[Phil Allison]

Iain Cherches Complete ASS "

Stevens and Billington say, on their website:

If for example a 5k Ohm resistive volume control were to be employed in a
passive control unit the source would be required to drive all the time a
quite severe load of 5k Ohm. If combined with a 1k Ohm source impedance
the
worst case output impedance of the combination would be 1500 Ohm at -6db
attenuation, while at -20db attenuation the output impedance would still
be
around 540 Ohm. If combined with around 1nF load capacitance (easily found
in longer, high capacitance interconnects), this leads to a 0.3db
attenuation at 20kHz for a 20db attenuation setting, practically showing
the
absolute permissible limit for load capacitance. The worst-case
attenuation
at 20kHz almost reaches 1db!!!


** The authors have used very funny maths.

With a 1500 ohm source and 1nF load:

- 3dB is at 106 kHz
- 1dB is at 53 kHz
- 0.25dB is at 26 kHz


With a 540 ohm source and 1 nF load:

- 3dB is at 295 kHz
- 1dB is at 147 kHz
- 0.25dB is at 74 kHz


Ordinary co-ax cable ( ie RG59) has only 22 pF per foot - good, low
capacitance RCA leads have similar values.

Takes a monster ** 50 foot long ** lead to create 1 nF.


**Conclusion:

The dB figures are false.

The reasoning is totally false.

What else would you expect from a marketing blurb ?



........ Phil

[Arny Krueger]

"Phil Allison" wrote in message

"Eeysore"

For a 10k pot , that's 2.5 kohms.

I've known ppl detect as little as -0.3 dB @ 20kHz so
let's say -0.1 dB shall we
?

That -0.1dB @ 20kHz will be caused by a tiny *38pF* of
load capacitance !


** You have shifted a decimal point.

A 2500 ohms source with a 38 pF load gives:

- 3dB at 1.68 Mhz.
- 1dB at 840 kHz
- 0.25dB at 420 kHz
- 0.1 dB at 230 kHz


Using a 10 kohms pot allows up to a 800 pF output cable
to be used, with only 0.25dB loss at 20kHz - worst
case.

No doubt your intent is to show a very conservative configuration, Phil.
You've done that! In fact droops of 1 dB @ 20KHz are still innocious. In
fact it is common for power amplifiers with output inductors to size them
for 0.5 to 1 dB droop with minimum rated load.

[Eeyore]

Phil Allison wrote:

"Eeysore"

For a 10k pot , that's 2.5 kohms.

I've known ppl detect as little as -0.3 dB @ 20kHz so let's say -0.1 dB
shall we
?

That -0.1dB @ 20kHz will be caused by a tiny *38pF* of load capacitance !

** You have shifted a decimal point.

I did worse than that. I wasn't fully awake.


A 2500 ohms source with a 38 pF load gives:

- 3dB at 1.68 Mhz.
- 1dB at 840 kHz
- 0.25dB at 420 kHz
- 0.1 dB at 230 kHz

Using a 10 kohms pot allows up to a 800 pF output cable to be used, with
only 0.25dB loss at 20kHz - worst case.

Sounds right.

To use my -0.1dB criterion that gives a C of 480pF or about 15 feet of typical
cable.

Graham

[Ian Iveson]

Eeyore wrote

Ian Iveson wrote:

It is not difficult, given short cables, to match a CD player with
a
valve amp via a potentiometer, because there can be a x100
difference
between source impedance and power amp input impedance, thus
allowing
for 2 10-to-1 ratios. For example, a 10k pot should be just OK with
1k
source and 100k load. In my view a resistor is preferable to an
inductor if both work OK.

There is NO *MATCHING* involved whatever.

It's purely a voltage transfer process. Voltage transfer works just
fine with
*ANY* practical ratio of impedances where the load impedance is
typically ~ =
10 times the source impedance for minimal loading ( 1dB ).

Matching refers a situation where 2 equal values of source and load
resistance
/ impedance are used to maximise *power* transfer as opposed to
voltage
transfer.

Maximum power transfer (impedance matching) is not a requirement for
good
performance in modern equipment and would indeed tend to degrade
performance in
modern equipment were it practiced.

Please use the correct terms.

What would you consider to be the correct term, in the context of my
post?

The way I have used the term "matching" is common.

Clearly I have not used it in the sense of matching for optimum power
transfer. I have used it in the sense of matching for compatibility,
to preserve linearity. Considering I spelled this out in terms of the
10-1 rule of thumb, there is no fear of confusion.

