Hi,
Can somebody walk me thru the calculation of an LC circuit where the
capacitor is tapped down on the coil? I see this often done for
bandspreading purposes.

Tnx,
Bill WX4A

Howdy,

I'm guessing that it can be solved like this...
Consider the autotransformer action of the tapped inductor.
Then divide the tap capacitor (C2) value by the square of the
turns ratio (N) before adding it to the primary capacitance
of the parallel tuned circuit (C1.)

F=2Pi*sqrt(L(C1+C2/N^2))

I found a rigorous solution in chapter 8 of Alternating
Current Circuits by K.Y. Tang but it's too messy to type.


73,
Grumpy



exray wrote in :

Hi,
Can somebody walk me thru the calculation of an LC circuit where the
capacitor is tapped down on the coil? I see this often done for
bandspreading purposes.

Tnx,
Bill WX4A

Thanks Grump...gonna save that for my files but I think I need to
clarify my question. One cap only...not one across the tank and another
on the tap although thats an interesting idea. Maybe this schematic
will help.

http://www.antiqueradios.org/gazette/pix/sw3-1sch.jpg

-Bill



Grumpy The Mule wrote:
Howdy,

I'm guessing that it can be solved like this...
Consider the autotransformer action of the tapped inductor.
Then divide the tap capacitor (C2) value by the square of the
turns ratio (N) before adding it to the primary capacitance
of the parallel tuned circuit (C1.)

F=2Pi*sqrt(L(C1+C2/N^2))

I found a rigorous solution in chapter 8 of Alternating
Current Circuits by K.Y. Tang but it's too messy to type.


73,
Grumpy



exray wrote in :

Hi,
Can somebody walk me thru the calculation of an LC circuit where the
capacitor is tapped down on the coil? I see this often done for
bandspreading purposes.

Tnx,
Bill WX4A

Howdy,

Oh, ok! I have seen a tapped down band spread capacitor
circuit so I assumed...

I wonder why they did that? Oh, looking at it the another
way, the grid impedance loading the parallel tuned circuit
is transformed by the turns ratio. Probably to provide
some voltage gain from that stage.

If the capacitance of the grid is negligible, then the
resonance is determined by the inductance from the
portion of the winding paralleled by the variable capacitor.

If you want to include the grid capacitance in parallel
with the variable capacitor, multiply it by the turns
ratio. Otherwise the solution is essentially the same.


73,
Grumpy



exray wrote in
:

Thanks Grump...gonna save that for my files but I think I need to
clarify my question. One cap only...not one across the tank and
another on the tap although thats an interesting idea. Maybe this
schematic will help.

http://www.antiqueradios.org/gazette/pix/sw3-1sch.jpg

Grumpy The Mule wrote:


If the capacitance of the grid is negligible, then the
resonance is determined by the inductance from the
portion of the winding paralleled by the variable capacitor.

So what about the 'top' portion of the coil being effectively in series
with the tank? Wouldn't that affect the tank values?

Tnx?

Howdy,


Maybe not. Imagine the top portion disconnected, it would
have very little effect. It's no more than a small capacitor
swamped out by the tuning capacitance.

I think if the grid capacitance times the square of the turns
ratio is small compared to the tuning capacitance, it can be
ignored. Because there are other strays that will require
a bit of pruning in any case.

The most significant effect would be when the tuning capacitor
is at its minimum value. So it's here that it may effect the
design of the inductor, requiring a bit less inductance for the
desired upper band limit compared to the other tuned winding.

There are no padding capacitors on that schematic. So I suspect
there are seperate tuning capacitors for each tuned winding on
the transformer. Because of this I believe it's not that
critical. You could measure the grid capacitance with the
tube mounted in its socket on the chassis. Then add that
capacitance times the turns ratio to the value of the tuning
capacitor when you calculate the necessary inductance value.

Or build the coil so you can stretch it (slip a few windings
further apart) a little once the thing is running. Then secure
them with a bit of wax.


73,
Grumpy



exray wrote in :

Grumpy The Mule wrote:


If the capacitance of the grid is negligible, then the
resonance is determined by the inductance from the
portion of the winding paralleled by the variable capacitor.

So what about the 'top' portion of the coil being effectively in series
with the tank? Wouldn't that affect the tank values?

Tnx?

On Nov 9, 8:36*am, Grumpy The Mule wrote:
Howdy,

I'm guessing that it can be solved like this...
Consider the autotransformer action of the tapped inductor.
Then divide the tap capacitor (C2) value by the square of the
turns ratio (N) before adding it to the primary capacitance
of the parallel tuned circuit (C1.)

