Harsh treble in active tone stack

[shaun]

Hi All,
I recently put something together from the remains of a Stromberg-Carlson unit. I managed to keep both transformers, plus the speaker. (Even at 60+ years old, I like the sound of old alnico. But perhaps that's part of my current problem.)

The schematic was cobbled together from projects and tutorials I found online. It has an EF86 pre with an "Aikido" CF, an active tone stack I found off Angelfire, and a simple PP, cathode biased 6V6 output. I was pretty amazed when the unit ran well with only a few minor tweeks - it usually takes 6 months of cursing and swearing:).

As you may notice, I have inputs for line level phono, bluetooth, and...yes, guitar. There is a trade off of course; it's not a great guitar amp - the focus was more toward getting a clean and undistorted audio sound.

The tone stack sounds considerably better than the typical TB Fender design I've used in the past - it's sort of rounder and warmer. Plus, I seem to have little problem getting a big fat bass. But as I often find in my builds, the treble is on the harsh side, especially as volume increases. It may be that the treble signal distorts further as the volume increases.

I installed a 290pF cap in the TS, as opposed to the original design, which had a 100pF. But with the 100pF, there was very little change in tone when the treble pot was adjusted, so I experimented and settled on the 290pF.

My goal is to understand how to create a smooth and creamy treble tone. I've read a lot on how to tame harsh treble in guitar amps (remove the bright cap, for eg.), but I'd like to better understand how treble filtering works, what creates harshness, and how to resolve it.

I know this is a tall order because, well, doesn't everyone seek the holy grail of smooth treble tones? And I am a little worried that my madcap schematic might get a few laughs. But hey, at least I'm trying:).

As always, thanks for any help.

[pdf64]

Maybe the signal is clipping before the volume control, ie it’s like one big multi part input stage, that when the tone controls are wound up, just has too much gain?
Does turning the instrument’s vol control down a bit get rid of the nasty? If so, consider putting the vol control earlier in the signal chain.

[jjasilli]

Very unusual topography.  I don't understand why the TS output is feeding the inputs of both PI tubes.  Maybe try disconnecting the "output" of the treble pot lug3??? 

[sluckey]

Very unusual topography.  I don't understand why the TS output is feeding the inputs of both PI tubes.  Maybe try disconnecting the "output" of the treble pot lug3???

It's not feeding both grids although I can see why a casual look may seem that it is. The TS output connects to pin 2 of the tube which is just a normal gain stage. The output of the plate connects through the volume pot and on to the cathodyne PI. But the output is also fed back to the TS. This feedback loop is what makes the TS active.

[2deaf]

Probably nothing to do with harsh treble, but those cathodynes always seem to work better when the grid is isolated from DC ground by a capacitor.

That master volume is particularly bogus.  It doesn't do anything until the pot is nearly at the end of its sweep and it throws the cathodyne off balance at low settings.

[PRR]

The cathode follower is not needed; bypass it for now. The cathodyne is real messed up but it is my bedtime. "Creamy" may be many different things.....

[DummyLoad]

see attached. visual of 2def's statement of the cathodyne's bias problem


--pete

[DummyLoad]

a suggested fix. the problem may still persist, but at least phase inverter will operate properly.


--pete

[Joel in Texas]

You’re getting good advice on possible circuit-based fixes to the “harsh treble” problem.

But I think something worth exploring more, can be found in your original post:

I recently put something together from the remains of a Stromberg-Carlson unit. I managed to keep both transformers, plus the speaker. (Even at 60+ years old, I like the sound of old alnico. But perhaps that's part of my current problem.)

Plug the amp into a different guitar speaker cab. It should have a speaker driver that you know delivers the kind of treble response you like. See if it makes things better.

That old alnico speaker might be great for guitar, or it might be voiced better for full-range applications.  Old alnico speakers are not uniformly voiced for “good” guitar tone, and in my experience with examples from old radios, stereos, and p.a. systems, many sound too harsh in the treble range for my own guitar tone tastes.

So before going crazy on the circuit, you might try a simple speaker swap.