Thanks anyway.

Ian

[Ian Iveson]

Iain wrote

1.Does anyone have experience of such transformers?
2. Are they actually better than a passive stepped attenuator
controller?


No. I would be interested to hear from folk who have compared the
two
sensibly.

I am planning to do just that, when time permits.

So in most circumstances this is a question about whether a
transformer is
better than an active preamp such as your own. In which case the
issue is
distortion, presumably.

I am told that the tracking accuracy between channels is superior,
althought with most stepped attenuators at betetr than 1% I wonder
if
the improvement is audible.

Stepped attenuators can be trimmed, surely?

I don't think it's a question of distortion either,and besides, the
Sowter has a THD of 0.03% at +20dBu at 50Hz.

But what is it for a 2V input at, say 20Hz, or whatever is the lowest
frequency it may encounter?

There may be some folk who would prefer transformer coupling in any
case,
possibly because they object to using capacitors in the signal path,
but
it's a while since anyone owned up to this apparent fetish, and I
forget
how they argue their case.

One thing I wonder about. It is common to use a capacitor in series
at the
input of a valve power amp, and this may be used to limit the LF
response.
If you use a transformer then arguably you don't need the series
cap for
its dc-blocking role. So it may seem that you can rely on the
transformer
also for the role of limiting LF. Would that be wise? I am
concerned about
the possibility of ensuing distortion.

Most matchning and input transformers are not designed to have
DC on them.

Oh. OK. I have a couple here and they aren't gapped. But this seems a
bit of a nonsequitur to me anyway, since I hadn't intended to imply
that they are used with DC. Only that they have by their nature no DC
output, and therefore obviate the need for a DC blocking cap on the
following input.

That's what got me to wonder about the wisdom of using a transformer
to define the lower limit of bandwidth. Signals of lower frequency
will be distorted, and the distortion products will be in band.

So if you use a transformer as an attenuator, do you ensure that it is
not subject to frequencies below its quoted bandwidth? How? Just
wondering.

cheers, Ian

[Eeyore]

Phil Allison wrote:

"Eeysore ****ing Pommy **** "

For a 10k pot , that's 2.5 kohms.

I've known ppl detect as little as -0.3 dB @ 20kHz so let's say -0.1 dB
shall we ?

That -0.1dB @ 20kHz will be caused by a tiny *38pF* of load capacitance !

** You have shifted a decimal point - ****HEAD !

A 2500 ohms source with a 38 pF load gives:

- 3dB at 1.68 Mhz.
- 1dB at 840 kHz
- 0.25dB at 420 kHz
- 0.1 dB at 230 kHz

Using a 10 kohms pot allows up to a 800 pF output cable to be used, with
only 0.25dB loss at 20kHz - worst case.

Why do you keep re-posting your correction ?

Graham

[Eeyore]

Arny Krueger wrote:

In fact droops of 1 dB @ 20KHz are still innocious.

Innocuous my arse !


In fact it is common for power amplifiers with output inductors to size them
for 0.5 to 1 dB droop with minimum rated load.

Typical output inductor 6uH j0.75 ohms at 20kHz

Droop due to the inductor into 8 ohms (more relevant to hi-fi ) ~ 0.04 dB.


Graham

[Eeyore]

flipper wrote:

Eeyore wrote:
west wrote:
"Eeyore" wrote
Iain Churches wrote:

2. Are they actually better than a passive stepped attenuator
controller?

Better ?

Explain what better is first !

As I've discoverd, in the tube audio world, "better" has no correlation
with measured technical accuracy / performance is is merely a subjective
matter.

I fail to see thefore how "better" can be anything other than a personal
opinion.

Why are you always trying to pick fly **** out of pepper? Maybe you should
frequent alt.lawyers. Let's stop this bs. I know what he means. I'd like to
see you post something in his language. Food for thought.

"Better" has no definable meaning.

That's the question he's asking: for an explanation or definition of
why it's supposedly 'better', or by what 'definition one would claim
it's 'better', and why one would use it.

We do finally seem to have established that point thank you. Some clarity never hurt.



E.g. one often hears here that SET's are better than other amplifiers types yet
they are demonstrably hugely technically inferior in almost every respect. If
this is to determine what 'better' means then better = worse.

How about asking a question that has a possible meaningful answer instead ?

He did. You just want to play word games rather than deal with the
intent of the question.

Determining the real intent of the OP's question is hardly an example of word games.
For those you'd need to engage Mr Joot's attention.

