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TubeCad Error re: LTP?

24.27.41.140

Posted on June 10, 2021 at 23:09:26
Triode_Kingdom
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The image below is from John Broskie's TubeCad magazine. Am I correct in thinking this is an error? Doesn't the CCS force equal signal currents, thus equal but opposite output amplitudes? I'm assuming this is a "perfect" CCS.













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RE: TubeCad Error re: LTP?, posted on June 11, 2021 at 02:56:46
cpotl
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I agree with you.

If the plate currents are I1 and I2, then I1 + I2 = I0, with I0 being the constant current through the CCS. So with equal plate resistors R, this means I1 R + I2 R = I0 R, and so (V0 - V1) + (V0 - V2) = I0 R where V0 is the V+ voltage, meaning the plate voltages V1 and V2 obey

V1 + V2 = 2 V0 - I0 R = constant

So if V1 increases, V2 must decrease by exactly the same amount, and vice versa.

An unusual mistake by TubeCad! Does he elaborate on why he says what he does? Do you have a link to his article?

 

it *IS* a 12AU7..., posted on June 11, 2021 at 03:06:24
PakProtector
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and with a CCS deserving the name, equal plate resistors are going to deliver equal output voltages. 12AU7 or not. Their relationship to the input is however in question...6CG7 and 6SN7 will do a bit better...heh-heh-heh

Mr. Broski has written a lot; if this is the first error, I'd say he is doing pretty well. At least he's not talking about replacing the leads of the chokes he happens to like using today.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: TubeCad Error re: LTP?, posted on June 11, 2021 at 04:43:47
dave slagle
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I fail to see the error. The closer the CCS is to ideal the closer the two plate resistors would need to be in value and in theory a perfect CCS with identical tubes would result in the need for matched resistors.

dave

 

RE: TubeCad Error re: LTP?, posted on June 11, 2021 at 05:11:07
cpotl
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"I fail to see the error. The closer the CCS is to ideal the closer the two plate resistors would need to be in value and in theory a perfect CCS with identical tubes would result in the need for matched resistors."

The error is that John Broskie is saying the two resistors should *not* be the same; he says that the left-hand one should be smaller.

By the way, I think the resistors should still be of equal value even if the two tubes are not matched? As long as there is no grid current, the current through the CCS must end up going through the two plate resistors, and when one plate current goes down the other must go up by an equal amount. This is true whether the tubes are matched or not.

This assumes there is no load on the outputs. If the outputs are loaded, things could get a bit more complicated. I wonder if that might be what he is driving at.

 

Link, posted on June 11, 2021 at 06:23:26
Triode_Kingdom
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Sorry, should have put this in the original post...

Oh, and notice that he repeatedly refers to "paraphrase" phase splitters. Must have been an off day!



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RE: TubeCad Error re: LTP?, posted on June 11, 2021 at 07:13:11
Triode_Kingdom
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"a perfect CCS with identical tubes would result in the need for matched resistors."

First, the article says the opposite - that even with the CCS, the resistors must be different values. That's the error.

Second, the tubes don't need to be identical. If the CCS is perfect, the AC signal current will be equal and opposite in the two triodes regardless. Only the presence of grid current can change that. One could be a 807 and the other an acorn tube.











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RE: Link, posted on June 11, 2021 at 08:27:39
Tre'
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John is a smart guy but we all make mistakes.

On UltraPath caps he rightly states that any ripple in the PS will get to the signal path and then goes about modifying the circuit searching for a fix. The fix is, of course, eliminating the ripply from the PS in the first place but he never goes there.

Like I said, we all make mistakes.

Tre'

Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: Link, posted on June 11, 2021 at 10:22:04
cellailuca
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By the way, if there is a grid resistor (instead of the grounded grid), depending on its value, the output becomes unbalanced with increasing frequency.


Omnes feriunt, ultima necat.

 

Not so much a mistake, but a different way, posted on June 11, 2021 at 15:10:41
Chip647
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Thank god for John though. What an active mind, I really envy his energy.