[2deaf]

see attached. visual of 2def's statement of the cathodyne's bias problem

It's the same thing as a volt meter providing the grid with a DC path to ground. 

 

[shaun]

"So before going crazy on the circuit, you might try a simple speaker swap."
Yes indeed, JFT. Very good idea.

"The output of the plate connects through the volume pot and on to the cathodyne PI. But the output is also fed back to the TS. This feedback loop is what makes the TS active."
Thanks for the light-bulb moment, Sluckey!

2deaf's idea, as clarified by DummyLoad, really helped me understand what is going on with the DC. I did wonder whether the feedback line could ignore a coupling cap, and I assumed it was something to do with the way the first half of the 12AT7 is part of the tone stack. In any case, I'll include another cap as suggested.

"That master volume is particularly bogus.  It doesn't do anything until the pot is nearly at the end of its sweep and it throws the cathodyne off balance at low settings."
Thank you, 2deaf. I realize the volume pot is in an unusual place. I just didn't realize how unusual - or that it would throw the cathodyne off balance, which makes sense now that you point it out. It's interesting how light illuminates the darkness. The darkness being my current abilities.     I'll move the vol pot elsewhere.

Regarding your suggestion, PRR, to remove the cathode follower, I've read so much about how the EF86 needs a CF when used as a preamp that I'm a little confused. I have used an EF86 pre in the past without a CF, and it seemed to go into a Fender style TB without a problem, but I also thought that, with a TS like the one I've used here, it required some impedance matching. But I learn at the feet of the masters, so I will bypass the CF and see what happens. Experiments are at least half the fun.

Thank you all!

Man, the things I don't know about amp circuitry.

[PRR]

> remove the cathode follower

There will be a difference. Big enough to justify a whole other dual-tube? I suspect if your dog bypassed it while you were sleeping, you wouldn't be sure anything changed.

[jjasilli]

Very unusual topography.  I don't understand why the TS output is feeding the inputs of both PI tubes.  Maybe try disconnecting the "output" of the treble pot lug3???

It's not feeding both grids although I can see why a casual look may seem that it is. The TS output connects to pin 2 of the tube which is just a normal gain stage. The output of the plate connects through the volume pot and on to the cathodyne PI. But the output is also fed back to the TS. This feedback loop is what makes the TS active.

Yikes.  I thought that there might be NFB but that it was occurring after the tone recovery stage > the cathodyne.  Didn't realize the negative signal was fed back into the TS. It would seem that the NFB back into the TS is of higher voltage than the signal in the TS.  I seriously don't understand this circuit.

[PRR]

> It would seem that the NFB back into the TS is of higher voltage than the signal in the TS.  I seriously don't understand this circuit.

12AT7(A) is a voltage gain stage wrapped in a Baxandall tone network. Knobs on "5" it gives near unity gain.

12AT7(B) is a cathodyne, unity gain to each of two outputs, one inverted.

EF86 is a voltage amplifier, voltage gain likely about 100.

6V6es probably need 20V signal to get full juice.

So what we have is an input sensitivity near 200mV. Normal guitar amps have 20mV sensitivity. This will be "hard work" to play loud.

We probably do not want to try to raise gain in the EF86 stage because it is already about to overload on hard strumming. (200mV hard strum in times 100 is 20V out, and there may not be 50V peak plate swing unless hand biased.) And too much gain before the first volume control just encourages front-end distortion which can't be controlled by any knob.

So I suspect either there is special thinking, or it has to be built and experimented with to clarify where it really wants to go.

[jjasilli]

OK, so it seems the circuit's performance is incomprehensible from the schematic, and you'd need to profile it empirically with test frequencies or a spectrum analyzer; then tweak from there.  (Gotta love those '50's designers).  A noble project. 


OTOH: 


* since you seem to like the Baxandall TS:  maybe rebuild a conventional Baxandall TS > conventional cathodyne PI. 
* per Merlin & various past threads on this Forum, there's ways to tame a cathodyne PI circuit (which I can't remember just now & am work overloaded), but maybe someone can chime in

[shaun]

Well, the funny thing in all this is that it's the best sounding amplifier I've built so far. Which is not saying much:). This is my 15th build, and I'm still in the dark about a lot of stuff. I'm a monkey-see-monkey do kinda guy, but I'm learning slowly through experience, and from this forum. And making lotsa mistakes along the way.

