WE417A is too much expensive for my budget. 😀
Wrong calculation, I need the grid at about 1000 ºK, not 500 ºK.
Anyway, if pin to socket resistance is your concern, you just can solder the pins. 😉
Silly question: How do you to distinguish, e.g. grid primary emission from gas grid emission?
You are kidding, right?
Let me google that for you
OK, like in high school
Jo(Mo)≈55[A/(cm²ºK²)](1000ºK)²exp{-4.15 eV/[(8.617x10⁻⁵eV/ºK)1000 ºK]}
≈ 6.7 x 10⁻¹⁴ A/cm²
Happy? 😛😀
Thanks for your kind words. 🙂
Your calculations seem wrong, what is c?
Last edited:
Thanks for running the check, I found an arithmetic error, ooops.
I now get:
J(o) = 0.013uA/cm2 for BaO (activated) over Au substrate
J(o) = 4.23 uA/cm2 for BaO/Ba over Ni-Mb substrate
Similar conclusion, I think.
BTW, c means 'circa' or 'roughly', not a symbol.
So go do for a frame grid tube within your budget then. You'll find that other tubes are even less feasible for electric grid heating.
Grids operate near 500K, not 1000K. A gold plated grid will not take 1000 K. But you can look up resistivity at 1000K if you like. I gave you a ref.
Perhaps.
You will still have the problem that the one end of the grid will be at a different temperature to the other, as conduction losses will not be even.
There's no such thing as a gas fee tube, but in good tubes, the effects of gas is negligible.
Grid primary emission current and gas grid secondary emission currents sum together. This means that for both the standard method and your proposed electric heating method, telling the two currents apart is not directly possible.
However, the presence of gas increases gm above the normal value. You often see the effect of this on a tube characteristic meter if you test a NOS (especially octal) tube that has long been in storage - the gm is often intially abnormally high and slowly decreases back to normal as the getter flash warms up and deals with the gas (or, down to the true low value if the tube has poor emission). A restorer's trick is to bake a tube that has become gassy in storage in an oven at 150 C for 30 minutes or so. That brings the gm down quite effectively. So, first verify your tube is not gassy, then do the grid emission test.
No, I wasn't kidding. My point is that you do not fully understand how to apply the equations you keep repeating. If you define each term then your errors will become more apparent, perhaps even to yourself.
In a previous post (and in another thread) I took an educated guess as to wht you thought terms meant, and included that guess in my post. You did not correct it, so I have to assume my guess as to your meaning was correct - unfortunately that meaning is wrong.
That would be better practice, but not what I asked for, as you know perfectly well.
It isn't just what people do in high school though. Not inline though, best done in 2 columns, one with the numbers, one with the dimensions. Engineers do it on the job. It uncovers errors. Dimensional analysis combined with the metric system is the greatest error stopper around.
Not with your woffle, no.
Last edited:
I dug out some 6P1P-EVs.
I have my granddaughters SE amp with two of them in it that has been running for several years.
I also have samples that have been run for only a few minutes in testing them.
Within reason, I can run them as long as necessary.
I can provide two sets with samples of test to evaluate, 1 hr run time(?), 100+hrs run time.
I'll look at samples tomorrow at work on the stereoscope to see if I can verify gold grids and screens. Otherwise I'll sacrifice one for dissection.
I have my granddaughters SE amp with two of them in it that has been running for several years.
I also have samples that have been run for only a few minutes in testing them.
Within reason, I can run them as long as necessary.
I can provide two sets with samples of test to evaluate, 1 hr run time(?), 100+hrs run time.
I'll look at samples tomorrow at work on the stereoscope to see if I can verify gold grids and screens. Otherwise I'll sacrifice one for dissection.
Hello SY,
I have been poking ariund diyaudio.com and found refreshes to emai. I have not found the actual tab to click on. I am on the road this week and attempting this on my cell phone. SY if you will please send me a email and I will respone with what is in mind regarding run in of these tubes.
Thanks DT
I have been poking ariund diyaudio.com and found refreshes to emai. I have not found the actual tab to click on. I am on the road this week and attempting this on my cell phone. SY if you will please send me a email and I will respone with what is in mind regarding run in of these tubes.
Thanks DT
I'll send it to the one you registered with (just in case it was a throwaway that you'll need to check).
Yes and no, in that order. 😀
Yes, the bottom end of the grid should be hotter than the top end, nevertheless, due to negligible resistance difference; the temperature gradient should be very low.
No, if we talking about the grid, the main mechanism of heat transmission to the exterior is conduction, and it favors the bottom end.
As I said before, nickel is a good heat conductor, and in steady state, all the grid will be at about the same temperature, not too relevant for the experiment by the way.
No, it can be proved that the number of electrons into the stream is a function of current density, the standard method produce a current orders of magnitude greater than in the proposed experiment, then for a given electron energy, the probability of producing a gas ion is almost negligible; even more, if you maintain the unused cathode at a tiny negative voltage, due to the small grid cross section, almost all gas ions will end up at the cathode.
