0.1pF three-terminal capacitor

According to Kibble and Rayner's "Coaxial AC Bridges" ISBN-0-85274-389-0) this is a Zickner capacitor.

Googling on "Zickner capacitor formula" threw up just one reference

which is to Springer Verlags "Landolt-Boernstein" which in turn gives two references to Zickner G., both in Arch. Elektrotech (Berlin), one to volume 38 page 1 in 1944, and the other to a joint paper with H.Hoyer in 1957, from page 271.

I know rthat you've got access to academic liibraries, so you should be able to take it from there - I could follow it up myself, but it wouldn't be all that easy.

---------- Bill Sloman, Nijmegen

Such a device is used in the Finnegan ITDS mass spectrometer to measure the high voltage. In that device, the output transformer produces a 1 MHz crystal controlled, amplitude modulated 0-30,000 V PP drive voltage.

The auto-transformer is nominally an 8" coil of aluminum wire with the windings spaced 1/2" apart about 12" - 18" tall. All is enclosed in an aluminum cube with a coax input, a high voltage ceramic feedthrough for output, and a nominally 1" diameter hole that has a 1" square plate set outside of it on 1" standoffs. Some type of "copper ball" about 3" in diameter can be moved up and down in the coil to "tune" it.

The connector to this plate matched the schematic to a capacitor in the feedback section of the high voltage control, but it took me quite some time to grok what was happening.

W> Is there an easy way to calculate the size of the hole necessary to

I doubt there is an easy way, short of having a finite element analysis program handy. I assume that air is the dielectric, throughout. But things like spacing between input and output plates and grounded layer (I also assume that the spacing between top plate and ground will be a lot bigger than between bottom plate and ground) and sharpness of hole edges will come into play. I think you will find that the result is a lot more predictable (tolerant of small assumption errors) if the sharp edges of the hole are replaced with 45 degree chamfers of about 1/3 of plate thickness, or best of all, a

This would make quite a calculus boundary problem for Mathcad to work on.

Don't know if this is applicable, but

drive #############################################

################# ######## ################## gnd to opamp gnd

is the usual way to do it, as it minimizes fringing.

The idea is to keep the gap in the lower electrode, the thing separating gnd from the opamp node, very small.

John

"Winfield Hill" schreef in bericht news: snipped-for-privacy@drn.newsguy.com...

Is that a homework question?

Hello Winfield,

That looks like some major Maxwell job. But why not take the unscientific road? Drill several times into double sided copper clad with a flat bit but don't penetrate the lower side layer. Use bits of various diameter. Then place the electrode up top and find the closest to 0.1pF, adjust height of top plate until it is just right.

You could also drill just one hole and have at it with a Dremel until

Regards, Joerg

Agilent HSFF or other 3D Field Solver would make short work of the calculation. Assuming you don't have a 3D Field $olver (or you wouldn't have asked the question) and you don't want to increase the hole size until you reach 0.1pf, you could try the freeware FastCap software from:

supplemental user manual (main manual included in program download):

A search will turn up considerable info on FastCap; I haven't used it but have heard it is a pain to use compared to commercial field solvers and that it provides quite accurate results if you can figure it out.

As starting point I'd expect the coupling capacitor to be area of the hole divided by the distance of the plates. The intermediate GND layer just increases the current in the top electrode. If the hole diameter is in the order of the distance I'd expect boundary effects, they may vanish when the diameter is hundred times the distance.

For a more accurate result, a field solver should be used.

Rene

Yes, but I don't have that choice.

OK, then you need a field solver or else a good scale model. You should have access to a good field solver, what with the MIT connections and all. I've used ATLC for stuff like this; it's a pain to use, but it's free. I think it may do 3-t caps now.

John

Mr Larkin's scheme would seem to be far easier to calculate needed size for the pickup. But i am guessing you need the given configuration to help protect the poor opamp from possible arcs... I think Don Lancaster has a scheme that uses PostScript as a field solver that might be useful.

In any case, start with Mr Larkin's scheme and determine the approximate size of the pickup patch (keep it round for symmetry and ease of calculation). Now keep the hole in the ground plane the same size and move the pickup patch/plate toward the preferred position; assume the ground plane is of zero thickness for first cut. I am guessing that this approach may allow either a simpler solution or easier way for a guesstimate. Certainly, the hole must get larger as the pickup moves away; perhaps a x^-2 (field) effect. I am guessing hole thickness is almost a second-order effect, as the hole will probably wind up being more than twice the plate separation.

Right. We have an excellent FEA program, which gives nice answers to this problem; but I'm looking for ways deal with it analytically. Hopefully there's a simple approach. :>)

My guess is that there is no analytical, as in exact closed-form math solution, way to do this, at least that human intelligence could provide. Certainly not simple. The boundary conditions are just too nasty, given that the hole won't be large compared to the plate thickness given your stated dims.

Incidentally, what hole size actually gets you 0.1 pF?

John

And 83,000 years of leisure time.

John

The FEA says 0.7 inches. Whoa! Too big by far. It must be time for plan B. Well, heck, what was plan B anyway?

If you had two parallel plates each 0.7" dia, 0.25" apart, nothing between, ignoring fringing, c would be about 0.35 pf. So the shielding of the center plate reduces c by about 3.5:1. Given the geometry, that doesn't sound unreasonable.

OK, we all want to know, what's this for?

John

Hello John,

There should be, but it'll go back to the roots. Meaning the Maxwell equations, two large pots of coffee and a stash of aspirins.

Regards, Joerg

How do you measure a 3-terminal capacitance with an oscillator?

John

Hello Winfield,

Just shoot a hole in there and measure?

Regards, Joerg