John Stewart wrote:

BFoelsch wrote:

"John Stewart" wrote in message
...
For the real deal look at
http://www.bonavolta.ch/hobby/en/audio/split.htm

JLS

See ABSE for a differential pair implemented with dissimilar tubes.
As long as the cathode return for the LTP is a real current source & the
tubes are
not overloaded the outputs will be equal for equal loads.

Problems arise as the weaker tube runs into overload.

Real problems with drift in a DC amplifier.

JLS

If the current source is perfect, the 6SN7 can be replaced by a
cathode-anode short and the output will still be perfectly balanced (AC
wise, of course), will it not?

An interesting concept. The current in the tubes follow the 3/2 power law,
while a short (piece) of wire would be linear, I guess. That should make for
more discussion.

Cheers, John Stewart

Ian Iverson made the assertion that one tube could be replaced
by a resistor equal to the static resistance of one triode.
He suggested a damp carrot be used, and still you'd get equal
inverted output, but gain of about 0.35 from input to each anode..
I did the math on the circuit arrangement on the follow up post.
If the "damp carrot" resistance is well bypassed by a an electro cap,
the circuit works as a normal CPI, with the same gain.
I cannot think of any use for such a creature as a Turnip-Carrot PI.


Patrick Turnip.

John Stewart wrote:

John Stewart wrote:

BFoelsch wrote:

"John Stewart" wrote in message
...
For the real deal look at
http://www.bonavolta.ch/hobby/en/audio/split.htm

JLS

See ABSE for a differential pair implemented with dissimilar tubes.
As long as the cathode return for the LTP is a real current source & the
tubes are
not overloaded the outputs will be equal for equal loads.

Problems arise as the weaker tube runs into overload.

Real problems with drift in a DC amplifier.

JLS

If the current source is perfect, the 6SN7 can be replaced by a
cathode-anode short and the output will still be perfectly balanced (AC
wise, of course), will it not?

An interesting concept. The current in the tubes follow the 3/2 power law,
while a short (piece) of wire would be linear, I guess. That should make for
more discussion.

Cheers, John Stewart

Thinking about that a little further, I believe there would be no gain at all.
The current in the 6SL7 cathode circuit is held constant, as though connected to
a very large value resistor. So I guess you get massive cathode degeneration &
no output at the plate at all.

Here is a copy of my reply to Ian :-
----------------------------
Consider the use of a single 1/2 of an 6SN7 triode with RL = 50k, IaQ = 2
mA,
and cathode taken to CCS tail with 4 mA.
Then we could have a carrot with 100k resistance, and another RL of 50k
to the B+, in this case let it be +300v, so with 2 mA in each RL
the anode and 50k-100k junction are both sitting on +200v.

If there is a -50v signal at the 50k-100k junction, there is a signal
current of
1 ma in the 50k, so at the cathode of the tube, bottom of the 100k carrot,
there must be a -150v signal.
Then since there is no current change in the CCS, there must be a
1 mA change in the tube, and the anode voltage change is +50v.
The load seen by the tube is 50k plus 150k effectively, so gain is
say 19 for a 6SN7, so the Vgk needed to produce the 200v change across the
tube
is 10.52 v, so the whole grid input voltage to give both the +/-50v output
voltages
at each 50k is -160.52v.

If the 100k was bypassed with a large electro, then the
situation is quite different, but then similar to a concertina
phase inverter, because the cathode load has been moved so
it is working with a differently biased load, but a load equal to the anode
load
of this lone triode.
So then the circuit is a true concertina phase inverter,
which needs about -55.2 v of input to make +50v at the anode, and -50v at
the "carrot" output.

It is not in any way able to work as a differential amplifier.

Patrick Turner.
-----------------------------------






JLS

Actually, with only 5 ma in the CC source, all of that would run thru the 27K that
was in the 6SN7 plate circuit & pull the 6SL7 cathode up so that the 6SL7 would be
cut off. To get the 6SL7 turned on you would need something like 11 ma from the CC
source to get the 6SL7 biased on. That would get it's cathode down to +3 volts so
conduction could begin. JLS

Something to ponder. Yet another phase splitter at ABSE. JLS

Not of any interest on RAT of course, but the early McIntosh SS power
amplifier, the MC2100, used this technique. A high-compliance current source
(3K6 with 80 V) in the collector circuit of a BJT.

You can argue all you want about gain and transconductance, KCL is hard to
overcome.

"John Stewart" wrote in message
...
Actually, with only 5 ma in the CC source, all of that would run thru the
27K that
was in the 6SN7 plate circuit & pull the 6SL7 cathode up so that the 6SL7
would be
cut off. To get the 6SL7 turned on you would need something like 11 ma
from the CC
source to get the 6SL7 biased on. That would get it's cathode down to +3
volts so
conduction could begin. JLS

Something to ponder. Yet another phase splitter at ABSE.
JLS

"Chris Hornbeck" wrote in message
...
On Sat, 26 Mar 2005 06:01:31 GMT, "Ian Iveson"
wrote:

All OK up to he

So the voltage at the cathodes varies until both
valves pass the same current. This is possible because the valve
receiving the signal at its grid perceives the consequent cathode
voltage as negative feedback.

