I've noticed that the various tunable inductors (and transformers) in a metal
can are much better shielded than I might have initially guessed, based upon
the premise that most all of the flux from the coil in concentrated inside the
coil which is 'visible' through the hole in the can (so that the thing can be
tuned!). In thinking about this, I've pretty much convinced myself that the
shielding works as well as it does because electrically it still "looks"
pretty much contiguous at the frequencies you're typically operating the coil
at (e.g., tens of MHz for a 1/4" hole), in a similar manner to how perforated
enclosures make an effective shield so long as the holes are small enough
(...and how you can even calculate the attenuation based on "waveguide below
cutoff" formulas).

Is that correct?

If so, it seems that if you tune the slug so far "out" that it protrudes from
the top of the can, you're probably starting to seriously degrade the
shielding... right?

Thanks,
---Joel Kolstad

Thanks for the responses, Tom and John -- it makes more sense now.

I do wonder how much of the shield could be removed before the shielding
effectiveness would drop significantly? I believe that some transformers are
shielded just by having a large metal band that covers perhaps 2/3 of the
transformer's area, right? ...and presumably they're still quite efficient.

---Joel

In a solenoidal, helically wound coil, most of the magnetic flux is aligned
with the center of the coil and flows along the axis of the coil.
Relatively little flux is outside the coil because of cancellation effects.
For a single turn, current in different parts of that turn tend to partially
cancel flux in regions that are outside that turn.

To see this, draw a circle of wire and show a current at one point flowing
in one direction (out of the page). At the corresponding point
diametrically opposite, that same current flows in the opposite direction
(into the page). Inside the turn the two magnetic fluxes that encircle the
wire add in-phase, but outside the turn these same fluxes cancel in opposite
phase. This effect is more pronounced in a coil of small diameter.

Coils of that kind are pretty well self-shielding for regions beyond the
*side* of the coil.

But if a copper disc is placed near the *end* of the coil a large
circulating current is induced around the edge of the disk and a flux is
created that opposes the flux inside the coil (Lenz's law) and a significant
reduction of the inductance of the coil can occur. This effect has been used
to "tune" coils.

The completely enclosed metal shield helps to create a closed path for
leakage current that makes the shield much more effective.

The shield around the coil can also reduce capacitive coupling to adjacent
circuitry. This is often important.

Bill W0IYH

"Joel Kolstad" wrote in message
...
I've noticed that the various tunable inductors (and transformers) in a
metal
can are much better shielded than I might have initially guessed, based
upon
the premise that most all of the flux from the coil in concentrated inside
the
coil which is 'visible' through the hole in the can (so that the thing can
be
tuned!). In thinking about this, I've pretty much convinced myself that
the
shielding works as well as it does because electrically it still "looks"
pretty much contiguous at the frequencies you're typically operating the
coil
at (e.g., tens of MHz for a 1/4" hole), in a similar manner to how
perforated
enclosures make an effective shield so long as the holes are small enough
(...and how you can even calculate the attenuation based on "waveguide
below
cutoff" formulas).

Is that correct?

If so, it seems that if you tune the slug so far "out" that it protrudes
from
the top of the can, you're probably starting to seriously degrade the
shielding... right?

Thanks,
---Joel Kolstad

"William E. Sabin" wrote in message
news:Roybe.17552$WI3.12208@attbi_s71...
In a solenoidal, helically wound coil, most of the magnetic flux is
aligned with the center of the coil and flows along the axis of the coil.
Relatively little flux is outside the coil because of cancellation
effects. For a single turn, current in different parts of that turn tend
to partially cancel flux in regions that are outside that turn.

To see this, draw a circle of wire and show a current at one point flowing
in one direction (out of the page). At the corresponding point
diametrically opposite, that same current flows in the opposite direction
(into the page). Inside the turn the two magnetic fluxes that encircle
the wire add in-phase, but outside the turn these same fluxes cancel in
opposite phase. This effect is more pronounced in a coil of small
diameter.

Coils of that kind are pretty well self-shielding for regions beyond the
*side* of the coil.

The flux at the center of the coil is highly concentrated. This flux then
leaves one end of the coil and returns to the other end via an external
path, just as it is supposed to do, but at a much smaller value of flux
*density* in the external path. This external flux can induce "eddy"
currents in a shield that can decrease coil inductance and Q, usually only a
small amount if the shield is not too "close".

This is in addition to the things mentioned before.

Bill W0IYH

The flux at the center of the coil is highly concentrated. This flux then
leaves one end of the coil and returns to the other end via an external
path, just as it is supposed to do, but at a much smaller value of flux
*density* in the external path. This external flux can induce "eddy"
currents in a shield that can decrease coil inductance and Q, usually only
a small amount if the shield is not too "close".
=====================
The above sparks-off my following question:

If an antenna matching unit (some prefer calling it a tuner) has air
/ceramic/polystyrene wound inductors ,to which extent will a metal cabinet
affect the Q of the said inductors ?
I have read somewhere that to maintain the best possible Q , the distance
between the inductor(s) and other metal parts should be not less than the
diameter of the inductor. Does this make sense ?

TIA for any response

Frank GM0CSZ / KN6WH

"Pipex News Server" wrote in message
...
The flux at the center of the coil is highly concentrated. This flux then
leaves one end of the coil and returns to the other end via an external
path, just as it is supposed to do, but at a much smaller value of flux
*density* in the external path. This external flux can induce "eddy"
currents in a shield that can decrease coil inductance and Q, usually
only a small amount if the shield is not too "close".
=====================
The above sparks-off my following question:

If an antenna matching unit (some prefer calling it a tuner) has air
/ceramic/polystyrene wound inductors ,to which extent will a metal
cabinet affect the Q of the said inductors ?
I have read somewhere that to maintain the best possible Q , the distance
between the inductor(s) and other metal parts should be not less than the
diameter of the inductor. Does this make sense ?