Good power supplies can be expensive. I appreciate his Aikido noise injection concept. I bought one of his boards a while back, (very nice board) I have yet to build it, probably because I like over-designing power supplies.

 

RE: Not so much a mistake, but a different way, posted on June 11, 2021 at 17:19:50
Tre'
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I could be off base and I'm sure in a commercial application, where the money needs to be watched, what I'm about to says won't make sense but what you call an over-designing power supply, I call a properly designed power supply.

Having said that, I agree with you. John has a sharp mind.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

Tube Cad Error??, posted on June 11, 2021 at 19:08:12
gusser
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I posted a similar question a couple of weeks ago. I find Tube CAD buggy. I can run the exact same tube type and circuit two days in a row and get different results.

Spice and my test bench always agree within component tolerances.

Just this week I was working on a 6DJ8 SRPP and when I ran "Make It So" it kept pushing the B+ up to 660 volts. That is ridiculous as the 6DJ8 is a low plate voltage tube (in tube terms).

I also don't like how you can't lock parameters like the B+. If you give it an impossible parameter list it should just error out versus inserting ridiculous values.

But hey, for $10 I got more than my money's worth. I know it is not certified to run on Windows 7 but it also says no 64bit. I have a 32bit Win 7 install? Perhaps there are more bugs with Win7 than just 64bit versions? I may try dusting off and old Win XP machine I have.

 

RE: TubeCad Error re: LTP?, posted on June 12, 2021 at 08:14:59
dave slagle
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If you assume a "perfect" CCS then I agree but it is you who added the "perfect" part and not broskie.

how close the "average" ccs is to the ideal and how closely matched the outputs need to be is open to interpretation. Then add in that broskie used the term "slightly different" and he gets a pass on this one from me. It is really easy to take any accurate statement and add a an additional caveat to it and make it appear inaccurate and that is what I see happening here.

dave

 

RE: TubeCad Error re: LTP?, posted on June 12, 2021 at 08:55:46
cpotl
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Well, that seems to be the most plausible explanation, certainly. And it is true that if the CCS is not perfect, as in the case in his previous schematic with a resistor connecting the cathodes to a negative supply rail, the left-hand anode resistor needs to be smaller than the right-hand one in order to achieve balanced outputs.

It would have been better if he had said that he was assuming an imperfect CCS, I feel, in that case.

 

Give me a break, posted on June 12, 2021 at 10:15:37
Triode_Kingdom
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You have certain areas of expertise that I respect, but It's truly tiresome to have to play trivial pursuit with your sub-atomic nit-picking. The impedance of a CCS should be in the range of megohms, and that obviates the need for different value anode resistors. Broskie doesn't say the CCS is imperfect. The text is in error unless there's a real world explanation other than "everything is subjective."


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Rosenblit Also Got This Wrong, posted on June 13, 2021 at 08:34:04
Triode_Kingdom
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According to Bruce Rozenblit ...

"If the current through V1 were identical to that through V2, then every increase in current flow through Rk caused by V1 would be met with a corresponding and equal decrease in current flow in Rk caused by V2. If the current is increased and decreased by the same amount at the same time, the net change is zero, and no signal would be present across Rk to drive V2. The only way the circuit can work is when the currents through the two tubes are unequal by a small amount." (Bold font is mine.)

Absolutely not true. The first sentence - on which every subsequent conclusion is based - is fundamentally incorrect. The two equal/opposite currents in each tube do not simply meet and cancel each other at Rk as Rozenblit suggests. Rather, the current flow between the two cathodes "seesaws" back and forth. This action occurs regardless of whether the currents are identical (such as when a CCS is used). When a tail resistor is used instead of a CCS, each triode attempts to create current flow independently according to its gain and drive level, and the difference passes through the tail resistor.

I'll just add that the error should have been caught when that last conclusion was put to paper. If it were true, it should be possible to null the output of V2 by moderate adjustment of the anode resistor ratio. The output of V2 should also null when currents are equalized by means of a CCS in the tail. Of course, neither scenario occurs, because the basic hypotheses regarding the necessity of unequal currents is wrong.



















