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In Reply to: RE: Broskie is correct on this one. posted by Triode_Kingdom on June 15, 2021 at 09:30:13
OK. If you used a resistor instead of a constant-current source, the two outputs will not be perfectly balanced if both triodes have the same value anode resistors (although balance will be pretty good). The difference in gain between the two outputs is given by:
A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))]
(Rk is the total cathode resistance in the case of a biasing scheme for the circuit)
So this all comes down to how effective the CCS is. You can look at the CCS as being very similar to a resistor tied to a very high negative voltage, but no CCS is perfect in practice, so a variance will exist. The lower the mu of the tube, the harder it will be to get rid of the offset.
I might be hung up on nomenclature- since 'perfect' doesn't exist I simply can't accept that the CCS is perfect (especially since most of them I see are really terrible).
Follow Ups:
Either both are grounded grid, or neither are, or they act like something in the middle...but they are the same.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"Either both are grounded grid, or neither are, or they act like something in the middle...but they are the same."
Well, no. V1 is driven at the grid and consumes no signal current or power from the input. V2 is driven at the cathode and consumes signal current and power. The power consumed by the cathode is fed through the tube and sums with the power available at the anode. I know it might seem like the same thing is happening inside the two tubes but as black boxes, they function differently.
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Buy Chinese. Bury freedom.
So if there is signal on the grounded grid, it becomes a different circuit?
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"So if there is signal on the grounded grid, it becomes a different circuit?"
What exactly are you asking Douglas? We both know a grounded grid amplifier behaves differently than a grounded cathode. Yes, the element selected for the application of drive matters.
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Buy Chinese. Bury freedom.
The point I am trying to make is that this grounded cathode amp y'all insist on labeling it as is not a grounded cathode amp anymore, and can not be examined as such since it is tied up in the LTP circuit.Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Edits: 06/16/21
My mention of a grounded cathode amplifier was only meant to describe the transformation of V2 if its cathode was grounded and it was driven at the grid. Isn't that what you meant when you asked, "if there is signal on the grounded grid, it becomes a different circuit?" V1 of the LTP is clearly a cathode follower when discussing drive for V2. I have never referred to it as a grounded cathode.
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Buy Chinese. Bury freedom.
Edits: 06/16/21
I thought I saw justification for different output presented with grounded grid to grounded cathode gain.
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"I thought I saw justification for different output presented with grounded grid to grounded cathode gain."Sorry, it's not clear to me what you mean by that. Nevertheless, based on your discussion with cpotl, I believe you're saying that other than the difference in gain between grounded cathode and grounded grid, there's no significant effect on the circuit when V2 is driven at its grid instead of its cathode. You're saying that the mechanism is the same.
"It comes down to there being no difference in reaction if the grids are driven balanced or if one side is fed no signal."
When V2 is driven at its grid, the circuit becomes a balanced, differential amplifier, and many things change. Primarily, rather than inverting the phase between the two triodes, the circuit merely repeats the two-phased signal presented at the grids, inverting both of them. The circuit is no longer actually a phase splitter. As a result, the AC signal voltage at the common cathodes falls to near zero, and this in turn, removes most of the current imbalance between the two triodes previously caused by current diverted through Rk. All in all, this is a very different scenario from the original purpose and function of the LTP.
--------------------------
Buy Chinese. Bury freedom.
Edits: 06/16/21
Couple of things, the grounded grid amplification factor of mu+1 leaves it requiring that side to have higher gain even with balanced resistors.
That never happens.
So the basic issue seems to be that treating it as a grounded cathode and a grounded grid seems to be improper. I don't see treating it that way when the diff amp model actually seems to give the correct output predictions with grid No.2 signal counted as zero.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"So the basic issue seems to be that treating it as a grounded cathode and a grounded grid seems to be improper. I don't see treating it that way when the diff amp model actually seems to give the correct output predictions with grid No.2 signal counted as zero."In Crowhurst's Fig. 626 it is manifest that V1 is driving into the cathode of V2, and since the grid of V2 is grounded, that means that V2 is by definition a grounded-grid amplifier. I don't see that there is any room for that statement to be incorrect.
It may be that one could find some other words to describe the same situation, but they must, logically speaking, be equivalent words describing exactly the same situation.
