[Andre Jute]

On Oct 26, 7:09 am, Eeyore
wrote:
Andre Jute wrote:
Eeyore wrote:
John Byrns wrote:
Patrick Turner wrote:

Cancelation of even order harmonics occurs in amps working in class AB
during that part of the wave forms which are in class A, ie, the bits
either side of the zero crossing.

But once each tube moves into cut off, nothing is cancelled.

Patrick, I'm surprised to hear you say this. What are you trying to
tell us, that the even order harmonics are only cancelled during those
parts of the cycle when both tubes are conducting, but that the even
order distortion components reappear during those parts of the cycle
when only one tube is conducting? If you actually believe that you
should go back to the books and study the theory of harmonic distortion
more carefully. I hope you didn't get this notion from the RDH4, I
haven't read the RDH4's harmonic distortion explanation, but if this is
what it says I have just lost any respect I had for the book. In a
perfectly balanced PP amplifier the even order harmonic distortion is
completely cancelled even when the tubes are cut off for parts of the
cycle.

I'd love to know how that happens. There's no cancellation of ANYTHING once one
side has ceased conducting !

Graham

Holy ****! Did I say yet that Poopie is ignorant and incompetent?

Nah, nobody can be that stupid and uninformed about tube basics.

Poopie must be cracking a joke. For the first time in his life.

Good on yer, cobber! If you can't be smart and informed, at least you
can try to be a clown, give people a giggle.

As an alleged 'wordsmith' you of all people ought to understand what cancellation
means. Apparently it went right over your head though.

Only to be expected from an ignorant non-technical ****wit.

Graham

Yeah, Poopie, you're the man when it comes to cancellations: your
definition of Class A, okay until then, cancelled out when you added
the superfluous words "under any signal condition".

Unsigned out of contempt

[Andre Jute]

On Oct 26, 7:21 am, Eeyore
wrote:
Andre Jute wrote:
Poopie Stevenson aka Eeyore wrote:
Andre Jute wrote:
Eeyore wrote:
Andre Jute wrote:

All of that follows logically from Poopie's absurd redefinition of Class A as
a Class in which "the output device(s)never cease conducting *under
any signal condition*," (emphasis added). It's ludicrous.

It's actually the only accurate definition.

I've already demonstrated several times that your words "under
any signal condition" make your definition grossly inaccurate.

But you're an ignorant **** and what you say is a load of ********.

Even when I'm right?

You're NOT right. Your ignorance is simply confusing you.

HOWEVER, to keep you happy I am happy to modify to modify my definition for clarity. I
already posted this once but I suppose you like to argue more than anything.

"the output device(s)never cease conducting *under any signal condition* within the rated
specification".

To be honest, to have to explicitly state "within the rated specification" is really a
case of pandering to fools, which certainly describes YOU, Jootikins.

Graham

Make up your mind, Poopie. Either Class A operation is possible only
with limited signal or it is possible, as you erroneously claim,
"under any signal condition". One or the other, not both.

There are about 500 messages across two threads in which you shilly-
shally about this gross error you, Pearce and Krueger have made. Which
is it?

Andre Jute
....who knows his own mind

[Andre Jute]

On Oct 26, 9:36 am, Eeyore
wrote:
John Byrns wrote:
Eeyore wrote:
John Byrns wrote:
(Don Pearce) wrote:
Eeyore wrote:

I'd love to know how that happens. There's no cancellation of ANYTHING
once one side has ceased conducting !

Because if you add an even harmonic to a signal, you have to make it
asymmetrical. You always get a peak coinciding with a trough on one
half cycle, followed by a peak coinciding with a peak on the next. If
you modify the signal to remove any asymmetry, you must by definition
remove the even harmonics.

Finally a man who understands the theory!

But it's not by ** CANCELLATION ** in the case of AB operation beyond the
crossover point. That's my issue with the description.

It does have that effect but the use of the word *cancellation* is wong IMHO.
There should be another way to describe it.

Cancelation is the right word,

********.

the two tubes, even when they, "operate beyond the crossover point",

When one tube has ceased conducting, there's NOTHING TO CANCEL, you IGNORANT
****WITTED ****.

CANCELLATION IS THE *** WRONG WORD ***.

In fact it's ADDITION of waveforms, not cancellation.

Graham

But d'y'see, dear old Poopster, the total harmonic distortion is less,
so something has been subtracted from the result. The net sense is the
same as cancellation. A little algebra, if you can handle it, would
help.

Or a little, a very little sensitivity to the English language would
help you arrive at the same conclusion just from reading the
explanations from Patrick and Pearcey and even Arny getting it right
for once.

Andre Jute
The anti-pedant

[Andre Jute]

On Oct 26, 10:06 am, Eeyore
wrote:
Patrick Turner wrote:
One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Graham

Here's a basic you should learn off by heart, Poopie:

Class A operation requires a limited signal so that the device(s) are
never driven to any point where they can stop conducting.

Andre Jute
Always ready to help a newbie

[Patrick Turner]

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

This also means that once the Ia travels below 10% of the idle value,
the gm of the tube cutting off
has diminished to such a low value the other tube turning on harder is
providing virtually all the Ichange x Vchange
across the available load, and is the only device coupled through only
1/2 the OPT primary
to the load, so the RL seen by this tube turning on hard has reduced to
1/2 its class A load,
or 1/4 of the nominal RL a-a, and in this case its 1.25k.

The load is the same as that for a class B amp.

Isn't 1.25k too low a load for getting maximum power from a KT88 in
triode mode, even in class B?

No.

If the RL a-a = 5k, then the class B load is 1.25k,
and if Ea = 500V, then max Ia at grid current is about 220mA.
If you run AB2, you get a heck of a lot more Ia up to around 350mA.
KT88 ca easily make 500mA, depending on loads etc.
One can get 140W from a pair in AB2 in tetrode.

But I was asking about the best load for a class B triode amp, is 1.25k
too low for a KT-88? I guess I will have to see if I can find the
triode plate curves for the KT-88, or maybe I can substitute the 6550
curves.

There is no best load for a class B triode amp. Class B is a horrid way
to build any amp.

Maybe you meant low bias class AB.

Do the load line analysis, or have a look at my website pages and print
out a set of curves for 6550
which are virtually the same as KT88.

http://www.turneraudio.com.au/loadma...p-triodes.html



What is all this talk about ARC's anyone else's rules that would keep
you from handing out free copies of their abominable concoctious junk,
assuming you drew the schematic your self? You mentioned this same
issue in connection with the ManleyLabs amplifier you modified, my
understanding is that they only have protection for schematics they have
drawn, if you draw your own schematic of the same circuit, they have no
rights with regard to it. Any Lawyers out there care to comment?

I fell OK about just letting folks know what they could do
to rebuild a Manley or ARC or start from scratch and use the schematic I
will be posting
at my site.
There is nothing I can gain by posting a copy of the original schematic
these companies use.

I was not talking about the original schematic drawn by these companies,
I was talking about a schematic of the same circuit that you or anyone
else may have drawn, it is my understanding that there is nothing to
prevent you from legally posting such a schematic, illustrating the same
circuit as the company circuit, you just can't post the schematic drawn
by the company.

I have zero reason or time available to re-draw anyone else's original
schematic,
and have no wish to disturb the minds of the ppl who work in prestigious
US audio companies
any more than I may have. But I doubt they are aware of my existance.
What I have to say is aimed really at those with a really keen interest
in such matters AND
who understand such things AND who can read a schematic AND understand
the effects of layouts,
AND who have time to use a soldering iron.

Maybe only 2.69 people in the world are actually interested....

Many companies do NOT like ppl posting copies of their schematics on the
web,
and I have no intention of offending them by doing so.

I am not suggesting that you should do it, but it is my understanding
that they have no say in your posting a schematic you drew of their
circuit.

It would be ungentlemanly for me to copy out and post a schematic of
theirs without their consent
especially if basically I was doing it to tell everyone what a POS it
was.

Its better for them, me, and the public if I merely leave out POS
descriptions,
and say "Here's an alternative that works better than the original..."
Then anyone really keen can focus in on it, and maybe do the same thing
for themeselves,
or hire me or somebody else to do similar. I am only trying to get other
maker's gear
to stop smoking and sing better.

I have now accumulated several schematics used in a range of PP amps and
they are worth publishing
at my website when I have time because they are fine tested designs for
anyone to try.

Of course you are going to offend them by doing that, and they
may retaliate by denying you access to replacement parts.

I am free to post alternative schematics used in the cases of their amps
though,

You are also free to post your rendition of the schematic for their
original circuit.

The schematic I have come up with for some of these amps is totally
mine,
and to use my design instad of the original meant removal of 80% of the
parts and tracks on the board and starting
all over again.

As I understand the situation they only have rights
to and control over their drawing of the original circuit, you are free
to create and distribute copies of a new drawing of the circuit that was
drawn by you.

I may have that right as you suggest, but I don't feel its right to copy
their schematic
out slightly differently in appearance and post it.

The intellectual content IS THEIRS, and remains theirs even after I have
drawn it up
myself.

So if anyone wants to see really what I am on about, they have to find
their own copy of the original schematic.
To get that you have to own one of their amps and be able to quote a
serial number.

I am not in the mood to be seen to publically question all these
companies might do.
I need only say what I have done in response to being presented with
samples of their amps that
had bad smoking habits.

In general, its my personal opinion that major US companies have
forgotten how to
build simple fine amplifiers, and have drifted to complexity, weight,
size,
high cost, and lots of do-dahs and bells and whistles that do nothing
for the sound.
Meanwhile, in general, there is an appalling lack of respect for good
biasing methods
of output tubes.
Their engineers seem to have misplaced optimism about reliability in
power amps.
There is never any active protection. But ****e happens anyway.....

I shouldn't ever have to be telephoned by someone saying to me "My nice
new
brand XXX tube amp was "fixed" elsewhere, but pharqued up again a
fortnight later
and the sound went really bad, and it blows fuses..."

But most of my last 12 months work was with ppl having to cope with
results of so called engineers.

I respect engineers in general, but so often its a dumb apprentice who
is used to design the amp
in way too little time. Sales are down, and engineers cost serious
money.
Engineers are professionals, and unlike tradesmen like myself they put
an extra faerking zero on the prices they charge.
Companies only hire them if the cost can be justified by the sales
figures.

And sales by US majors are probably falling as ppl turn to chinese crap
imports.

I make no apology for my cynicism.

Patrick Turner.



Regards,

John Byrns

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

[Patrick Turner]

Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

But Graham, once the tubes stop cancelling each other's even order
distortions
when they move from class A to AB, their non linear current behaviour is
utterly
attrocious, and each tube only 1/2 amplicates the signal.

Yet the VOLTAGE outcome across the OPT primary is substantially linear.

So some would say that by means of the SUMMING action of the OPT, there
is cancelling going on.
I am simply saying the summing action merely obstructs the gross non
linearity of currents
from being current in the OPT secondary.

Its Mysterious, this whole simple business.