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RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 09:10:38
cpotl
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Part of the problem is caused by the lack of clarity in his writing, I think.

For example, when he says "If the current through V1 were identical to that through V2...", he must, I suppose, be talking of the quiescent (no-signal) situation. (There never is any question of the currents through V1 and V2 being the same when there is a signal; if they were, then the signal voltages on the two anodes would be identical and in phase, which of course is not at all what happens.)

He then, as you point out, erronenously goes on to say "...every increase in current flow through Rk caused by V1 would be met with a corresponding and equal decrease in current flow in Rk caused by V2." This *would* be true in the case of an ideal constant current source, but it is not true for an Rk connected at its bottom end to some (finite) negative voltage source. (I suppose that must be what his schematic has?)

He then says "If the current is increased and decreased by the same amount at the same time, the net change is zero, and no signal would be present across Rk to drive V2." Of course, since his premise is false, the conclusion is false.

I think it is quite instructive to think about the situation where the negative voltage, call it Vneg, to which the bottom end of Rk is tied, is adjustable. Of course to get a given desired quiescent current through Rk, one has to adjust the value of Rk in step with the adjustment of Vneg. The ideal constant-current source is achieved in the limit where Vneg goes to minus infinity (and of course Rk correspondingly has to go to infinity, so as to keep the quiescent current through Rk at the desired value).

With that understanding, it is clear that as Vneg is made more and more negative, so that Rk is correspondingly getting larger and larger, the sum of the two cathode currents when there is a signal is getting to be more and more nearly constant (and equal to the quiescent current through Rk). In other words, the signal *current* through Rk is getting smaller and smaller as Vneg gets bigger and bigger. Despite the signal current getting smaller and smaller, the signal *voltage* across Rk, that is to say, the signal voltage on the cathodes, is not getting smaller and smaller. This is because the value of Rk is being adjusted to be bigger and bigger as Vneg is made bigger. In the limit that Vneg is sent to minus infinity, the "signal current" through Rk (which is now approaching infinite resistance) will be going to zero, but the signal voltage across Rk will be reaching its asymptotic value, which is roughly one half of the input signal voltage. In this limit the two tube currents do indeed see-saw up and down in exact anti-phase.

He then says "The only way the circuit can work is when the currents through the two tubes are unequal by a small amount." Does he mean the quiescent currents through the two tubes? If not, what does he mean? He certainly, presumably, does not mean the actual currents through the two tubes when there is a signal. As mentioned above, there never was any possibility of the tube currents, including signal, being equal. Rather, the expectation is that when one tube current rises above its quiescent value the other tube's current falls below its quiescent value. Or, when he says "when the currents through the two tubes are unequal by a small amount," is he referring to the *change* in current in one tube being nearly equal (and actually opposite) to the change in current in the other tube? It really ends up being an exercise in trying to understand the psychology of the writer!

Anyway, I'm sure you and I would agree on the facts. For an ideal constant-current source, the two outputs will have exactly equal and opposite (i.e. anti-phase) audio signals, if the two anode resistors are identical in value. When one instead has a cathode resistor Rk connected to a finite-voltage (negative) source, the two signals will not be of equal magnitude if the anode resistors are equal. The output signal from the left-hand anode would be larger than that from the right-hand anode, and so one would need to make the left-hand anode resistor appropriately smaller in order to balance the outputs. Of course, as Vneg is made more and more negative, the imbalance for equal anode resistors would become less and less.

 

RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 11:47:53
PakProtector
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It helps me to think of the LTP as a composite. It is like a grounded cathode amp for the input stage, and a grounded grid for the second. They are joined( ideally ) at the cathode node and taken to ground by a current regulator.

It does not matter what the idle current is, or how it is shared( until a lack of voltage outptu somewhere comes into play). Raise the grid of the input stage and the current through that stage increases. The cathode follows it, driving the second half towards cut off.

There is some solid math out there to describe the behaviour when the cathode load is resistive.

