I was looking at a schematic for a McIntosh C22 and noticed that there is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the above
method?

r




--
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

On Wed, 03 Dec 2003 22:40:20 -0000, Rich Andrews
wrote:

I was looking at a schematic for a McIntosh C22 and noticed that there is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the above
method?

The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

Chris Hornbeck
"That is my Theory, and what it is too."
Anne Elk

Chris Hornbeck wrote

"The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

Perhaps it is worth noting that the cathode will have a low output
resistance so the connection to the cable screen will not trouble
it.

If the voltage on one plate of a capacitor follows the other, then
there is no capacitance. Hence, as you say, high impedance.

Is the 36V a red herring? Is that a signal or a DC bias voltage?

cheers, Ian

"Ian Iveson" wrote in newsAwzb.11387
:

Chris Hornbeck wrote

"The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

Perhaps it is worth noting that the cathode will have a low output
resistance so the connection to the cable screen will not trouble
it.

If the voltage on one plate of a capacitor follows the other, then
there is no capacitance. Hence, as you say, high impedance.

Is the 36V a red herring? Is that a signal or a DC bias voltage?

cheers, Ian




I forgot a bit of information. The preamp Roger is talking about is the
C26.

Bootstrapping? Is that a British term like torch or did I fall asleep
during lecture again?

I cannot answer the question about the red herring. I can tell you this
much; when the issue arose in the design stage, his boss told him to try
connecting the shield to the emitter and it resolved the issue. His boss
at the time was Sidney Cordeman who designed the C22 and a few other
rather cool trinkets at McIntosh. You can view a schematic for the C22 at
http://www.berners.ch/McIntosh/Downloads/C22_ser.zip. If you look at V4a
or V6a you will see the shield connected to the cathode.

I thought about the voltage differential between the center and
shield/screen and how that might affect capacitance, and to be honest, I
can't see how the capacitance of the cable would change. If the
capacitance were to change due to voltage, then one would have a varactor,
correct?

What I do see is that the leakage would be reduced. My thinking says that
if there is a small ac signal on the shield and it is phase with the
center conductor, then any apparent leakage would be reduced by virtue
that the signal on the shield would leak back to the center. Thus, the
losses through the cable would be minimized. In other words, leakage is a
2 way street. Am I looking at things right or should I get some glasses?
(:)



r


--
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

Chris Hornbeck wrote:

On Wed, 03 Dec 2003 22:40:20 -0000, Rich Andrews
wrote:

I was looking at a schematic for a McIntosh C22 and noticed that there is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the above
method?

The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

I think that the driven sheild is often referred too as a
"Faraday Shield". Cheers, JLS



Chris Hornbeck
"That is my Theory, and what it is too."
Anne Elk

John Stewart wrote in news:3FCF123B.383B0026
@sympatico.ca:

Chris Hornbeck wrote:

On Wed, 03 Dec 2003 22:40:20 -0000, Rich Andrews
wrote:

I was looking at a schematic for a McIntosh C22 and noticed that there
is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this
explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded
wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the
emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the
above
method?

The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

I think that the driven sheild is often referred too as a
"Faraday Shield". Cheers, JLS



Isn't a Faraday shield also an enclosure to protect one from things like
lightning.

r


--
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

"Rich Andrews" wrote in message
.44...
John Stewart wrote in news:3FCF123B.383B0026
@sympatico.ca:

Chris Hornbeck wrote:

On Wed, 03 Dec 2003 22:40:20 -0000, Rich Andrews
wrote:

I was looking at a schematic for a McIntosh C22 and noticed that there
is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this
explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded
wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the
emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the
above
method?

The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

I think that the driven sheild is often referred too as a
"Faraday Shield". Cheers, JLS



Isn't a Faraday shield also an enclosure to protect one from things like
lightning.