Graham

[Eeyore]

Ian Iveson wrote:

Eeyore wrote
Ian Iveson wrote:

It is not difficult, given short cables, to match a CD player with
a valve amp via a potentiometer, because there can be a x100
difference between source impedance and power amp input impedance, thus
allowing for 2 10-to-1 ratios. For example, a 10k pot should be just OK
with 1k source and 100k load. In my view a resistor is preferable to an
inductor if both work OK.

There is NO *MATCHING* involved whatever.

It's purely a voltage transfer process. Voltage transfer works just
fine with *ANY* practical ratio of impedances where the load impedance is
typically ~ = 10 times the source impedance for minimal loading ( 1dB ).

Matching refers a situation where 2 equal values of source and load
resistance / impedance are used to maximise *power* transfer as opposed to
voltage transfer.

Maximum power transfer (impedance matching) is not a requirement for
good performance in modern equipment and would indeed tend to degrade
performance in modern equipment were it practiced.

Please use the correct terms.

What would you consider to be the correct term, in the context of my
post?

The way I have used the term "matching" is common.

And like many things that are 'common' it is completely wrong.


Clearly I have not used it in the sense of matching for optimum power
transfer. I have used it in the sense of matching for compatibility,
to preserve linearity. Considering I spelled this out in terms of the
10-1 rule of thumb, there is no fear of confusion.

That's not *matching* though. At least not to my way of thinking.

The point being that in a voltage transfer interface there is nothing to match.
That's one of the many attractions it has.

In the pro-audio world if 'matching' came into the conversation it would
invariably be in the context of 'level matching' such as when interconnecting
professional and consumer equipment with different operating levels. Impedances
would never normally merit a second thought, or even a first one !

Graham

[Phil Allison]

"Eeysore ****ing Lying Pommy **** From Hell "


Genetic, autistic ****wits like Graham Stevenson are not human.

Just sub human Zombies with malicious intent.

Ought to be baited like any other vermin.



........ Phil

[Phil Allison]

"Eeysore ****ing Lying Pommy **** From Hell "


Genetic, autistic ****wits like Graham Stevenson are not human.

Just sub human Zombies with malicious intent.

Ought to be baited like any other vermin.




........ Phil

[John Byrns]

In article ,
"Ian Iveson" wrote:

One thing I wonder about. It is common to use a capacitor in series at
the input of a valve power amp, and this may be used to limit the LF
response. If you use a transformer then arguably you don't need the
series cap for its dc-blocking role. So it may seem that you can rely
on the transformer also for the role of limiting LF. Would that be
wise? I am concerned about the possibility of ensuing distortion.

I don't see why distortion would become any more of a problem below the
lowest frequency the transformer the transformer is rated for, assuming
the LF signal level is no higher than the signal level the transformer
is designed for in its specified range. If the shunt inductance of the
transformer is being used in this way to limit the LF response it
implies that a finite source impedance is being used to drive the
transformer and if you go through the math I think you will find that LF
signals take the transformer no closer to saturation than do signals at
the transformers rated LF limit. I'm not sure if this argument hold's
for actual DC though?


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article i,
"Iain Churches" wrote:

"west" wrote in message news:zIL9i.212$ng.61@trnddc07...

"Eeyore" wrote in message
...


west wrote:

BTW Krueger states that any pot over 10K is "unwise." If that's the
case,
then why do so many commercial tube amps have a 100K-250K input pot?

They're not connected to a long length of cable are they ?

Have you completely missed the point ?

Possibly. How much do you consider a cable which is long? I though Phil
said
that RG58 (or I mentioned RG6) has negligable capacitance in reasonable
lengths.

I was looking at a typical commercially-built passive controller yesterday.
It had cables of 1m50 on both input and output, and a 100k DACT
stepped attenuator.

What does "1m50" mean? It doesn't make sense any way I parse it.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Eeyore]

John Byrns wrote:

if you go through the math I think you will find that LF
signals take the transformer no closer to saturation than do signals at
the transformers rated LF limit. I'm not sure if this argument hold's
for actual DC though?

Uh ???

Can someone unscramble that ?

Graham

[John Byrns]

In article ,
Eeyore wrote:

John Byrns wrote:

if you go through the math I think you will find that LF
signals take the transformer no closer to saturation than do signals at
the transformers rated LF limit. I'm not sure if this argument hold's
for actual DC though?

Uh ???

Can someone unscramble that ?