Gain plate to plate is about u, gain for a single plate to ground is half of u. From the time when a high gain pentode drove a split load( a fairly easy load ), another stage ahead of it was usually employed( the 5-20 comes to mind). Laurent and the 6AN8/7199 made things slightly lower in parts count.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 15:30:56
Triode_Kingdom
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"Laurent and the 6AN8/7199 made things slightly lower in parts count."

I had a private discussion recently with another member regarding the pentode-triode topology. I had planned to use it for a series of amplifiers I'm building, but halfway through the first design, realized it wouldn't have sufficient gain. in fact, the goal of driving an amplifier to full output with roughly 1V input places distinct limitations on the use of this circuit.

For example, it works in the Dynaco ST-70 because UL operation reduces the amount of global NFB needed. Rather than the typical 20 dB or more, the ST-70 applies about 13 dB. This reduces the driver gain requirement by at least 7 dB in comparison to a straight pentode amplifier. The 7199 also works well when driving higher-sensitivity output types, such as the 7591, which typically bias with -20V or so on the grids. However, driving a 6L6 variant with -30V or -35V on the grids to full output isn't possible, because that requires more gain than the 150 or so that's available with this circuit.

This limitation wasn't obvious to me until my first project in the series of amplifiers I'm currently building was well underway. I eventually had to backtrack and revise the driver circuit from a cathodyne to a medium-mu triode driving a LTP. That shares the gain structure between two stages and brings much more total gain to the table.

You probably know all about this, but I haven't designed a lot of push-pull pentode amps from scratch. The large numbers of commercial amplifiers that use the triode-pentode topology created an assumption on my part that wasn't true. I'll be more careful next time. :)








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RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 16:56:22
AmadeusMozart
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Well there must be something wrong with my LtSpice since it confirms Brosky and Rozenblit.

AM

 

RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 18:10:48
Triode_Kingdom
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I used it too, to study the current flows and AC voltages. Put a CCS under the tail to force equal currents through the tubes. If V2 is still working, Rosenblit is wrong.






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RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 18:32:22
cpotl
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"Well there must be something wrong with my LtSpice since it confirms Brosky and Rozenblit."

Are you saying that you used a CCS supplying the cathodes, and found that you needed to make the left-hand anode resistor smaller than the right-hand one, in order to get balanced outputs? (That is what was said in the Broskie article) This cannot be possible.

I also checked using LTSpice, and confirmed that in the simulation the two anode resistors need to be equal in order to get balanced outputs. This agrees with theory, and it agrees with what TK has said.

Of course, if you instead use a cathode resistor connected to a (finite) negative supply voltage at its bottom end, then you *will* find that you need to make the left-hand anode resistor smaller than the right-hand one, in order to achieve balance. This is because the cathode supply is not acting as a perfect constant-current source.

 

It's Even Worse than What I Posted..., posted on June 13, 2021 at 18:59:43
Triode_Kingdom
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I left one thing out of the original post because it's so confusing. As I said, Rozenblit stated that the only way the circuit can work is when the currents through both tubes are unequal by a small amount. It needs to be pointed out that he's not talking about some molecular-level minutiae. Here's the next thing he says:

"The only way to make the currents unequal is to make RL1 and RL2 unequal. The required difference is on the order of 10-20%, with RL2 always the larger of the two."

This statement (plus his previous assertion) tells me he has totally failed to grasp the operation of the circuit. The need to make the resistors unequal is of course because we need to convert the unequal currents to balanced differential output voltages. But the unequal currents aren't necessary to the function of the circuit, they're a detrimental side effect of using a resistor in the tail while driving one triode at its grid and the other at its cathode.

One thing interesting about all this is that Broskie has referred to Rozenblit's work in several places within TubeCad. I do wonder if they're working off a shared misconception in this one area.





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RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 20:11:53
elblanco
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page 137 might help.

 

RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 20:32:54
elblanco
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isnt it because gain of v1 is common cathode

mu * RL / Rp + RL

and v2 is grounded grid which is

(mu + 1)RL / Rp + RL

CCS or no CCS

 

RE: Rosenblit Also Got This Wrong, posted on June 13, 2021 at 20:39:07
elblanco
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and cathode (v2) Gm = (mu +1)/Rp

 

Owww!, posted on June 13, 2021 at 21:13:17
Triode_Kingdom
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Well, THAT made my brain hurt! Excellent text, thanks for the link!








