I don't know if anyone other than you and I are interested in this so any
uninterested parties please hit your delete key and I apologize.
You are right Gary, there is no physical d.c. connection. There is a
physical rf connection however, as the bypass looks like a near short to rf .
There two different ways to look at the short: 1> there are many pF of
capacitance between the bypass plate and the body of the cavity. How does that
go? .224 X surface area of one plate divided by the spacing of the plate
further divided by velocity factor of the insulator. So, if you have a bypass
plate say 16 square inches, with a teflon insulator .005" thick, that
would be 16 X .224 / .005 /.68 = 1054 pF. At 1296 MHz this value of
capacitance = .116 ohms. Not exactly a short circuit but pretty close. This
method
gives an approximation but is not as precise as the following. The reason is
because this method always results in -j values and ignores the skin
effects.
2> if the width of the bypass plate is made 1/4 wavelength, and open
circuit, the plate acts as a quarter wave transformer and reflects a short from
the open end. Either way, the rf sees a low impedance path across the teflon
insulator. The width has to be exactly 90 degrees to reflect a short
circuit but again, it is still a pretty low number. Considering that the
spacing
between the bypass and the cavity body is so small, the characteristic
impedance of 1/4 wavelength "line" is low, on the order of a couple of ohms.
With an impedance so low, the bandwidth of the bypass is pretty wide. The
reactance of such a "line" is Z0 times tangent of line length in degrees.
Going to half frequency, the length becomes 45 degrees. The reactance then is
only 2 ohms(-j) and going to 1.5 X frequency, the reactance is also 2 ohms
but +j. At frequencies between these two, the reactance is lowered until it
reaches essentially zero at design frequency. There are inherent losses in
any transmission line and this prevents reaching an actual zero ohm
reflection. This is one of the reasons quarter wave open stubs don't have
infinite loss at the design frequency.(there also other reasons).
Hope this is not more than you wanted to know!
73,
Gerald K5GW
In a message dated 11/20/2009 4:56:21 P.M. Central Standard Time,
gpatterson53@hotmail.com writes:
Thanks so much for your reply. I have scoured the internet for info on
this and there is little to none.
I need to reread your response carefully but I still have a disconnect.
The n6ca cavity has the fingerstock anode plate (where the 2c39 plate
plugs in) , then a teflon sheet and then the outside surface of the cavity.
There is no physical connection between the 2c39 plate and the cavity and
therefore no "skin effect surface" to facilitate rf traveling as you
described.
does that make any sense. Thanks 73 Gary
____________________________________
From: TexasRF@aol.com
Date: Fri, 20 Nov 2009 13:41:37 -0500
Subject: Re: [Amps] trying to understand 1296 cavity amp
To: gpatterson53@hotmail.com; amps@contesting.com
Gary,
Without getting too deep into cavity theory, the rf is generated between
the plate and grid in that tube. The rf flows on the cavity inner surfaces
only. The rf will not pass through any of the metal parts.
Following this reasoning, any rf present at the plate flows from the tube
contact ring into the plate bypass fingerstock and out and on to the inner
surface of the bypass plate. At this point, the rf sees a very low impedance
and exits the bypass plate and on to the inner surface of the cavity upper
wall. The rf continues on to the outer wall of the cavity and sees a
short circuit there. The short circuit causes a phase shift of 180 degrees and
sends the rf back toward the tube in reverse order from above.
When every thing is in resonance, the returning energy is in phase at the
tube plate connection.
The same action occurs on the lower side of the cavity from the grid to
the cavity wall except there is no bypass capacitor plate in the rf flow path.
There is heavy coupling between the upper and lower cavity walls as they
are very close in terms of wavelengths.
The concept of ground essentially disappears inside the cavity. There is
rf everywhere inside and ground is meaningless. Also, the action performs a
balun function perfectly so that the output coax, which does have a ground
reference, can couple to the rf inside the cavity.
There is a standing wave inside the cavity due to the short at the outer
wall this is nearly infinite. The loss in the cavity is the limiting factor
in this vswr. The inherent high Q of cavities is related to this high vswr.
Pretty cool huh?
73,
Gerald K5GW
In a message dated 11/20/2009 9:48:31 A.M. Central Standard Time,
gpatterson53@hotmail.com writes:
I have a n6ca 2c39 1296 cavity amp and have trouble understanding how it
works. The 2c39 plate has a large teflon sheet capacitor to ground. I dont
understand how any rf gets into the cavity when the plate is dead shorted
with this capacitor. any help?
gary
w4af
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