But any AB amp can be made to be a class A amp if the load value is
simply raised high enough
to prevent cut off occuring, which I define as being the reduction in Ia
to less than 1/10 of
the idle current for each tube.
Cut off could also be described as being where distortion in current
waves in each tube
exceeds 5% to 10%.

Patrick Turner.


Graham

[Patrick Turner]

Eeyore wrote:

Patrick Turner wrote:

But the SUM of the joint action of each tube in
class AB with very non linear currents
gives a linear voltage outcome.

And should not be confused with genuine CANCELLATION of distortion by Class A
push-pull operation.

Agreed, and see my last post.

Patrick Turner.

Graham

[Patrick Turner]

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

I should not mention whether or not I should be worried about your
levels of understanding.

I just let ppl decide for themseleves.

But I do know what is going on in each output tube of an AB pair,
milisecond by milisecond,
electron by electron.

Patrick Turner.





Regards,

John Byrns

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

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

This also means that once the Ia travels below 10% of the idle value,
the gm of the tube cutting off
has diminished to such a low value the other tube turning on harder
is
providing virtually all the Ichange x Vchange
across the available load, and is the only device coupled through
only
1/2 the OPT primary
to the load, so the RL seen by this tube turning on hard has reduced
to
1/2 its class A load,
or 1/4 of the nominal RL a-a, and in this case its 1.25k.

The load is the same as that for a class B amp.

Isn't 1.25k too low a load for getting maximum power from a KT88 in
triode mode, even in class B?

No.

If the RL a-a = 5k, then the class B load is 1.25k,
and if Ea = 500V, then max Ia at grid current is about 220mA.
If you run AB2, you get a heck of a lot more Ia up to around 350mA.
KT88 ca easily make 500mA, depending on loads etc.
One can get 140W from a pair in AB2 in tetrode.

But I was asking about the best load for a class B triode amp, is 1.25k
too low for a KT-88? I guess I will have to see if I can find the
triode plate curves for the KT-88, or maybe I can substitute the 6550
curves.

There is no best load for a class B triode amp. Class B is a horrid way
to build any amp.

That is certainly a broad brush stroke, McIntosh did a nice business in
what were essentially class B amplifiers. Many of the older readers
here enjoyed Rock & Roll music during their teen years delivered via the
class B amplifier at the local AM radio station, without "horrid"
results.

Maybe you meant low bias class AB.

No, I actually meant class B.

Do the load line analysis, or have a look at my website pages and print
out a set of curves for 6550
which are virtually the same as KT88.

http://www.turneraudio.com.au/loadma...p-triodes.html

I did the analysis as I earlier had said that I would. 1.25k does seem
to be a reasonable class B load for the KT-88/6550 just as you said. I
asked the question because I have not built any amps with this tube and
am not familiar with it beyond the fact that the Quadraplex VTRs at the
Television Station where I worked as a youth had a couple dozen 6550s in
each VTR. The class B load of 1.25k seemed low to me relative to your
2.5k class A load, but I made two erroneous assumptions in asking that
question. First I didn't realize that ra for the KT-88 is as low as it
is, and second I didn't take into consideration that your class A load
line is dissipation limited rather than voltage limited as with the
class B load line.


Regards,

John Byrns

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

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs! I am having trouble
reconciling this with your claim that cancelation can't occur. Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

Please explain.


Regards,

John Byrns

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

[Doug Bannard]

"John Byrns" wrote in message
...
In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that
in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs! I am having trouble
reconciling this with your claim that cancelation can't occur. Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

Please explain.


Regards,

John Byrns

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

John:

I think that the problem that Patrick may be having with this concept is
more one of semantics than anything else. It's hard to intuitively view
"cancellation" of even harmonic terms if both tubes are not conducting at
the same time. Let's view it with a very small amount of math thrown in.
Sometimes intuition does not take one far enough.

Take a pure Class B amplifier. Each tube or transistor conducts for 180
degrees (forget about any crossover effects). The current in each half of
the OPT primary, neglecting any magnetizing current is a half sine. Do a
Fourier expansion on this half sine and it's full of harmonics...all even
ones. But as long as the two tubes are well matched, the two half primary
currents are of identical shape and add in the OPT secondary to produce a
full sine, symmetrical about its zero voltage axis.
The even harmonics are gone. Is this cancellation? Well I guess you might
call it that.

Basically any waveform symmetrical about its zero voltage axis will contain
no even harmonics...even a symmetrical square wave. Similarly, if we
symmetrically clip the tops off both the positive and negative half cycles
of our sine we do not see any even harmonic content either for the same
reason. The ideal push-pull output stage using matched devices is not going
to produce even order distortion regardless of its operating class, whether
it be Class A, Class AB or Class B.

If in any doubt, see pages 300-301 of RDH4, or any university level math
text covering Fourier analysis.

Best Regards : Doug Bannard, P. Eng.

[Multi-grid]

On Oct 28, 2:29 am, "Doug Bannard" wrote:
"John Byrns" wrote in message

...





In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that
in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs! I am having trouble
reconciling this with your claim that cancelation can't occur. Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

Please explain.

Regards,

John Byrns

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

John:

I think that the problem that Patrick may be having with this concept is
more one of semantics than anything else. It's hard to intuitively view
"cancellation" of even harmonic terms if both tubes are not conducting at
the same time. Let's view it with a very small amount of math thrown in.
Sometimes intuition does not take one far enough.

Take a pure Class B amplifier. Each tube or transistor conducts for 180
degrees (forget about any crossover effects). The current in each half of
the OPT primary, neglecting any magnetizing current is a half sine. Do a
Fourier expansion on this half sine and it's full of harmonics...all even
ones. But as long as the two tubes are well matched, the two half primary
currents are of identical shape and add in the OPT secondary to produce a
full sine, symmetrical about its zero voltage axis.
The even harmonics are gone. Is this cancellation? Well I guess you might
call it that.

Basically any waveform symmetrical about its zero voltage axis will contain
no even harmonics...even a symmetrical square wave. Similarly, if we
symmetrically clip the tops off both the positive and negative half cycles
of our sine we do not see any even harmonic content either for the same
reason. The ideal push-pull output stage using matched devices is not going
to produce even order distortion regardless of its operating class, whether
it be Class A, Class AB or Class B.

If in any doubt, see pages 300-301 of RDH4, or any university level math
text covering Fourier analysis.

Best Regards : Doug Bannard, P. Eng.- Hide quoted text -

- Show quoted text -

That looks like not created v. cancelled from here...
cheers,
Douglas

[Andre Jute]

On Oct 27, 4:29 pm, "Doug Bannard" wrote:
"John Byrns" wrote in message
{snip]
I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs! I am having trouble
reconciling this with your claim that cancelation can't occur. Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

Please explain.

Regards,

John Byrns

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

John:

I think that the problem that Patrick may be having with this concept is
more one of semantics than anything else. It's hard to intuitively view
"cancellation" of even harmonic terms if both tubes are not conducting at
the same time. Let's view it with a very small amount of math thrown in.
Sometimes intuition does not take one far enough.

Take a pure Class B amplifier. Each tube or transistor conducts for 180
degrees (forget about any crossover effects). The current in each half of
the OPT primary, neglecting any magnetizing current is a half sine. Do a
Fourier expansion on this half sine and it's full of harmonics...all even
ones. But as long as the two tubes are well matched, the two half primary
currents are of identical shape and add in the OPT secondary to produce a
full sine, symmetrical about its zero voltage axis.
The even harmonics are gone. Is this cancellation? Well I guess you might
call it that.

Well, I'd call it cancellation. I didn't want to give my take on this
and get the subthread invaded by the usual fartcatchers who follow me
around before John received a sound answer, and yours is a good and
complete answer; thanks Doug from me too. I don't see how anyone but a
pedant can object to the "removal result" of the combination of a
positive and a negative half-cycle being called a cancellation; it is
an "addition" only in a mathematical sense (and it goes against the
grain!). But Patrick, clinging so tightly to the formal definition,
had me wondering whether there was something else that made my
commonsense view wrong.

Basically any waveform symmetrical about its zero voltage axis will contain
no even harmonics...even a symmetrical square wave. Similarly, if we
symmetrically clip the tops off both the positive and negative half cycles
of our sine we do not see any even harmonic content either for the same
reason. The ideal push-pull output stage using matched devices is not going
to produce even order distortion regardless of its operating class, whether
it be Class A, Class AB or Class B.

If in any doubt, see pages 300-301 of RDH4, or any university level math
text covering Fourier analysis.

Nah. We don't need to resort to such desperate measures while we have
you and Patrick to explain things to us.

Best Regards : Doug Bannard, P. Eng.

Thanks again for restoring my faith in the simple things...

Andre Jute
There are more things in heaven and earth, Horatio -- Will the Shake
Your first take on a problem is usually the right one -- John Z
DeLorean, when we were young

[Patrick Turner]

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

No, because I endorse what many books tell us such as RDH4.

I refuse to spoon feed ppl about the basics.

But simple use of the brain about what happens inside each tube of a PP
pair
in class A, AB or B or C etc reveals to most minds what happens.

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Once one tube is well and truly cut right off, the other is conducting
delivery
of all the power through only 1/2 the OPT primary, and its as if the
other cut off tube is unplugged
for that part of the wave.

Each output tube takes a turn at current delivery by means of turn on
harder to produce
voltage change which then SUM but without cancelation of current.
The cancelation thinge is when the tubes are in class A and the 2H is
cancelled, similarly
to in any LTP. The Class A Wiulliamson has a common cathode resistance
which is unbypassed
to assist the cancellation, but ideally, a CCS should be used, or even a
choke for the cathode
resistance. One may argue whether having each cathode with separate RC
bypass networks
is better than a shared cathode impedance/resistance.
But the individual cathode RC or fixed bias does allow class AB
operation, wheras the
common cathode R/Z does not.

Work out the 2H current in each tube while in class A of each tube and
see how such currents are applied across the primary.
The reason 2H is low in class A is that the 2H currents are the same
phase at each end of the
OPT primary which cannot produce voltage in the load if the current is
applied in common mode to
both ends of the load.


Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs!

Class B and C amps have a summing action of their turn on currents which
pull the voltage
in oposite directio. There is huge distortion in the Class B device
current,
but the total action produces a linear voltage outcome.
a pair of complementary pnp and npn SS devices do exactly the same
thing.

But its not unusual that the amount of V swing achieved in each
direction by each class B device is not equal so
some 2H will allways appear in the output.
But usually, the 3H dominates H distortion products.


I am having trouble
reconciling this with your claim that cancelation can't occur.

You have to see the distinction between each tube in class A "sharing
the load"
and the summing actions once cut off has occurred.
One could say correctly that the severe current distortions of each
device in class B
are cancelled by means of the summing. In effect they are.

In class A, what one tube does with the load current affect the other
tube.
If you have one tube with high gm and the other with a low gm, then
the amount of class A power produced in each tube varies.
This in effect is because the load seen by each tube working as an SE
tube
varies, and where you have RLa-a = 10k, then the class A load of each
tube = 5k in theory.
Tubes ain't any more perfect than I am, and you will find that perhaps
one tube "sees" 4.5k,
and the other sees 5.5k. Careful measurements with 10 ohm current
sensing R in each anode or cathode circuit
will tell you about the current flow in each tube.


Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

A sine wave symetrically clipped usually has a pile of odd numbered H,
and not much 2H. Some 2H is usually there though.

Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping
progesses from mild to severe.

Patrick Turner.

Please explain.

Regards,

John Byrns

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

[John Byrns]

In article . com,
Multi-grid wrote:

On Oct 28, 2:29 am, "Doug Bannard" wrote:
"John Byrns" wrote in message

...

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that
in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs! I am having trouble
reconciling this with your claim that cancelation can't occur. Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

Please explain.


John:

I think that the problem that Patrick may be having with this concept is
more one of semantics than anything else. It's hard to intuitively view
"cancellation" of even harmonic terms if both tubes are not conducting at
the same time. Let's view it with a very small amount of math thrown in.
Sometimes intuition does not take one far enough.

Take a pure Class B amplifier. Each tube or transistor conducts for 180
degrees (forget about any crossover effects). The current in each half of
the OPT primary, neglecting any magnetizing current is a half sine. Do a
Fourier expansion on this half sine and it's full of harmonics...all even
ones. But as long as the two tubes are well matched, the two half primary
currents are of identical shape and add in the OPT secondary to produce a
full sine, symmetrical about its zero voltage axis.
The even harmonics are gone. Is this cancellation? Well I guess you might
call it that.

Basically any waveform symmetrical about its zero voltage axis will contain
no even harmonics...even a symmetrical square wave. Similarly, if we
symmetrically clip the tops off both the positive and negative half cycles
of our sine we do not see any even harmonic content either for the same
reason. The ideal push-pull output stage using matched devices is not going
to produce even order distortion regardless of its operating class, whether
it be Class A, Class AB or Class B.

If in any doubt, see pages 300-301 of RDH4, or any university level math
text covering Fourier analysis.

Best Regards : Doug Bannard, P. Eng.


That looks like not created v. cancelled from here...

It wouldn't look like "not created" if you used a current probe to
observe the anode current waveforms of each of the two PP output tubes.


Regards,

John Byrns

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

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

No, because I endorse what many books tell us such as RDH4.

Yes, and if we read, understand, and apply what the RDH4 tells us, we
understand that even harmonic cancellation does occur in class AB
amplifiers, even when the tubes are cutoff for part of the cycle. That
is what this discussion is all about, Multi-grid, at least I think
Multi-grid was the first, made the claim that even harmonic cancellation
doesn't occur when the tubes in a class AB amplifier are cut off, then
both you and Eeyore jumped in to back up his claim.

Have you read pages 300 & 301 of the RDH4 as Doug Bannard, not to be
confused with the Multi-grid Doug, suggested? There is not a lot there,
but it is a start and hopefully might lead you to follow up by pursuing
some more complete references on Fourier analysis and the theory behind
it.

I refuse to spoon feed ppl about the basics.

But simple use of the brain about what happens inside each tube of a PP
pair
in class A, AB or B or C etc reveals to most minds what happens.

Simple use of the brain reveals that the cathode current of each tube of
a class A, AB or B amplifier contains even order harmonic currents,
among others. What simple use of the brain apparently doesn't reveal to
some is the reason these even harmonic currents are canceled in the
output of a PP amplifier. If the two PP sides of the amplifier are
identical, then the cancellation will be complete, if the to sides are
not identical, for any one of a multitude of reasons, including but not
limited to, mismatched tubes then the cancellation will only be partial.

I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Once one tube is well and truly cut right off, the other is conducting
delivery
of all the power through only 1/2 the OPT primary, and its as if the
other cut off tube is unplugged
for that part of the wave.

Yes, that is obvious, but not especially relevant.

Each output tube takes a turn at current delivery by means of turn on
harder to produce
voltage change which then SUM but without cancelation of current.

It is not necessary for each tube to be continuously delivering current
for the cancellation of the even order harmonic currents from the two
tubes to cancel, it is only necessary that the two currents exhibit the
required symmetry with respect to one another. The harmonic currents
flow even when the tube is cutoff, it's just that all the currents,
including the 2H current, happen to add to zero for the period the tube
is cutoff.

The cancelation thinge is when the tubes are in class A and the 2H is
cancelled, similarly
to in any LTP.

"The cancellation thingy" is also when the tubes are in class AB or even
class B, where the 2H is also canceled.

The Class A Wiulliamson has a common cathode resistance
which is unbypassed
to assist the cancellation, but ideally, a CCS should be used, or even a
choke for the cathode
resistance. One may argue whether having each cathode with separate RC
bypass networks
is better than a shared cathode impedance/resistance.
But the individual cathode RC or fixed bias does allow class AB
operation, wheras the
common cathode R/Z does not.

This is a diversion into a side issue but let me ask, why does a common
cathode RC not allow class AB? Granted fixed bias provides far better
class AB operation than does cathode bias, but given that we are using
cathode bias I fail to see how a common cathode RC would prevent class
AB operation anymore than separate cathode RCs do? What separate
cathode RCs do is reduce to some extent the effect of mismatched tubes
in the two sides of the PP circuit, but given well matched tubes a
common RC does not prevent class AB operation.

Work out the 2H current in each tube while in class A of each tube and
see how such currents are applied across the primary.
The reason 2H is low in class A is that the 2H currents are the same
phase at each end of the
OPT primary which cannot produce voltage in the load if the current is
applied in common mode to
both ends of the load.

The same is also true of the even order harmonic currents in class AB
and class B amplifiers!

I am beginning to get a glimmer of what your problem is here, it appears
that you are not correctly identifying the even order harmonic currents
that are applied to the output transformer primary in common mode. You
are looking at the total current waveform from each tube, which is
causing you to become confused. When the tube is cut off, the even
order harmonic currents continue to flow through the tube, its just that
the DC, fundamental, and all the harmonics together sum to zero during
the time the tube is cutoff, however this does not imply that the
current of any particular harmonic is aero during the time of cutoff,
hence cancellation goes on as if the tube had not cutoff.

Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs!

Class B and C amps have a summing action of their turn on currents which
pull the voltage
in oposite directio.

It sounds like you are describing the push pull action of any class of
amplifier?

There is huge distortion in the Class B device
current,
but the total action produces a linear voltage outcome.

And that is exactly the point, which you persist in trying to deny.
Although odd order non-linearity does remain in class A, AB, and B
amplifiers.

a pair of complementary pnp and npn SS devices do exactly the same
thing.

But its not unusual that the amount of V swing achieved in each
direction by each class B device is not equal so
some 2H will allways appear in the output.
But usually, the 3H dominates H distortion products.

You are simply stating that in the real world, devices, drive signals
and other parameters of the two sides of a PP amplifier may not be
perfectly matched, and as a result the even harmonic cancellation will
be incomplete, but partial cancellation still occurs reducing the even
order harmonic distortion, even when one or the other tube is cutoff.

I am having trouble
reconciling this with your claim that cancelation can't occur.

You have to see the distinction between each tube in class A "sharing
the load"
and the summing actions once cut off has occurred.
One could say correctly that the severe current distortions of each
device in class B
are cancelled by means of the summing. In effect they are.

In class A, what one tube does with the load current affect the other
tube.
If you have one tube with high gm and the other with a low gm, then
the amount of class A power produced in each tube varies.
This in effect is because the load seen by each tube working as an SE
tube
varies, and where you have RLa-a = 10k, then the class A load of each
tube = 5k in theory.
Tubes ain't any more perfect than I am, and you will find that perhaps
one tube "sees" 4.5k,
and the other sees 5.5k. Careful measurements with 10 ohm current
sensing R in each anode or cathode circuit
will tell you about the current flow in each tube.

Not only will the load seen by each of the two tubes in a class A
amplifier vary depending on the relative gm of the two tubes, but even
with two perfectly matched tubes the relative load each sees will vary
over the complete audio cycle due to the dynamic changes in the tubes
characteristics over the cycle.

Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

A sine wave symetrically clipped usually has a pile of odd numbered H,
and not much 2H. Some 2H is usually there though.

There is no 2H if the clipping is symmetrical.

Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping
progesses from mild to severe.

I'm not sure what you think this experiment proves? All it proves is
that a voltage amplifier or driver stage is producing some 2H at high
levels of drive, or that the output stage begins clipping asymmetrically
at high output levels due to some circuit feature or imperfection, I
don't see what it has to do with the issue of even order harmonic
cancellation? If one wanted to go to the trouble I have little doubt
that given some effort one could design a class AB amplifier that
wouldn't exhibit the problems you are alluding to. You are simply
dragging a Red Herring across the path with this issue which is
unrelated to the issue of even order harmonic cancellation in class AB
amplifiers when the tubes cutoff.

And these sorts of problems aren't restricted to class AB amplifiers,
there are class A PP amplifiers that show a rise in even order,
including 2H, distortion even before clipping sets in.


Regards,

John Byrns

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

[John Byrns]

In article ,
flipper wrote:

On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner
wrote:

In class A, what one tube does with the load current affect the other
tube.

Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are
essentially voltage controlled current sources and will source that
current regardless of what the 'other' is doing.

Yes, but what the other tube is doing will affect load seen by the first
tube, which will affect the voltage on its anode, if not its anode
current.

I think all this is just going to further confuse poor old Patrick, we
should probably let him alone so he can get back to his shed to wind
some output transformers and catch up with the backlog of Big US amps
awaiting modifications. While he working on that, his sub conscious can
work on the cutoff problem and perhaps an Epiphany will eventually come.


Regards,

John Byrns

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

[John Byrns]

In article ,
flipper wrote:

On Sun, 28 Oct 2007 21:40:26 GMT, John Byrns
wrote:

In article ,
flipper wrote:

On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner
wrote:

In class A, what one tube does with the load current affect the other
tube.

Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are
essentially voltage controlled current sources and will source that
current regardless of what the 'other' is doing.

Yes, but what the other tube is doing will affect load seen by the first
tube, which will affect the voltage on its anode, if not its anode
current.

Of course, but Patrick was speaking of what happens 'in the tube' and
(ideal) pentode conduction is independent of anode voltage

The load change will affect voltage but as long as everything is
'identical' (which is never the 'real world' case, of course) both
sides will see and do the same thing so, again, even harmonics are
canceled. Odd harmonics are another matter.

The point here, though, was that 'tube interaction', and 'mutual'
conduction, isn't the 'canceling' mechanism. It's the OPT summing two
equal but opposite waveforms and it's inherent to what even harmonics
'are'. With both halves 'equal but opposite' they just plain don't
exist no matter 'how' you got there and how many you started with,
because they 'cancel'. Even if it's just equal but opposite 1ns blips,
no even order harmonics.


I think all this is just going to further confuse poor old Patrick, we
should probably let him alone so he can get back to his shed to wind
some output transformers and catch up with the backlog of Big US amps
awaiting modifications. While he working on that, his sub conscious can
work on the cutoff problem and perhaps an Epiphany will eventually come.

