FAQ - HV feedthroughs: Theory, off the shelf parts, and fully custom designs

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Liam David
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FAQ - HV feedthroughs: Theory, off the shelf parts, and fully custom designs

Post by Liam David »

IN PROGRESS

Quite some time ago I promised to write an FAQ on feedthrough design. This report summarizes things I've learned from constructing and using a variety of feedthroughs and will hopefully help others make informed engineering decisions.

The Fundamentals

Getting a high voltage into a grounded metal vacuum chamber is not trivial. Applying voltages to conductors sets up charge differentials that fundamentally do not want to exist. These charges (electrons) will jump out of surfaces, travel through vacuum, move along surfaces, punch through dielectrics (insulators), or otherwise leak away to neutralize. Leakage currents less than, say, a few tens of microamps are usually not a problem. Anything more than that, and you're getting dangerously close to breakdown or arcing conditions.

Underlying Physics

Before stating the fundamental engineering design criteria, I'll give an overview of the physics effects driving them. I would encourage those unfamiliar with these concepts to give this section at least a cursory read. I'm going to assume as prior knowledge on e.g. dielectric strength of an insulator, or that you should make the air-side of the feedthrough sufficiently long to avoid arcs.

Pesky Electrons
Within the realm of standard fusors, there are three processes that generate these free electrons, namely thermionic emission, field emission, and ion-induced secondary electrons.

In a metal, valence electrons are generally free to move and thus form a spatially homogenous electron "sea". The energies of these electrons are not all the same nor are they described by e.g. a Maxwell-Boltzmann distribution like particles in a gas. Instead, due to the Pauli exclusion principle, they "stack" in energy up to the Fermi energy. It's important to stress that this is an energy distribution at every point in the metal, rather than a distribution of electrons within the metal.

One might think that electrons can just "fall off" the top of the energy stack and move out of the metal, but this is prevented by the material's work function. It's essentially an additional potential energy barrier above the Fermi energy. As an electron attempts to escape the metal, it induces a positive charge in the metal which pulls it back.

Thermionic emission can overcome this work function using temperature. At temperatures greater than absolute zero, you find some small fraction of electrons with energies greater than the Fermi energy. Some are just high enough to fall over the work function barrier and escape the metal. The higher the temperature, the higher the thermionic current (which is why fusors with hot cathodes are easier to ignite). In the plot below, the shaded yellow section represents the electrons that can escape since they have energies above the Fermi energy (E_f) + the work function (W).

Fermi Energy.png

Field emission can overcome this work function by sort-of ignoring it. Even when you don't, but especially when you do apply a negative voltage to the metal (thus generating an electric field), electrons can quantum tunnel through the work function barrier and out into space! This is analogous (but temporally reversed) to how fusion is greatly enhanced by quantum tunneling through the Coulomb barrier. An electric field on the metal's surface lowers the height of the work function barrier and increases tunneling. A field also enhances thermionic emission in this way; by lowering the height of the work function barrier, lower-energy temperature-excited electrons can climb out.

Here phi=W is the work function.
Here phi=W is the work function.

Secondary electrons are generated when an ion or electron plows into the metal, giving fewer than one (probabilistically) or many electrons enough energy to pass over the work function barrier. There are many intricacies that I will not discuss here. Secondary electrons do help sustain a glow discharge fusor, but you do not want them in your feedthrough.

Where do electrons that are liberated from metals go? To ground either directly, or after collisions with background gas, or after discharging from an insulator that's in the way. Within a feedthrough, you want any electrons that are liberated to go directly to ground.

Any multiplication leads to a multiplication in leakage current. Thus, generally speakings, lower-pressure systems will be more resistant to breakdown, which agrees with our understanding of Paschen's law.

Charged insulators present problems because they can lead to sparking and potentially insulator destruction. Electrons will also move along the surfaces of insulators, especially those with large electric fields tangent to the surfaces, which can cause breakdown. If there are too many hydrocarbons on the insulator surface or in the residual vacuum gas, this can create conductive carbon deposits. If too many electrons build up, they can puncture through the insulator entirely, at best rendering it useful only for significantly lower voltages and potentially ruining it entirely.

