Coaxial capacitor calculation (not coaxial cable as capacitors)
Duncan Clark
Does anyone know of how to design a coaxial capacitor i.e. two aluminium
tubes, one within the other with say polystyrene as dielectric for use
in gamma matches?. How does one calculate the capacitance if one knows
the diameters of the tubing and spacing? It's the standard formula for
flat plate capacitors but how does one calculate using hollow tubes?
Many thanks
Duncan
G4ELJ
--
A distraction is only a distraction if you pay attention to it.
Duncan Clark
GeneSys Ltd.
http://www.dnamp.com
Tel: +44(0)1252376288
FAX: +44(0)8701640382
Reg Edwards
COAXPAIR.exe from my website.
It supplies an exact analysis of performance of coaxial
structures. For other than air-spaced lines (which I
suggest you use) you will have to enter the value of the
insulant's velocity factor. For polythene VF = 0.665.
Please yourself about what impedance you terminate the line
with.
Amongst the many computed characteristics is the one you
need - picofarads per unit length for given dimensions.
Downloads in a few seconds. It is not zipped up. Can be
run immediately. User friendly.
--
******************************
Reg, G4FGQ For free software go to:-
http://www.btinternet.com/~g4fgq.regp
******************************
Duncan Clark <Dun...@nospam.demon.co.uk> wrote in article
<BZY4nCAR...@genesys.demon.co.uk>...
mun...@ibm.net
anyway. And, my spell checker don't like it even though it is a real word.
--
Crazy George
Roy Lewallen
C = 2*pi*epsilon*L / ln(r1/r2)
where
epsilon = erel * epsilon0
erel = dielectric constant (relative permittivity) of the dielectric
(2.56 for polystyrene)
epsilon0 = permittivity of free space = 8.854 * 10^-12 coul^2/nt-m^2
L = length of the conductors
r1 = outer radius of inner conductor
r2 = inner radius of outer conductor
If L, r1, and r2 are in meters, you can use the value of epsilon0
shown above, and the answer will be in farads. This formula doesn't
account for fringing at the ends, so a short, fat capacitor might have
a noticeably higher C than this predicts. The capacitor also must be
short in terms of wavelength for the apparent capacitance to be what
the formula predicts.
As Reg so frequently says, you don't need a computer program for this.
Roy Lewallen, W7EL
Roy Lewallen
Roy Lewallen
mun...@ibm.net
out of date.
--
Tell Bill Gates. Its in one of his abortions.
--
Crazy George
Reg Edwards
pop up.
But you are quoting me out of context. And as far as I can
recall I have only said it once. Clearly it made some
impact. A computer program (or pocket calculator) is
needed when you cannot, or are too lazy, or do not have the
time to work it out in your head.
99.99 percent of radio amateurs do not have the time to
attend a physics course which includes inernational
standards and units of measurements. And many do not have
the confidence to enter such values as -
epsilon0 = permittivity of free space = 8.854 * 10^-12
coul^2/nt-m^2
in their pocket calculators, which may not even have
Logs-to-the-base e facilities. In the US you are still on
feet and inches.
And what on earth is coul^2/nt-m^2 anyway ?
By the time one has sorted out what one is actually doing
with your formula and gained sufficient confidence to use
it, a dedicated computer program will have sorted out
everyone's requirements for a gamma-match capacitor for the
next 100 years. And a vast range of other requirements too.
It will also allow experimentation on a 'what-if ?' basis.
The formula you give, of course, is perfectly correct and
perhaps the more scientifically minded-readers of this
newsgroup will record it their technical notebooks. Radio
amateurs will find it valuable to maintain a technical
notebook alongside their station log books. Mine has been
maintained for at least 20 years. Never regretted the
effort. Probably, Roy, you keep something similar
yourself.
--
******************************
Reg, G4FGQ For free software go to:-
http://www.btinternet.com/~g4fgq.regp
******************************
Roy Lewallen <w7...@teleport.com> wrote in article
<36DDBD61...@teleport.com>...
Reg Edwards
But here's a multiple-choice-answer question : -
An insulant is -
(A) A device for attaching artificial flies to
salmon-fishing hooks ?
(B) A birth control pill appropriate to Gays ?
(C) The generic name of a broad group of materials used
for constructing electrical insulators and supports ?