I don't have the analytical chops to do spectrum analysis, and much of the math - other than the basics such as cathode biasing - turns my brain to mush. My oscilloscope is a vintage Heathkit model that I suspect needs a reboot. But they say Rome wasn't built in a day, so maybe I'll figure out the harder math at some point.

For now, I will relocate the volume, probably to just after the input selector, and I'll add the extra 0.22 cap to block DC. See how that sounds.

Thank you for the great feedback.

[shooter]

This is my 15th build

I did 18 "prototypes", the 18th got converted to my 1st build 
after that is got easier, my brain exploded less and the bulb kept lighting and the amps kept sounding sweeter!
so hang in there, you're on the right train

[shaun]

Thanks Shooter! Yes, my "builds" are mostly spaghetti. But one day!

[jjasilli]

Should have occurred to me earlier, possible simple fixes:


*  100K grid stopper on the signal input lug into the cathodyne tube. (I think this is the Merlin thing)
*  snubber cap across the cathodyne outputs: try different values 10pF - 120pF

[2deaf]

The gain of a cathodyne is approximately one so that the Miller effect isn't a consideration.  The lack of a Miller effect means that high frequency roll-off won't occur until the grid stopper resistor gets absolutely huge.  Even a 1M grid stopper on a cathodyne has no effect on guitar frequencies.

Another thing, which is not immediately obvious, is that both outputs of a balanced cathodyne have the same output impedance and this impedance is very low.  One of the ramifications of this is that a cross-line master volume will not work properly.  It also means that the cathodyne has no problem driving capacitive loads while maintaining a wide bandwidth.  This neuters cross-line treble cut controls and high frequency roll-off with the typical fixed capacitors across the outputs of a cathodyne.

   

[pdf64]

...
Another thing, which is not immediately obvious, is that both outputs of a balanced cathodyne have the same output impedance and this impedance is very low...

Can you walk me through why the plate output impedance is very low, rather than being high, eg nearly equal to the plate load resistance?

ie why doesn't Aiken's common cathode triode stage analysis apply?
From http://www.aikenamps.com/index.php/designing-common-cathode-triode-amplifiers
Output Impedance (output taken from the plate, unbypassed cathode)

The internal plate resistance will increase if there is negative feedback due to an unbypassed cathode resistor, and the voltage gain of the stage will decrease as well. Assuming a cathode resistor of 1.5k + 56k, the resistance seen looking into the plate when the cathode resistor is unbypassed is:
ra'(unbypassed Rk) = ra + (mu + 1)*Rk
                              = 62.5k + (101)*57.5k
                              = 5.87M
Therefore, the output impedance will be:
R = ra' || Rp
   = 5.87M || 56k
   = 55.5k

I guess it may be due to a combination of the loads being equal on both cathode and plate, and plate current = cathode current?
But I can't quite get it straight in my head 

[shaun]

Thanks, 2deaf. Great tutorial that I will use the next time I try to install an MV.

Thanks, jjasilli. Both good suggestions. I'll try the snubber first. Should be a fun experiment.

[shooter]

Can you walk me through

I'm lurking with pencil in hand, brain might be out to lunch though 

[PRR]

> why the plate output impedance is very low

I think that is incomplete, and may mislead someone.

One oddity is that loading one side affects the other side, sometimes severely. Heavy load on cathode side will INcrease gain to plate side. While I don't care to go that road, I can see how it might go goofy.

[2deaf]

Can you walk me through why the plate output impedance is very low, rather than being high, eg nearly equal to the plate load resistance?

Yeah well..you know..man...it's kinda like......If you're asking me to explain something to you, then I must have said something that was wrong.  I'll tell you one thing, though, and I've known this for a long time:  Those balanced cathodynes sho' 'nuff behave as though they have equal, low output impedances.  The 570 ohm output impedance I used for my example was found experimentally. 

Maybe where I erred was when I claimed that the plate had a low output impedance when in reality it has an apparent/virtual/equivalent/effective/Hoodoo output impedance.  In either event, my example is still valid.