Triple Junctions
A consequence of introducing insulators into electrostatic systems is that they can enhance fields, especially at boundaries. I don't have a good intuitive explanation for why (I'm not sure one exists), but it is a consequence of the laws of electrostatics that boundaries between 3 or more materials of different electric permittivity (e.g. a conductor, insulator, and vacuum) generate very high electric fields. This is especially true when the low permittivity material (vacuum, in this example) has an acute angle at the boundary. Theoretically, the field diverges to infinity at a perfect boundary, but physical boundaries aren't quite that bad. If such a junction exists on a cathode, it can cause high local electron emission, insulator charging, and all kinds of breakdown issues. Avoid them like the plague.

Consider this example from a non-final version of my custom cube chamber. I have an alumina insulator sealing to an aluminum chamber using a Viton gasket.

triple junction.PNG

There are 3 junction types present: vacuum-Viton-alumina, vacuum-Viton-aluminum, vacuum-alumina-aluminum. All three enhance the field, shown by the magenta coloring. You may notice that some junctions lack a field enhancement. This is because I used some specific geometries in the chamber to shield them from the high voltage, which is something I will discuss below.

Thus: There are three fundamental design criteria in engineering a feedthrough system:
1. Minimizing surface electric fields.
2. Eliminating or shielding triple junctions.
3. Preventing electrons on the surfaces of dielectrics from building up and traveling.


How do we achieve this? The rest of this post will address points 1 and 2. I'll cover point 3 at a later date.

Feedthrough Engineering

First I'll discuss minimizing surface electric fields. At the highest level, you want your surfaces to be as smooth, clean, and cold as possible. Any tiny bumps, protrusions, scratches, etc., or crud on the surface, will locally enhance the electric field and hugely increase field emission. This applies to electrodes in air as well, hence the idea of spheres, toroids, and other forms of "field control". All feedthrough parts should be thoroughly cleaned, ideally with a strong detergent and an ultrasonic cleaner, although most times you can get away with just IPA or the like.

Unless there is some significant spark or arc event, and ignoring metal deposition on the ceramic, the withstand voltage of a feedthrough will increase with time. There are a variety of mechanisms contributing to this, but the primary is the heating and obliteration of small imperfections in the metal by field emission-induced sparks. If the sparks are too powerful, however, they blow craters in the metal and make things worse. This process is called conditioning and is performed by slowly ramping the voltage over a long time, potentially hours.

It will come as no surprise that physically larger feedthroughs can generally withstand higher voltages. This is because electric fields are measured in volts/meter, so for a given voltage, the electric field and thus all the unpleasantries discussed above are reduced for bigger components. Compare, for example, the common commercial 30kV and 100kV feedthroughs:

https://mpfpi.com/shop/power-feedthroug ... 0234-7-cf/
https://mpfpi.com/shop/power-feedthroug ... at-flange/

Illustrative Example
Here is a simple example showing how to optimize the diameter of a bare HV stalk, which should get across the idea of optimizing geometry for a given feedthrough size. Very few situations can be solved analytically, so after establishing some basic dimensions, the best tool for designing a feedthrough system is FEMM or another capable electrostatic solver. FEMM, for example, has 2D and axisymmetric solvers and can handle dielectrics.

Most of us have to mount our feedthroughs on a 2.75" conflat or similarly-sized flange. Let's approximate the electric field on the HV stalk (which is our negative surface, and thus the surface of interest regarding electron emission) using a naive method. The internal radius is ~17.4mm and assuming a stalk diameter of ~1/4", we have a field at 50kV applied of E ~ 50kV/(14mm) ~ 3.6e6 V/m. I'll state here that the experimental threshold field for which problems can really start is ~ 1.5e7 V/m, so we seem to have a safety factor of ~ 5. However, this calculation is only valid for infinite parallel plates. Any other geometry with a minimum electrode separation equal to this parallel-plate distance will have a stronger field. In other words, any electrode curvature leads to field enhancement somewhere. I'm not sure there's a name for it, but I call it geometric enhancement.

