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0.1pF three-terminal capacitor

T

Terry Given

Hi Joerg,
Hello Terry,



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

So: Whiskey in honour of Maxwell, or Vodka for Schelkunoff?

I suspect stabbing oneself with a pencil will be far more fun than this
analysis.


Cheers
Terry
 
J

John Larkin

John Larkin wrote...

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
 
J

Joerg

Hello Terry,
So: Whiskey in honour of Maxwell, or Vodka for Schelkunoff?

I'd go for a screwdriver. It contains oranges (West) and Vodka (East).
I suspect stabbing oneself with a pencil will be far more fun than this
analysis.

That's why I'd simply measure it with a resonant circuit, checking the
oscillator frequency with and without the little cap. Of course, in
academic circles that might be considered cheating ;-)

Regards, Joerg
 
T

Terry Given

Joerg said:
Hello Terry,



I'd go for a screwdriver. It contains oranges (West) and Vodka (East).

A synergistic harmony. May lead to dancing badly though.
That's why I'd simply measure it with a resonant circuit, checking the
oscillator frequency with and without the little cap. Of course, in
academic circles that might be considered cheating ;-)

Regards, Joerg

I'd go for a 100:1 scale model using tinfoil, cardboard and glue. Bugger
all use analytically though.

Seriously, I'd do a bunch of parametric sweeps with FEA, and curve-fit.
If a curve-fit is good enough for Grover, its good enough for me :)

I am curious why Win needs an analytic approach though - I'm sure its
not just masochism. I guess he needs to build a variety of different
ones. Still, if anyone here is up to the task, its Win.

Cheers
Terry
 
J

Joerg

Hello Terry,
I am curious why Win needs an analytic approach though - I'm sure its
not just masochism. I guess he needs to build a variety of different
ones. ...

Maybe to torment his students a little? We were bombarded with Maxwell
for two semesters until it came out of our ears.
... Still, if anyone here is up to the task, its Win.

That's for sure. Normally you'd have to have a long beard for that
though and IIRC he had shaved that off.

Regards, Joerg
 
T

Terry Given

Joerg said:
Hello Terry,



Maybe to torment his students a little? We were bombarded with Maxwell
for two semesters until it came out of our ears.

ditto. I especially liked analysing the 2nd order butterworth filter
using maxwells equations. Talk about cracking a nut with a nuke. But
afterwards it proved a great example of just why we simplify whenever
possible. Yay for electrically small circuits.

That guy would happily chalk up 5-deep nested integrals, from memory.
Amazingly smart, yet paid surprisingly little.
That's for sure. Normally you'd have to have a long beard for that
though and IIRC he had shaved that off.

LOL :)
Regards, Joerg

Cheers
Terry
 
J

John Larkin

Hello Terry,


I'd go for a screwdriver. It contains oranges (West) and Vodka (East).


That's why I'd simply measure it with a resonant circuit, checking the
oscillator frequency with and without the little cap. Of course, in
academic circles that might be considered cheating ;-)

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

John
 
R

Robert Baer

John said:
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
I beg to differ; the hole will be at least 2 times the thicknes; i am
guessing ruoghly 3-4 times the thickness.
 
R

Robert Baer

Winfield said:
John Larkin wrote...



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?
That is not too different than my wild guess given earlier.
 
T

Tony Williams

Winfield Hill said:
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?

Plan B is to see if you can move the mechanical design to
the classic guarded ring construction, which is calculable.

___________|__________
|______________________|
__________ _ ___ _ __________
0v|__________||_||_|_||_||__________|0v
/ |
/ |___Connection to SJ(at 0v)
/
Insulating ring

Might even be a pcb layout......
 
F

Fred Bloggs

Winfield said:
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. :>)

Does your FEA plot show the electric flux lines emanating from the top
plate above the hole passing through the hole, and then splitting, with
some ending on the bottom of the middle plate and others, closer to hole
centerline, terminating on the bottom plate? This will be the only type
of field that would yield zero work for a vertical path integration of
the E field from bottom plate to middle plate, and if that is the case,
then the equivalent "electrostatic hole" should be smaller than the
physical hole in so far as using a simple parallel plate capacitance
formulation. I'll look for an analytical solution but it will most
likely have to be an infinite series of hyperbolic sines and cosines...I
don't see why the bottom plate cannot be reduced to a sheet charge
density so that the coupling capacitance between it and the top plate
can be defined as the ratio of the integral of bottom plate total sheet
charge to V.
 