Graham, I am surprised that someone of your supposed technical knowledge
doesn't get the point immediately! Please keep in mind the part of my
post that you clipped, which was that the OP was talking about using a
transformer to limit the LF response of a system, and my claim that if
that were the case it would assume a finite source impedance driving the
transformer. Do the math, think about it a bit, and if you have a
normal number of neurons you should be able to figure it out.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Eeyore]

John Byrns wrote:


What does "1m50" mean? It doesn't make sense any way I parse it.

1.50 metres (59")

Ever seen a 2.2 microfarad cap marked 2u2 ?

Graham

[Eeyore]

John Byrns wrote:

In article ,
Eeyore wrote:

John Byrns wrote:

if you go through the math I think you will find that LF
signals take the transformer no closer to saturation than do signals at
the transformers rated LF limit. I'm not sure if this argument hold's
for actual DC though?

Uh ???

Can someone unscramble that ?

Graham, I am surprised that someone of your supposed technical knowledge
doesn't get the point immediately! Please keep in mind the part of my
post that you clipped, which was that the OP was talking about using a
transformer to limit the LF response of a system, and my claim that if
that were the case it would assume a finite source impedance driving the
transformer. Do the math, think about it a bit, and if you have a
normal number of neurons you should be able to figure it out.

You assertion still doesn't help the above quoted sentences of yours make any
sense though.

Graham

[Iain Churches]

"John Byrns" wrote in message
...

What does "1m50" mean? It doesn't make sense any way I parse it.

Sorry John. That's how we write one metre and fifty centimetres or 150cms
here in Scandinavia. We also use the comma and not the point (full stop/dot)
as
a decimal marker. That might have confused you even mo-)


Iain

[John Byrns]

In article ,
Eeyore wrote:

John Byrns wrote:


What does "1m50" mean? It doesn't make sense any way I parse it.

1.50 metres (59")

Ever seen a 2.2 microfarad cap marked 2u2 ?

Yes, frequently, and that was exactly my problem, I thought Iain was
talking about the cable capacitance and I was trying to decode it as
such, and it just didn't make sense in that context.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
"Iain Churches" wrote:

"John Byrns" wrote in message
...

What does "1m50" mean? It doesn't make sense any way I parse it.

Sorry John. That's how we write one metre and fifty centimetres or 150cms
here in Scandinavia.

Thanks, I thought you were talking about the cable capacitance where
"1m50" doesn't make sense and/or is a little extreme.

We also use the comma and not the point (full stop/dot)
as
a decimal marker. That might have confused you even mo-)

No, I am not even aware of comma/point usage in that context. I learned
my lesson on that as a small boy reading British radio magazines, or was
it German?


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Iain Churches]

"John Byrns" wrote in message
...
In article ,
"Iain Churches" wrote:

"John Byrns" wrote in message
...

What does "1m50" mean? It doesn't make sense any way I parse it.

Sorry John. That's how we write one metre and fifty centimetres or 150cms
here in Scandinavia.

Thanks, I thought you were talking about the cable capacitance where
"1m50" doesn't make sense and/or is a little extreme.

We also use the comma and not the point (full stop/dot)
as
a decimal marker. That might have confused you even mo-)

No, I am not even aware of comma/point usage in that context. I learned
my lesson on that as a small boy reading British radio magazines, or was
it German?

Almost certainly German. Here we use the point as a designator to mark
thousands, so 10. 000 is ten thousand, as opposed to 10,000 in the UK
(and also in the US?)

One has to be *very* carefully when entering bids on e-Bay:-)))

Regards
Iain


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[robert casey]

Most people seem to pick 100k.


For a passive control unit ? That would be exceptionally silly, but given the
audiophoolery that routinely exists surprises me not one jot.

At its -6dB position, a 100k volume control presnts an output resistance of 25k
to the world. A relatively modest cable capacitance of ~ 300 pF will cause a
-3dB low pass filter effect at ~ 20kHz.

Now that'll be audible.


Isn't that why volume controls are usually located inside the audio
amplifier? The input jack feeds thru a selector switch to the top of
the pot. The wiper of the pot feeds (usually via a cap) directly to a
tube grid, without the capacitance to ground you'd get with a long patch
cord. Or interconnect if you prefer...

[Eeyore]

John Byrns wrote:

In article ,
"Iain Churches" wrote:

"John Byrns" wrote in message
...

What does "1m50" mean? It doesn't make sense any way I parse it.