Well, let's not get too smug about it because there could/should be
'some' interaction with triodes so his description has intuitive merit
in that case. But it's a special case due to plate feedback in triodes
(which could/should affect all harmonics to some extent) and not PP,
per see.

I may not have been very clear, but my second sentence was referring to
the issue of whether or not cutoff makes even order cancelation
impossible. I consider the interaction question I referred to in the
first sentence to be only a side issue to the original question of
cutoff defeating even order cancelation.


Regards,

John Byrns

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

[Patrick Turner]

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Eeyore wrote:

Patrick Turner wrote:

One cannot have distortion cancelling by one tube cancelling that in
another when one is cut off.

THANK YOU !

Basics do matter.

Indeed they do, but neither Patrick, myself, or anyone else is correct
on every issue. In this case Patrick has vigorously asserted that this
view, which he holds in common with you, is true, but he has failed to
even attempt an argument that might demonstrate its truth. Patrick is
an extremely skilled and talented fellow in the practical aspects of
tube amp design and construction, but he has a very limited
understanding of what is going on behind the scenes in the theory of
tube amp operation.

I think my website might indicate that your are not quite right about my
levels of understanding.

Can you point out where on your website it explains why "One cannot have
distortion cancelling by one tube cancelling that in another when one is
cut off"?

No, because I endorse what many books tell us such as RDH4.

Yes, and if we read, understand, and apply what the RDH4 tells us, we
understand that even harmonic cancellation does occur in class AB
amplifiers, even when the tubes are cutoff for part of the cycle. That
is what this discussion is all about, Multi-grid, at least I think
Multi-grid was the first, made the claim that even harmonic cancellation
doesn't occur when the tubes in a class AB amplifier are cut off, then
both you and Eeyore jumped in to back up his claim.

I have lost track of whatever it was that Multigrid might have said.

But once one tube cuts off, the part of the cycle handled by the other
tube alone
generates a substantially linear output voltage when summed with the
other tube's efforts. IMHO, the summing action of tubes or other devices
in class B does not include the cancelation of harmonics
because of the common mode application of the near identical even
harmonics
of the same phase of both tubes across the whole primary.
The class B action means the current wave of the device is a series
of 1/2 sine waves with a flat part and many harmonics are present
if you ever wanted to filter them out and quantify them.
But these harmonics don't appear in the load. Fact.
There isn't mutual cancellation like in class A.
But there is crossover distortion as one device turns off, and the other
turns on.

Class B is like the two guys sawing the log with the long bush saw,
and each guy only pulls the saw from centre position his way and pushes
it back
to centre then lets go. The mens' current production is severely
distorted, and jerky,
stop'start in fact, but they manage to saw the log with the saw
travelling
one way and the other like an output voltage due to the summed
applications of the
efforts, ie, currents.

Once one device cuts off, no current flows, so there are no distortion
currents produced
by that device.

So the only relevant issue is what each tube does when it turns on, and
has the load
power under its control.


Have you read pages 300 & 301 of the RDH4 as Doug Bannard, not to be
confused with the Multi-grid Doug, suggested?

Page 299 through to pg 303 is about Fourier Series and Harmonics.

There is so much high falooting math I don't understand very much, and
i don't need to. The pages do not contain a large range of
down to earth simple current wave forms of PP amps with the
list and typical amplitudes of the harmonics present.


There is not a lot there,
but it is a start and hopefully might lead you to follow up by pursuing
some more complete references on Fourier analysis and the theory behind
it.

I do not need to become fully familiar with Fourier analysis.
I am not formally university educated, and don'y undertsand the math.
I make amops and do whatever relevant hands on measuring and analysis
that I need to do
without putting on the garb of the white coated boffin.
I do know that were I to measure the harmonics of a typical AB PP amps
current waves,
I would find a shirtfull and bootfull of harmonics when the amp is well
into class AB.
But a very small fraction of them find their way into the secondary of
an OPT,
or into a direct connected load of complementary pnp and npn devices.


I refuse to spoon feed ppl about the basics.

But simple use of the brain about what happens inside each tube of a PP
pair
in class A, AB or B or C etc reveals to most minds what happens.

Simple use of the brain reveals that the cathode current of each tube of
a class A, AB or B amplifier contains even order harmonic currents,
among others. What simple use of the brain apparently doesn't reveal to
some is the reason these even harmonic currents are canceled in the
output of a PP amplifier.

In a class B amp, the output current produced by each output tube is
substantially
linear with input voltage for each 1/2 wave of the sine wave.
Its tuned off for the other 1/2.

Its mate does the same substantially linear job on the other 1/2 of the
wave not handled by
the first.

Its a simple case of each device substantially linearly amplifying each
1/2 wave.
The Fourier Analysis might describe things with math, and any given
repeating waveform can be expressed in terms of various amplitudes of
sine wave harmonics
with amplitudes and phase relationships required to give the original
un-analysed non sine wave.
The quantified harmonics are simply not relevant in the class B case.

It could be argued cancelation does take place, but I see things simply
as a
arched 1/2 wave being a linear effort with ZERO effort made upon the
other 1/2 of the wave.

Now were you to analyse and quantify and draw the graphs for the
fundememental
and harmonics of EACH class B device, you could congratulate yourself on
your wizardry,
but just getting a PC spectral analysis program to do it all for you
could be
less prone to errors after days of work doing the RDH4 math "by hand".

After doing the graphs of the current waves displaying them so they
represent the currents
flowing in/out of each end of the OPT primary, perhaps you'd find that
lo and behold,
when the harmonic waves are all viewed, it is seen they applied in
common mode to each
end of the load and thus cannot appear as load voltage current since the
amplitude and phase of the harmonics is the same at each end of the
load, ie,
each end of the OPT.

But I see no need to do this labourious and unecessary math and graphing
task which may satisfy
the lofty smartarse acedemics.
RDH4 does not have too many full descriptions to simply illustrate what
I have been saying.
So what use are the smartarse acedemics? have they earned their dinner?
I don't see they have
unless they make easy for lesser mortals to understand the concepts.
So the Fourier analysis is merely confusing.
I see the tubes simply turning on and off to do various parts of the
wave, and while 'on',
they have a linear control of output current, but while off, they have
no control.

In some SS amps, there are two crossover regions as the output voltage
lifts its own rail voltages
as in class H, all very complicated in Fourier terms but all you need to
know is the
timing sequence of what happens during wave cycles.
The Soundcraftsmen amp was class H.


If the two PP sides of the amplifier are
identical, then the cancellation will be complete, if the to sides are
not identical, for any one of a multitude of reasons, including but not
limited to, mismatched tubes then the cancellation will only be partial.

The perfectly complementary action of two devices in class B never ever
occurs.
Crossover distortion of some kind spoils the attempt at perfection.

Where you have EL34 on one side and 6L6 on the other of a PP AB amp,
expect a large amplitude DIFFERENCE in the applied commonality of
harmonic currents applied
to each end of the OPT, so you'd get considerably more resultant THD
than if you had a matched pair
of the same tube type number.

But in common sense terms, each of the pair of tubes are just switching
on and off,
and while off, each tube is oblivious to its Fourier productions. Its
bloody turned off,
to there ain't nothin going on. If a tube could talk, it'd tell you
I am NOT making any harmonics while I is turned off, OK.
You might answer the cheeky brat of a tube by saying but all your
harmonics
add to a flat line of zero current while you are turned off,
and don't argue, because you can't argue with that guy Fourier.
So all these many harmonics are there, but appear not to be there for
1/2 a wave....


I admit I do not understand this notion that even order distortion can't
be canceled when one tube is cut off, and I would really appreciate an
explanation from someone that understands why cancellation should fail
to occur?

Once one tube is well and truly cut right off, the other is conducting
delivery
of all the power through only 1/2 the OPT primary, and its as if the
other cut off tube is unplugged
for that part of the wave.

Yes, that is obvious, but not especially relevant.

We cannot remove tubes and plug them back in fast enough to demonstrate
the common sense notion...

Each output tube takes a turn at current delivery by means of turn on
harder to produce
voltage change which then SUM but without cancelation of current.

It is not necessary for each tube to be continuously delivering current
for the cancellation of the even order harmonic currents from the two
tubes to cancel, it is only necessary that the two currents exhibit the
required symmetry with respect to one another. The harmonic currents
flow even when the tube is cutoff, it's just that all the currents,
including the 2H current, happen to add to zero for the period the tube
is cutoff.

See the above comments.

The cancelation thinge is when the tubes are in class A and the 2H is
cancelled, similarly
to in any LTP.

"The cancellation thingy" is also when the tubes are in class AB or even
class B, where the 2H is also canceled.

One could prove almost anything with Fourier math....

Its easier to see the mutually cancelling 2H of each tube in a class A
amp.

But while tubes are cut right off, summed distortion currents in that
tube = zero.

I have to say this does not mean the distortion currents are NOT
PRESENT.
They must be, but all summ to zero amps for the flat line of the tube
when cut off.
If you filter out harmonics of the current wave of a single tube of the
clas B amp,
you'll see the plethera of harmonics, and these are all sine wave
signals
which flow continuously, and without cutting off, but which when added
will give you zero current for
1/2 the cycle wave at fundememtal F.



The Class A Wiulliamson has a common cathode resistance
which is unbypassed
to assist the cancellation, but ideally, a CCS should be used, or even a
choke for the cathode
resistance. One may argue whether having each cathode with separate RC
bypass networks
is better than a shared cathode impedance/resistance.
But the individual cathode RC or fixed bias does allow class AB
operation, wheras the
common cathode R/Z does not.

This is a diversion into a side issue but let me ask, why does a common
cathode RC not allow class AB?

Because after one tube cuts off, the common Rk is the Rk for one tube
only,
and the Ra of the tube rises acording to (µ + 1) x Rk, and severe 3H
distortion results.



Granted fixed bias provides far better
class AB operation than does cathode bias, but given that we are using
cathode bias I fail to see how a common cathode RC would prevent class
AB operation anymore than separate cathode RCs do?

Well, the C bypassing the common Rk does improve AB working
considerably.
Quad-II has 180 ohms plus 40uF between cathode FB winding and 0V.
So for music at well below clipping the biasing stays fixed.
But with a sine wave the biasing voltage Ek rises dramatically in AB
and is as bad as separate RC to each cathode.

What separate
cathode RCs do is reduce to some extent the effect of mismatched tubes
in the two sides of the PP circuit, but given well matched tubes a
common RC does not prevent class AB operation.

True.

Its the low ZC that counts. If the C is ommitted, class AB is horrid.

The separate RC give far better self regulation of bias than a common RC
does.



Work out the 2H current in each tube while in class A of each tube and
see how such currents are applied across the primary.
The reason 2H is low in class A is that the 2H currents are the same
phase at each end of the
OPT primary which cannot produce voltage in the load if the current is
applied in common mode to
both ends of the load.

The same is also true of the even order harmonic currents in class AB
and class B amplifiers!

Well yes, but they are less easy to see, as i explained above.

In effect we agree then, but practically speaking a tube cut off has
zero
harmonic distortion currents for the time its cut off.