F=2Pi*sqrt(L(C1+C2/N^2))

I found a rigorous solution in chapter 8 of Alternating
Current Circuits by K.Y. Tang but it's too messy to type.

73,
Grumpy

exray wrote :

Hi,
Can somebody walk me thru the calculation of an LC circuit where the
capacitor is tapped down on the coil? *I see this often done for
bandspreading purposes.

Tnx,
Bill WX4A



One thing to be a bit careful about is including the coupling between
the coil sections. Since I don't know how those particular coils were
designed, I can't say for sure, but in general the coupling between
pieces of the coil isn't as high as you might think.

You can make a good estimate for typical HF single-layer air-core
solenoid coils just using your favorite coil calculation. For
example, consider a coil that's one inch diameter, two inches long, 20
turns per inch, and tapped at the 30th turn (1.5 inches) up from the
bottom. Then the whole coil is about 16.32uH, the 1.5" part is about
11.54uH, and the top 0.5" is about 2.63uH. If the coupling were
perfect between the sections, I believe the inductance of the whole
would be about 11.54uH + 2.63uH + 2*sqrt(11.54*2.63)uH = 25.19uH. At
16.32uH for the whole coil, the implied coupling coefficient between
those two sections is only about 0.20, and you need to be careful to
not think of the tapped coil as simply a transformer with a 3:1 turns
ratio, with implied close coupling between the sections. (This also
illustrates why you can short out turns of a tank coil without totally
killing the net inductance...)

I trust if I've hosed the calculation too badly, someone will point
out the error of my ways. ;-)

Cheers,
Tom

Howdy,

That's an excellent point. I expect that the sections
of the coil in question will not have very good coupling.
I hadn't considered it! I calculate the same inductance
values.

Bah! In my previous posts the "(sqrtL*C)" should be "(1/sqrtL*C)"
Heh... with a pencil and paper the formula turned out right.
I noticed this when I calculated the resonance of the two
values 16.32uH and 11.54uH with an arbitrary value of 50pF
just to verify how I thought things should be.

So from this I gather that the extension of the coil
connected to the grid circuit will have even less
effect on the tuning than I expected.

The Easy Teenage NewYork method of solving this would be
to put the problem into SPICE with the estimated coupling
coefficent of 0.2.

I've never liked shorting turns to reduce an inductance.
Seems like an avoidable source of some losses no matter
how you slice it. But for some long coils, like antennas,
it works ok. I guess.



K7ITM wrote in
:

One thing to be a bit careful about is including the coupling between
the coil sections. Since I don't know how those particular coils were
designed, I can't say for sure, but in general the coupling between
pieces of the coil isn't as high as you might think.

You can make a good estimate for typical HF single-layer air-core
solenoid coils just using your favorite coil calculation. For
example, consider a coil that's one inch diameter, two inches long, 20
turns per inch, and tapped at the 30th turn (1.5 inches) up from the
bottom. Then the whole coil is about 16.32uH, the 1.5" part is about
11.54uH, and the top 0.5" is about 2.63uH. If the coupling were
perfect between the sections, I believe the inductance of the whole
would be about 11.54uH + 2.63uH + 2*sqrt(11.54*2.63)uH = 25.19uH. At
16.32uH for the whole coil, the implied coupling coefficient between
those two sections is only about 0.20, and you need to be careful to
not think of the tapped coil as simply a transformer with a 3:1 turns
ratio, with implied close coupling between the sections. (This also
illustrates why you can short out turns of a tank coil without totally
killing the net inductance...)

I trust if I've hosed the calculation too badly, someone will point
out the error of my ways. ;-)

Cheers,
Tom

Hmmm, looks to me like that tapped tank is the oscillator set up for
the super regeneration...

denny ... k8do

On Nov 10, 3:30*pm, Denny wrote:
Hmmm, looks to me like that tapped tank is the oscillator set up for
the super regeneration...

denny ... *k8do

And a second thought is see the same thing on the second stage, where
the top portion of that tank is in series between the tank and the
control grid... Redraw that circuit in your head so that the top
portion of coil is bent over at a right angle to the tank... I wonder
if the grid to cathode capacity is too high for comfort and the
series section of coil is for taming that... We do that for
amplifiers on 10 meters where the plate capacity is too high for
comfortable tuning... A small series coil between the plate and the
tank calms that down...