No, it's 4 metres of razor wire to protect you from a daft old **** with
white hair, in a frock coat, called Michael Faraday
jim


--
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

On Thu, 04 Dec 2003 06:19:52 -0000, Rich Andrews
wrote:

Bootstrapping? Is that a British term like torch or did I fall asleep
during lecture again?

Just an old term for positive feedback.

What I do see is that the leakage would be reduced. My thinking says that
if there is a small ac signal on the shield and it is phase with the
center conductor, then any apparent leakage would be reduced by virtue
that the signal on the shield would leak back to the center. Thus, the
losses through the cable would be minimized. In other words, leakage is a
2 way street. Am I looking at things right or should I get some glasses?

You may have better luck thinking about capacitive reactance
rather than capacitance. Those raised on opamps may have luck
drawing the circuit as an opamp, with the cables capacitance
as a lumped-sum capacitor between inverting and non-inverting
inputs.

It's a crude analogy but may help get the point across.

Good fortune,

Chris Hornbeck
"That is my Theory, and what it is too."
Anne Elk

Hi Rich,

When the shield is connected to ground, the signal voltage must charge and
discharge the cable capacitance. But since a bipolar transistor has pretty
constant base-emitter voltage, connecting the shield to the emitter will
almost eliminate the AC voltage between the signal and the cable shield,
effectively reducing the effect of its capacitance. The DC voltage
differential is inconsequential. But the emitter of this circuit has a low
output impedance, so the shield maintains its low impedance to ground, and
works well as an electrostatic shield. This would have been important in
the C26 due to its rather high interstage impedances. Probably the best
circuit analogy to this technique would be a cascode input stage.

I actually think that the C26 is quite underrated as a preamp . . . I know
this is heresy on this NG, but I much prefer them to a C20 or even a C22 . .
.. they definately seem to hold their performance very well over the years,
too. I have one that I modified years ago (tone control elimination,
high-quality volume pot, local power supply bypassing, etc.) that still
sounds excellent, even compared to some pretty high-priced stuff. Measured
performance is also outstanding, of course a little 3H, but pretty far down
and perfectly consistant with regard to frequency and volume setting. The
C26 has a very low output impedance, making it suitable for long cable runs
to the amp(s). You can still pick one up in excellent condition for less
than $500.00, and pair it with a tube amp of your choice, and have a
fantastic-sounding system and the classic McIntosh look as well. The front
glass is even still available as a new part from Mac.

BTW, a Faraday shield as I understand it is something used for
electromagnetic shielding, not electrostatic shielding. These are commonly
found inside line and microphone input transformers. Perhaps some of the
magnetics gurus can correct this if I've oversimplified . ..

Regards,

Kirk Patton

"Rich Andrews" wrote in message
.44...
I was looking at a schematic for a McIntosh C22 and noticed that there is
a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this
explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded wires
to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the
emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the above
method?

r




--
Nothing beats the bandwidth of a station wagon filled with DLT tapes.

"Kirk Patton"

BTW, a Faraday shield as I understand it is something used for
electromagnetic shielding, not electrostatic shielding.


** Other way around.

A Faraday shield is typically a box made of wire mesh.


These are commonly found inside line and microphone input transformers.


** If they were INSIDE then they could protect the unit from what comes
from outside.

Low level audio transformers often have a "mu-metal" magnetic shield.
This is fitted around the whole OUTSIDE of the item - the CRTs inside a
scope have a similar shield fitted to protect the electron beam from
deflection by external hum fields.





........... Phil

Rich Andrews wrote

Bootstrapping? Is that a British term like torch or did I fall
asleep
during lecture again?

Dunno. It has heaps of applications and when I look in a dictionary
I just get confused. No idea of etymology. I blame the French, who
have never made sense.

Let's consider two cases common in valve amplifiers.

First, a simple cathode follower with, say 1M grid resistor to
ground. The input resistance will be, more or less, 1M, right? Let's
say you are driving from a low impedance source, and 1V will result
in 1uA current through the resistor.