In any case, whatever words one uses to describe the circuit, there can really only be one set of correct equations that can be written down for the small-signal analysis, and so only one correct prediction for the two signal outputs. The physics, after all, is what it is, and all that is really being done is to apply Kirchoff's law to the currents at each node, employ Ohm's law to relate potential differences across resistors to the currents flowing through them, and employ the small-signal analysis of how the tubes behave under small perturbations around the quiescent conditions. All of these are universally agreed.
There shouldn't really be room for there to be any differences of opinion about what the correct system of equations to describe the circuit should be. And it is a very simply circuit. I think that analysis of Crowhurst's, after correcting the obvious algebraic mistake, is valid.
By the way, what do you mean when you say "the diff amp model actually seems to give the correct output predictions with grid No.2 signal counted as zero"? Correct as judged how?
Edits: 06/18/21
With a grid grounded in a diff amp, that section is not presenting a gain of mu+1 as it would if the circuit could be treated as a grounded cathode and a grounded grid. The balance is dependent on the magnitude of the tail load.The usual implementation of a diff amp third stage in a Williamson is going to behave quite differently vs one with a hard, unsaturated CCS in its tail. The signal at the cathode node will be larger, and the plate output will be different.
If it were acting like a separate grounded grid stage, an LTP built around a pair of 6C19Pi running a hard CCS would show higher gain in the section with a grounded grid since mu/mu+1 is near the minimum practically possible. IOW, if treating the grounded grid as separate were proper, a minimum mu circuit would show the grounded grid section with higher gain. That never happens.
As you said, examining the circuit as a whole is required, and it will not show that the grounded grid side has a delivered amplification factor any different from the other half. The interaction between mu, gm, plate loads and the cathode load show the imbalance generating causes.
Without the equations in front of me, increases to mu and gm will both shrink the imbalance. Differences in the plate loads can work towards correcting this imbalance, and increasing Rk shrinks imbalance...
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Edits: 06/18/21
"As you said, examining the circuit as a whole is required, and it will not show that the grounded grid side has a delivered amplification factor any different from the other half. The interaction between mu, gm, plate loads and the cathode load show the imbalance generating causes."One needs to be sure that disagreements are not merely about how the words are being used. In terms of Crowhurst's equations, if the amplification factors are called A1 and A2, and defined to be the changes in the plate voltages of tube 1 or tube 2 respectively, divided by the input voltage on the grid of tube 1, then they can be read off from his equations (92) and (94) respectively (since his Ep1 and Ep2 are defined to be the changes in plate voltages on tube 1 and tube 2 in response to putting a 1 volt input on the grid of tube 1). That is to say,
A1 = Ep1 (in eqn (92)), and A2 = Ep2 (in eqn (94)).
(Actually, there is a further typo, in his eqn (92); there should be a plus sign after Rp1 in the denominator of his expression.)
Using the corrected eqn (96) that I had previously given, these equations then say it all. That is, all the signal voltages, namely Ep1 at plate 1, Ep2 at plate 2, and Eck at the common cathode point, are all expressed in terms of the input voltage on grid 1, the resistors RL1, RL2, Rk, and the characteristics mu1 and Rp1 for tube 1 and mu2 and Rp2 for tube 2.
Crowhurst is, of course, precisely worrying about the fact that tube 1 is driving into the impedance corresponding to the paralleling of the cathode resistor Rk and the input impedance of the grounded grid amplifier that it is driving. That is the basis of his calculation. So he has taken everything properly into account.
By the way, if we assume RL1 = RL2 = RL, and assume identical tubes so that Rp1 = Rp2 = Rp and mu1 = mu2 = mu, the resulting formula for A1/A2 agrees exactly with the one Ralph gave a few days ago, namely
A1/A2= 1 + (RL + Rp)/((1+mu) Rk).
If the tubes are taken to be different in their characteristics (unequal mu and Rp), but the two anode loads are still set equal, RL1 = RL2 = RL, then the ratio of the amplification factors is given by
A1/A2 = 1 + (RL + Rp2)/((1+mu2) Rk).
So the mu and plate resistance of tube 1 do not enter in the expression for the ratio of the amplification factors.
Edits: 06/18/21 06/18/21 06/18/21 06/18/21
What in the hell is eqn 92 and 94?...and feel free to put up the corrected version.
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"What in the hell is eqn 92 and 94?"
Equations in the book by Crowhurst that has been under discussion in this thread. I put the relevant pages in a post on this thread a few days ago, with my corrections indicated on the 4th of the posted pages.