I am beginning to get a glimmer of what your problem is here, it appears
that you are not correctly identifying the even order harmonic currents
that are applied to the output transformer primary in common mode. You
are looking at the total current waveform from each tube, which is
causing you to become confused. When the tube is cut off, the even
order harmonic currents continue to flow through the tube, its just that
the DC, fundamental, and all the harmonics together sum to zero during
the time the tube is cutoff, however this does not imply that the
current of any particular harmonic is aero during the time of cutoff,
hence cancellation goes on as if the tube had not cutoff.

I explained this above.

The humble man in the street can be forgiven if he doubts any currents
of any kind flow
when the DC and harmonic currents are ZERO when the tube is cut off.



Why I don't understand this is because even order distortion is canceled
in class B amplifiers, where one or the other tube is cutoff over almost
the entire cycle, yet cancellation still occurs!

Class B and C amps have a summing action of their turn on currents which
pull the voltage
in oposite directio.

It sounds like you are describing the push pull action of any class of
amplifier?

There is huge distortion in the Class B device
current,
but the total action produces a linear voltage outcome.

And that is exactly the point, which you persist in trying to deny.
Although odd order non-linearity does remain in class A, AB, and B
amplifiers.

a pair of complementary pnp and npn SS devices do exactly the same
thing.

But its not unusual that the amount of V swing achieved in each
direction by each class B device is not equal so
some 2H will allways appear in the output.
But usually, the 3H dominates H distortion products.

You are simply stating that in the real world, devices, drive signals
and other parameters of the two sides of a PP amplifier may not be
perfectly matched, and as a result the even harmonic cancellation will
be incomplete, but partial cancellation still occurs reducing the even
order harmonic distortion, even when one or the other tube is cutoff.

The even order artifacts in PP are generated when the two halves of the
sine wave fundemental
are amplified to a different amplitude; ie, the transfer curve of each
tube isn't the same...


I am having trouble
reconciling this with your claim that cancelation can't occur.

You have to see the distinction between each tube in class A "sharing
the load"
and the summing actions once cut off has occurred.
One could say correctly that the severe current distortions of each
device in class B
are cancelled by means of the summing. In effect they are.

In class A, what one tube does with the load current affect the other
tube.
If you have one tube with high gm and the other with a low gm, then
the amount of class A power produced in each tube varies.
This in effect is because the load seen by each tube working as an SE
tube
varies, and where you have RLa-a = 10k, then the class A load of each
tube = 5k in theory.
Tubes ain't any more perfect than I am, and you will find that perhaps
one tube "sees" 4.5k,
and the other sees 5.5k. Careful measurements with 10 ohm current
sensing R in each anode or cathode circuit
will tell you about the current flow in each tube.

Not only will the load seen by each of the two tubes in a class A
amplifier vary depending on the relative gm of the two tubes, but even
with two perfectly matched tubes the relative load each sees will vary
over the complete audio cycle due to the dynamic changes in the tubes
characteristics over the cycle.

And you will get resultant intermodulations....

But class A PP remains basically more linear process than SE...

Also the
related observation that there is no even order distortion in a sine
wave which has had both the tops and bottoms of the sine wave
symmetrically clipped off.

A sine wave symetrically clipped usually has a pile of odd numbered H,
and not much 2H. Some 2H is usually there though.

There is no 2H if the clipping is symmetrical.

But only if the clipping is symetrical.
In many amps it is not perfectly symetrical, and 2H remains substantial.



Try measuring one of your AB amps at 1W, 3.2W, 10W, 32W and as clipping
progesses from mild to severe.

I'm not sure what you think this experiment proves?

Depends what you measure.

If its the amplitude of each harmonic one will get some surprising
results showing varying levels of each harmonic as outpt voltage is
increased.

For any given amp, its very difficult to predict and or calculate the
rate of rise in harmonics with output voltage.

Better you merely assume you'll get THD, and then measure it to know how
much.

Some empiricle assumptions can be made, but to know, you must measure.


All it proves is
that a voltage amplifier or driver stage is producing some 2H at high
levels of drive, or that the output stage begins clipping asymmetrically
at high output levels due to some circuit feature or imperfection, I
don't see what it has to do with the issue of even order harmonic
cancellation? If one wanted to go to the trouble I have little doubt
that given some effort one could design a class AB amplifier that
wouldn't exhibit the problems you are alluding to. You are simply
dragging a Red Herring across the path with this issue which is
unrelated to the issue of even order harmonic cancellation in class AB
amplifiers when the tubes cutoff.

And these sorts of problems aren't restricted to class AB amplifiers,
there are class A PP amplifiers that show a rise in even order,
including 2H, distortion even before clipping sets in.

Don't worry, be happy.

Patrick Turner.



Regards,

John Byrns

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

[Patrick Turner]

flipper wrote:

On Sun, 28 Oct 2007 13:12:57 GMT, Patrick Turner
wrote:

snip.

I mentioned...

In class A, what one tube does with the load current affect the other
tube.

You said...

Not with pentodes (or any SS device) it doesn't. (Ideal) Pentodes are
essentially voltage controlled current sources and will source that
current regardless of what the 'other' is doing.

But the pentode (or tetrode) gm could be different, and thus if so,
current turn on/off
is different despite the same amplitude of grid input signals and total
summed output voltage being the same for each tube.
So the load each pentode sees is also different.
It does not follow that just because you have PP pentodes instead of
some other devices
such as triodes, bjts, or mosfets, that in PP class A the pentodes will
always be experiencing
the same load as each other.

But even with un-matched pentodes, the amount of class A 2H cancelling
is so substantial
that the 3H is still usually much more than any 2H left due to poor tube
matching.
That one PP output tube never sees an exactly equal load to the other is
of negligible consequence
because the THD in a couple of watts from a pair of tubes capable of 30
watts of class A will be inherently low.


Patrick Turner.

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
Patrick Turner wrote:

No, because I endorse what many books tell us such as RDH4.

Yes, and if we read, understand, and apply what the RDH4 tells us, we
understand that even harmonic cancellation does occur in class AB
amplifiers, even when the tubes are cutoff for part of the cycle. That
is what this discussion is all about, Multi-grid, at least I think
Multi-grid was the first, made the claim that even harmonic cancellation
doesn't occur when the tubes in a class AB amplifier are cut off, then
both you and Eeyore jumped in to back up his claim.

I have lost track of whatever it was that Multigrid might have said.

Multi-grid said something to the effect that 2H cancellation can't occur
in a class AB amplifier when one one or the other of the tubes is
cutoff, the same as you are, or at least were previously saying.

But once one tube cuts off, the part of the cycle handled by the other
tube alone
generates a substantially linear output voltage when summed with the
other tube's efforts. IMHO, the summing action of tubes or other devices
in class B does not include the cancelation of harmonics
because of the common mode application of the near identical even
harmonics
of the same phase of both tubes across the whole primary.

But that is just saying in different words that 2H cancellation does
occur even while one or the other tube is cutoff, "common mode
application of the near identical even harmonics of the same phase of
both tubes across the whole primary" is simply the mechanism for
accomplishing the 2H cancellation.

We are saying the same thing in different words.

The class B action means the current wave of the device is a series
of 1/2 sine waves with a flat part and many harmonics are present
if you ever wanted to filter them out and quantify them.
But these harmonics don't appear in the load. Fact.

Exactly, that is the whole point of 2H and even harmonic cancellation in
a PP stage, the odd harmonics do appear in the load however because they
appear across the output transformer as a differential signal, not a
common mode signal.

There isn't mutual cancellation like in class A.
But there is crossover distortion as one device turns off, and the other
turns on.

I don't see the distinction between class A and class B operation, the
2H and even harmonic cancellation mechanism is the same for both classes
of operation, the only difference is the amplitude of the harmonics
generated. In both cases the 2H and even harmonics appear at the output
transformer as a common mode signal and are canceled in the load by the
action of the transformer.

Once one device cuts off, no current flows, so there are no distortion
currents produced
by that device.

That does not follow, all that follows is that the sum of the DC
current, the fundamental current, and the harmonic distortion currents
sum to zero during the interval that the tube is cut off. That is not
the same thing as saying there are no distortion currents flowing while
the tube is cut off.

Mr. Fourier explains this for us.

[Big Snip]

But I see no need to do this labourious and unecessary math and graphing
task which may satisfy
the lofty smartarse acedemics.

Exactly, there is no need for you to do it because the academics have
all ready done it for you, so you already know that the 2H and other
even order harmonics will cancel in a PP stage, be it class A, class B,
or anywhere in between, or even if it is PP class C.

RDH4 does not have too many full descriptions to simply illustrate what
I have been saying.
So what use are the smartarse acedemics? have they earned their dinner?

Certainly they have, you are beginning to sound a little like Ian.

The cancelation thinge is when the tubes are in class A and the 2H is
cancelled, similarly
to in any LTP.

"The cancellation thingy" is also when the tubes are in class AB or even
class B, where the 2H is also canceled.

One could prove almost anything with Fourier math....

Its easier to see the mutually cancelling 2H of each tube in a class A
amp.

To me it is easier to see the mutually canceling 2H of each tube in a
class B amp because the amplitudes of the harmonics are larger and as a
result are easier to observe.

But while tubes are cut right off, summed distortion currents in that
tube = zero.

I have to say this does not mean the distortion currents are NOT
PRESENT.
They must be, but all summ to zero amps for the flat line of the tube
when cut off.
If you filter out harmonics of the current wave of a single tube of the
clas B amp,
you'll see the plethera of harmonics, and these are all sine wave
signals
which flow continuously, and without cutting off, but which when added
will give you zero current for
1/2 the cycle wave at fundememtal F.

Exactly, I think you have got it now, the even order distortion currents
cancel even while the tubes are cut off in either class AB or class B.

Work out the 2H current in each tube while in class A of each tube and
see how such currents are applied across the primary.
The reason 2H is low in class A is that the 2H currents are the same
phase at each end of the
OPT primary which cannot produce voltage in the load if the current is
applied in common mode to
both ends of the load.

The same is also true of the even order harmonic currents in class AB
and class B amplifiers!

Well yes, but they are less easy to see, as i explained above.

Your reasoning on this seems backwards, I think we both agree that the
harmonic currents in a class AB stage are larger than those in a class A
stage, I have always found it easier to see large things, within reason,
than to see small things.

In effect we agree then, but practically speaking a tube cut off has
zero
harmonic distortion currents for the time its cut off.

No, a tube cutoff does not have zero harmonic distortion currents for
the time it is cut off, what happens is that the harmonic distortion
currents, along with the fundamental current, and the DC bias current
sum to zero over that interval, which is a completely different thing
than saying the distortion currents are zero during that interval.

You even said this yourself a few paragraphs up from here where you said
"I have to say this does not mean the distortion currents are NOT
PRESENT. They must be, but all summ to zero amps for the flat line of
the tube when cut off."


Regards,

John Byrns

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

[Andre Jute]

On Oct 29, 2:14 pm, Patrick Turner wrote:

I have lost track of whatever it was that Multigrid might have said.

I expect so has Cuddles Multi-grid. He's a troll. He says whatever
comes into his head; it doesn't have to make sense, it just has to
stir up someone.

