M.I.T. Press 1959.
Short version:
The number of electrons emitted by a given coated cathode at a given temperature is not homogeneous or constant. I call this the Little Gopher Effect.
Tongue in Cheek
DT
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http://www.google.com.ar/url?sa=t&r...R-BBTzwUFLvCL5R7fC5YFBQ&bvm=bv.99261572,d.Y2I
Sure, I born confused. 😀
You are kidding, right? 😕
If not, and I say with all due respect, you do not understand neither thermionic emission nor Schottky effect.
If you do not apply a voltage between anode and cathode, it only exist an electron cloud around the cathode, and this is almost useless.
Maybe you have another way to make your valves work, enlighten us please.
Please, say to your little gopher, that to obtain saturation current it must be applied an external acceleration voltage between anode and cathode.
Agree, but this elementary Electrodynamics was already discussed, on post#73.
Agree, that’s why the electric field is that big there.
Bingo! You got it! You used the linear superposition principle.
No, it is not the voltage drop, but the electric field E on its surface, they are related, but not the same thing.
Note that, at the interface, there is a huge discontinuity on the normal component of electric field D.
Agree, but at saturation, the electric field inside the oxide cathode cannot exist as it is, without the external electric field, which in turn cannot exist as it is, without an applied voltage between anode and cathode.
The expression for saturation current
Js = Jo exp [(α √ Va ) / ( k T )]
Was used by me, and all over the world, and it is valid for any kind of emitter material.
The other one
Js = Jo exp{e √[e E / (4 π ε0)] / ( k T )}
It is also valid for any kind of emitter material, provided that E is the electric field at its surface, for semiconductors or insulators, E have a more complex theoretical form, by the way.
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That's why it was so hard to take stable measurements with my poor 5R4GY.
Little Gopher Effect. 😀
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Not according to Hermann/Wagener.
Hermann/Wagener The Oxide-Coated Cathode 1951 Vol1 sec 12.1 p 100 not only makes it plain that BaO evaporates significantly, but publishes rate measurements versus temperature in fig 54. These rates exceed those of Ba atoms by perhaps an order of magnitude at likely temperatures, as set out in Hermann/Wagener The Oxide-Coated cathode vol 2 sec 31 pp261.
That's about what Morgan got. He was kind enough to get me the rest of the dimensions, now I just need 15 minutes of quiet time with my books. 😀
I've measured this for a 6AL5; it more closely resembles the Shockley equation with a resistor in parallel.
I can't be certain what the significance of that is (plate leakage, measurement error?), but if we take n = 1 (ideality coefficient), the temperature measures as 678 K.
Tim
Nice! This seems like the complete plot Ia vs Va, note the saturation Io at about 0.5 V, from there you enter into Schottky effect region.
That puzzles me is Ln(Ia+Is), Ln(Ia) vs Va, with retarding potentials, must be a straight line.
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Hi,
The question one should ask one self is why gold sputtered grids on modern valves?
Big or small signal.
The answer is?
Thank you, 😉
The question one should ask one self is why gold sputtered grids on modern valves?
Big or small signal.
The answer is?
Thank you, 😉
The blue one? 😀
I do not think that ion bombardment is the issue, because the grid cross section is too small for that.
I tend to think more on contamination by cathode evaporation.
See Hensley E B, Conduction mechanism in oxide-coated cathodes, Journal of Applied Physics Vol 27(3) March 1956, pp286-290.
From the synopsis: "...The high temperature conductivity is a property of the electron gas in the cathode pores."
The first sentence in the first paragraph is:
"Prior to 1949 it was generally believed that the conductivity of oxide-caoted cathodes could be adequately accounted for on the basis of the alkaline earth oxides being n-type semiconductors."
After some description of experimental apparatus, Hensley goes on in detail why this is not so. Authors in earlier years did not devise sufficiently accurate lab techniques.
True. However, tubes are normally operated so that anode current less than 40% of the emission possible (power tubes) and usually quite a bit less than that (low level stages). This causes a space charge buildup between the cathode and grid. This space charge smooths things out greatly, so the noise in the anode current is dominated by shot effect.
Over the life of the tube, the oxide layer usually becomes more an more porous, having less and less oxide. This does not necessarily mean evaporation of whose BaO molecules, although that will occur to a slight extent. The Ba get separated from it's O, deep within the matrix. The Ba atom can then migrate to the surface and evaporate. The O of course just diffuses through the pore stuctrure and joins the tube gas, ultimately to be captured by the getter flash.
Herrman/Wagener is a very old text complied from stuff published in the 1930's. Much of it refuted by later publications.
Yeah right, Popilin. Turns out that is a site listed by MacAfee as malignant. MacAfee will not aloow my PC to acess it.
No. Obviously what goes on inside a cathode coating cannot be affected by an electric field that is not within it. Duh.
You've realy answered this yourself. That cathode makes available an electron cloud around it. In the absence of any anode or grid field, electroncs get continually thrown out of the cathode and fall back again. Somewhat like a lot of schoolboys in hard hats continually throwing rock up into the air.