Just for the purpose of recording the result, I'm attaching a set of jpegs of the relevant pages in Crowhurst's book discussing the long-tailed pair, with what I believe are the proper corrections on the fourth page of the extract. I kept the characteristics of the two tubes (mu, and plate impedance) different, so that one can see that in eqn (97), the condition on the ratio of plate loads for equal output signals, implies that the characteristics of tube 1 are immaterial. I think my sums were done properly, but I'll be happy to be corrected if anyone disagrees!
Apologies for the way this has come out; I was wanting to attach a single pdf with the 6 pages, but I couldn't find a way to do that! Did I miss some obvious way to do it?
If you have a printer/scanner then print the 6 pages out and then you can scan it into one pdf (if it has an automatic feed).
"If you have a printer/scanner then print the 6 pages out and then you can scan it into one pdf (if it has an automatic feed)."
That wasn't the problem; I started out with a single pdf for the six pages, in fact. But I found that the AA website wouldn't let me post a pdf, so I couldn't see any option but to turn it into six png files. But also, I was hoping to attach it as a clickable link or thumbnail, but there didn't seem to be any way to do that, either. The options for including attachments seem to be very limited. Is there some way I'm not seeing, to do this?
So with one grid driven the balance is delivered by the magnitude of the cathode resistor( be it a resistor or a ccs). With two grids driven, the balance is STILL driven by Rk. The Rk still delivers balance when both are driven...a stage like used in a Williamson will have poorer balance than one rigged with either a longer resistor, or a current reg.
Nowhere does that equation presented by Ralph which I believe to represent the ratio between common cathode amplification factor and grounded grid amplification factor( delivered in circuit that is ) come into play. The second half does not have a mu+1 vs mu delta.
The LTP is no longer a separate set of common cathode and grounded grid amplifiers. They're tied together and the gain equations applied to them separately does not apply that way. The drive voltage is entirely a function of Rk when run SE:PP and the amplification does not magically change if you put some signal into the other grid.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Is the way he depicts the two tubes (ignoring for now the cathode resistor).
We can clearly see that the first tube is driving the 2nd via its cathode.
We also know that the greater the mu of a cathode follower, the less signal voltage loss there will be on the cathode.
So in the scenario where only one tube is driven by the grid, we can see that the other tube will always have less output, as a function of its mu.
The better the CCS becomes, the more this difference is reduced. If both sides are driven by opposite signals, it becomes no concern. Since our circuits are all balanced anyway that is why we use matched resistors.
Put another way, if you're driving this circuit single-ended, there will be a slight difference at the outputs, especially if your CCS is questionable.
If the tube with no signal on its grid actually has more gain, it would work to compensate for an imperfect tail load. The circuit works the other way...
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
If the tube with no signal on its grid actually has more gain, it would work to compensate for an imperfect tail load.
Emphasis added.
It might help. But at that point IMO/IME its all about slop.
The grounded grid has more gain than a grounded cathode.And yet the section with the grid grounded *NEVER has more gain.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Edits: 06/16/21
The point here is that the input side will have voltage loss delivering the signal to the other half that does not have its grid driven.
GG has more gain. Not in this case due to less input. But
So then, with a large Rk, where is this input going?
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
I remember some time back simming the input section of the EAR 509 cct to get my head around it. It uses DC coupled cascaded LTPs; equal anode resistors. The first has a very long tail to a neg supply.
The outputs of the first were ever so slightly different but the second evened that up so the result was the same from the 2nd anodes.
Best not discuss what happens after those stages ;)
I have built with cascaded LTP's too. They do a very fine job IME if the gain is needed. Since I like a medium-mu linestage, two-stage amps have done quite well. The latest are LTP's for balanced drive to the active cross-over.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"So if there is signal on the grounded grid, it becomes a different circuit?"
No. The mode in which each tube is operating is the same whether there is a signal or not. The fact that tube 2 is operating in grounded-grid mode, while tube 1 is not, becomes evident when one considers a small-signal analysis.
"Either both are grounded grid, or neither are, or they act like something in the middle...but they are the same."Why do you say that? Whatever the other faults in Crowhurst's discussion, the principle of his analysis, based on his redrawing the schematic as in his Fig. 626, looks OK. (I was unhappy when I first encountered his Fig. 626 because he appeared, erroneously, to have taken the bottom end of Rk to ground rather than to a negative supply rail. But actually, for the small-signal analysis he is making, it doesn't change anything. All he needs for his analysis is that the bottom end of Rk is held at a constant voltage with respect to ground.)