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RE: Owww!, posted on June 14, 2021 at 03:08:39
cpotl
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As I said before, the main problem with the Rozenblit article is the lack of clarity and precision in his writing. That sentence of his that you highlighted, "The only way the circuit can work is when the currents through the two tubes are unequal by a small amount," is an excellent example of this.

First of all, when he says "the currents through the two tubes" he doesn't literally mean the total currents flowing in the two tubes. He means the *changes* in currents in the two tubes when the signal is present. And his sign conventions are such that if an *increase* in the current in tube 1 counts as a positive "change in current," then at the same time the *decrease* in current in tube 2 counts as a positive "change in current." That point is made much clearer in Crowhurst, because he gives equations.

The other misleading thing about Rozenblitz's sentence is that when he says " The only way the circuit can work is when the currents through the two tubes are unequal by a small amount," he makes it sound as if this is something that the user is going to have to arrange in order to be able to make the circuit work. And this is not the case. It would have been much clearer if he had said:

"It is a fact that the current-changes in the two tubes are unequal, and this is why the circuit works."

Crowhurst is much clearer about things, largely because he gives equations. However, even he makes it a bit confusing, I think, when he says "In order to get any drive to V2 at all, it is necessary to have unequal values of I1 and I2...". Again, this makes it sound almost as if this is something the user is going to have to arrange. Whereas in fact the true situation is that I1 and I2 *are* unequal, and this is why V2 is getting driven.

Of course, all of the above applies to the case of a cathode resistor Rk whose bottom end is connected to a (finite) negative voltage source Vneg. In the limit where one approaches an ideal constant-current source (which could be viewed as Vneg goes to minus infinity and at the same time Rk goes to infinity so as to keep the quiescent cathode current at the desired value), the tube V2 still gets driven even though the changes in the two tube currents *do* now become equal. That is because it takes only an infinitesimal change in current to change the voltage dropped across an infinite resistance by a non-zero amount.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 03:13:16
PakProtector
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There are ways to turn up the gain of the pentode section that don't have to raise the output impedance. These do require silicon current regulators, which I do recall are contrary to the religious teachings you subscribe to...LOL

Another reason to miss the participation of Mssr's Pimm and Kelly.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

Dave..., posted on June 14, 2021 at 03:20:24
PakProtector
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I do believe that the amount of compensation in terms of resistor value suggested by Mr. Broski precludes that conclusion. IIRC, one of us posted the suggested/required plate load resistor compensation to be in the neighborhood of 15-20%. That is for sure suggesting that the ccs is about as good as a tail resistor 2/3 to 3/4 the value of the plate resistors.

With even a single 10M45 the circuit needs resistors as closely balanced as you can select( until the high frequency degradation of the 10M45 drops its effective resistance ).

I'd drag Mr. Broski in here by preference if I could. My bet is he'd recognize the mistake gracefully.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 05:58:11
PakProtector
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The ccs is not to force equal currents through the tubes. It is to enforce its quantity. For reasonably matched sections( say a 5687 ), the CCS is not going to do anything to enhance the match. It will enforce that the sum of the two cathodes remains constant...
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 06:00:35
PakProtector
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Its cathode is not grounded. Not even close. On that basis your equation set needs attention.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 06:12:32
elblanco
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okay CCS.

still doesnt change the fact that gain is slightly higher with grounded grid.

No?

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 06:19:05
PakProtector
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Those two bits are symmetrical. neither cares where the signal comes from because each is loaded identically.

IOW, try and show how the bit with the grid grounded is actually different from the one we are applying signal to.

I say both are the same in this particular LTP circuit.

Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 06:24:19
cpotl
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"The ccs is not to force equal currents through the tubes."

In the very confusing article by Rozenblit, when he speaks of "the currents through the two tubes" he actually means the *changes* in the currents through the two tubes. In other words the quantities that Crowhurst calls I1 and I2. And the CCS *does* force Crowhurst's currents I1 and I2 to be equal.