Andre Jute
"You don't need global feedback to build a good-sounding amplifier."
-- Henry Pasternack

[Andre Jute]

If it walks like a cancellation, if it talks like a cancellation, if
it sounds like a cancellation, if it causes less total secind harmonic
like a cancellation, if it *cancels* like a cancellation, it *is* a
cancellation.

I don't care who calls it an addition. That's just algebraic sleight
of hand performed inside the halves of a transformer: a cancellation
by any other name.

Quack. Eeeh. Deeeh.

Andre Jute
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review

[Patrick Turner]

I don't see the distinction between class A and class B operation, the
2H and even harmonic cancellation mechanism is the same for both classes
of operation, the only difference is the amplitude of the harmonics
generated. In both cases the 2H and even harmonics appear at the output
transformer as a common mode signal and are canceled in the load by the
action of the transformer.

Fair enough...

Once one device cuts off, no current flows, so there are no distortion
currents produced
by that device.

That does not follow, all that follows is that the sum of the DC
current, the fundamental current, and the harmonic distortion currents
sum to zero during the interval that the tube is cut off. That is not
the same thing as saying there are no distortion currents flowing while
the tube is cut off.

Mr. Fourier explains this for us.

But for ordinary ppl its difficult for them to see that
a number of harmonics flowing continually can sum to give a straight
line signal
with zero current at all.

[Big Snip]

But I see no need to do this labourious and unecessary math and graphing
task which may satisfy
the lofty smartarse acedemics.

Exactly, there is no need for you to do it because the academics have
all ready done it for you, so you already know that the 2H and other
even order harmonics will cancel in a PP stage, be it class A, class B,
or anywhere in between, or even if it is PP class C.

RDH4 does not have too many full descriptions to simply illustrate what
I have been saying.
So what use are the smartarse acedemics? have they earned their dinner?

Certainly they have, you are beginning to sound a little like Ian.

Hmm, some might say that the ordinary man would only feed them porridge
for dinner....

Me?, like Ian? heaven forbid!


The cancelation thinge is when the tubes are in class A and the 2H is
cancelled, similarly
to in any LTP.

"The cancellation thingy" is also when the tubes are in class AB or even
class B, where the 2H is also canceled.

One could prove almost anything with Fourier math....

Its easier to see the mutually cancelling 2H of each tube in a class A
amp.

To me it is easier to see the mutually canceling 2H of each tube in a
class B amp because the amplitudes of the harmonics are larger and as a
result are easier to observe.

Well they would be. I built a tunable filter with Q = 50
for all H between 1kHz and 10kHz, maybe I will connect to a cathode one
day
when an typical class AB signal is passing.


But while tubes are cut right off, summed distortion currents in that
tube = zero.

I have to say this does not mean the distortion currents are NOT
PRESENT.
They must be, but all summ to zero amps for the flat line of the tube
when cut off.
If you filter out harmonics of the current wave of a single tube of the
clas B amp,
you'll see the plethera of harmonics, and these are all sine wave
signals
which flow continuously, and without cutting off, but which when added
will give you zero current for
1/2 the cycle wave at fundememtal F.

Exactly, I think you have got it now, the even order distortion currents
cancel even while the tubes are cut off in either class AB or class B.

Its not obvious to most though. A cut off tube has no currents, period.


Work out the 2H current in each tube while in class A of each tube and
see how such currents are applied across the primary.
The reason 2H is low in class A is that the 2H currents are the same
phase at each end of the
OPT primary which cannot produce voltage in the load if the current is
applied in common mode to
both ends of the load.

The same is also true of the even order harmonic currents in class AB
and class B amplifiers!

Well yes, but they are less easy to see, as i explained above.

Your reasoning on this seems backwards, I think we both agree that the
harmonic currents in a class AB stage are larger than those in a class A
stage, I have always found it easier to see large things, within reason,
than to see small things.

In effect we agree then, but practically speaking a tube cut off has
zero
harmonic distortion currents for the time its cut off.

No, a tube cutoff does not have zero harmonic distortion currents for
the time it is cut off, what happens is that the harmonic distortion
currents, along with the fundamental current, and the DC bias current
sum to zero over that interval, which is a completely different thing
than saying the distortion currents are zero during that interval.

You even said this yourself a few paragraphs up from here where you said
"I have to say this does not mean the distortion currents are NOT
PRESENT. They must be, but all summ to zero amps for the flat line of
the tube when cut off."

OK, I am just showing the difference between commonsense perceptions and
theoretical perceptions involving Fourier analysis.

Complementary pairs of transistors which operate close to class B and
spend all their lives
each only half amplifying music create the same perception dilemnas.
One should be able to filter then measure the many harmonic currents in
one of the 0.22 ohm emitter resistors
and get a string of harmonics in a filter which will add up to zero
current for 1/2 the wave.

Then when you measure the other 0.22 ohm R there should be a similar set
of harmonic currents
present which must shunt those of the other bjt, giving a net current
applied to the
SINGLE output terminal which must sum substantially close to zero,
leaving only crossover distortions.

Its never explained this way in the books though.

There are never any graphs pf all the harmonics shown in correct to
scale amplitudes and phase.

Patrick Turner




Regards,

John Byrns

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

[Patrick Turner]

There is no best load for a class B triode amp. Class B is a horrid way
to build any amp.

That is certainly a broad brush stroke, McIntosh did a nice business in
what were essentially class B amplifiers. Many of the older readers
here enjoyed Rock & Roll music during their teen years delivered via the
class B amplifier at the local AM radio station, without "horrid"
results.

The crock and troll was so full of distortions and screaming gits that
using low distortion
radio gear was wasted on them.


Maybe you meant low bias class AB.

No, I actually meant class B.

Well, class B is where corssover distortion is not pretty....

Do the load line analysis, or have a look at my website pages and print
out a set of curves for 6550
which are virtually the same as KT88.

http://www.turneraudio.com.au/loadma...p-triodes.html

I did the analysis as I earlier had said that I would. 1.25k does seem
to be a reasonable class B load for the KT-88/6550 just as you said.

But only where Ea isn't too high.

But KT88 were run with up to Ea = 800V, in beam mode, and with fixed
bias and for
140 watts and working class AB2, and large PO is available.
TT21 were ideal at this. I forget the load value used.


I asked the question because I have not built any amps with this tube and
am not familiar with it beyond the fact that the Quadraplex VTRs at the
Television Station where I worked as a youth had a couple dozen 6550s in
each VTR. The class B load of 1.25k seemed low to me relative to your
2.5k class A load, but I made two erroneous assumptions in asking that
question. First I didn't realize that ra for the KT-88 is as low as it
is, and second I didn't take into consideration that your class A load
line is dissipation limited rather than voltage limited as with the
class B load line.

5k is a typical RLa-a for a pair of KT88.

I have seen many with 3.2k.
When wound up to full power with a sine wave they might smoke.

Even 4k was used with Quad-II for a pair of KT66.

So when the amp works AB, the B part of the operation means each tube
sees only 1k
on alternate wave crests.

I quite like Ea = 500V and RLa-a = 8k for KT88/6550.

This means the B load is 2k min, and distortion in UL or CFB isn't too
much.

The change in load from the class A load of 1/2 RLa-a to class B load of
1/4 RLa-a
means that there is a dynamic gain reduction on each wave peak leading
to
3H and other odd H, 5H especially. Class AB beam and pentode amps are
worst, triode AB amps are better,
and make the best AB amps. UL is between the two with regard to
harmonics
generated by the cut off and load changes.


Patrick Turner.




Regards,

John Byrns

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

[Patrick Turner]

Andre Jute wrote:

If it walks like a cancellation, if it talks like a cancellation, if
it sounds like a cancellation, if it causes less total secind harmonic
like a cancellation, if it *cancels* like a cancellation, it *is* a
cancellation.

I don't care who calls it an addition. That's just algebraic sleight
of hand performed inside the halves of a transformer: a cancellation
by any other name.

Politicians can sum and cancel almost any economic strategems,
mostly leaving behind them an odd order of broken promises,
and gaining no realization they were ever wrong.

Of some engineers, similar could be said, but for the man in the street
unused to Fourier, a cut off tube is as silent, harmless, dead, and
as distortion current free as a dead politician. Even better than a dead
cat.

We have a Federal Election in a few weeks and the BS is really flowing
well,
and all sorts of arguments happening.

The Liberal pollies can't see that reducing wages and putting up
interest rates really hurts ppl
all the while saying they are not doing either.

The pollies would tell us that income and expenditure currents of many
types flow and sum together to cancel
flows and cut offs in the next person along to give a single wonderful
income and fundemental wellbeing
to all. A Mr Furious dreamed up the theory of frequent political
expedients.

I heard that a candidate Blue Party candidate, Mr Phark Yutu is looking
to get a large vote.


Anyway, back at the amplifier, too much switching and stop start
amplifying is dreadful for the
music. The less we rely on Fourrier, the better.

Patrick Turner.










Quack. Eeeh. Deeeh.

Andre Jute
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review

[Rudy]

"John Byrns" wrote in message
...
: In article .com,
: Andre Jute wrote:
:
: On Oct 25, 7:40 am, Eeyore
: wrote:
:
: I'd love to know how that happens. There's no cancellation of ANYTHING
once
: one
: side has ceased conducting !
:
: Graham
:
: Holy ****! Did I say yet that Poopie is ignorant and incompetent?
:
: Nah, nobody can be that stupid and uninformed about tube basics.
:
: There are a lot of "stupid and uninformed" people around, there are at
: least three people involved in this discussion that have expressed this
: same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore.
:
:
: Regards,
:
: John Byrns

it seems the error made is in using the wrong mental model, that is
seeing the transformer primary as a resistance.
sure, PP operating on a pure R load would cease to
cancel when one side would be cutoff, as it would no longer
partake in the transfer curve for PP !
with a transformer, however, the transfer curve is a function
of the tight coupling of the primary halves, so always PP ;-)
(not the current, but the core flux is what matters)

Rudy

[Patrick Turner]

Rudy wrote:

"John Byrns" wrote in message
...
: In article .com,
: Andre Jute wrote:
:
: On Oct 25, 7:40 am, Eeyore
: wrote:
:
: I'd love to know how that happens. There's no cancellation of ANYTHING
once
: one
: side has ceased conducting !
:
: Graham
:
: Holy ****! Did I say yet that Poopie is ignorant and incompetent?
:
: Nah, nobody can be that stupid and uninformed about tube basics.
:
: There are a lot of "stupid and uninformed" people around, there are at
: least three people involved in this discussion that have expressed this
: same belief as Eeyore, they are Multi-grid, Patrick Turner, and Eeyore.
:
:
: Regards,
:
: John Byrns

it seems the error made is in using the wrong mental model, that is
seeing the transformer primary as a resistance.

To the vacuum tubes, RLa-a IS a resistance.

One R set up set up with a CT so any signal voltage applied between the
end and the CT
is applied in opposite phase to the other end, courtesy of the action of
the transformer.

Its RL a-a.