Anyway, Crowhurst makes it look pretty convincing that V2 is a grounded grid, while V1 is not. Do you see any flaw in his strategy for the analysis?
Edits: 06/15/21
I see plenty wrong with it. It comes down to there being no difference in reaction if the grids are driven balanced or if one side is fed no signal. The output is different in magnitude, as expected since the input is also different.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"I see plenty wrong with it. It comes down to there being no difference in reaction if the grids are driven balanced or if one side is fed no signal. The output is different in magnitude, as expected since the input is also different."
I'm not sure I understand what you are saying. Crowhurst presents a specific set of calculations which (apart from the error he made!) provide a formula for the required ratio of anode load resistors in order to achieve a balanced output. If you think his analysis is incorrect, can you provide an alternative derivation of an expression for the ratio RL1/RL2 of anode resistors that will produce a balanced output? What is your formula?
Don't quite feel like coming up with the equations. I am quite happy with the conclusion that the usual input side can not be considered a grounded cathode amp.
I suspect the proof of this will resemble the one presented that show the plate of an equally loaded split-load PI has the same low output Z as its cathode.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
"The difference in gain between the two outputs is given by: A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))]"
This is consistent with the answer Crowhurst would have got if he hadn't made an algebraic error. He has transcribed his formula from Chapter 12 for the input impedance of a grounded grid amplifier incorrectly. The correct formula from Chapter 12 would say Rin = (RL2 + Rp2)/(1 + mu2), but first of all this has ended up in the garbled form appearing in the line above eqn (96), and then he has evidently interpreted this, incorrectly, as (RL2 + Rp2)/mu2. This gives him a wrong formula (96) for Rx, and then in turn this gives him a wrong formula (97) for the required ratio RL1/RL2 of anode load resistors for getting balanced outputs.
His formula (97) clearly cannot be right because if you take the limit when Rk goes to infinity (the ideal CCS limit), it gives RL1/RL2 = 1+ 1/mu, which would say that the anode resistors should be unequal even in the ideal CCS limit. As has been discussed extensively already, this is simply impossible.
Correcting his calculation, one instead gets
RL1/RL2 = (1 + mu) Rk/ [RL2 + Rp2 + (1 + mu) Rk],
and this does indeed imply RL1/RL2 becomes 1 in the large Rk CCS limit. It is also consistent with Ralph's formula (his (A1/A2)^{-1} is the corrected expression for RL1/RL2 above).
Out of curiousity I tried the repeating the (corrected) Crowhurst calculation in the case where one does not make any assumptions about the two tubes being the same. So I allowed them to have different mu values mu1 and mu2, and different internal plate resistances Rp1 and Rp2. If I did my sums correctly the required ratio RL1/RL2 for getting balanced outputs is now
RL1/RL2 = (1 + mu2) Rk/ [RL2 + Rp2 + (1 + mu2) Rk]
In other words, the characteristics of tube 1 do not enter at all into the formula for the ratio of load resistors one needs in order to get balanced outputs! This seems a bit remarkable, but I think it is correct; at least, if Crowhurst's other formulae are correct. It happens because Crowhurst's eqns (92) and (93) imply Ep1 = RL1 Ek/Rx, and so when one equates Ep1 to Ep2 and solves for RL1, the Rp1 and mu1 parameters characterising tube 1 do not enter.
Having followed this, it sounds as though trying to be too idealistic about the elements in a ltp means the imbalance in the circuit will be swamped by the reality of tolerances in these elements.
This will lead to distortion profiles for each type of phase splitter and so a preference for each type sonically.
More power to those who advocate set topologies ;)
I am very fond of a floating paraphase inverter that is adjustable. Rather than making the "perfect" inverter with a "perfect balanced output".
The advantages of the adjustable floating paraphase inverter is that you can adjust symetry with output tubes that have either more or less different amplification factors.
I measured this when I built an Audio Note clone that had a floating parapahese inverter followed by a6SN7 before driving the output tubes.
when the inverter was adjusted for "perfect symmetry" the end result with "perfect (within 1%) matched 6SN7 followed by "perfect matched" output tubes was more than twice as large as when I simply adjusted the inverter for best output.
When using less than perfect matched 6SN7's or output tubes the difference became larger although a hefty dosis of negative feedback attempts to correct this. But then IMD started to shoot up (the result of GNFB).
The moral of this post? Start viewing a design in its totality rather than splitting a design up and ending up with a worse result.
