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[Philip Hardcastle]

Since eV are a measure of energy, inserting a 6eV source inserts a tiny bit of energy.  It seems obvious to me that virtually nothing happens.

Then you have not brought your intellect to the task.


What is the state of charge of the 6eV mesh when it comes into contact with a metal of 5eV work function, what is the chemical potential of the 5eV metal when it reaches equilibrium without the mesh?


I think you need to see the mesh as a region that is more negative than the collector, this being so it has a few effects, the one that is well understood, and the reason for S3 grids in pentodes, is that it suppresses secondary electron emissions from the collector (plate).


The other effect is that the presence of an electric charge on the mesh in proximity to the collector will cause the chemical potential at the 5eV metal surface to be lowered, but at the same time the charge of the whole collector rises, that is to say the chemical potential at the left hand surface of the 5eV metal must therefore rise, and if so it means that the equilibrium of the left hand side is disturbed, electron must flow from the left hand side 5eV, this then means that the left hand side 1eV chemical potential must rise from the equilibrium point it would have absent the grid on the right hand side becoming charged, and of course the increase of chemical potential in the 1eV on the left hand side translates to a rise in chemical potential on the right hand side 1eV surface, so it (the RHS emitter) more readily emits electrons into the RHS gap.


Now we know that the grid has a charge higher than the RHS 5eV metal surface, so we know that hot tail electrons from the RHS 1eV surface has to do work to get past the mesh, so hot tail electrons present in the population are selectively (by dint of their own energy) allowed to reach the 5eV RHS surface only after they have cooled (done work) getting past the charged grid of the 6eV mesh.


I suppose this is what you mean by it is obvious that virtually nothing happens?

[MarkE]

Then you have not brought your intellect to the task.


What is the state of charge of the 6eV mesh when it comes into contact with a metal of 5eV work function, what is the chemical potential of the 5eV metal when it reaches equilibrium without the mesh?

If you introduce 1E-18J, which is what ~6eV is then unless you are at an atomic scale, which as soon as you say the word "mesh" you aren't, you haven't done much of anything.  Are you certain you are saying what you intend to say?  Are you saying that you want to introduce 6eV into the system or a material with a 6eV work function?  The latter would make more sense but still doesn't get you anywhere.

I think you need to see the mesh as a region that is more negative than the collector, this being so it has a few effects, the one that is well understood, and the reason for S3 grids in pentodes, is that it suppresses secondary electron emissions from the collector (plate). 


The other effect is that the presence of an electric charge on the mesh in proximity to the collector will cause the chemical potential at the 5eV metal surface to be lowered, but at the same time the charge of the whole collector rises, that is to say the chemical potential at the left hand surface of the 5eV metal must therefore rise, and if so it means that the equilibrium of the left hand side is disturbed, electron must flow from the left hand side 5eV, this then means that the left hand side 1eV chemical potential must rise from the equilibrium point it would have absent the grid on the right hand side becoming charged, and of course the increase of chemical potential in the 1eV on the left hand side translates to a rise in chemical potential on the right hand side 1eV surface, so it (the RHS emitter) more readily emits electrons into the RHS gap.


Now we know that the grid has a charge higher than the RHS 5eV metal surface, so we know that hot tail electrons from the RHS 1eV surface has to do work to get past the mesh, so hot tail electrons present in the population are selectively (by dint of their own energy) allowed to reach the 5eV RHS surface only after they have cooled (done work) getting past the charged grid of the 6eV mesh.


I suppose this is what you mean by it is obvious that virtually nothing happens?

Really, nothing happens.  We can and do put metals with different work functions near each other all the time and they sit there happily in thermodynamic equilibrium all day long even though there is an internal electric field that results from the difference in work functions.

[Philip Hardcastle]

What I thought was obvious, and I accept your statement as being confused on the issue, was that it was the work functions of the materials. The chemical potentials will of course come to equilibrium in a circuit with no persistent current (the diagram shown but without the mesh). Clearly a circuit producing a persistent current must, ipso facto, have a non uniform chemical potential of end surfaces.


As to your saying nothing happens that simply goes against the proved use of a suppressor grid in a pentode tube.


Sorry MarkE but if you cannot see it from that diagram then I cannot make you see it, no amount of argument from me is going to sway your view.

[MarkE]

What I thought was obvious, and I accept your statement as being confused on the issue, was that it was the work functions of the materials. The chemical potentials will of course come to equilibrium in a circuit with no persistent current (the diagram shown but without the mesh). Clearly a circuit producing a persistent current must, ipso facto, have a non uniform chemical potential of end surfaces.


As to your saying nothing happens that simply goes against the proved use of a suppressor grid in a pentode tube.


Sorry MarkE but if you cannot see it from that diagram then I cannot make you see it, no amount of argument from me is going to sway your view.

Mr. Hardcastle pentodes do not spontaneously drive energy into external circuits.  External power supplies supply operating power to pentodes.  If you think differently, then you can devise and publish an experiment that you think will show otherwise.

[Philip Hardcastle]

Mr. Hardcastle pentodes do not spontaneously drive energy into external circuits.  External power supplies supply operating power to pentodes.  If you think differently, then you can devise and publish an experiment that you think will show otherwise.

I have all the proof I need backed up by others doing the same experiment, all you are showing in your statement is that you are prejudiced by the view that the 2nd is absolute.


MarkE, with due respect because I can see you are a professional in science, get a pentode, wire it as shown in my $10 experiment, heat it in a fancy oven and measure it with all the care you can, then tell me what you know.