Re: 0.1pF three-terminal capacitor

John Larkin wrote: 
On 10 Aug 2005 09:40:13 -0700, Winfield Hill
<Winfield_member@xxxxxxxxxxx> wrote:


John Larkin wrote...

Winfield Hill wrote:

John  Larkin wrote...

Winfield Hill wrote:


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.

Yes, but I don't have that choice.

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.  :>)




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.
.

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