W

Winfield Hill

Tony Williams wrote...
Plan B is to see if you can move the mechanical design to
the classic guarded ring construction, which is calculable.

___________|__________
|______________________|
__________ _ ___ _ __________
0v|__________||_||_|_||_||__________|0v
/ |
/ |___Connection to SJ(at 0v)
/
Insulating ring

Might even be a pcb layout......

Not a pcb in my case, but yes that's a good plan B. If necessary
I could project the 0.1pF electrode plug into my chamber, closer
to the pickup electrode:

.. insulating signal
.. ring | outside world
.. 0V __________ \_ _|_ _ __________ _
.. XXXXXX || || || || 0V shield,
.. XXX_________||_|| ||_||_________ _ 0.125 thick
.. |___|
.. __________________________________ _ 0.125 space
.. XXXXXX
.. XXXX______________________________ _ 0.125 ring
.. / |
.. / |___Connection to SJ(at 0v)
.. pickup electrode

Apparently at these scales fringing-fields have a *huge* effect.
Sloggett, et.al., J Physics A, 19, p2725 (1986), has an accurate
formula with fringing-field correction terms (see a.b.s.e.). **

Using the terms in their equation 36, the correction factor is a
huge 3.466 times! for disks spaced with a separation of half the
disc diameter (considered a fairly-distant separation, enhancing
the fringing-field effects). Wow! Actually, I suspect a typo in
the equation; so perhaps the correct ratio is 2.447 times. Hmm.

At any rate these big numbers raise my eyebrows, and I'll try to
check them out with some bench measurements later today, and Mike
here at the Institute will probably look at them with his FEA.

Bartlett & Corle, J Physics A, 18, p1337 (1985) and another fellow,
J Cooke (1958) have the brute-force numerical calculations that the
Sloggett team used to evaluate the various formulas in their paper.

** In a followup article Sloggett points out their formula is the
same as one published by Shaw in 1970, Phys Fluids, 13, p1935.

BTW, the 0.1pF capacitor at issue is the feedback element for a
sensitive five-axis capacitive position-measuring system.
 
W

Winfield Hill

Fred Bloggs wrote...
My guess would by HV voltage-current converter at frequency....

Good guess, but for those one needs 25kV of insulation. I use
old color-TV rectifier tubes, 1B3, etc., IIRC, for capacitance
dividers to measuring and servo-control the high rf voltage.
 
F

Fred Bartoli

Winfield Hill said:
John Larkin wrote...

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?

Maybe plug some dielectric material (Al203 or something) into the
intermediate plate hole.
 
J

Joerg

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

If I understood correctly Win needs to determine the capacitance between
drive electrode and opamp input, through the hole. So, why not rig up an
oscillator where a small part of the LC capacitance is just that
capacitance? Measure the frequency with and without the capacitance
connected. The oscillator would need to be a diff version or at least
the LC needs to be symmetrical to ground to accommodate the third
terminal (GND). Amidon #12 iron powder might be suitable to transfer the
LC from symmetrical to non symmetrical.

Or simply feed a signal from a known impedance into the drive electrode,
terminate the other side and see what comes through. I think an
oscillator would be more accurate.

Regards, Joerg
 
J

John Larkin

I beg to differ; the hole will be at least 2 times the thicknes; i am
guessing ruoghly 3-4 times the thickness.

As Win says, the hole will be about 0.7" diameter, a bit under 3x the
platc thickness.

To simplify the math enough to even consider an analytic solution, I'd
consider "large" to be 100:1 maybe. 3-4 isn't "large" in this context.

John
 
F

Fred Bloggs

Winfield said:
Apparently at these scales fringing-fields have a *huge* effect.
Sloggett, et.al., J Physics A, 19, p2725 (1986), has an accurate
formula with fringing-field correction terms (see a.b.s.e.). **

Adobe 6.0 keeps saying the file is damaged and cannot be repaired.
 
W

Winfield Hill

Fred Bloggs wrote...
Adobe 6.0 keeps saying the file is damaged and cannot be repaired.

Testing the posting just now - it downloads and displays fine.
But it's a two-part post, because of the binary file's length.
Perhaps your viewing program didn't successfully download the
second part to complete the file? Would you like me to repost
it to see if that fixes the problem?

I'm think you'd be interested in the article.
 
J

Jim Thompson

Adobe 6.0 keeps saying the file is damaged and cannot be repaired.

Doesn't that frost you? Adobe is getting as bad as M$oft for
generating version incompatibilities :-(

...Jim Thompson
 
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