AM
"The advantages of the adjustable floating paraphase inverter is that you can adjust symetry with output tubes that have either more or less different amplification factors."It's possible to adjust balance with any phase inverter, including the cathodyne. The LTP inverters in the amp I'm building now will have adjustable balance. The circuit is similar to the Citation V, although I'm using different tube types and component values. It's also worth nothing that imperfect balance here is largely compensated when NFB is used and the PI is included in the loop. Bandwidth is important for this purpose, and the LTP is excellent in that regard.
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Buy Chinese. Bury freedom.
Edits: 06/17/21
Looks good,, just be aware that different tubes and/or output transformer are highly likely to upset the feedback loop and the frequency compensation in the driver. (A lot of) testing will be require to ensure stability over different frequency ranges and output levels.
Personally I would be keen to add some resistors in series with the screens.
That's the schematic of the HK Citation V, just an example of the PI balance circuit in an LTP (and the same topology as the amp I'm building). I'm using HK transformers, but not from the Citation. Screens in each channel will be driven by a VR tube in series with B+, similar to the Leslie 122 amplifier. This will set the screens at 105V below B+, so just about any 6L6 variant can be used.
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Buy Chinese. Bury freedom.
For all the worry about balance, a lot of the stuff I have built of late runs a pentode LTP. Screen current is not going through the loads, but most certainly is passing through a cascode MOSFET current regulator at the cathodes. Phase-to-phase balance is not to be despised...LOL
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Two schematics from the RCA 26 manual, basically identical except for using a different output tube (and adjusting the B+ voltage).
To continue on my earler remark of having to adjust feedback loop and frequency compensation: Check the values of the feedback capacitor and the addition of frequency compensation in the phase splitter.
In the second schematic I like the way that the bias voltage is derived - an automatic adjustment when B+ comes on and there cannot be a run away because of a fault in the bias circuit.
This is what you have to figure out when not using the same components - especially transformers are troublesome, just any transformer which supposedly has the same impedance can result in total different behavior. This is due to interwinding capacitance and leakage inductance between primary and secondary. personally I would never ever consider trying to build a clone of the Williamson amplifier as it is only marginally stable and needed a very specially constructed OPT.
AM
"In the second schematic I like the way that the bias voltage is derived - an automatic adjustment when B+ comes on and there cannot be a run away because of a fault in the bias circuit."I do like the fact that bias tracks B+ regarding variations due to line voltage. However, this circuit brings the same liabilities to an AB1 amplifier as self bias. Namely, increased amplifier output and anode current causes bias voltage to increase, and that moves the amplifier closer to Class B. Small amplifiers operating near Class A may not suffer the effects, but large amplifiers biased in true AB1 will show crossover distortion at high output levels. This distortion will persist into quiet musical passages due to the time constant of the bias circuit, degrading fidelity. In addition, this circuit is essentially a voltage divider with the negative bias voltage being subtracted from available B+. That may not be a disadvantage, depending on the transformer used, but it has to be taken into account during initial design.
"I would never ever consider trying to build a clone of the Williamson amplifier as it is only marginally stable and needed a very specially constructed OPT."
The most significant issue with the Williamson IMO is the inclusion of too many stages and too many coupling caps. The Citation V topology is much cleaner, due in part to the simplification of the gain structure made possible by driving the outputs in pentode mode, rather than triode. Only two coupling caps are necessary, and there is significant headroom in every stage ahead of the outputs.
--------------------------
Buy Chinese. Bury freedom.
Edits: 06/18/21
And many rate the floating paraphrase for it's sound. From those that like the single ended distortion characteristics but have been persuaded by going PP. WE124 amplifier being a case of this.Anecdotal of course. No science involved in this post.
Edits: 06/16/21
The advantage here is that a differential amplifier has only a cubic non-linearity to express, rather than the higher distortion quadratic non-linearity of a single-ended circuit. In a nutshell if you have differential circuits from input to output (and power tubes can be differential too) then distortion does not compound as much from stage to stage as much since even orders are canceled.That leaves the 3rd as the dominant distortion product and it gets the same treatment as the 2nd (IOW nearly inaudible). But is has the same valuable property of masking the higher orders, which won't be as high amplitude to start with.
So you wind up with a circuit that sounds just as smooth and organic as SET, but with inherently lower distortion, so more detailed and smoother than SET. This is easy to hear- its not subtle. Easy to measure too, and confirms the math on paper- when all three happen at the same time its very real.
Edits: 06/15/21
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