This transformed single R appears to the tubes somewhat variably.
In class B, while a tube is conducting, the load on the tube is 1/4
RLa-a,
and when both conduct in class A the load on each tube is 1/2RLa-a.

The secondary R load is always reflected to the two tubes as RLa-a, with
a CT.
The Ra of the tubes seen at the secondary varies.

Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR =
1,000:1
In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT.
In class B, only one tube is connected via 1/2 the primary, so only one
Ra = 1.2k
and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't
operating
because one tube is cut off, so turn ratio is halved, Z ratio Ra in
class B = 1.5k/6 = 250,
so Ra at the output = 1.5k/250 = 6 ohms.

The Rout of the amplifier at the sec is Ra transformed, and notice the
change in Rout
between A and AB. Its responsible for major 3H, 5H and other
distortions.



sure, PP operating on a pure R load would cease to
cancel when one side would be cutoff, as it would no longer
partake in the transfer curve for PP !
with a transformer, however, the transfer curve is a function
of the tight coupling of the primary halves, so always PP ;-)
(not the current, but the core flux is what matters)

If one runs a class B PP amp with only ONE output tube,
you will get a badly distorted signal like the one you'd get
with a pure un-magnetically coupled R between B+ and the anode,
only at each end of the OPT pri there would be two phases of the same
distorted signal.
The load with one tube and OPT is the same as with a single R taken to a
suitably higher B+
than used with an OPT.

Fourier could describe what harmonics flow all the time despite cut off.

Furious arguemnts have followed about him.

Patrick Turner.


Rudy

[Rudy]

"Patrick Turner" wrote in message
...
:
:
: Rudy wrote:
:
: "John Byrns" wrote in message
: ...
: : In article .com,
: : Andre Jute wrote:
: :
: : On Oct 25, 7:40 am, Eeyore
: : wrote:
: :
: : I'd love to know how that happens. There's no cancellation of
ANYTHING
: once
: : one
: : side has ceased conducting !
: :
: : Graham
: :
: : Holy ****! Did I say yet that Poopie is ignorant and incompetent?
: :
: : Nah, nobody can be that stupid and uninformed about tube basics.
: :
: : There are a lot of "stupid and uninformed" people around, there are at
: : least three people involved in this discussion that have expressed
this
: : same belief as Eeyore, they are Multi-grid, Patrick Turner, and
Eeyore.
: :
: :
: : Regards,
: :
: : John Byrns
:
: it seems the error made is in using the wrong mental model, that is
: seeing the transformer primary as a resistance.
:
: To the vacuum tubes, RLa-a IS a resistance.

for the purpose of loadline analyses, that is a fair approximation.
for analyzing the actual working mechanism, it lacks accounting for
the element of time, as the transformer actually stores energy
comparable to a fast rotating heavy object storing rotational impulse
so in case a tube is cutoff, it's plate will happily follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded, perhaps

the output result will just be more or less symmetrically compressed,
3H, 5H, etc.

this is distinctly different from a transistor PP, mostly apparent
when you think of asymmetrical signals (music) driving it ;-)

the load each PP tube sees is not suddenly changed at cutoff,
it is in fact only balanced somewhat is a small region around the
bias point, as with larger swing the dynamic plate resistance
will vary more between the tubes, 3H will start to appear

a measure of 'true class A' power of an AB pp, meaning
the 3H would be low without nfb, would be the region where
current is below twice the bias and cutoff AND dynamic
plate R of the output tubes doesn't change more than
"X" percent.

.... to be discussed,
heh,

Rudy

:
: One R set up set up with a CT so any signal voltage applied between the
: end and the CT
: is applied in opposite phase to the other end, courtesy of the action of
: the transformer.
:
: Its RL a-a.
:
: This transformed single R appears to the tubes somewhat variably.
: In class B, while a tube is conducting, the load on the tube is 1/4
: RLa-a,
: and when both conduct in class A the load on each tube is 1/2RLa-a.
:
: The secondary R load is always reflected to the two tubes as RLa-a, with
: a CT.
: The Ra of the tubes seen at the secondary varies.
:
: Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR =
: 1,000:1
: In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT.
: In class B, only one tube is connected via 1/2 the primary, so only one
: Ra = 1.2k
: and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't
: operating
: because one tube is cut off, so turn ratio is halved, Z ratio Ra in
: class B = 1.5k/6 = 250,
: so Ra at the output = 1.5k/250 = 6 ohms.
:
: The Rout of the amplifier at the sec is Ra transformed, and notice the
: change in Rout
: between A and AB. Its responsible for major 3H, 5H and other
: distortions.
:
:
:
: sure, PP operating on a pure R load would cease to
: cancel when one side would be cutoff, as it would no longer
: partake in the transfer curve for PP !
: with a transformer, however, the transfer curve is a function
: of the tight coupling of the primary halves, so always PP ;-)
: (not the current, but the core flux is what matters)
:
: If one runs a class B PP amp with only ONE output tube,
: you will get a badly distorted signal like the one you'd get
: with a pure un-magnetically coupled R between B+ and the anode,
: only at each end of the OPT pri there would be two phases of the same
: distorted signal.
: The load with one tube and OPT is the same as with a single R taken to a
: suitably higher B+
: than used with an OPT.
:
: Fourier could describe what harmonics flow all the time despite cut off.
:
: Furious arguemnts have followed about him.
:
: Patrick Turner.
:
:
: Rudy

[John Byrns]

In article ,
"Rudy" wrote:

for the purpose of loadline analyses, that is a fair approximation.
for analyzing the actual working mechanism, it lacks accounting for
the element of time, as the transformer actually stores energy
comparable to a fast rotating heavy object storing rotational impulse
so in case a tube is cutoff, it's plate will happily follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded, perhaps

Huh, wouldn't the transformer store energy more like a spring does,
after all the schematic symbols are very similar?

If "the transformer actually stores energy comparable to a fast rotating
heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a
serious loss of high frequencies?


Regards,

John Byrns

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

[Rudy]

"John Byrns" wrote in message
...
: In article ,
: "Rudy" wrote:
:
: for the purpose of loadline analyses, that is a fair approximation.
: for analyzing the actual working mechanism, it lacks accounting for
: the element of time, as the transformer actually stores energy
: comparable to a fast rotating heavy object storing rotational impulse
: so in case a tube is cutoff, it's plate will happily follow to wherever
: that winding will take it.
: (well, not so happy when certain V's are exceeded, perhaps
:
: Huh, wouldn't the transformer store energy more like a spring does,
: after all the schematic symbols are very similar?
:
: If "the transformer actually stores energy comparable to a fast rotating
: heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a
: serious loss of high frequencies?
:
:
: Regards,
:
: John Byrns
:
err, yes, i just wanted to emphasize the
'flux doesn't stop on a dime' aspect.
:-)
Rudy

[Multi-grid]

On Oct 31, 4:46 pm, John Byrns wrote:
In article ,

"Rudy" wrote:
for the purpose of loadline analyses, that is a fair approximation.
for analyzing the actual working mechanism, it lacks accounting for
the element of time, as the transformer actually stores energy
comparable to a fast rotating heavy object storing rotational impulse
so in case a tube is cutoff, it's plate will happily follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded, perhaps

Huh, wouldn't the transformer store energy more like a spring does,
after all the schematic symbols are very similar?

If "the transformer actually stores energy comparable to a fast rotating
heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a
serious loss of high frequencies?

Regards,

John Byrns

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

Yes it does. This is from the leakage inductance. Things come to a
complete stop when leakage L and capacitance resonate...
cheers,
Douglas

[John Byrns]

In article .com,
Multi-grid wrote:

On Oct 31, 4:46 pm, John Byrns wrote:
In article ,

"Rudy" wrote:
for the purpose of loadline analyses, that is a fair approximation.
for analyzing the actual working mechanism, it lacks accounting for
the element of time, as the transformer actually stores energy
comparable to a fast rotating heavy object storing rotational impulse
so in case a tube is cutoff, it's plate will happily follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded, perhaps

Huh, wouldn't the transformer store energy more like a spring does,
after all the schematic symbols are very similar?

If "the transformer actually stores energy comparable to a fast rotating
heavy object" wouldn't that cause a 6 dB/octave roll-off, resulting in a
serious loss of high frequencies?

Regards,

John Byrns

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

Yes it does. This is from the leakage inductance. Things come to a
complete stop when leakage L and capacitance resonate...

We specify a well designed and wound transformer with low leakage
inductance so that it doesn't cause high frequency roll-off problems
until well above the audio range. The impact of excess leakage
inductance is also minimized by using a pentode output tube, which given
your name I am sure you know all about.



Regards,

John Byrns

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

[Patrick Turner]

Rudy wrote:

"Patrick Turner" wrote in message
...
:
:
: Rudy wrote:
:
: "John Byrns" wrote in message
: ...
: : In article .com,
: : Andre Jute wrote:
: :
: : On Oct 25, 7:40 am, Eeyore
: : wrote:
: :
: : I'd love to know how that happens. There's no cancellation of
ANYTHING
: once
: : one
: : side has ceased conducting !
: :
: : Graham
: :
: : Holy ****! Did I say yet that Poopie is ignorant and incompetent?
: :
: : Nah, nobody can be that stupid and uninformed about tube basics.
: :
: : There are a lot of "stupid and uninformed" people around, there are at
: : least three people involved in this discussion that have expressed
this
: : same belief as Eeyore, they are Multi-grid, Patrick Turner, and
Eeyore.
: :
: :
: : Regards,
: :
: : John Byrns
:
: it seems the error made is in using the wrong mental model, that is
: seeing the transformer primary as a resistance.
:
: To the vacuum tubes, RLa-a IS a resistance.

for the purpose of loadline analyses, that is a fair approximation.
for analyzing the actual working mechanism, it lacks accounting for
the element of time, as the transformer actually stores energy
comparable to a fast rotating heavy object storing rotational impulse
so in case a tube is cutoff, it's plate will happily follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded, perhaps

The analogy is fine, as long as you remember there is no such thing as
momentum
in electronics.

the output result will just be more or less symmetrically compressed,
3H, 5H, etc.

this is distinctly different from a transistor PP, mostly apparent
when you think of asymmetrical signals (music) driving it ;-)

Asymetrical music signals will still have a dc content of zero,
and there's very nearly always equal energy each side of zero.

In a bjt output stage or a tube stage where the load is taken from the
junction of
the two output bjts or tubes, the Fourier waves are shunting one
another,
rather than being applied common mode as John B is so adament about.

If one device is omitted, the class B waves are like you get with an SE
tube
biased on the brink of cut off.



the load each PP tube sees is not suddenly changed at cutoff,
it is in fact only balanced somewhat is a small region around the
bias point, as with larger swing the dynamic plate resistance
will vary more between the tubes, 3H will start to appear

a measure of 'true class A' power of an AB pp, meaning
the 3H would be low without nfb, would be the region where
current is below twice the bias and cutoff AND dynamic
plate R of the output tubes doesn't change more than
"X" percent.

