Cecil Moore wrote:
On Oct 17, 6:42 am, Alejandro Lieber alejan...@Use-Author-Supplied-
Address.invalid wrote:
It appears to me that in the short circuited turns, a very big current
must be circulating, adding heat losses and lowering the Q of the circuit.

For a screwdriver antenna, the problem is solved by a conductive
sleeve over the outside of the shorted turns that keeps most of the RF
on the conductive sleeve instead of in the shorted turns of the coil.
--
73, Cecil, w5dxp.com

But, isn't that conductive sleeve itself a shorted turn? It's
conductive, coaxial with the rest of the inductor above the sleeve, so
the magnetic field certainly passes through it.

I think the real answer is that everything is a tradeoff.

On Oct 18, 11:04*am, Jim Lux wrote:
But, isn't that conductive sleeve itself a shorted turn?

Yes, but that particular low-loss shorted turn solves the problem that
needs solving. Nobody said it was a perfect solution.
--
73, Cecil, w5dxp.com

On Oct 18, 9:57*pm, Cecil Moore wrote:
On Oct 18, 11:04*am, Jim Lux wrote:

But, isn't that conductive sleeve itself a shorted turn?

Yes, but that particular low-loss shorted turn solves the problem that
needs solving. Nobody said it was a perfect solution.
--
73, Cecil, w5dxp.com

well, if its not the perfect solution then the problem is not
completely solved... so there must be a better solution to really
solve the problem.

On Oct 18, 5:13*pm, K1TTT wrote:
well, if its not the perfect solution then the problem is not
completely solved... so there must be a better solution to really
solve the problem.

The problem of transmission line losses can be solved with a perfect
lossless transmission line. Have you seen such or does reality force
us to settle for a reasonable non-perfect solution?
--
73, Cecil, w5dxp.com

K1TTT wrote:
On Oct 18, 9:57 pm, Cecil Moore wrote:
On Oct 18, 11:04 am, Jim Lux wrote:

But, isn't that conductive sleeve itself a shorted turn?
Yes, but that particular low-loss shorted turn solves the problem that
needs solving. Nobody said it was a perfect solution.
--
73, Cecil, w5dxp.com

well, if its not the perfect solution then the problem is not
completely solved... so there must be a better solution to really
solve the problem.

one could simply slit the tube (and finger stock at top). Consider that
the lower tube is serving two purposes: mechanical support and a movable
contact on the inductor.

On 10/18/10 12:04 PM, Jim Lux wrote:
Cecil Moore wrote:
On Oct 17, 6:42 am, Alejandro Lieber alejan...@Use-Author-Supplied-
Address.invalid wrote:
It appears to me that in the short circuited turns, a very big current
must be circulating, adding heat losses and lowering the Q of the
circuit.

For a screwdriver antenna, the problem is solved by a conductive
sleeve over the outside of the shorted turns that keeps most of the RF
on the conductive sleeve instead of in the shorted turns of the coil.
--
73, Cecil, w5dxp.com

But, isn't that conductive sleeve itself a shorted turn? It's
conductive, coaxial with the rest of the inductor above the sleeve, so
the magnetic field certainly passes through it.

Whic of course means that everything everywhere is a shorted turn. 8^)


I think the real answer is that everything is a tradeoff.

That's a pretty good answer.

Mike Coslo wrote:
On 10/18/10 12:04 PM, Jim Lux wrote:
Cecil Moore wrote:
On Oct 17, 6:42 am, Alejandro Lieber alejan...@Use-Author-Supplied-
Address.invalid wrote:
It appears to me that in the short circuited turns, a very big current
must be circulating, adding heat losses and lowering the Q of the
circuit.

For a screwdriver antenna, the problem is solved by a conductive
sleeve over the outside of the shorted turns that keeps most of the RF
on the conductive sleeve instead of in the shorted turns of the coil.
--
73, Cecil, w5dxp.com

But, isn't that conductive sleeve itself a shorted turn? It's
conductive, coaxial with the rest of the inductor above the sleeve, so
the magnetic field certainly passes through it.

Whic of course means that everything everywhere is a shorted turn. 8^)

Indeed, but some shorted turns intercept more flux than others.
Hopefully the field in the passenger seat is small enough that the power
dissipated in one's spouse's ring is fairly low.