Sorry John. That's how we write one metre and fifty centimetres or 150cms
here in Scandinavia.

Thanks, I thought you were talking about the cable capacitance where
"1m50" doesn't make sense and/or is a little extreme.

We also use the comma and not the point (full stop/dot)
as a decimal marker. That might have confused you even mo-)

No, I am not even aware of comma/point usage in that context. I learned
my lesson on that as a small boy reading British radio magazines, or was
it German?

That would have been German.

Brits use the period / full stop / point as the decimal separator like
yourselves.

I love the French equivalent for 'floating point'. It translates as 'flying
comma'. :~)

Graham

[Eeyore]

robert casey wrote:

Most people seem to pick 100k.

For a passive control unit ? That would be exceptionally silly, but given the
audiophoolery that routinely exists surprises me not one jot.

At its -6dB position, a 100k volume control presnts an output resistance of 25k
to the world. A relatively modest cable capacitance of ~ 300 pF will cause a -3dB
low pass filter effect at ~ 20kHz.

Now that'll be audible.

Isn't that why volume controls are usually located inside the audio
amplifier? The input jack feeds thru a selector switch to the top of
the pot. The wiper of the pot feeds (usually via a cap) directly to a
tube grid, without the capacitance to ground you'd get with a long patch
cord. Or interconnect if you prefer...

I thought volume controls are usually inside the amplifier because it's simply
convenient to put them there.

Graham

[Ian Iveson]

John said:

So it may seem that you can rely
on the transformer also for the role of limiting LF. Would that be
wise? I am concerned about the possibility of ensuing distortion.

I don't see why distortion would become any more of a problem below
the
lowest frequency the transformer the transformer is rated for,
assuming
the LF signal level is no higher than the signal level the
transformer
is designed for in its specified range. If the shunt inductance of
the
transformer is being used in this way to limit the LF response it
implies that a finite source impedance is being used to drive the
transformer and if you go through the math I think you will find
that LF
signals take the transformer no closer to saturation than do signals
at
the transformers rated LF limit. I'm not sure if this argument
hold's
for actual DC though?

What argument? You are asking me if my guess at your maths hold true
for DC? Dunno, haven't done the sums yet. Let's see...

No.

Actually, my guess at your maths doesn't hold true at all, as far as I
have calculated so far.

I can see a countervailing force...a restorative reaction...vaguely.
Voltage falls and frequency falls and you reckon the upshot remains
pretty much the same. But I don't think it's enough.

Maths on NG is really hard.

Take two constants, K = 2 x pi and B = N squared.

N is the turns ratio
f is the frequency
Vs is the source voltage
Rs is the source resistance
RL is the load resistance
Lp is the primary inductance

Adopt the following circuit model. Vs is applied to a voltage divider
comprising Rs and X(Lp)||B.RL

ip is the current through X(Lp), and is directly related to
magnetising current.

From some algebra I reckon that:

ip = Vs / {Rs + K.f.Lp.(1 + Rs / B.RL)}

Note I have no primary resistance.

Note then that, at DC, f = 0 and so ip = Vs / Rs

Your argument could be true if Rs was great enough compared to the
rest of the bottom line. But if the transformer is being used to limit
bandwidth, then we should know how big it is...ie at -6dB it will be
equal to X(Lp)||B.RL

For f to be insignificant wrt magnetising current, Rs would need to be
an order of magnitude greater than that, I would expect.

A bit more maths to do...but so far I don't think you are right.

I guess you have an easier solution. What is it?

And why, if as seems inevitable your maths has a term for frequency in
it, can't you find out whether it "holds for DC" by substituting zero
for f in the equation?

Incidentally, if I include the fact that Lp increases with ip
according to the BH curve for the core material, then the maths
becomes impossible, as we are short of an expression for the BH curve
and even if we had one the equation would end up a proper dogs dinner.

I still can't simulate a non-linear inductor...stuck for either an
expression or a way of getting spice to use a look-up table and
interpolate.

Isn't this already worked out in RDH4 or somewhere? Patrick?

cheers, Ian

[robert casey]

Isn't that why volume controls are usually located inside the audio
amplifier? The input jack feeds thru a selector switch to the top of
the pot. The wiper of the pot feeds (usually via a cap) directly to a
tube grid, without the capacitance to ground you'd get with a long patch
cord. Or interconnect if you prefer...


I thought volume controls are usually inside the amplifier because it's simply
convenient to put them there.