Now connect the resistor to the cathode instead of ground. Now 1V at
the grid will result in about 0.9V at the cathode end of the
resistor. Because you are driving from a low impedance source, and
this feedback path has 1M resistance, the voltage at the grid will
not be significantly different as a result of the change. Hence it
is perhaps misleading to talk in terms of feedback in the usual way.

However, if you consider the current flowing through that 1M, it is
reduced by a factor of 10. You have 1V on one side and 0.9V on the
other, giving a current of 0.1 / 1M = 100nA.

You can then say that the *effective* input resistance is 10M,
because 1V results in 100nA through 10M. As long as AC voltage on
the cathode is 0.9 times that on the grid, the resistor will
continue to behave like 10M, without any effect on linearity.

As far as I know, the resistor is then said to be bootstrapped.

Now consider the capacitance between grid and anode of that same
stage. Because the anode has a low resistance to ground there is no
signal there, and the capacitance will be whatever stated in the
valve data. The capacitance is defined by how much current is
required to create a particular voltage *difference* between the
plates. If one plate is constant, the other sees the rated
capacitance.

Now convert to a concertina by adding an anode resistor, and you
will have a gain of, say -15 between grid and anode. Now a change of
1V at the grid results in a change of -15 at the anode. The plates
see a 16V difference, and so require 16 times the current to follow
the voltage. In effect, the capacitance has been multiplied by 1 +
amplification factor. That is the Miller effect.

The opposite is also true. A capacitor from grid to cathode would
have an effective value of 0.9 times, etc.

What's a varactor? Does the capacitance change with voltage? Not the
same thing. In all the above cases, it is assumed that the signal at
one end is an exact replica of that at the other, as in the case of
your cable. Hence the voltage *difference* maintains a constant
ratio, so the effective capacitance is also constant.

However, as Rudy recently pointed out, if the gain varies with
voltage (as it does) then the miller capacitance does too, so it
becomes like a varactor to some degree. As will your cable
arrangement. There is bound to be some phase error and a bit of
distortion. Any difference between the signals will see the
capacitance, so it is a method of transmitting error. But it will be
tiny and lost at the low impedance cathode.

This is long because I didn't know the answer. Perhaps soon I will.

Oh, about the red herring. If it is DC it is not the significant
factor. It's the signal that counts.

You have the right idea with what you say of leakage. It depends on
the voltage difference. Everything does.

cheers, Ian


in message . 44...
"Ian Iveson" wrote in newsAwzb.11387
:

Chris Hornbeck wrote

"The cathode tends to "follow" the grid signal. To the extent
that
this is true, no voltage difference appears between them;
hence,
a high impedance. This is also called "bootstrapping".

Perhaps it is worth noting that the cathode will have a low
output
resistance so the connection to the cable screen will not
trouble
it.

If the voltage on one plate of a capacitor follows the other,
then
there is no capacitance. Hence, as you say, high impedance.

Is the 36V a red herring? Is that a signal or a DC bias voltage?

cheers, Ian




I forgot a bit of information. The preamp Roger is talking about
is the
C26.


I cannot answer the question about the red herring. I can tell you
this
much; when the issue arose in the design stage, his boss told him
to try
connecting the shield to the emitter and it resolved the issue.
His boss
at the time was Sidney Cordeman who designed the C22 and a few
other
rather cool trinkets at McIntosh. You can view a schematic for
the C22 at
http://www.berners.ch/McIntosh/Downloads/C22_ser.zip. If you look
at V4a
or V6a you will see the shield connected to the cathode.

I thought about the voltage differential between the center and
shield/screen and how that might affect capacitance, and to be
honest, I
can't see how the capacitance of the cable would change. If the
capacitance were to change due to voltage, then one would have a
varactor,
correct?

