0.1pF three-terminal capacitor

[Winfield Hill]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

.. drive electrode ---,
.. |
.. #######################
.. ,-- hole
.. #################### ########################-- GND
..
.. ##########################################
.. | \
.. opamp SJ at GND potential
.. measure ac current

 

[Frank Bemelman]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

Is that a homework question?

 

[[email protected]]

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

http://www.google.com/search?q=cach...er+capacitor+formula&hl=en&lr=lang_nl|lang_en

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.

 

[Aubrey McIntosh, Ph.D.]

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.

 

[John Popelish]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

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
1/16th inch radius on both sides of the hole. It is probably a lot
more predictable (first or second order change in capacitance with
hole size and plate spacings, versus higher orders for smaller holes)
is the hole is at least as big as the thickness of the ground plate.

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

 

[Joerg]

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
0.1pF is reached.

Regards, Joerg

 

[Glen Walpert]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

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:

http://www.fastfieldsolvers.com/

supplemental user manual (main manual included in program download):
http://rleweb.mit.edu/vlsi/codes/FastCapsuppl.pdf

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.

 

[Rene Tschaggelar]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

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

 

[John Larkin]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

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

 

[Robert Baer]

Is there an easy way to calculate the size of the hole necessary to
create a 0.1pF capacitor in the drawing below? The middle grounded
plate with the hole is 1/8" thick, and the plates are 1/8" apart.

. drive electrode ---,
. |
. #######################
. ,-- hole
. #################### ########################-- GND
.
. ##########################################
. | \
. opamp SJ at GND potential
. measure ac current

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.

 

[Winfield Hill]

John Larkin wrote...

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.

Yes, but I don't have that choice.

 

[John Larkin]

John Larkin wrote...

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

 

[Winfield Hill]

John Larkin wrote...

OK, then you need a field solver or else a good scale model.

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. :>)

 

[John Larkin]

John Larkin wrote...

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

 

[Joerg]

Hello John,

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.

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

 

[Terry Given]

John Larkin wrote...



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. :>)

Do an n-tuple (I just love that word of different geometry runs with
FEA, and curve-fit

I think Joerg underestimates the required caffeine dosage for an
analytic solution

Can you use, by analogy, Snellings formula for fringing flux?
(wild-assed guess)


Cheers
Terry

 

[John Larkin]

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.

And 83,000 years of leisure time.

John

 

[Winfield Hill]

John Larkin wrote...

Incidentally, what hole size actually gets you 0.1 pF?

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?

 

[Joerg]

Hello Winfield,

... Well, heck, what was plan B anyway?

Just shoot a hole in there and measure?

Regards, Joerg

 

[Joerg]

Hello Terry,

I think Joerg underestimates the required caffeine dosage for an
analytic solution

I was just anticipating that when the frustration level has reached a
certain limit and reems of paper are full of integrals one would
gradually step over to booze ;-)

Regards, Joerg