You're either yanking my chain, or your model is flawed.
Although inclined to give you the benefit of the doubt and
believe the former, I'll respond to the latter.

Don't overthink this. It's not anything new and it's not even
interesting. The division of voltages across the diff-pair
of transconductance devices is purely algebraic and inversely
proportional to their transconductances. Period. No magic,
no feedback. It's a simple voltage divider.

So, if the division of voltages is inversely proportional to
transconductance, and gain is directly proportional to transconductance,
then the transconductance drops out, no?

"Chris Hornbeck" wrote

All OK up to he

Cool. So you agree that unmatched valves produce equal output
voltages in a differential pair with matched loads and a CCS at its
tail?

So the voltage at the cathodes varies until both
valves pass the same current. This is possible because the valve
receiving the signal at its grid perceives the consequent cathode
voltage as negative feedback.

You're either yanking my chain, or your model is flawed.
Although inclined to give you the benefit of the doubt and
believe the former, I'll respond to the latter.

Whatever takes your fancy. But if you put these two paragraphs
together...

A voltage appears across a pair of series'd transconductance
devices' inputs. The voltage is divided between the pair in
inverse
linear proportion to their transconductances.

Each device's share of the voltage generates an output
current that is the product of its share of the voltage
and its transconductance.

....it amounts to...

So the voltage at the cathodes varies until both
valves pass the same current.

That is, the voltage is divided such that currents are equal. What I
said explains how that happens.

Don't overthink this. It's not anything new and it's not even
interesting. The division of voltages across the diff-pair
of transconductance devices is purely algebraic and inversely
proportional to their transconductances. Period. No magic,
no feedback. It's a simple voltage divider.

Now you are wandering off track again. Perhaps you could explain,
for example, how come there is no feedback?

The gain of each device is the simple linear product of its
transconductance and its load impedance.

Open loop gain, yes. But they are not open loop.

Guys, there just isn't any more to it than this. No handwaving
is necessary or allowed.

Er, you are the one waving hands. You have made no reference to the
specific circuit we are discussing, but just repeat fundamentals
without explaining how they apply in this case. If you do try and
explain in terms of the actual circuit, you may find that we agree.

But not if your fundamentalism denies the simple truth about
balance.

cheers, Ian

On Sat, 26 Mar 2005 10:26:06 -0500, "BFoelsch"
wrote:

So, if the division of voltages is inversely proportional to
transconductance, and gain is directly proportional to transconductance,
then the transconductance drops out, no?

Yes, for the current outputs of the transconductance devices.
In special cases, including matched output loading, infinite
internal impedance and zero external impedance, mismatched
transconductances can give matched voltage outputs.

But these are special cases, not a general rule.

Chris Hornbeck
"Excuse me, since when is getting paid for it not USING the property?"
-Bob Olhsson

On Sat, 26 Mar 2005 15:39:43 GMT, "Ian Iveson"
wrote:


Now you are wandering off track again. Perhaps you could explain,
for example, how come there is no feedback?

What feedback? You *are* yanking my chain, aren't you?

Chris Hornbeck
"Excuse me, since when is getting paid for it not USING the property?"
-Bob Olhsson

"Patrick Turner" wrote

Ian Iverson made the assertion that one tube could be replaced
by a resistor equal to the static resistance of one triode.
He suggested a damp carrot be used, and still you'd get equal
inverted output, but gain of about 0.35 from input to each anode..
I did the math on the circuit arrangement on the follow up post.
If the "damp carrot" resistance is well bypassed by a an electro
cap,
the circuit works as a normal CPI, with the same gain.
I cannot think of any use for such a creature as a Turnip-Carrot
PI.


I V E S O N

Two syllables, please.

http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF

cheers, Ian

ps figures are static values, graphs are both with 1V pk signal at
1k.

zoom to actual size and don't complain you can't read it

"Chris Hornbeck" wrote

What feedback? You *are* yanking my chain, aren't you?

Well, at least I know what you're missing now. Signal input to
triode is Vgk. Vk varies in the case of the LTP.

cheers, Ian

pps carrots are actually quite dangerous in HV circuits so please
don't try this at home.

Morrisons carrots, with slight trace of broccoli, have poor
conductance but they might be dead because one of them is a bit
furry. I bought them because they were on special offer with the
broccoli, which I ate. Difference in cooking time would have meant
cutting the carrots into little strips and I couldn't be arsed so
they were abandoned in the bottom of the fridge.