Also a good reason, which doesn't conflict with my reason. Sometimes
things work out together like that.

[Eeyore]

flipper wrote:

Eeyore wrote:

I thought volume controls are usually inside the amplifier because it's simply
convenient to put them there.

Depends on who's convenience you're talking about and it might have
been more 'convenient' for the 50's TV viewer to have a 20 foot wired
remote.

50's wives would never have stood for such a thing.

Graham

[Iain Churches]

"Eeyore" wrote in message
...


robert casey wrote:

Most people seem to pick 100k.

For a passive control unit ? That would be exceptionally silly, but
given the
audiophoolery that routinely exists surprises me not one jot.

At its -6dB position, a 100k volume control presnts an output
resistance of 25k
to the world. A relatively modest cable capacitance of ~ 300 pF will
cause a -3dB
low pass filter effect at ~ 20kHz.

Now that'll be audible.

Isn't that why volume controls are usually located inside the audio
amplifier? The input jack feeds thru a selector switch to the top of
the pot. The wiper of the pot feeds (usually via a cap) directly to a
tube grid, without the capacitance to ground you'd get with a long patch
cord. Or interconnect if you prefer...

I thought volume controls are usually inside the amplifier because it's
simply
convenient to put them there.

In these days of high level sources, no preamp is actually
required, so one can mount a pair stepped attenuators in
the power amp chassis, and feed the CD player direct.

http://www.kolumbus.fi/iain.churches...-30%20SA01.jpg



Iain



Graham

[Iain Churches]

"Eeyore" wrote in message
...


Iain Churches wrote:

"Eeyore" wrote

Load capacitance is what messes up passive attenuators using pots.

OK. Now we are getting to the crux of the matter.
Can you give us a formula and some shirtcuff calculations, Graham?

Assuming the source impedance is lowish compared to the pot value, the
maximum
output impedance / resistance of a potentiometer volume control is 1/4 the
pots'
value.

For a 10k pot , that's 2.5 kohms.


If we want to be confident that the pot's resistance and the load
capacitance of
the output cable's not introducing any colouration we need to choose a
'minus
dB' @ say 20kHz that's small enough to be inaudible.

I've known ppl detect as little as -0.3 dB @ 20kHz so let's say -0.1 dB
shall we
?

Doing the sums elsewhere.......

Could you do a typical calculation here? It would be useful to know how to
calculate a pot and cable combination.

Best regards
Iain

[Phil Allison]

"Iain Churches"

Doing the sums elsewhere.......

Could you do a typical calculation here? It would be useful to know how
to
calculate a pot and cable combination.


** It is sooooo simple.

The highest output resistance for a pot or step attenuator is when the level
is at -6dB.

For a 10 kohm pot, when the wiper is set exactly half way - you get 5 kohms
to ground and 5 kohms in series with the source. The two 5 kohms are
effectively in parallel for a low impedance source & so sum electrically to
2.5 kohms.

he formula for capacitive reactance is:

Xc = 1 / ( 2.pi.F.C )

Plus, when the resistance and Xc are the same number, attenuation is 3dB.

For a simple 6 dB per octave low pass filter you can apply a very good rule
of thumb.

If the - 3dB frequency is at F, then at

F/2 the attenuation is 1dB from flat & at

F/4 the attenuation is 0.25dB from flat.


Example:

Say the cable capacitance is known to be 800pF,

Then at 80 kHz

Xc = 1 ( 2 x 3.14 x 8exp4 x 8exp-10 )

= 2590 ohms

= near enough to 2.5 kohms

So, the attenuation will be 3 dB at 80 kHz, worst case.

It follows from the rules of thumb that

at 40 kHz it will be 1 dB and

at 20 kHz it will be 0.25dB.



....... Phil

[Chris Hornbeck]

On Fri, 8 Jun 2007 08:21:08 +0300, "Iain Churches"
wrote:

Could you do a typical calculation here? It would be useful to know how to
calculate a pot and cable combination.

Two useful simplifications (subject to correction and
embellishment I'd hope):

Assume source impedance to be negligibly small and load
impedance to be capacitive only. Output resistance is at
maximum at 6dB attenuation, where it's 1/4 of the ladder's
resistance. For example, a 10K ohm ladder has a maximum
output resistance of 2K5 ohms.

Calculate a -3db frequency where output resistance equals
1/ 2 pi F C , where C is loading capacitance. This works
out as:

F-sub-3dB = reciprocal 2 pi (output resistance) (loading C)

Divide the 3dB number by four to get a 1/4dB number and
assume that that is good enough! (Or fudge from here. Arf.)