It's only been in the past few years that I read anything about cathodyne output impedance.  I didn't particularly buy into some of what was said.  It struck me as though an unexpected phenomenon was observed and an after-the-fact explanation was concocted.

From RDH4:  "The equivalent source impedance that determines the high frequency attenuation for each channel due to shunt capacitance is given by
R_o = r_pR_L / [r_p + R_L(u + 2)]
and is of the order of 1000 ohms.

       

[jjasilli]

Merlin: Designing Tube Preamps for Guitar and Bass, 2009 Edition, Ch 7 The Cathodyne PI, p. 161, ff: assuming both outputs are equally loaded (and how would they not be??? I ask) . . . "The fact is that not only is the output impedance equal at both outputs, but also very low in value (usually less than 1k Ohm) has puzzled many authors, yet is true [and beyond the scope of this book. . .]"


However, I don't see why the low output impedance would defeat a snubber cap.

[DummyLoad]

why do you need a white CF to drive the TS? the pentode with ~56K Z-out would work just fine. i converted white CF (V2) to single gain stage and as TS driver - should have better sensitivity. if you want to keep it cleaner, sub V2 with 12AU7 and re-bias for 12AU7. how i would add a MV. please see attached.

respectfully,

--pete

[shooter]

when in reality it has an apparent/virtual/equivalent/effective/Hoodoo output impedance.

I can confirm by experimentation there are oddity's and they manifest as Merlin shows.  these are from my last build.
Big G1 R "fixed", I left calculating output Independence left to chance    


as to "why drive a TS" with a White CF", probably the same reason I used a DCCF to drive a cathodyne....the circuits "looked cool", never seen them stitched together, wonder how they sound. 

[PRR]

...assuming both outputs are equally loaded (and how would they not be??? I ask) . . .

In guitar amps: when the output stage clips, one side or the other becomes near-shorted as the grid tries to go positive.

[shaun]

why do you need a white CF to drive the TS? the pentode with ~56K Z-out would work just fine. i converted white CF (V2) to single gain stage and as TS driver - should have better sensitivity. if you want to keep it cleaner, sub V2 with 12AU7 and re-bias for 12AU7. how i would add a MV. please see attached.
respectfully,
--pete

Thanks Pete. PRR pointed out early in the thread that the CF in the preamp was superfluous. I am relatively a newbie and am still trying to understand impedance matching between stages, so I was going on what I'd read about EF86 V1s needing a CF, especially going into a more complicated TS, which would need to "see" the low impedance signal a CF provides. Obviously, I need to fully understand the math if I want to avoid the inclusion of unnecessary CFs!

I will study your examples closely and try to get a handle on the math. I certainly like the idea of an extra gain stage.

I think it's going a bit far to add me as a "designer' on the schem:). But I appreciate the thought.

Much gratitude - this incredible thread is teaching me some great stuff I may never figure out otherwise.

[pdf64]

...assuming both outputs are equally loaded (and how would they not be??? I ask) . . .

In guitar amps: when the output stage clips, one side or the other becomes near-shorted as the grid tries to go positive.

Also, would a Type 3 / crossline master vol, or similar non 0V referenced load such as plate to cathode snubber cap, equally load the cathodyne's outputs?

Another thing is that a low output impedance doesn't equate to the circuit being able to maintain good signal linearity at high outputs into what would apparently be a reasonable impedance bridge, eg 10 x its output impedance.
Its capability to sink and source current is a different characteristic.

[jjasilli]

FWIW, Merlin goes on to say (in the book I quoted above), that very large grid stoppers on the power tubes (signal input), stabilize the cathodyne PI. 


Also, it seems impossible to overly generalize about cathodynes, because they are either AC coupled or DC coupled to their driving stage -- circuit configurations and alternatives seem to depend upon which type of coupling is employed. 

[PRR]

why do you need a white CF to drive the TS?

It's not a White. Maybe it was supposed to be?

There's no sense resistor in the upper plate. So the lower triode is nominally an active resistor. The capacitor up to B+ "could" bleed ripple in counter-clockle "Akido" style. Or may be a remnant of a WCF idea.