TIA for any response

Frank GM0CSZ / KN6WH

I prefer the term "tuner" because the unit very seldom actually "matches"
impedances. Instead, it "transforms" the impedance at the sending end of
the transmission line, coax or whatever, to the 50 ohms resistance that the
transmitter is usually (these days) designed for. The term "tuner" is common
usage and it's OK.

The coil in a CLC tee-type tuner can have a Q as high as 400, and stray
coupling to the metal cabinet or ground plane can easily cut the Q in half.
What effect that has depends on the load impedance of the antenna
feedpoint. If the load is highly reactive (high X, low R) the coil can get
quite hot.

I believe it is important that the open ends of the coil should be at least
one coil diameter away from any metal surface. The sides of the coil are
less critical, but the mechanical design should do a pretty reasonable job
of reducing that stray coupling to a low value also.

Stray capacitance from coil to ground or elsewhere can be a problem
sometimes.

I have been thinking about my previous inputs to this thread, and I am not
entirely satisfied with them. I will try to improve them sometime today.

Bill W0IYH

William E. Sabin wrote:
....

The coil in a CLC tee-type tuner can have a Q as high as 400, and stray
coupling to the metal cabinet or ground plane can easily cut the Q in half.
What effect that has depends on the load impedance of the antenna
feedpoint. If the load is highly reactive (high X, low R) the coil can get
quite hot.

I believe it is important that the open ends of the coil should be at least
one coil diameter away from any metal surface. The sides of the coil are
less critical, but the mechanical design should do a pretty reasonable job
of reducing that stray coupling to a low value also.

....
Hi Bill,

In practical tuner applications the unused part of the coil is usually
shorted. The reasoning is that it prevents the generation of high RF
voltages. Shorting part of the coil should ruin the Q quite a bit.
Any comment on this practice?

Thanks, Peter

You certainly have your mind strolling down paths I have wandered...
Yes, it does seems so--is that found to be true when carried into actual
construction practices--well, yes and no, but not to any extent which poses
a "real" problem...

Or, simply put, I no longer lose sleep over the fact my L-Tuner shorts turns
when tuning the coil...

Regards,
John

Some years ago I found that leaving the unused turns open-circuit caused
very serious interactions with the remaining part of the coil at certain
frequencies, and actually made the tuner untunable at those frequencies. I
was sure that the open turns and the capacitances involved with those turns
was resonating, causing "suckouts". If the unused turns were shorted, the
spurious resonances in the coil disappeared, a well as I could determine.

Assume the shorting bar is perfect (lossless). Then:

The equivalent circuit just for the coil alone is the used part with its
resistance, magnetically coupled to the shorted part with its resistance.
The coupling from the shorted part to the used part adds an inductive
reactance plus some resistance in series with the used part. The shorted
turns have nearly the same ratio of inductive reactance per inch of coil
length to resistance per inch of coil length as the rest of the coil, so the
Q should not be degraded by this coupling. In other words the net loss is
the same as that of the entire coil operating alone. The perfect short does
not add power loss into the coil.

Assume the shorting bar is not perfect (not lossless). Then:

Some additional power loss is added to the shorting bar and the Q should
decrease.

Bill W0IYH

"Peter Orban" wrote in message
...


William E. Sabin wrote:
...

The coil in a CLC tee-type tuner can have a Q as high as 400, and stray
coupling to the metal cabinet or ground plane can easily cut the Q in
half. What effect that has depends on the load impedance of the antenna
feedpoint. If the load is highly reactive (high X, low R) the coil can
get quite hot.

I believe it is important that the open ends of the coil should be at
least one coil diameter away from any metal surface. The sides of the
coil are less critical, but the mechanical design should do a pretty
reasonable job of reducing that stray coupling to a low value also.

...
Hi Bill,

In practical tuner applications the unused part of the coil is usually
shorted. The reasoning is that it prevents the generation of high RF
voltages. Shorting part of the coil should ruin the Q quite a bit.
Any comment on this practice?

Thanks, Peter

"William E. Sabin" wrote in message
news:9rSbe.21106$WI3.17494@attbi_s71...

Assume the shorting bar is perfect (lossless). Then:

The equivalent circuit just for the coil alone is the used part with its
resistance, magnetically coupled to the shorted part with its resistance.
The coupling from the shorted part to the used part adds an inductive
reactance plus some resistance in series with the used part. The shorted
turns have nearly the same ratio of inductive reactance per inch of coil
length to resistance per inch of coil length as the rest of the coil, so
the Q should not be degraded by this coupling. In other words the net
loss is the same as that of the entire coil operating alone. The perfect
short does not add power loss into the coil.

Assume the shorting bar is not perfect (not lossless). Then:

Some additional power loss is added to the shorting bar and the Q should
decrease.

Bill W0IYH

A followup to my previous input:

I connected an old roller coil with short circuited turns to my ancient
Boonton 260A Q meter.

Over most of the inductor range at 10 MHz the Q remained "fairly" constant.
At the low values of inductance the Q (about 250) dropped about 15%. I
suspect changes in shape factor of the unshorted turns are involved. The
winding pitch of the coil is greatly reduced at low L values to try to
maintain shape factor and improve Q. At low L values, inreasing the
frequency to 15 MHz helped to restore most of the Q.

The detailed analysis of all that is going on in the roller coil is a
time-consuming chore that I don't want to get into. I believe my previous
explanation is OK under ideal textbook conditions (Kraus "Electromagnetics"
4th edition page 235, Eq 2) but of course that is just an approximation to
the actual coil, as could be expected.

On to other things.

Bill W0IYH