Anyway, 95k required a slice 1cm diameter 2mm thick sandwiched
between two 1p pieces, previously cleaned with brown sauce and
rinsed. Carrot is not a well-behaved load, taking a second or two to
settle, like a big cap and resistor in series. Possibly
characteristics are similar to a cap with polarising dielectric.
Maybe it's a semiconductor. Perhaps a grid could be threaded between
the anode and cathode?

The other problem is long-term stability. A bonsai carrot might be
OK. Wouldn't need replacing, just an occasional trim and watering.
Multi-terminal carrots can be grown if you put some stones in the
soil.

cheers, Ian

On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson"
wrote:

"Chris Hornbeck" wrote

What feedback? You *are* yanking my chain, aren't you?

Well, at least I know what you're missing now. Signal input to
triode is Vgk. Vk varies in the case of the LTP.

A valve's cathode looks to the outside world like a
resistor equal to the reciprocal of its transconductance.
It has all the characteristics of a real resistor, both
from the viewpoint of an imaginary idealized
tranconductance "deamon" inside the valve and from the
viewpoint of the external world. It even has the
noise characteristics of a real resistor of appropriate
temperature.

IOW, each valve sees each other's cathode as a resistor,
and neither sees the (idealized) CCS. It's too big. The resistors
make a simple voltage divider. There is *no* feedback involved.

To put it another way, if the grounded grid valve were replaced
by a resistor equal to 1/Gm the signal voltage appearing at
the cathodes/CCS junction would not change. There is *no*
feedback involved. And there is no such thing as "internal
feedback in triodes" either. That's whack.

Chris Hornbeck

So you have to fit what you know about transconductance around that
fact, not vice versa. Ohms law is not tap dancing.

Isn't the key to the scheme is that the current from the
constant current source (sink actually) is split between the
two triodes? One triode has the input signal applied to its
grid, the other triode's grid connected to a non varying bias
voltage (say ground). Adjust the input signal on the first
triode and make it draw say 1ma more current from the CCS.
But the only way that can happen is if the other triode decreases
its current by 1ma. Use the same value resistors in both plate
circuits, and you'd see inversion from one plate to the other.

At least one John has unravelled the conundrum, potentially. The CCS
works by varying its output voltage in order to achieve the correct
constant current. So the voltage at the cathodes varies until both
valves pass the same current. This is possible because the valve
receiving the signal at its grid perceives the consequent cathode
voltage as negative feedback.

One thing we have missed out, BTW, is the capacitance to the grids.
Current lost to or gained from the grids from or to anodes and
cathodes is not accounted for so far.

A few picofarads. Not of significance at audio frequencies.

"robert casey" wrote

Isn't the key to the scheme is that the current from the
constant current source (sink actually) is split between the
two triodes? One triode has the input signal applied to its
grid, the other triode's grid connected to a non varying bias
voltage (say ground). Adjust the input signal on the first
triode and make it draw say 1ma more current from the CCS.
But the only way that can happen is if the other triode decreases
its current by 1ma. Use the same value resistors in both plate
circuits, and you'd see inversion from one plate to the other.

Yup, we are all but one agreed on this. That's why I said ohm's law
isn't tap dancing.

...Not of significance at audio frequencies.

Famous last words!

You are probably right. Even so, in the presence of current
source/sink there is a need to be careful because of high impedance.
A few tens of pF can become significant.

cheers, Ian

"Chris Hornbeck" wrote

[below in full]

You are not applying yourself to the matter in hand, still. You are
boxing with strange shadows, setting up cardboard villains and
chopping them down. I don't like to watch.

The feedback halves the gain.

I've posted a picture. Others have posted circuits. Either make your
refutation clear and specific or please stop pestering.

cheers, Ian

in message ...
On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson"
wrote:

"Chris Hornbeck" wrote

What feedback? You *are* yanking my chain, aren't you?

Well, at least I know what you're missing now. Signal input to
triode is Vgk. Vk varies in the case of the LTP.

A valve's cathode looks to the outside world like a
resistor equal to the reciprocal of its transconductance.
It has all the characteristics of a real resistor, both
from the viewpoint of an imaginary idealized
tranconductance "deamon" inside the valve and from the
viewpoint of the external world. It even has the
noise characteristics of a real resistor of appropriate
temperature.

IOW, each valve sees each other's cathode as a resistor,
and neither sees the (idealized) CCS. It's too big. The resistors
make a simple voltage divider. There is *no* feedback involved.

To put it another way, if the grounded grid valve were replaced
by a resistor equal to 1/Gm the signal voltage appearing at
the cathodes/CCS junction would not change. There is *no*
feedback involved. And there is no such thing as "internal
feedback in triodes" either. That's whack.

Chris Hornbeck

On Sun, 27 Mar 2005 03:06:33 GMT, "Ian Iveson"
wrote:


...Not of significance at audio frequencies.