The actual electric field on the stalk surface is given by the expression
field_eq.PNG
field_eq.PNG (3.49 KiB) Viewed 5699 times
where V is the absolute voltage, and a and b are the stalk and tube radii. This results fundamentally from the cylindrical geometry. We obtain an actual field of ~ 9.3e6 V/m, or nearly 3x higher!

We should also examine the equation this equation as a function of a. We plot it and find that there's a minimum!

Minimum normalized to 1
Minimum normalized to 1

This minimum occurs at dE/da = 0, from which we obtain that a = b/e, where e = 2.718... is Euler's number. Thus, for a 17.4 mm radius tube, the optimal stalk diameter is 2*6.4mm = 12.8mm. The field is 7.8e6 V/m.

The overall shape can be intuited by considering the trade-off between the stalk curvature and gap size. A bigger gap reduces the field until the stalk becomes too small and the high curvature increases the field again.

We also note that the function is relatively flat around the minimum, meaning that even significant deviations from this ideal radius don't hurt too much. This explains why we can often get away with using a variety of stalk diameters.

This equation is also all that's needed to demonstrate why a 2-stage feedthrough can potentially support higher voltages than a single-stage feedthrough. I'll leave that as an exercise to the reader.

If we introduce some dielectric like an alumina insulator, the situation becomes more complicated. Within the scope of this post, the only reasons one would add an insulator to the stalk, or between the stalk and chamber, are to help prevent plasma formation and to stop the generation of secondary electrons. This is most helpful at high pressures and low voltages where a glow discharge can form on the stalk. Electric fields from the stalk induce opposing dipoles in the insulator, pushing the field out of the insulator and increasing your surface fields. See the addendum for more details. For all reasonably well-made neutron-producing fusors, I would advise against an insulated stalk.

While I recommend leaving the HV stalk bare, it is still important to understand how a dielectric between the stalk and chamber modifies the fields. I'll discuss this in the next section.

Example Commercial Feedthrough
By far the most common feedthrough used on fusors is the standard 30kV model linked above. Keep it clean, throw a toroid on top, and use low-profile bolts, and it'll support up to 70kV, far more than most here will ever need. Immersed in oil, such a feedthrough could potentially support 90kV.

On the vacuum side, the stock stalk diameter is 2.4mm which gives a surface field of ~1.6e7 V/m, much larger than the ideal 7.8e6 V/m. We should increase the stalk diameter outside the ceramic bit that protrudes into the chamber to something near 12.8mm. When the stalk recedes into the ceramic, its optimum diameter will change since dielectrics expel fields. I'll spare you the derivation, but for a stalk radius a, tube radius b, and dielectric inner and outer radii r_1, r_2, the fields are:
field_eq_dielectric.PNG
field_eq_dielectric.PNG (9.42 KiB) Viewed 5693 times
where
F.PNG
We see that for epsilon = 1, corresponding to no insulator, the radii r_1 and r_2 become unimportant and we recover the equation from earlier (with the leading 1/a replaced by 1/r). Minimizing the stalk surface field gives the equation
amin.PNG
from which we recover a = b/e with the substitution epsilon = 1. Plugging in values for the 30kV feedthrough, and using epsilon = 10 for alumina, we obtain a diameter of 8.3mm. Technically, the inside of the ceramic tapers, and the tube diameter reduces to 1" to mate with the ceramic. I'll leave the math to the reader, but the result looks something like:

feedthrough.PNG

The tapers between the different diameters require simulations to optimize and mechanically there is nothing keeping the stalk centered, but the basic idea is there. I'm not suggesting that everyone's fusor should use this stalk design. It may be a starting point for someone wishing to really push their feedthrough far beyond its rating and is just meant to give the equations and illustrate the process.

Making examples of my previous feedthroughs
Most of what I said in viewtopic.php?p=85822 remains valid.







Addendum
To show how dielectrics expel fields, consider the following plot:

comparison.png

The vertical lines denote the stalk radius, dielectric inner and outer radii, and tube radius. The blue curve is the electric field with the dielectric absent. The red curve is the electric field with the dielectric present. We see that everywhere outside the insulator, including on the stalk surface, the red curve is higher. Inside the insulator, the field is greatly reduced.
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