 

RE: Dave..., posted on June 14, 2021 at 06:28:09
cpotl
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I agree with you. In the context of the Broskie article it really makes no sense to suppose that when he speaks of a CCS, he is actually talking of a very poor CCS that doesn't really deserve the name CCS.

It must be a slip-up on his part.

 

RE: Dave..., posted on June 14, 2021 at 06:55:00
dave slagle
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I initially simmed this with a cascoded 10m45 since I already had the circuit built and indeed the values needed to be matched which seemed to confirm TK's original criticism. Then I went to a single lowly 10m45 and found for a 6SN7 with a 30,000Ω resistor on the left tube I needed a 30,775Ω to match the outputs from 100Hz to 1kHz. This combined with broskies use of the term "slightly" got me into this discussion. If this is picking nits then so be it but there is probably an equal chance of the same conversation happening if his underlying assumption was the mythical ideal CCS. If you look at his audience, I suspect a good majority of them fall into the group who think a single 10M45 is close enough to the ideal which makes me not take issue with his statement and I still feel adding the assumption of the ideal and then calling it in error is more akin to a cheap shot than a valid criticism.

dave

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 07:17:02
elblanco
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but the tube section on the right has different characteristics due to transconductance being higher. Grounded grid Gm is higher. they are essentially two different tubes.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 07:26:38
PakProtector
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That is what I am trying to get you to see; the one is no more a grounded grid deserving special treatment than the other. I asked you to point to the reason that one is indeed as you describe, and thus deserving of the special consideration.

The nominal 'input' section is electrically identical to the one you refer to as grounded grid.

cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Dave..., posted on June 14, 2021 at 07:38:08
cpotl
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I don't think it a question of "cheap shots" or anything like that at all.

Personally, I have an extremely high opinion of John Broskie, and I love the clarity and the precision of his articles. They are also very pedagogical, which is great.

Anyone can make the occasional mistake, though, and it is not at all a "cheap shot" to draw attention to the very exceptional slip-up if it occurs.

As I said, I like the clear and pedagogical style of his articles. But in an article on the long-tailed pair, the key pedagogical distinction to be made between the case of a cathode resistor connected to a negative voltage source, versus the case of the CCS, is that in the former one needs unequal anode resistors whereas in the latter case the anode resistors should be equal. One could certainly go on, after having made this clear pedagogical point, to discuss what happens when the CCS is not giving a truly constant current. But it just doesn't seem to fit with his usual clear style of writing for him to be assuming when he first introduces the CCS that it isn't a true CCS.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 08:01:06
PakProtector
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Now you are adding specific words to what they wrote. Ones that while useful go a long way to forward the idea that these folks were brilliant writers about the subject. It is irrelevant how brilliant they are, it does not change how a LTP works...LOL

Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Rosenblit Also Got This Wrong, posted on June 14, 2021 at 08:10:34
cpotl
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"Now you are adding specific words to what they wrote. Ones that while useful go a long way to forward the idea that these folks were brilliant writers about the subject."

I didn't think I was promoting the idea that Rozenblit wrote brilliantly on this subject; quite the opposite!

All I was saying is that having seen how Crowhurst describes it, and he makes himself clear by actually writing down some relevant equations, one can then gain some inkling as to what Rozenblit was trying, in his confusing style, to say.

But I agree with you, that what really matters is what is actually happening.

 

Another Text - Long Tailed Pair, posted on June 14, 2021 at 08:19:38
Triode_Kingdom
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Looks interesting, a different approach than Crowhurst, haven't had time to read through it yet. The LTP text begins on p. 61.







