Scott,
Capacitors made from G-10 material (dielectric constant 4.8 and 1/16"
thick) have about 17pF per square inch of capacitance and a breakdown
voltage of about 50kV not including edge effects. The breakdown voltage
at the edges is considerably less and one should allow about 1/8"/kV of
dielectric to extend out past the plate conductors (copper clad). This
makes it hard to trim pc board capacitors for high voltage applications.
Instead of just trimming the board edge to reduce capacitance, one must
peal up or mill away the copper clad while leaving the dielectric
intact.
The dissipation factor for G-10 is about 2% at 1MHz, which means it's
got a Q of about 50 at 1MHz. Much lower loss capacitors can be made of
teflon board material which commonly have dielectric losses of only
about 0.1% (Q=1000). Teflon is more expensive and somewhat harder to
find but it's often available at surplus electronic parts dealers for a
song. The formula for calculating capacitance of other board materials
is:
C = 0.2249 * k * A / d (picofarads/in.)
where:
k is the dielectric constant of the material
A is the plate area of the capacitor (in)
d is the board thickness (in)
One other note. The equivalent circuit of this capacitor can be modeled
as a series RLC combination. The magnitude of R is equal to the
capacitive reactance at a given frequency multiplied by the dissipation
factor. So if Xc is -j100 ohms, then R=2 ohms if the dissipation factor
is 2%. Also, the inductance is roughly about 30nH per inch of capacitor
radius it turns out. If lower series inductance is needed, one can stack
boards (connecting alternating layers at the edges) and thereby reduce
area (and inductance) for a given capacitance.
Hope this helps.
Dave
K0QE
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