In practice, if you can drive a 10K ohm ladder, you're
golden, as I believe Arny has already said. Not everything
can drive 10K ohms well, so the numbers do still matter,
but the ability is a useful idealized goal.

Much thanks, as always,

Chris Hornbeck

[Iain Churches]

"Phil Allison" wrote in message
...

"Iain Churches"

Doing the sums elsewhere.......

Could you do a typical calculation here? It would be useful to know how
to
calculate a pot and cable combination.


** It is sooooo simple.

The highest output resistance for a pot or step attenuator is when the
level is at -6dB.

For a 10 kohm pot, when the wiper is set exactly half way - you get 5
kohms to ground and 5 kohms in series with the source. The two 5 kohms are
effectively in parallel for a low impedance source & so sum electrically
to 2.5 kohms.

he formula for capacitive reactance is:

Xc = 1 / ( 2.pi.F.C )

Plus, when the resistance and Xc are the same number, attenuation is 3dB.

For a simple 6 dB per octave low pass filter you can apply a very good
rule of thumb.

If the - 3dB frequency is at F, then at

F/2 the attenuation is 1dB from flat & at

F/4 the attenuation is 0.25dB from flat.


Example:

Say the cable capacitance is known to be 800pF,

Then at 80 kHz

Xc = 1 ( 2 x 3.14 x 8exp4 x 8exp-10 )

= 2590 ohms

= near enough to 2.5 kohms

So, the attenuation will be 3 dB at 80 kHz, worst case.

It follows from the rules of thumb that

at 40 kHz it will be 1 dB and

at 20 kHz it will be 0.25dB.


Phil. It's a lot simpler than I imagined. I have
never needed to calculate that kind of thing,
even though the formula was familiar.

Thanks for taking the time to explain it.
Iain

[Eeyore]

Iain Churches wrote:

"Eeyore" wrote in message

I thought volume controls are usually inside the amplifier because it's
simply convenient to put them there.

In these days of high level sources, no preamp is actually
required, so one can mount a pair stepped attenuators in
the power amp chassis, and feed the CD player direct.

http://www.kolumbus.fi/iain.churches...-30%20SA01.jpg

For many decades I didn't own a preamp. Everything of mine had line level outs
anyway (including my turntable).

The current one only qualifies as a preamp since it has a phono input but in
reality it's just a source switcher the way I use it.

Graham

[Eeyore]

Iain Churches wrote:

"Eeyore" wrote

I've known ppl detect as little as -0.3 dB @ 20kHz so let's say -0.1 dB
shall we ?

Doing the sums elsewhere.......

Could you do a typical calculation here? It would be useful to know how to
calculate a pot and cable combination.

I messed up last time round. I post a new thread.

Graham

[Arny Krueger]

"Eeyore" wrote in
message
flipper wrote:

Eeyore wrote:

I thought volume controls are usually inside the
amplifier because it's simply convenient to put them
there.

Depends on who's convenience you're talking about and it
might have been more 'convenient' for the 50's TV viewer
to have a 20 foot wired remote.

50's wives would never have stood for such a thing.

The first commercial wireless TV remote went on the market in 1955, and sold
well. 50's wives arguably had wireless remotes at their disposal.

[Arny Krueger]

"Iain Churches" wrote in message
i.fi

In these days of high level sources, no preamp is actually
required, so one can mount a pair stepped attenuators in
the power amp chassis, and feed the CD player direct.

Somebody needs to clue Iain in on the existence of 2-channel stepped
attenuators, and the 50-year old technology about how to wire in a stepped
or continuously-variable balance control.

http://www.kolumbus.fi/iain.churches...-30%20SA01.jpg

Butt-ugly amplfier if there ever was one.

[Arny Krueger]

"John Byrns" wrote in message

In article
, "Ian
Iveson" wrote:

One thing I wonder about. It is common to use a
capacitor in series at the input of a valve power amp,
and this may be used to limit the LF response. If you
use a transformer then arguably you don't need the
series cap for its dc-blocking role. So it may seem that
you can rely on the transformer also for the role of
limiting LF. Would that be wise? I am concerned about
the possibility of ensuing distortion.

I don't see why distortion would become any more of a
problem below the lowest frequency the transformer the
transformer is rated for, assuming the LF signal level is
no higher than the signal level the transformer is
designed for in its specified range.