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HA!, posted on June 14, 2021 at 09:53:37
PakProtector
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I can read the 10Mxx spec sheets...LOL Don't mistake me for thinking that beast is a good CCS used singly. Your spice results about parallel what I discovered whilst running longer and longer resistor tails( at higher and higher voltages ). The chip's specs are basically honest.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Dave..., posted on June 14, 2021 at 09:58:38
PakProtector
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Actually it is a matter of covering everything that can possibly count. And then explaining it thoroughly without missing anything. Pretty simple...LOL

Douglas
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Crowhurst et al..., posted on June 14, 2021 at 11:17:55
Triode_Kingdom
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Well, I believe the confusion in Crowhurst's paper lies in a certain difficulty he may have experienced in translating what he knew and formulated into a plain English explanation. His text must be read very carefully in order not to misinterpret his meaning. That's not what I see when reading the hypothesis by Broskie or Rozenblit. Their descriptions of the circuit seem flawed by a fundamental conceptual error. No matter how many times I review, it simply isn't possible to reconcile what they're saying with my understanding of the circuit, which includes results obtained in SPICE. :(






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RE: Crowhurst et al..., posted on June 14, 2021 at 11:28:21
elblanco
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just for giggles...-

separate the two tube halves, using a CCS for each one and couple the cathodes together with a cap.

 

RE: Crowhurst et al..., posted on June 14, 2021 at 13:08:55
cpotl
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Crowhurst's schematics leave a little bit to be desired. In his first schematic, Fig. 624, I was assuming that the connection to the grid of V2 should be understood to be at ground potential, and that the "non-cathode end" of Rk is connected to the supply written as something not quite readable, but looks like maybe -B0, which is presumably indicating some negative voltage (relative to ground). This would be a standard type of LTP. (It would have been nice if a ground symbol had been attached to that line coming from the grid of V2, though!)

But then he says he is going to redraw the schematic in the form of Fig. 626. But now, in Fig. 626, the "non-cathode end" of Rk is connected to ground, which is also where the grid of V2 is connected, of course.

So it appears that his Fig. 626 is not merely a redrawing of Fig. 624; he has done away with the negative supply for the bottom end of the tail of the LTP.

Unless, of course, the line from the "non-cathode end" of Rk in Fig. 624 is actually supposed to have a blob at the point where it crosses the line from the grid of V2. In which case the slightly unreadable voltage that could be -B0 is actually just denoting the negative end of the B+ supply, and is intended to be synonymous with ground.

I don't think that is a very likely scenario, though. I think the whole essence of a long-tailed pair, in its traditional form, is that the "non-cathode end" of Rk should be connected to a negative voltage (i.e. negative with respect to ground)? As an LTP, it would really work very poorly indeed if Rk were connected to ground at its bottom end.

But then, what is going on with Fig. 626? It is clearly then not simply a redrawing of Fig. 624.

 

Current Change, posted on June 14, 2021 at 15:32:51
Triode_Kingdom
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"Of course, all of the above applies to the case of a cathode resistor Rk whose bottom end is connected to a (finite) negative voltage source Vneg. In the limit where one approaches an ideal constant-current source (which could be viewed as Vneg goes to minus infinity and at the same time Rk goes to infinity so as to keep the quiescent cathode current at the desired value), the tube V2 still gets driven even though the changes in the two tube currents *do* now become equal. That is because it takes only an infinitesimal change in current to change the voltage dropped across an infinite resistance by a non-zero amount. "

I believe you're referring to the current through Rk, and that is clearly equal to (I1-I2). However, this is not responsible for creating the AC voltage that drives V2. That voltage derives from the V1 cathode. In fact, I'm bothered by Crowhurst's statement that, "This change is the input to V2 and, if Rk is very large, we can have (I1-I2) very small and still get some drive into V2." This implies that the drive for V2 is created only by (I1-I2) flowing in Rk. However, the drive into V2 is in fact provided by V1 as a voltage source. It only takes a few minutes with SPICE to demonstrate that the AC voltage at the cathodes remains relatively unchanged as Rk approaches infinity and the tube currents become equal.













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RE: Current Change, posted on June 14, 2021 at 16:06:56
cpotl
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"This implies that the drive for V2 is created only by (I1-I2) flowing in Rk. However, the drive into V2 is in fact provided by V1 as a voltage source."

