? K is grounded, for sure, but grid is not.
Well, it is mechanically related with a fixed distance to cathode and there is a flow of electrons...
If the Grid is left open the Tube becomes a Diode and the grid potential will "float" to a voltage between cathode and Anode determined by the geometry.
As the grid is typically close to the electron emitting cathode and far from the electron attracting anode, it would be a few Volt above the cathode.
Now, let's connect the grid to a capacitor.
The grid potential will initially cause current to flow, the grid emits negatively charged big fat lazy green electrons.
These will charge the capacitor NEGATIVELY, with respect to the Grid's "open circuit" potential.
Eventually the grid will become sufficiently negative to pinch off the current in the Tube.
This means less electrons reach the grid and the charge in the capacitor falls allowing more current to flow and the capacitor potential will rise allowing more current.
Basically the circuit works as a result of the parasitic grid current and given non-gassy tubes will reach an equilibrium with the grid a few 100mV lower than the cathode and will remain there stable.
My basic and naive knowledge tells me that a continuous flow of electrons reaching an electrode would result in ever increasing negative voltage if there is no discharge path.
But there is a discharge path. Through the dielectrics.
Yes, this circuit fundamentally requires parasitics to work, it actually only works on parasitics, but it works fine and stable with good quality tubes (which makes it less ideal in the 21st century).
This type of stable operation indicates there is actually a discharge path.
Of course is. Air has finite resistance.
Actually, my view on it is that once the grid has reached a negative potential high enough to reduce the flow of electrons to a trickle, grid current decreases to a point where leakage becomes sufficient to compensate the remaining flow of electrons hitting the grid. However I fail to model and quantize this discharge path. It is very hard for me to admit taht a whole industry relied on uncontrolled parameters.
The parameter is NOT uncontrolled. It's quite simple, the tube works close to Vgc = 0V. In this region the grid emission will provide a small negative bias when the grid is connected to a capacitor. Note that the negative will be relative to the cathode, NOT ground.
Thus the cathode is pulled up by appx. the diode operation current and the grid is stabilised by the grid-emission at a few 100mV below the cathode voltage.
Without a discharge path, the voltage is continuous increasing. The Van de Graaf generator voltage would increase infinitely if there was no air ionization for discharge.
Incorrect, eventually the grid stops emitting enough electrons to overcome the losses in the dielectric and this happens before the Anode Current is pinched off completely.
Curves are the results of experiments, they show observations, they don't explain why.
Considering the number of circuts that operate this way, I can just acknowledge it works.
Of course it does.
I fail to see how the operation near zero Vgk helps understanding how grid voltage is regulated.
Because only near 0Vgc is there enough grid emission for the whole thing to work... Otherwise the Grid emits insufficient electrons to have this mechanism function.
And that is main parameter here, which IS CONTROLLED (by inherent design).
It does mean that circuit must be designed explicitly for a SPECIFIC tube and needs reasonably narrow tolerance, so it is not something to randomly "roll any tube that fits the base" into.
Thor