Distortion is more of a problem at low frequencies because of core
saturation effects:

http://www.rane.com/note159.html

"Core saturation happens when the magnetic field in the core reaches its
maximum possible density, which is what happens when the applied voltage
polarity remains the same for too long."

Obviously, the applied voltage polarity remains the same for a longer time
at lower frequencies than higher frequencies.

"Saturation has nothing to do with power delivery: the onset of saturation
depends only on the voltage waveform applied to the primary."

[Arny Krueger]

"Eeyore" wrote in
message
Arny Krueger wrote:

In fact droops of 1 dB @ 20KHz are still innocious.

Innocuous my arse !

http://www.pcavtech.com/soundcards/t...LevelMatch.gif

Shows that octave-wide dip at 20 KHz can be almost 3 dB without being
audible.

http://www.pcabx.com/technical/dips_pips_tips/index.htm

Has a downloadable sample illustrating a - 3 dB first order butterworth
roll-off.

In fact it is common for power amplifiers with output
inductors to size them for 0.5 to 1 dB droop with
minimum rated load.

Typical output inductor 6uH j0.75 ohms at 20kHz

Droop due to the inductor into 8 ohms (more relevant to
hi-fi ) ~ 0.04 dB.

I've measured the droop - the amp's own roll-off must be a major
contributor.

[John Byrns]

In article ,
"Arny Krueger" wrote:

"John Byrns" wrote in message

In article
, "Ian
Iveson" wrote:

One thing I wonder about. It is common to use a
capacitor in series at the input of a valve power amp,
and this may be used to limit the LF response. If you
use a transformer then arguably you don't need the
series cap for its dc-blocking role. So it may seem that
you can rely on the transformer also for the role of
limiting LF. Would that be wise? I am concerned about
the possibility of ensuing distortion.

I don't see why distortion would become any more of a
problem below the lowest frequency the transformer the
transformer is rated for, assuming the LF signal level is
no higher than the signal level the transformer is
designed for in its specified range.

Distortion is more of a problem at low frequencies because of core
saturation effects:

http://www.rane.com/note159.html

"Core saturation happens when the magnetic field in the core reaches its
maximum possible density, which is what happens when the applied voltage
polarity remains the same for too long."

Obviously, the applied voltage polarity remains the same for a longer time
at lower frequencies than higher frequencies.

"Saturation has nothing to do with power delivery: the onset of saturation
depends only on the voltage waveform applied to the primary."

That quote should be corrected to state that saturation depends on
frequency as well as voltage. The reason distortion (saturation) would
not be a problem when a transformer is being used to limit the LF
response as the OP was asking about, is because what we are talking
about then is a low pass filter where the source resistance driving the
transformer primary operates in conjunction with the transformer's
primary inductance to roll of the LF response below a specific point.
This roll off of the primary voltage at lower frequencies prevents the
core from going into saturation (distortion) at lower frequencies as
would happen if the primary voltage were held constant as the frequency
decreases.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Arny Krueger]

"John Byrns" wrote in message

In article ,
"Arny Krueger" wrote:

"John Byrns" wrote in message

In article
, "Ian
Iveson" wrote:

One thing I wonder about. It is common to use a
capacitor in series at the input of a valve power amp,
and this may be used to limit the LF response. If you
use a transformer then arguably you don't need the
series cap for its dc-blocking role. So it may seem
that you can rely on the transformer also for the role
of limiting LF. Would that be wise? I am concerned
about the possibility of ensuing distortion.

I don't see why distortion would become any more of a
problem below the lowest frequency the transformer the
transformer is rated for, assuming the LF signal level
is no higher than the signal level the transformer is
designed for in its specified range.

Distortion is more of a problem at low frequencies
because of core saturation effects:

http://www.rane.com/note159.html

"Core saturation happens when the magnetic field in the
core reaches its maximum possible density, which is what
happens when the applied voltage polarity remains the
same for too long."

Obviously, the applied voltage polarity remains the same
for a longer time at lower frequencies than higher
frequencies.

"Saturation has nothing to do with power delivery: the
onset of saturation depends only on the voltage waveform
applied to the primary."

That quote should be corrected to state that saturation
depends on frequency as well as voltage.

Why? the previous quote already said that once!

The reason
distortion (saturation) would not be a problem when a
transformer is being used to limit the LF response as the
OP was asking about,

It is the frequency and amplitude of the applied voltage that counts. When a
transformer is being used as a high pass filter, its primary is receiving
the applied voltage.