On 17/10/10 22:42, Alejandro Lieber wrote:
Since I built my first 80meter/40meter 6aq5 + 6DQ6 transmitter with pi
output in 1972, when I want to vary the inductance of a coil in a
tunner, or loading coil in an antenna, I just short circuit some turns.
I see that this is the usual practice everywhere.

My question is why do we not just leave the turns open circuited instead
of short circuiting them.

It appears to me that in the short circuited turns, a very big current
must be circulating, adding heat losses and lowering the Q of the circuit.



Only fairly basic AC circuit theory is needed to analyse the effect of
the shorted turns.

If you have a air cored solenoid inductor of n turns, and short m turns
at one end, you can treat that as two independent inductors of n-m and m
turns with some flux coupling factor k. The mutual inductance can be
calculated, and a T equivalent of Ln Lm-n Rn Rm-n M elements constructed
and solved. k of course depends on coil construction and n and m, a
value can be determined by measurement of the reactance of the
combination. (You might be surprised at how low k is.)

Essentially, when the power lost in the shorted turns is low (due to the
combination of low k and low R), then the technique works fine.

We (hams) have some pretty inadequate word based explanations for some
of these kind of things when there are simple quantitative solutions at
hand.

An example is the traditional explanation of link coupling ratios. See
http://vk1od.net/tx/concept/lctr.htm for a quantitative explanation
using the same techniques as suggested above.

BTW, the solenoid inside an aluminium tube is a case of an inductor
surrounded by a single shorted turn... but if R in that turn is very
low, then little heat is generated in the tube.

We also sometimes use a movable shorted turn to adjust an inductor... a
brass or preferably silver plated brass slug was often used in VHF / UHF
tuned circuits.

Owen

On 18/10/10 07:24, Owen wrote:
....
Only fairly basic AC circuit theory is needed to analyse the effect of
the shorted turns.

If you have a air cored solenoid inductor of n turns, and short m turns
at one end, you can treat that as two independent inductors of n-m and m
turns with some flux coupling factor k. The mutual inductance can be
calculated, and a T equivalent of Ln Lm-n Rn Rm-n M elements constructed
and solved. k of course depends on coil construction and n and m, a
value can be determined by measurement of the reactance of the
combination. (You might be surprised at how low k is.)

To give a mental arithmetic example...

Assuming ideal inductors for the moment...

Suppose you had an air cored inductor, that when you measure the
inductance of the first half of the inductor (other terminal open) you
get 10µH. You now measure the whole inductor and get 30µH. We can
calculate that M=5µH.

Now forming a T equivalent of the inductor with one half shorted,
L=10+5+(5//(10+5))=18.75µH. Notably, the current in the s/c is
3.75/15=25% of the current in the other section, so losses are about 6%
of that in the other section... not usually a big issue.

That is, if I got the maths right on the fly!

Now, real inductors have some distributed capacitance which changes this
as you approach the inductor's self resonance frequency.

This works ok because mutual inductance is lowish. Increase k by
introducing a magnetic core material for instance, and the situation
changes.

You might see the technique applied to powdered iron core inductors. Not
necessarily a good idea, but it 'works' for some because they are low
permeability powders and flux leakage is high (ie k is not nearly 1).

Owen

Owen wrote in news:vtKuo.256$tk4.180
@viwinnwfe02.internal.bigpond.com:

Well, I didn't get the maths right, there was a sign error in the formula
below. Here is what it should have read.

Suppose you had an air cored inductor, that when you measure the
inductance of the first half of the inductor (other terminal open) you
get 10µH. You now measure the whole inductor and get 30µH. We can
calculate that M=5µH.

Now forming a T equivalent of the inductor with one half shorted,
L=10-5+(5//(10-5))=7.5µH. Notably, the current in the s/c is
50% of the current in the other section, so losses are about 25%
of that in the other section... not usually a big issue.

Of course, the situation depends on the tapping point, and is much worse
when you short just one turn... but we don't usually do that. The example
has a fairly high coupling factor k (for an air cored coil), and losses
are lower for lower k.

I am working on a note that expands on this.

Owen