I guess we agree that pentode with 56k load will drive a tone network which does not come below 120k. Oh it may not be "dead flat" as in utterly buffered High Fidelity design, but that's not the job.

[jjasilli]

...assuming both outputs are equally loaded (and how would they not be??? I ask) . . .

In guitar amps: when the output stage clips, one side or the other becomes near-shorted as the grid tries to go positive.

Also, would a Type 3 / crossline master vol, or similar non 0V referenced load such as plate to cathode snubber cap, equally load the cathodyne's outputs?

Another thing is that a low output impedance doesn't equate to the circuit being able to maintain good signal linearity at high outputs into what would apparently be a reasonable impedance bridge, eg 10 x its output impedance.
Its capability to sink and source current is a different characteristic.

I don't understand how a snubber cap can be considered a load.  It's a short circuit, which is the "opposite" of a load. But it's a only partial short circuit for a fraction of the AC bandwidth.  And there's 2 leg's to the AC, 180 degrees out of phase, so they cancel out totally, or drop off at the knee, rather than actually short-circuiting.  I think this is is not a load issue; it's a phase issue. 


According to Merlin, cathodynes have great signal linearity which is why they're used in hi-fi, until you get to overdrive when their performance degrades, absent corrective measures .  But you can't be so general.  You need to distinguish between the power tubes being overdriven by the cathodyne, and/or the cathodyne itself being overdriven; and whether the cathodyne is AC or DC coupled.  At which point we're beyond the scope of even Merlin's book, which is too deep into the weeds for me.  Merlin cites further scholarly sources for those interested.   

[HotBluePlates]

I don't understand how a snubber cap can be considered a load. ...

Anything other than ∞Ω is "a load."  0Ω is an "extreme load" because the tube cannot sustain its output voltage into the short-circuit.

You could look at "loading" in terms of "ability to maintain output voltage."  If the load is removed and a voltage suddenly appears at the output, then the load was "heavy."

OR you could think in terms of a voltage divider (since we're talking about voltage amplifier stages here).

    - Load Impedance 100x Source Impedance:  Output Voltage (Mostly) Unchanged (99% applied voltage) = Very Light Load
    - Load Impedance 10x Source Impedance:  Output Voltage Somewhat Lower (91% applied voltage) = Light Load
    - Load Impedance 1x Source Impedance:  Output Voltage Halved (50% applied voltage) = Significant Load
...
    - Load Impedance 1/100th Source Impedance:  Output Voltage Almost Zero (1% applied voltage) = Near-Short Circuit

[HotBluePlates]

...assuming both outputs are equally loaded (and how would they not be??? I ask) . . .

In guitar amps: when the output stage clips, one side or the other becomes near-shorted as the grid tries to go positive.

I don't understand how a snubber cap can be considered a load.  It's a short circuit, which is the "opposite" of a load. But it's a only partial short circuit for a fraction of the AC bandwidth.  ...  I think this is is not a load issue; it's a phase issue. 
...

PRR and pdf64 are talking about different things (or at least, from different directions).

PRR is talking about the output tubes distorting (when the grid-drive equals/exceeds the bias voltage), at which point the output tube grid stops looking like ∞Ω.  Instead, it draws grid-current, and may effectively be <1kΩ while the grid-voltage is near/above cathode-voltage.

[jjasilli]

Hotblue, good points!  So, 0Ω is not the absence of a value, but is a value = a very heavy, perhaps the heaviest possible, load (unless there is such a thing as a load less than -0-).  But this does resolve the points under discussion.

For the cathodyne's signal outputs, the snubber cap presents as a no load = ∞Ω's to the 2 lo signal bands remaining in the main signal path. The identical but opposite phase hi frequency bands see a 0Ω path through the cap, but they cancel one another, due to opposing phase; not because they are bled to ground and back into the circuit (per the typical loaded circuit).  For AC operation in the hi frequency bands, there is no whole or partial short circuit.  I.e., none of the bled away signal reaches "ground" and gets recirculated. Instead, it vanishes from the circuit inside the cap.  So:

* Do cathodyne outputs see the snubber as a load for the hi frequencies?  I still think not. Bui if Yes, is such load significant enough to affect the performance of the cathodyne?