Hmm, a little hard to follow you, but while a tube is cut off there
is *no load* connected to it, unless in other words we say the Fourier
currents all add to zero current for
1/2 a wave and are flowing in tube and 1/2 the OPT, but are conveniently
summing to 0.0amps dc or ac for
that 1/2 cycle.

Usually 3H is HIGH witout NFB and unacceptably high when tubes are
biased for class B at low levels,
then as PO increases the % reduces, then inceases again near clipping.

This was a problem with early SS amps; the early designers didn't know
how to deal with
the SS distortions at low levels where tube amps at the same level were
much freer from distortions,
and spectra was benign.

The early designers could not afford to have hot idling class AB stages
because germanium transistors
went phut all too easily when they got hot.
So they were forced int making all their crummy creations using class B
working,
and with token idle current.

Bean counters made damn well sure OPTs were damn well eliminated from
the SS amps
during the transition from tubes to SS. For awhile, input transformers
on SS amps were used because of the
appallingly low base input resistance of the early power transistors,
and besides,
there were lots of out of work OPT winders available to wind simpler
cheap IPT for SS
to fill them in until permanent retrenchment, retirement or the sack.

The appalling design ideas carried over into the days of early silicon
bjts which turned out
to be much more rugged than the germanium types.

But SS heralded the use of class B, and the use of ever increasing
amounts of NFB,
so that we have amps with 40dB of local series voltage NFB in the form
of emitter follower connection,
and another 60dB of global NFB, to crossover distortion is difficult to
identify as a spurious artifact
amoungst the macro 3H and 2H which dominate the THD spectra just below
clipping,
but which rarely exceed 0.005% if the designer knows his stuff.

In effct, many VAS stages in SS amps are merely a gain bjt with CC
collector load to power
a relaetively high impedance base input impedance of darligton output
stage device arrays.
The Single Ended VAS stage often has gain = 5,000x, and THD of this
stage is say 5% at say
25Vrms, mainly 2H and 3H, very much like a pentode's spectra. The VAS
stage THD well exceeds the
output stage THD because of the emitter follower NFB.
The 60dB of global NFB reduces the 5% of VAS thd to 0.005%.

Because tubes have lower gain when used in audio circuits than available
with bjts,
and because of stability reasons, a large amount of NFB cannot be
applied, with a total of 40 dB being about the limit.



Patrick Turner.



... to be discussed,
heh,

Rudy

:
: One R set up set up with a CT so any signal voltage applied between the
: end and the CT
: is applied in opposite phase to the other end, courtesy of the action of
: the transformer.
:
: Its RL a-a.
:
: This transformed single R appears to the tubes somewhat variably.
: In class B, while a tube is conducting, the load on the tube is 1/4
: RLa-a,
: and when both conduct in class A the load on each tube is 1/2RLa-a.
:
: The secondary R load is always reflected to the two tubes as RLa-a, with
: a CT.
: The Ra of the tubes seen at the secondary varies.
:
: Say you have Ra = 1.2k for EL34 in triode, and 6k:6 ohms OPT. ZR =
: 1,000:1
: In class A, Ra-a = 2.4k and is transformed to 2.4ohms by the OPT.
: In class B, only one tube is connected via 1/2 the primary, so only one
: Ra = 1.2k
: and OPT Z ratio has changed to 1.5k:6, because 1/2 the OPT primary isn't
: operating
: because one tube is cut off, so turn ratio is halved, Z ratio Ra in
: class B = 1.5k/6 = 250,
: so Ra at the output = 1.5k/250 = 6 ohms.
:
: The Rout of the amplifier at the sec is Ra transformed, and notice the
: change in Rout
: between A and AB. Its responsible for major 3H, 5H and other
: distortions.
:
:
:
: sure, PP operating on a pure R load would cease to
: cancel when one side would be cutoff, as it would no longer
: partake in the transfer curve for PP !
: with a transformer, however, the transfer curve is a function
: of the tight coupling of the primary halves, so always PP ;-)
: (not the current, but the core flux is what matters)
:
: If one runs a class B PP amp with only ONE output tube,
: you will get a badly distorted signal like the one you'd get
: with a pure un-magnetically coupled R between B+ and the anode,
: only at each end of the OPT pri there would be two phases of the same
: distorted signal.
: The load with one tube and OPT is the same as with a single R taken to a
: suitably higher B+
: than used with an OPT.
:
: Fourier could describe what harmonics flow all the time despite cut off.
:
: Furious arguemnts have followed about him.
:
: Patrick Turner.
:
:
: Rudy

[Ian Iveson]

John Byrns wrote:

Rudy wrote:

for the purpose of loadline analyses, that is a fair
approximation.
for analyzing the actual working mechanism, it lacks
accounting for
the element of time, as the transformer actually stores
energy
comparable to a fast rotating heavy object storing
rotational impulse
so in case a tube is cutoff, it's plate will happily
follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded,
perhaps

: Huh, wouldn't the transformer store energy more like a
spring does,
: after all the schematic symbols are very similar?

A capacitor is analogous to a spring. Don't be fooled by the
symbols. A resistor is analogous to a damper, incidentally.

: If "the transformer actually stores energy comparable to
a fast rotating
: heavy object" wouldn't that cause a 6 dB/octave
roll-off, resulting in a
: serious loss of high frequencies?

Low frequencies, considering the mass is in shunt. Yes, it
does.

The concept of momentum in electronics is a bit interesting.
Is LI concerved?

If I have a perfect inductor passing current in a loop, and
switch a second, equal, inductor into the circuit in series
with the first, will the current be halved? If instead I
double the inductance by inserting a core into the first
inductor, will the current be also halved? In both cases
large voltages would accompany the change, analogous to
collision forces.

Ian

[John Byrns]

In article ,
"Ian Iveson" wrote:

John Byrns wrote:

Rudy wrote:

for the purpose of loadline analyses, that is a fair
approximation.
for analyzing the actual working mechanism, it lacks
accounting for
the element of time, as the transformer actually stores
energy
comparable to a fast rotating heavy object storing
rotational impulse
so in case a tube is cutoff, it's plate will happily
follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded,
perhaps

: Huh, wouldn't the transformer store energy more like a
spring does,
: after all the schematic symbols are very similar?

A capacitor is analogous to a spring. Don't be fooled by the
symbols. A resistor is analogous to a damper, incidentally.

It depends on which dual you are using, an inductor can also be
analogous to a spring and a capacitor to a mass.


Regards,

John Byrns

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

[Don Pearce]

On Thu, 01 Nov 2007 12:26:01 -0500, John Byrns
wrote:

In article ,
"Ian Iveson" wrote:

John Byrns wrote:

Rudy wrote:

for the purpose of loadline analyses, that is a fair
approximation.
for analyzing the actual working mechanism, it lacks
accounting for
the element of time, as the transformer actually stores
energy
comparable to a fast rotating heavy object storing
rotational impulse
so in case a tube is cutoff, it's plate will happily
follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded,
perhaps

: Huh, wouldn't the transformer store energy more like a
spring does,
: after all the schematic symbols are very similar?

A capacitor is analogous to a spring. Don't be fooled by the
symbols. A resistor is analogous to a damper, incidentally.

It depends on which dual you are using, an inductor can also be
analogous to a spring and a capacitor to a mass.


Regards,

John Byrns

I haven't given this any thought, but my intuition says that when you
give an inductor a kick with a voltage, it is reluctant to get moving
(current flow), which makes it more like the mass. How would you
intuit that the other way?

d

--
Pearce Consulting
http://www.pearce.uk.com

[John Byrns]

In article ,
(Don Pearce) wrote:

On Thu, 01 Nov 2007 12:26:01 -0500, John Byrns
wrote:

In article ,
"Ian Iveson" wrote:

John Byrns wrote:

Rudy wrote:

for the purpose of loadline analyses, that is a fair
approximation.
for analyzing the actual working mechanism, it lacks
accounting for
the element of time, as the transformer actually stores
energy
comparable to a fast rotating heavy object storing
rotational impulse
so in case a tube is cutoff, it's plate will happily
follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded,
perhaps

: Huh, wouldn't the transformer store energy more like a
spring does,
: after all the schematic symbols are very similar?

A capacitor is analogous to a spring. Don't be fooled by the
symbols. A resistor is analogous to a damper, incidentally.

It depends on which dual you are using, an inductor can also be
analogous to a spring and a capacitor to a mass.


Regards,

John Byrns

I haven't given this any thought, but my intuition says that when you
give an inductor a kick with a voltage, it is reluctant to get moving
(current flow), which makes it more like the mass. How would you
intuit that the other way?

It all depends on whether you are using the Force-voltage analogy or the
Force-current analogy.

Similarly you can redraw a purely electrical circuit so that all the
capacitors become inductors, the inductors become capacitors, the
resistors become conductances, voltages become currents, and currents
become voltages.


Regards,

John Byrns

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

[Don Pearce]

On Thu, 01 Nov 2007 13:40:47 -0500, John Byrns
wrote:

In article ,
(Don Pearce) wrote:

On Thu, 01 Nov 2007 12:26:01 -0500, John Byrns
wrote:

In article ,
"Ian Iveson" wrote:

John Byrns wrote:

Rudy wrote:

for the purpose of loadline analyses, that is a fair
approximation.
for analyzing the actual working mechanism, it lacks
accounting for
the element of time, as the transformer actually stores
energy
comparable to a fast rotating heavy object storing
rotational impulse
so in case a tube is cutoff, it's plate will happily
follow to wherever
that winding will take it.
(well, not so happy when certain V's are exceeded,
perhaps

: Huh, wouldn't the transformer store energy more like a
spring does,
: after all the schematic symbols are very similar?

A capacitor is analogous to a spring. Don't be fooled by the
symbols. A resistor is analogous to a damper, incidentally.

It depends on which dual you are using, an inductor can also be
analogous to a spring and a capacitor to a mass.


Regards,

John Byrns

I haven't given this any thought, but my intuition says that when you
give an inductor a kick with a voltage, it is reluctant to get moving
(current flow), which makes it more like the mass. How would you
intuit that the other way?

It all depends on whether you are using the Force-voltage analogy or the
Force-current analogy.

Similarly you can redraw a purely electrical circuit so that all the
capacitors become inductors, the inductors become capacitors, the
resistors become conductances, voltages become currents, and currents
become voltages.

Thought you might say that! No, I've never bought into the notion of
current causing voltage, it is always the other way round in my head.

You apply a voltage, and a current results.

d

--
Pearce Consulting
http://www.pearce.uk.com

[Ian Iveson]

John said:

A capacitor is analogous to a spring. Don't be fooled by
the
symbols. A resistor is analogous to a damper,
incidentally.

It depends on which dual you are using, an inductor can
also be
analogous to a spring and a capacitor to a mass.

And a mass to a spring and a spring to a mass?

It is more intuitive surely to consider voltage to be an
analogue of mechanical force.

Whichever way round you think, I suppose. The analogy is of
limited value unless you are better at analysing complex
mechanical networks than you are at electronics, or you're a
mechanical engineer into analogue computing. The equations
are the same in the end.

cheers, Ian