What I do see is that the leakage would be reduced. My thinking
says that
if there is a small ac signal on the shield and it is phase with
the
center conductor, then any apparent leakage would be reduced by
virtue
that the signal on the shield would leak back to the center. Thus,
the
losses through the cable would be minimized. In other words,
leakage is a
2 way street. Am I looking at things right or should I get some
glasses?
(:)

On Thu, 4 Dec 2003 22:46:58 -0000, "Ian Iveson"
wrote:

Bootstrapping?
Dunno. It has heaps of applications and when I look in a dictionary
I just get confused. No idea of etymology.

Comes from "pulling oneself up with one's bootstraps".
Even the visual analogy is dated, I guess.
It's just plain old positive feedback, nothing special.

Chris Hornbeck
"That is my Theory, and what it is too."
Anne Elk

"Chris Hornbeck" wrote in message
...
On Thu, 4 Dec 2003 22:46:58 -0000, "Ian Iveson"
wrote:

Bootstrapping?
Dunno. It has heaps of applications and when I look in a dictionary
I just get confused. No idea of etymology.

Comes from "pulling oneself up with one's bootstraps".
Even the visual analogy is dated, I guess.
It's just plain old positive feedback, nothing special.


** With one proviso - the gain is always less than unity. More than that
and you have an oscillator.

I once had the bright idea of making a condenser mic pre-amp by
bootstrapping a 22 Mohm resistor to give 500 Mohms effective impedance at
the gate of a FET.

The result was a pre-amp with good frequency response BUT an horrendous
background noise level produced by the fact that noise was being injected
into the mic capacitor / FET gate via the 22 Mohms aided by the positive
feedback.

Those 500 or 1000 Mohm resistors you need for such a pre-amp are still
hard to come by .....




.......... Phil

Rich Andrews wrote:

John Stewart wrote in news:3FCF123B.383B0026
@sympatico.ca:

Chris Hornbeck wrote:

On Wed, 03 Dec 2003 22:40:20 -0000, Rich Andrews
wrote:

I was looking at a schematic for a McIntosh C22 and noticed that there
is a
bit of shielded wire connecting some stages together and the shield is
connected to the cathode of the following stage. I found this
explaination
on Roger Russell's site:

"Cable capacity is a problem at high frequencies for some shielded
wires to
the loudness and balance controls. I use a driven shield. Instead of
connecting the shield to ground, it is connected to 36 volts at the
emitter
of the following amplifier stage."

Can someone explain to me how cable capacitance is reduced using the
above
method?

The cathode tends to "follow" the grid signal. To the extent that
this is true, no voltage difference appears between them; hence,
a high impedance. This is also called "bootstrapping".

I think that the driven sheild is often referred too as a
"Faraday Shield". Cheers, JLS



Isn't a Faraday shield also an enclosure to protect one from things like
lightning.

Most cars work well in that function. But a convertible
with the roof retracted is not recommended!!!! Cheers, JLS

Nothing beats the bandwidth of a station wagon filled with DLT tapes.

On Fri, 5 Dec 2003 11:20:52 +1100, "Phil Allison" wrote:



** With one proviso - the gain is always less than unity. More than that
and you have an oscillator.

I once had the bright idea of making a condenser mic pre-amp by
bootstrapping a 22 Mohm resistor to give 500 Mohms effective impedance at
the gate of a FET.

The result was a pre-amp with good frequency response BUT an horrendous
background noise level produced by the fact that noise was being injected
into the mic capacitor / FET gate via the 22 Mohms aided by the positive
feedback.

Those 500 or 1000 Mohm resistors you need for such a pre-amp are still
hard to come by .....


I have some super high value resistors surplus, from high voltage (7kv) power supplies. They look like a barber pole, I
guess thats a carbon strip on the side. I bet they would be noisy as hell?

Why would you need such a high value?

"Bob Flint" wrote in message
...
On Fri, 5 Dec 2003 11:20:52 +1100, "Phil Allison"
wrote:



** With one proviso - the gain is always less than unity. More than
that
and you have an oscillator.

I once had the bright idea of making a condenser mic pre-amp by
bootstrapping a 22 Mohm resistor to give 500 Mohms effective impedance at
the gate of a FET.

