bksaif12...
Thanks for accepting my response as an answer. I was an
Electronics Technician with a specialty in radar in the
Navy.
I'll repost it here in the Answer box for the sake of future
readers, and then respond to your request for clarification.
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I'm not sure of the significance of your formulae, since
you don't define 'a', but I assume they have something
to do with the area of the plates?
Without using formulae, but simply perception and logic,
it's not hard to see that the area of conjunction of the
two sets of plates is about ten times greater when the
plates are fully meshed than when they are unmeshed.
This is the source of the greater capacitance.
Keep in mind that the image is only an illustration, and
does not necessarily represent a true picture of what
the plates would look like when meshed at the minimum
(50pf) capacitance. In reality, they might not be 1/3rd
of the way meshed, as shown in the image, but, perhaps,
1/10th of the way meshed. Fully meshing them increases
the conjunction by a factor of 10, since they are then
in conjunction over 10 times the surface area of the
plates, vs 1/10th of that surface area at the low end.
Proving this with mathematics would require a complex
set of equations based on the exact dimensions of the
actual plates, but, again, it's pretty obvious simply
on a visual basis.
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As for designing a variable capacitor that would provide
an increase of 100-fold, the approach would be similar,
I would think. I'm not an electronics engineer, but my
approach, using the logic above, would be that we need
the minimum capacitance to be 1/100th of the capacitance
when the plates are in full conjunction, and we need that
capacitance to be in a usable range.
So if, e.g., we want the maximum capacitance to be 5000uf,
and the minimum to be 50uf, we need to design the plates
such that they produce 5000uf capacitance when they are
fully meshed. This might involve using a double gang
approach, where there are two sets of plates (or more)
on the shaft. This creates more plate area and therefore
more capacitance. We could also simply make the plates
larger in size to increase the area, or we could do both,
if necessary.
We can also create less distance between the plates, which
will increase the value of the capacitance. The limit to
this will depend on the voltage we're using, as we don't
want sparking between the plates, and a larger voltage
could produce this.
Once we have the maximum capacitance worked out, then we
need to make the capacitor adjustable in a way that is
usable. The gear ratio in the illustration you provided
is too direct to be usable, and would cause large changes
too quickly to make an adjustment down to 50uf. So we'd
need to use a gear ratio that allowed for much finer
adjustment of the value.
Theoretically, even the capacitor in the illustration you
provided could decrease to 1/100th of the maximum value,
but, as illustrated, it appears to have some mechanical
"stops" in place, which keep it from decreasing below
about 1/3rd of the full plate conjunction - plus the
gear ratio is unusable.
Changing the mechanical stops so the minimal conjunction
of the plates produces a conjunction equal to 1/100th of
their total surface area, and changing the gears to allow
for a fine-tuning of the incremental values should do it.
A good discussion of the relationship between capacitance,
the distance between the plates, and the area of the plates
can be found on this page, from James Maxwell's page at the
Physics Department at the University of California at Davis:
http://maxwell.ucdavis.edu/~electro/dc_circuits/capacitance.html
Please do not rate this answer until you are satisfied that
the answer cannot be improved upon by way of a dialog
established through the "Request for Clarification" process.
sublime1-ga
Searches done, via Google:
capacitance "distance between the plates"
://www.google.com/search?q=capacitance+%22distance+between+the+plates%22 |