No prizes offered. Your spell-checker is about 100 years
out of date.
--
******************************
Reg, G4FGQ For free software go to:-
http://www.btinternet.com/~g4fgq.regp
******************************
mun...@ibm.net wrote in article
<36dd...@news1.us.ibm.net>...
at...@imap1.asu.edu
snip
: 99.99 percent of radio amateurs do not have the time to
: attend a physics course which includes inernational
: standards and units of measurements. And many do not have
: the confidence to enter such values as -
: epsilon0 = permittivity of free space = 8.854 * 10^-12
: coul^2/nt-m^2
: in their pocket calculators, which may not even have
: Logs-to-the-base e facilities. In the US you are still on
: feet and inches.
: And what on earth is coul^2/nt-m^2 anyway ?
snip
Reg,
This is why most physicists use Gaussian units instead of those funny
SI ones that have extra unnecessary quantities like Coulombs etc.
Capacitance is measured in centimeters in Gaussian units. You
immediately know that a capacitance hat made of a spherical copper
toilet float has a capacitance given by its radius in centimeters.
It's also clear then that if you double all the dimensions of
a capacitor, its capacitance doubles..
However, if you need to talk to people who use Farads, you have to know
the conversion factor which is about 9 centimeters = 10 picofarads.
73 Kevin w9...@ptolemy.la.asu.edu
Roy Lewallen
system, the unit of permittivity is coul^2/nt-m^2. With epsilon0 in
these units (that is, using the numerical value I gave) and the
capacitor length in meters, the equation I posted gives capacitance in
farads. For people who can't enter 8.854 * 10^-12 on a pocket
calculator, who have a computer but not a $10 scientific calculator,
or who can't handle dimensions in meters and capacitances in farads, a
computer program, or a formula simplified for the user's particular
range of values (e.g., length in inches, capacitance in pF) would
indeed be necessary.
(*) In the rationalized MKSA system, the fundamental units are meters,
kilograms, and seconds, and the derived electrical unit is the
ampere.) Other units in the MKSA system include familiar ones like the
coulomb, henry, farad, ohm, siemens, volt, joule, watt, and so forth,
so it's a convenient system for us to use. If your calculations are
done with these units, the results come out in these units. For
example, if you divide volts by amperes, the result is in ohms. There
are other useful systems which produce more convenient values for some
types of calculation, like the one Kevin mentions (and which I'm not
familiar with). One other common system is the CGS
(centimeter-gram-second) system, in which the derived units include
the erg and dyne. There's the CGS electrostatic system, in which the
fundamental units are the centimeter, gram, and second, but the
derived current unit is the statcoulomb. In this system the
permittivity of free space (epsilon0) is 1, so permittivity is equal
to dielectric constant. (In this system you'd do Ohm's law
calculations in statvolts and statamperes, and the result would be in
statohms. When converting from units in this sytem to rationalized
MKSA units, you'd find the factor of 8.854 * 10^-12, or its square or
square root, to keep showing up.) And there's the CGS electromagnetic
system (CGS + abampere) in which the permeability of free space, mu0,
is 1, so the permeability equals what we rationalized MKSA users call
the relative permeability. There's the unrationalized MKSA system, in
which a number of electrical units (like permeability) differ by 4*pi
from the rationalized system. And a handful of others.
Thanks to Reg and Kevin for providing this opportunity to do a little
explaining. I agree with Reg that most amateurs don't have the time
(or inclination) to take a physics course, so maybe this will help
raise the awareness about units of measurement. For those amateurs
still in high school, consider taking a physics course. It should
cover the MKS or MKSA system and what it means. If more Americans did
this, we might some day be able to finally get rid of this abominable
system we inherited from the English (but who have had the sense to
abandon -- sort of, anyway -- when I was in England a couple of years
ago, the rental car's speedometer was calibrated in MPH).
Roy Lewallen, W7EL
William E. Sabin
Roy Lewallen wrote:
> Most electronics engineers use rationalized MKSA units(*).
And there is not a damned thing wrong with it. Not only that, I have known a
lot of fine physicists over the years who have done the same thing. If
someone wants to use some other method it's OK, of course, but there is no
particular "cachet" connected to it.
Bill W0IYH