The result was a pre-amp with good frequency response BUT an horrendous
background noise level produced by the fact that noise was being injected
into the mic capacitor / FET gate via the 22 Mohms aided by the positive
feedback.

Those 500 or 1000 Mohm resistors you need for such a pre-amp are still
hard to come by .....


I have some super high value resistors surplus, from high voltage (7kv)
power supplies. They look like a barber pole, I
guess thats a carbon strip on the side. I bet they would be noisy as hell?

Why would you need such a high value?



** Condenser mic capsules are simply a small capacitor of about 20 to 90
pF depending on its size. The load must not reduce the low frequency
utput - eg a capsule 20 pF and a 1000 Mohms load equates to - 3dB at 8
Hz .

PLUS the mic capacitance actually shunts the thermal noise of the load
resistor - so the higher the value the better.



........... Phil

On Fri, 5 Dec 2003 07:46:12 +1100, "Phil Allison" wrote:


"Kirk Patton"

BTW, a Faraday shield as I understand it is something used for
electromagnetic shielding, not electrostatic shielding.


** Other way around.

A Faraday shield is typically a box made of wire mesh.

I worked in a 2-way radio shop that had a wire mesh 'room' for working on low level RF stages in radios. It was referred
to as a Faraday cage, and was supposed to keep out RF transmissions so you could work on radios with the same frequency
as those transmitting outside in the city. It could also suppress your own transmitter tests. This type of radiation is
electromagnetic.

There is a copper shield in power transformers, it's referred to as a Faraday shield but I think it is there to cut
capacitance coupling between windings. Basically it's to get rid of high fr. line hash. It should be grounded... This is
considered electrostatic shielding I guess.

Phil is right, if you want magnetic shielding, you need MU metal, (or sometimes a mild steel cage, but they are not
reliable since they can re-radiate the field.) From what I've heard, physically damaging MU metal, say by hitting it
with a hammer, will destroy it's anti magnetic properties.

"Bob Flint" wrote in message
...
Phil is right, if you want magnetic shielding, you need MU metal,
(or sometimes a mild steel cage, but they are not
reliable since they can re-radiate the field.) From what I've heard,
physically damaging MU metal, say by hitting it
with a hammer, will destroy it's anti magnetic properties.

Several layers of well-used mild (or silicon steel, as if you could
find it in anything but transformer cutouts) will do just as good. And
yes, deforming mu-metal will change its properties; it has to be annealed
specially to work at its peak.

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

On Thu, 4 Dec 2003 22:00:03 -0600, "Tim Williams" wrote:

"Bob Flint" wrote in message
.. .
Phil is right, if you want magnetic shielding, you need MU metal,
(or sometimes a mild steel cage, but they are not
reliable since they can re-radiate the field.) From what I've heard,
physically damaging MU metal, say by hitting it
with a hammer, will destroy it's anti magnetic properties.

Several layers of well-used mild (or silicon steel, as if you could
find it in anything but transformer cutouts) will do just as good. And
yes, deforming mu-metal will change its properties; it has to be annealed
specially to work at its peak.

Tim

You know the 'bell' they put on transformers... I have some open frame types with no bell and I want to limit their
field... any suggestions on what type of metal I could use? I hadn't thought of old laminations... maybe there IS a use
for those old trannys of mine!!

"Bob Flint" wrote in message
...
You know the 'bell' they put on transformers... I have some open frame
types with no bell and I want to limit their field... any suggestions
on what type of metal I could use? I hadn't thought of old
laminations... maybe there IS a use for those old trannys of mine!!

Uh... if I had to do this I'd probably make a rectangular box for it,
ala potted transformers. Just use 20-30ga. mild steel, galvanized or
not depending on needs.

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms