All of the questions, I hope to answer with this
description. I have a
brand new National R-175A, as the plate choke, feeding into
a pair of 500 pf
doorknob caps. A 300 pf vacuum variable tune cap into the
B&W 852 tank
coil, then a 1300 pf vacuum variable with a padding cap to
get up to 1800 pf
on 80 meters. I have checked the tank coil using two
impedance bridges,
getting 6.9 and 7.1 uH against a B&W spec of 7.0 uH for the
80 meter
inductance. All was checked for shorts to ground. Each cap
was also
checked, and they can be varied from minimum to maximum
capacitance without
any shorts or other negative items being detected. >>
Chuck,
I will have 3,150 VDC no load, coming from a 5 Kw
distribution transformer
that was unwound down to 2150 VAC -- lots of fun. (That's
another story) I
am figuring about 2900 VDC under load, at 800 ma. So the
resulting plate
impedance, using the formula Rp = Ep/2*Ip gives around 1800
ohms. The
specific B&W 852 tank coil was selected, since it was
designed to work with
a low voltage high current system, which I believe I have.
Do the
calculations yourself and you will find that this results in
a design Q of
12, the good ol' magic number. I fully agree with the
observations made by
one of the poster, this tank coil would be bad news with a
4500 VDC plate
voltage and low current. However, I feel it has a high
probability of
working , within it's intended work envelope. We will see
soon, I hope!>>
You may agree with that post, but if you do you are agreeing
with bad advice. Even if you are off by a factor of four in
loaded Q the end result, if you have reasonable size and Q
components, is negligable. As such "bad news" is a highly
dramatic exaggeration.
Even the Q number "12" so many people worship is a number
pulled from someone's rear pocket, and in actual fact the
formula given for both plate impedance and tank Q with that
loadline are approximations.
The minimum Q you can use and have the network behave like a
pi is some number slightly over the square root of the
impedance ratios. Say 1800/50=36 and sqrt 36=6. Any Q over
SIX would transform impedances fine in this case. Below that
and the tank would "quit working".
The high end of Q, before the tank "quits working", is the
Q where circulating currents get so high something in the
tank overheats, or where the bandwidth is so narrow you
can't tolerate the tank selectivity. Other than that there
will be almost no change you would ever notice.
The only real problem you face with the B&W, other than it
might not make some people have a warm fuzzy feeling and be
in their happy place, is you might have to use a little more
C than normal.
After all that nice looking work, I'd do the minimal amount
possible to check for something **in the resonant path**
that is totally screwed up. The blocking cap cannot possibly
be an issue at this point, and neither should the choke. As
a matter of fact the first thing I would have done is
isolate the tank to ONLY the vacuum variables and the
inductor and then checked it. At this stage anything else
you did without doing that is wasting time and energy.
Read what Jim Brown wrote. You are being given good advice
on building a complex sound proof room, but the real problem
is a pot hole in the path between those three components or
you are doing something wrong with the grid dip meter.
73 Tom
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