Famous last words!

You are probably right. Even so, in the presence of current
source/sink there is a need to be careful because of high impedance.
A few tens of pF can become significant.

The impedance at the cathode of a valve is the reciprocal
of its transconductance, (typically a coupla K ohms -ish).
Until you decide to believe this, I'll shut up.

Thanks,

Chris Hornbeck

Ian Iveson wrote:

"Patrick Turner" wrote

Ian Iverson made the assertion that one tube could be replaced
by a resistor equal to the static resistance of one triode.
He suggested a damp carrot be used, and still you'd get equal
inverted output, but gain of about 0.35 from input to each anode..
I did the math on the circuit arrangement on the follow up post.
If the "damp carrot" resistance is well bypassed by a an electro
cap,
the circuit works as a normal CPI, with the same gain.
I cannot think of any use for such a creature as a Turnip-Carrot
PI.

I V E S O N

Sory.
Seems I put in an 'r' where it wasn't needed
Don't let it wurie you too much.



Two syllables, please.

http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF

So this circuit also confirms that if CCS and equal RL are used,
and gM, Ra and u are both quite differnt, you get equual VO?

Patrick Turner.



cheers, Ian

Chris Hornbeck wrote:

On Sat, 26 Mar 2005 20:26:08 GMT, "Ian Iveson"
wrote:

"Chris Hornbeck" wrote

What feedback? You *are* yanking my chain, aren't you?

Well, at least I know what you're missing now. Signal input to
triode is Vgk. Vk varies in the case of the LTP.

A valve's cathode looks to the outside world like a
resistor equal to the reciprocal of its transconductance.

Yes true, but only for the case of a CF, where Rkin = 1/gm
For where there is a load between the anode and B+,
Rkin = RL/A, where A is the gain of the tube with that RL.
And if there is a cathode resistor, values of Rkin have to be adjusted
for the
parallel R.


It has all the characteristics of a real resistor, both
from the viewpoint of an imaginary idealized
tranconductance "deamon" inside the valve and from the
viewpoint of the external world. It even has the
noise characteristics of a real resistor of appropriate
temperature.

IOW, each valve sees each other's cathode as a resistor,
and neither sees the (idealized) CCS. It's too big. The resistors
make a simple voltage divider. There is *no* feedback involved.

Hmm, but in an LTP, each tube has mutual negative current,
since the cathode of one tube is connected to the Rkin of the other.
So if Dn occurs in one tube it also must occur inverted at its cathode,
and be reduced
by series current NFB.
The output impedance from one anode of the LTP
is less than the output resistance of the same tube in a one tube common
cathode circuit with full
cathode bypassing.
But where the loads imposed on both the LTP anodes are the same,
they act as if the Ro was the same as a single tube in common cathode
mode.




To put it another way, if the grounded grid valve were replaced
by a resistor equal to 1/Gm the signal voltage appearing at
the cathodes/CCS junction would not change. There is *no*
feedback involved. And there is no such thing as "internal
feedback in triodes" either. That's whack.

NFB in triodes was defined by professor Child in
Terman's Radio Engineering in 1937.

I wouldn't dare say you are wrong, and would only ever
wonder what you would think about what the Professor Child wrote,
all those years ago.

Patrick Turner.



Chris Hornbeck

"Patrick Turner" wrote

I V E S O N

Sory.
Seems I put in an 'r' where it wasn't needed
Don't let it wurie you too much.

There has been a tendency for the runts of the family to add a
syllable. Several epochs past, Ive the French butcher became
entangled with a Scandinavian girl about 30 miles from here in
Wensleydale, and was never allowed to leave. There was never an
Iver, AFAIK.

http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF

So this circuit also confirms that if CCS and equal RL are used,
and gM, Ra and u are both quite differnt, you get equual VO?

Hope so.

It is also the circuit you were just describing, so you can compare.
Don't know where you got the 0.35 from but it's about right.
Incidentally, a signal to the resistor side's output results in a
gain of exactly 0.5 at the other output.

The value of the resistor doesn't matter for balance until it is so
small that the valve turns off. Then the circuit works via the
interelectrode capacitance to give small, in-phase and unequal
outputs.

The electrical life of a carrot is quite interesting. Resistance
seems to fall as it ages at first, as does its apparent capacitance.
Then the resistance rises again but the capacitance keeps falling.
Throughout this, the LTP remains balanced.

Anyway, more interesting questions are about how the output
impedances behave, and how it reacts to unequal loads and power
supply hum.

cheers, Ian

On Sat, 26 Mar 2005 20:13:22 GMT, "Ian Iveson"
wrote:

http://www.ivesonaudio.pwp.blueyonder.co.uk/carrot.GIF

Ian, I hadn't seen this before now. You're right
of course and I was wrong.

Thanks for the correction, and your patience.

Chris Hornbeck