*  Why would a cathodyne defeat the hi-killing purpose of the snubber?

[shooter]

So, 0Ω is not the absence of a value

my understanding of practical, and ohms law, 0 ohms in an "active circuit" equates to "attempted" infinite current anytime voltage is present, superconductivity aside 

[pdf64]

For the cathodyne's signal outputs, the snubber cap presents as a no load = ∞Ω's to the 2 lo signal bands remaining in the main signal path. The identical but opposite phase hi frequency bands see a 0Ω path through the cap, but they cancel one another, due to opposing phase; not because they are bled to ground and back into the circuit (per the typical loaded circuit).  For AC operation in the hi frequency bands, there is no whole or partial short circuit.  I.e., none of the bled away signal reaches "ground" and gets recirculated. Instead, it vanishes from the circuit inside the cap...

I'm not sure that's correct, rather each output provides a path to 0V for the other, ie via their output impedances.

...* Do cathodyne outputs see the snubber as a load for the hi frequencies?  I still think not. Bui if Yes, is such load significant enough to affect the performance of the cathodyne?

As per above, as frequency increases, each output will be the other's most significant load.

...Why would a cathodyne defeat the hi-killing purpose of the snubber?

Due to the very low output impedances, for the snubber to be useful at an appropriate frequency, the cap's value would have to be an order of magnitude or two larger than those normally used. But as I noted previously, output impedance is usually derived at small signal levels, whereas a phase splitter has to be able to operate at high signal levels; my concern is that the circuit's small signal characteristics may not apply. Consider that an op amp has very low output impedance, and can output almost a rail-to-rail signal, eg 30Vp-p when lightly loaded. But as the load impedance decreases, so must the max signal level.

[HotBluePlates]

Yeah well..you know..man...it's kinda like......If you're asking me to explain something to you, then I must have said something that was wrong.  ... balanced cathodynes sho' 'nuff behave as though they have equal, low output impedances.  ...

From RDH4 ...

Flip to Page 330 of RDH4, where the (cathodyne) phase splitter is discussed:  "The effective output impedance is different for the two output channels since [each uses a different form of feedback]."

... Do cathodyne outputs see the snubber as a load for the hi frequencies? ... is such load significant enough to affect the performance of the cathodyne?

I have to think about it simply, cause I'm not using Spice, nor will I be drawing out a bunch of equivalent-circuit diagrams & solving equations.  Replace the "snubber cap" with a direct connection, which it will be for some "high-enough frequency".

There are 56kΩ load resistors for this split-load inverter, and normally they'd be analyzed as a 56k+56k = 112kΩ load resistor (and then consider the effects of feedback).  If there is a piece of wire from plate to cathode, these are now 56k||56k = 28kΩ.  The B+ filter cap & ground are "the same" except for edge-cases of extremely high frequency & imperfect caps.

Would there not be "a loading effect" from the loadline rotating from 112kΩ to 28kΩ?  Towards less voltage output?

Hey, I'm on-board with saying opposing-polarity signals cancel (they do), but there's also some kind of loading effect reducing the voltage-output capability of the inverter.

...Why would a cathodyne defeat the hi-killing purpose of the snubber?

Due to the very low output impedances, for the snubber to be useful at an appropriate frequency, the cap's value would have to be an order of magnitude or two larger than those normally used. ...

The above is true.

I can tell you I put a Cut control on a physical amp at the split-load inverter.  I used a cap-to-plate and a cap-to-cathode running to my rheostat, as both have d.c. voltages to be blocked.  I had to use 0.047µF caps, with a 250kΩ pot to get the capacitors large enough to audibly shave treble.

The amp & this Cut control works quite well.   

[2deaf]

Flip to Page 330 of RDH4, where the (cathodyne) phase splitter is discussed:  "The effective output impedance is different for the two output channels since [each uses a different form of feedback]."

Keep reading.  The very next sentence identifies one notable exception where the high frequency roll-off due to capacitive loads is governed by equation 34a.