If there is an electric field (say imposed by a positive anode), the anode draws of some of the electrons, so less fall back to the cathode. Below a certain point, the anode electic field is insufficient to draw off all teh elctrons, and the anode current follows the three-halves power law: I proportional to V^3/2. Under such conditions a space charge (cloud of excess electrons) continues to exist around the cathode.
In a pure metal cathode, above the anode voltage required to draw off al the electrons, the anode current flat-lines at the Richardson-Dushman value.
In an oxide coated cathode in a tube with anode volatge below the saturation value, the net flow of cathode current causes a radial voltage drop across the oxide layer (Ohm;s Law: V=I.R). This impoves emission slighly thu schottly effect but have no noticeable effect on anode current, as anode current still follows the three-halves power law.
However, if in an oxide cathide tube, the anode voltage if above the satuation value, does cause a noticeable modest rise in anode current, and teh rise in anode current causes more radial volatge drop and thus more schottky effect.
There are two distinct regions in a diode with an oxide coated cathode: The region within the oxide layer, governed by temperature and schottky effect arising from the voltage gradient between the metal base an the oxide surface, and the region between the oxide surface and the anode, governed by the anode-imposed electric field, modified by the space charge.
You could, with two identical oxide cathodes, have then in two diodes, one with an anode 20 mm in diameter (ie. >> space charge distance from cathode), the other with an anode 200 mm in diameter. To get the same current, the second will need an anode voltage ten times as high. But as the anode current is the same, the net cathode current is the same, thus the volt drop across the oxide layer is the same. Therefore the amount of shottky effect is the same.
No, I did not!
That makes no sense. A field, by definition, must extend over some finite distance. It cannot be a point of line - that is a
Of course.
Well, obviously. So...?
See aslo Tomlinson T B, On the conductivity of oxide cathode coatings, Journal of Apllied Physics Vol 25(6) Jun 1955.
I quote from page 720" ...there is an emission current that flows between granules, the spaces or pores being filled with an electron cloud."
I quote from page 723: "The conduction paths are partly across the pores and partly though or around the granules. The complete picture is one of many diodes is series with many resistances, all in parallel with many other similar combiniations."
The axis is ln(Ia + Is), modeling its compliance with the Shockley equation, essentially. Is is about 7nA in this case (corresponding to the bottom tail). The exponential range spans about 5 decades. At the top, current flow transitions from diffusive and exponential, to ballistic and power law (ideally 1.5 power, but the published 6AL5 curve is closer to 1.29 plus an offset, and obviously, with an exponential tail such as this).
Tim
Just in order you can use your measurements, I take the audacity to give some suggestions.
i) Determination of cathode work function
From Richardson-Dushman equation
Jo = Ao T² exp ( - e φ / k T )
You can obtain e φ from the slope of the straight line
Ln ( Jo / T² ) = - ( e φ / k ) (1 / T) + Ln ( Ao )
ii) Determination of cathode temperature
Applying a retarding potential, Vr
Jr = Ao T² exp [ - e ( φ + Vr ) / ( k T )] = Jo exp ( - e Vr / k T )
Then you can obtain cathode temperature, T, from the slope of the straight line
Ln ( Jr ) = - ( e / k T ) Vr + Ln ( Jo )
iii) Determination of Jo
Applying an accelerating potential, Va, for a high value enough, current density reaches its saturation value, Js
Js = Jo exp [(α √ Va ) / ( k T )]
Then, you can obtain Jo from the straight line
Ln ( Js ) = [ α / ( k T )] √ Va + Ln ( Jo )
You will need more points than in your plot, with higher voltages.
If you use a picoammeter, just to avoid damage in the valve, you can measure with different values of heater voltage.
With the 5R4GY I used: 1 V, 1.25 V, 1.5 V, 1.75 V, and 2 V
I also used DOS TableCurve software (now in a windows version) to make all statistics.
www.te.kmutnb.ac.th/~msn/8890_CH05.pdf See page 24 of the pdf or it is page 376 of the book extract as follows:
The end result is that grids are forced to work at temperatures as high as 1500◦C. Their primary
and secondary emission must, however, be low. To prevent grid emission, high electron affinity must be
ensured throughout the life of the tube, even though it is impossible to prevent material evaporated from
the cathode from contaminating the grid surface.
In tubes with oxide cathodes, grids made of tungsten or molybdenum wire are coated with gold to
reduce primary emission caused by deposition. The maximumsafe operating temperature for gold plating,
however, is limited (about 550◦C). Special coatings have, therefore, been developed for high-temperature
applications that are effective in reducing grid emission. In tubes with thoriated tungsten cathodes, grids
made of tungsten or molybdenum are coated with proprietary compounds to reduce primary emission.
Primary grid emission is usually low in a thoriated tungsten cathode device. In the case of an oxide cathode,
however, free barium can evaporate fromthe cathode coating material and find its way to the control
and screen grids. The rate of evaporation is a function of cathode temperature and time. A grid contaminated
with barium will become another emitting surface. The hotter the grid, the greater the emissions.
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