Van de Graaff generator voltage measurement

[davekni]

P.S. I found the source of the positive field reading when I grounded the terminal after turning off the VDG. I feel kind of stupid now. For my self excited negative charging VDG, the negative roller is on top inside the terminal and the positive roller is on the bottom and not shielded. After turning off the VDG and grounding the terminal it shielded the negative roller but left the positive roller still unshielded giving a net positive field. I confirmed this by putting some grounded shielding around the bottom roller and the positive field disappeared.

Certainly not anything I would have thought about.  Great work figuring out this detail!

Given the difficulty of measuring VDG voltages, I'd say your results from three separate techniques are relatively close.  Impressive work measuring voltages.

[haversin]

I modeled my VDG terminal in FEMM to see what the E field magnitude looked like around it's surface. The terminal voltage was set 100kV. The results are shown below. It's seen that the max E field magnitude is about 1.5*10^6 V/m and occurs at the lower outside of the terminal  Assuming air breaks down at 3*10^6 V/m this predicts a max terminal voltage of only 200kV. I'm guessing the space charge due to corona is why the terminal can go higher than this. I haven't figured out how to model this space charge. Does anyone have any ideas? Uspring?
 

[Uspring]

I've used some heuristics in my calculations and split the air region in 2 parts: One with a field >= 30 kV/cm and one with < 30 kV/cm. For the first region I assumed, that the field stays pretty close to 30 kV/cm, since conduction increases rapidly above this threshold causing the space charge to redistribute itself so that the field remains close to it. For the other region we have no further ionisation and only charges of one polarity. The charge density there follows from the current and the speed of the charges. The speed of charges again can be calculated from the field. It is roughly proportional to the field, about 2e-4 (m/s)/(V/m). Note the continuity condition for the current, i.e. the current is the same through every surface around the top load.
The current thus uniquely determines the cloud charge distribution and therefore the field. Integrating over the field you get the voltage.The spherical symmetry that I've assumed for my calculation makes a simple solution feasible.

In your case, I see two possibilities, a full scale calculation or an attempt to simplify. For the first case, have a look here: https://publications.lib.chalmers.se/records/fulltext/188939/188939.pdf . That is a very well done work and quite thorough.
The other possibility is to use your FEMM calculation and assume some current density originating from the top load surface, where the field is above 30 kV/cm and initially assume, that the field is not changed by the current. The current will follow the field lines. The charges will spread out due to the field lines spreading out but will also be compressed along the path direction due to the lower speed. The ion speed can be calculated from the field. Then you can add this charge density into the FEMM calculation and see, how it will modify the field. That's a bit of a trial and error method, but you might get a feeling of how corona current affects the field. Ideally the current distribution you choose will cause a 30 kV/cm layer around the bottom of the top load.

I haven't put much thought into the happenings inside the >= 30 kV/cm region. One likely needs to consider spatially distributed ion sources.

[Uspring]

I should have put more thought into the happenings inside the >= 30 kV/cm region. 

Besides having a field of more than 30 kV/cm there is another requirement for corona, which I have neglected:

For corona to sustain a current, a sufficient number of free electrons, which start avalanches, have to be available. The avalanche starting from a single electron fizzles out, when it reaches a low field region. Avalanches become self sustaining, when an avalanche causes the next one. That is, when there are enough ionisation events to produce light, which in turn frees electrons on the surface of the top load, i.e. causes a photo effect. Then the next avalanche will start there.
This is a quite inefficient feedback, since about 100 million ionisation events in the avalanche are needed to start a new one. That implies, that the initial avalanche has to run for a certain distance to become large enough. I suspect, that the > 30kV/cm layer around the top load is just thick enough to sustain corona. The paper I quoted has a table of ionisation coefficients.
Shielding effects through space charges might not be relevant at the currents your VDG can provide.

[haversin]

I ran a simulation of the spherical symmetry case with your ion mobility of 2e-4 (m*m)/(v*s) and a terminal sphere radius of 0.1 m. E at the surface was set to 3.e6 v/m. Numerically integrating -E*dr from r  = 10m to the sphere radius 0.1m the terminal voltage with no current was 297 kV and with a 5 uA current the voltage was 335 kV. So the 5 uA current did make a difference. Going to the FEMM shape of the VDG terminal is much more complex.

[Uspring]

I chose a different outer radius (1m) in my calculation. The voltages without and with space charge were then 270 and 271.2 kV, respectively. The outer radius I chose is probably a bit too low, since your VDG is possibly higher than 1 m and the ceiling much further away. But I'd expect the field lines to bend to ground in not too far a distance and the ions to follow that, so I think 10m as an outer radius is a bit large.
For 10 m I get the same result as yours.

[haversin]

I'm glad were getting the same numbers! are you getting a closed form solution? I got a nasty first order nonlinear differential equation that I could only solve numerically. I agree that 10m is too large. The height to the top of the terminal is only 77 cm. Of course the actual geometry is far from spherically symmetric.

[Uspring]

There is an analytic expression for the E field:

E = sqrt(I/(6*π*μ*ε0)*(r^3-r0^3) + E0^2*r0^4)/r^2

I is the current, μ the ion mobility, r0 the terminal radius and E0 the field at r0.

Wolfram Alpha claims the integral of this to be a hypergeometric function. So there is an analytic solution even for the voltage, but considering my own knowledge, a useless one. But you can make some approximations, which simplify that.

Did you try to see the corona in the dark? It might be visible to a dark adapted eye.

[haversin]

There is an analytic expression for the E field:

E = sqrt(I/(6*π*μ*ε0)*(r^3-r0^3) + E0^2*r0^4)/r^2

I is the current, μ the ion mobility, r0 the terminal radius and E0 the field at r0.

Very nice! I guess that differential equation wasn't so nasty after all!

Did you try to see the corona in the dark? It might be visible to a dark adapted eye.

  While investigating the difference between the field mill readings of the VDG with and without the 25 cm diameter sphere on top I took photos of the corona with and without the sphere. They were exposed for 30 seconds in the dark at ISO 3200 see below. The terminal was illuminated briefly with a small flashlight so it could be seen in the photo.  To the eye the corona jets appear to be flickering randomly around the terminal the photos give the wrong impression that they are all happening at the same time. With the sphere on top the top jets disappeared and the bottom jets were much brighter and more frequent. After looking at these photos I thought the sphere on top might actually be lowering the voltage by increasing the E field at the bottom. However FEMM results showed that the sphere actually lowers the E field at the bottom see below.

[Uspring]

Very nice! I guess that differential equation wasn't so nasty after all!

Thank you!
The images are interesting. The assumption of spherically symmetric current distributions seems dubious now, even with nearly symmetric setups. I would have expected a glowing layer around the terminal such as in a glow lamp. The jet like structures in your photos indicate possibly quite large current densities, which might be affected by the correspondingly large space charges. So possibly space charges are back in the game even for 5uA currents.

The FEMM simulation shows illustratively, how the high field regions move downward to the bottom of the terminal when the sphere is added. I believe terminal voltage increases with the extra sphere.