[jjasilli]

For those of us who would rather have you do the math, what is the point your are making?

[2deaf]

However, I don't see why the low output impedance would defeat a snubber cap.

It's part of a low-pass filter and the low value has a dramatic effect on the -3dB cutoff frequency.  It causes a 120pF snubber to have a cutoff frequency of 1MHz.

The crossline resistor appears to be half as big and connected to ground for each output.  The crossline snubber capacitor appears to be twice as big and connected to ground for each output.

The reason that they appear to be connected to ground is because there is a point between the two phases where the voltage is zero referenced to ground.  I mean, if you started walking across that capacitor from -20V towards +20V, then somewhere along that walk you are going to find 0V.  That point is an apparent/virtual/equivalent/effective ground that can be treated as an actual ground for the purposes at hand.

The reason that the capacitor appears to be twice as big is that the current is still the same and one end is still at the same voltage, but the other end is connected to zero volts.  In order for the current to remain the same, the impedance must be cut in half.  Twice the capacitance has half the impedance.  This is all at some constant frequency.

     

[2deaf]

For those of us who would rather have you do the math, what is the point your are making?

Who?

[shaun]

Man, I feel like I'm sitting in a master class. The line of inquiry is sometimes over my head, but this thread is gold. Thank you!

[2deaf]

Can you walk me through why the plate output impedance is very low, rather than being high, eg nearly equal to the plate load resistance?

ie why doesn't Aiken's common cathode triode stage analysis apply?

I finally got around to drawing some more comic strips and one of them is for the output impedance of a cathodyne.  My explanation for the performance of a cathodyne can be applied to an unbypassed common cathode gain stage in order to derive the anode output impedance.  Aiken's common cathode triode stage analysis cannot be applied to a cathodyne and he probably never intended for it to be.

I think that the output impedance is just some Hoodoo thing that doesn't really exist but that comes in right handy in a lot of situations.  I also think that just as soon as you find something that contradicts a theory, that theory is proven invalid.  Or maybe those two are actually the same thought and the invalid theory comes in right handy sometimes with or without Little John the Conqueroo.

   

[PRR]

> more comic strips

Good. Though I would consider losing the bias parts and just feeding 100VDC to the grid.

> Aiken's common cathode triode stage analysis apply?

This is certainly not a "common cathode". The cathode is one output but self-jacks far from the input common.

There are at least three "output impedances" here. Plate, cathode, and plate-to-cathode. Each of them is affected by load.

AC-ground the plate, the cathode output voltage increases very slightly but the cathode impedance falls to half.

AC-ground the cathode, the plate output rises about 50X.

AC-connect the cathode to the plate output, my mind boggles.

The cathodyne works very well as long as both outputs are loaded moderately and THE SAME. Which is why it was popular in hi-fi. It don't stink in overdrive but other topologies go 'bad' with different flavors.

[pdf64]

> more comic strips
...The cathodyne works very well as long as both outputs are loaded moderately and THE SAME...

I still can't work out whether a plate-to-cathode snubber would act (ie at high frequencies) to load each output equally   

[2deaf]

Also, would a Type 3 / crossline master vol, or similar non 0V referenced load such as plate to cathode snubber cap, equally load the cathodyne's outputs?

Let's try it if the crossline device didn't load each output equally.  The ratio of the dissimilar impedances presented to the outputs will determine the ratio of the output voltages.  Now we put the same amount of impedance in parallel with each of the dissimilar impedances.  The ratio of one parallel combination to the other parallel combination is not the same as the original ratio between the dissimilar impedances alone.  If we try to use the original output voltages with the ratio of the parallel combinations, the currents will not add up.

It becomes apparent that the only time the currents are going to add up is when the impedances of the parallel combinations have the same ratio as the original singular impedances.  The plate and the cathode resistors are both the same, so the effective crossline impedance will have to be the same for each output in order for the parallel combinations to have the same ratio as the singular impedances.

Shorter answer:  If you're going to have the same current coming out of the anode as went into the cathode and the anode resistor is the same as the cathode resistor, then the effective impedance for each output from the crossline device must be the same.