Keit, you do not understand what is a field, if you also do not understand my posted equations, just do not read my posts. 😉
If you must repeat one equation over and over, no need for more than one clarification of the meaning of its terms, I did it here
http://www.diyaudio.com/forums/tubes-valves/276917-why-gold-grids-7.html#post4409001
If you need additional tubes, I have a pair of 6P1P-EVs from an amp that has been in use for four years. I also have NOS unused samples and can run samples for some period of time (you define how long) and provide them as a set.
I dissected one and confirmed it has gold grids.
I dissected one and confirmed it has gold grids.
Appreciate it. I'll let you know if I end up needing them.
This is the wonderful thing about this forum- we come up with an experiment to answer a question and people jump in to help out.
This is the wonderful thing about this forum- we come up with an experiment to answer a question and people jump in to help out.
You have it round the wrong way. The bottom of the grid will be cooler than the top, because it will be loosing heat via the leads to the pins. Unless you propose heating teh pins, which will most likely cause loss of vacuum.
Your test, as you proposed in your post #211, involves (after preheating to stabilise the gas level) heating the grid electrically to something like its normal temperature, aparently while disabling the cathode by turning of the heater. So the cathode will be colder than normal due to its conduction losses to the heater pins and the cathode pins. So the electrically heated grid will be loosing heat by radiation to the cool cathode. There will be additional heat loss be radiation near the ends of the tube, where the grid can "see" the micas.
That the grid will loose a lot of heat by radiation is easily understaood, remembering that:-
1) the grid surface area is of the same order of magnitude as the cathode;
(do not confuse grid thermal surface area with the apparent area offered to radial electron flow)
2) Much of same radiation paths are responsible for the grid running typically around 500 K during normal operation.
If the grid is uncoated, estimating the median grid temperature during normal operation and during your proposed test is relatively easy. Just monitor it's AC electrical resistance by the 4-terminal method.
Unfortunately, this will not be sufficient, as thermionic emission is so sharply dependent on temperature. Parts of the grid only slightly hotter will contribute to most of the emission. How is it known that these slightly hotter parts correspond or don't correspond to parts with the most Ba contamination?
I appreciate that the grid support pins offer significant conduction, but only a diffrence of only a few K will significantly unbalance emission.
It's hard to make sense of that. In normal operation, most ions get lost in the space charge between the cathode and the grid. Since the grid-anode electric field is presumably about the same, the fraction of ions hitting the grid vs those passing through will be about the same, not significantly affected by whether the cathode is a volt or two positive or negative to the grid.
I understand something about fields that it apears you do not: Nothing can be affected by a field unless it is in that field.
In your post #63, you said "..apply an acelerating potential Va...". But you did not say just where Va is applied.
You contended that, contrary to textbooks on thermionic emission, that schottky effect is significant. It is well known that schottky effect has no impact on pure metal cathodes (inc hot uncontaminated grid wires) and onlt a small effect with oxide cathodes. I'm still waiting for your defined terms numerical calculation supporting your contention. All we get from you is seemingly intentional misreading of what I said, plus red herrings.
You can see the difference between pure metal cathodes (no schottky effect) and oxide cathodes (small schottky effect) quite easily. Plot anode current vs anode voltage for a pure tungsten cathode diode, eg an A2087 noise diode and plot for a small oxide cathode diode such as a 6AL5, at lowish filament voltages. You'll find that with increasing anode voltage past the emission saturation point, the pure tungsten diode clearly assympotes to a horizontal line, whereas the oxide cathode diode assymtotes to a line somewhat slanted upwards. The upward slanting is not entirely due to schottky effect - the net flow of current causes resistive I^2xR heating in the oxide layer, not entirely balanced out by the surface cooling effect of loosing energetic electrons.
Last edited:
Yes. Plus the question is challenging to answer satisfactorily in its own right, justifying an exploration of a backwater of published material and understanding, which I find interesting in itself.
I've been reading more about gold chemistry. Very interesting stuff gold, think I'll start collecting it, SY is sure onto a winner here with 'send me yer gold' 😉 It has 5d106s1, but with a twist, check it out. Energy wise, it prefers to gain an electron to fill the 6s shell rather than lose the one it has, so readily forms Au- ions, in some respects like a halogen ie a non-metal. So Au readily forms compounds with certain metals such as Ba, eg BaAu2, which was the first Auride to be confirmed in 1938. It melts at about 750 deg C, apparently. As a non-metal its electronegativity approaches that of iodine, it wants to fill that 6s shell. Pt 5d96s1 has similar traits, I read.
So I'm still on that kick about Au surface chemisorption and dissociation of BaO, forming BaAu2 with liberation of O2, or even Barium Aurate which is stable and solid up to 500 deg C or so. Will be wonderful to find out experimentally.
Last edited:
- Status
- Not open for further replies.