I think it is in a sense a matter of convention, what one ascribes as being responsible for what. The important point is that there is an equation that relates the changes I1 and I2 in the anode currents, and the change (Delta Vk) in the voltage on the cathodes, namely

(I1 - I2) * Rk = (Delta Vk)

You could interpret this equation as saying that "if I know (Delta Vk) I can calculate I1 - I2." Or else you could interpret it as saying "If I know I1 - I2 I can calculate (Delta Vk)." I don't think it really matters whether you think of it as the right-hand side of the equation dictating what the left-hand side does, or the other way around.

The important point is that the whole system is described by a well-posed set of equations, which can be solved. (As, indeed, is being done by LTSpice.)

Yes, I agree that the AC voltage (represented by (Delta Vk) at the cathodes is relatively unchanged as one passes to the CCS limit in which Rk goes to infinity (with Vneg on the "tail" correspondingly going to minus infinity). That is how it comes about that (I1 - I2) goes to zero in the Rk --> infinity limit.

I suppose, by the way, that Crowhurst just made a mistake with his Fig. 626, and that he really intended to connect the bottom end of his Rk to a negative source voltage (his -B0 of Fig. 624, if that is what that smudgy symbol is), rather than having it connected to ground. I haven't had the time or patience to go through all his equations yet.

 

RE: Current Change, posted on June 14, 2021 at 16:52:41
Triode_Kingdom
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"(I1 - I2) * Rk = (Delta Vk)"

Isn't this telling us that if I1and I2 are equal, so that the difference is 0, there will be no AC drive voltage at the cathode?






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RE: Current Change, posted on June 14, 2021 at 17:33:08
cpotl
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" "(I1 - I2) * Rk = (Delta Vk)"

Isn't this telling us that if I1and I2 are equal, so that the difference is 0, there will be no AC drive voltage at the cathode? "

No, because Rk is tending to infinity at the same time as (I1 - I2) is going to zero. The rate at which Rk goes to infinity exactly balances against the rate at which (I1 - I2) goes to zero, so that the product of the two gives the finite and non-zero quantity (Delta Vk).

It is maybe clearer to view it as the equation (I1 - I2) = (Delta Vk)/Rk,
so that ones sees that as Rk goes to infinity, while holding (Delta Vk) approximately fixed (finite and non-zero), then it implies (I1 - I2) goes to zero. This is the CCS limit, where I1 and I2 become equal.

 

RE: Current Change, posted on June 14, 2021 at 22:18:40
Triode_Kingdom
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No, you've misunderstood it too, but what you said forced me to re-examine Crowhurst's meaning.

There is indeed no drive to V2 when the two currents are equal. Crowhurst confirms this meaning when he says, "If you look at Fig. 625-a, you will notice that to get any drive at all for V2 it is necessary to have unequal values of I1 and I2..."

Of course, we know that if a perfect CCS is used, I1 and I2 are theoretically equal. According to the above, this should mean there's no drive for V2. This is the same point in Rozenblit's statement that got my attention. So what's going on?

Well, Rozenblit DID get it wrong. He clearly intends this to mean that V2 will have no AC output if AC currents I1 and I2 are equal, and his statement in total confirms this. In the case of Crowhurst, however, the statement regarding the necessity of I1 and I2 being unequal isn't referring to AC current flows. It refers only to the instantaneous, simultaneous value of the two currents at a single point in time, a reference to the various static values along the see-saw diagram. It's just one step in the set of explanations that lead up to the total picture, and it can't be taken out of context.

Crowhurst himself explains the circuit in a way that almost completely obscures this issue. However, he's only saying that if both currents are, say, +2 mA, there will be no drive for V2. The same is true when the currents are both +3 mA, or both +4.5 mA. Drive is only created when they are different, such as +4 mA and -4 mA, or +2 mA and -2.5 mA. I believe this may be the source of confusion.



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RE: Current Change, posted on June 15, 2021 at 04:35:34
PakProtector
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I am not going through the olde math. It is clearly poorly presented.

The sum of the two sections current is a constant.
At idle the two sections do not need equal currents.
For the time being, treat the CCS as perfect...

Examine the performance; driven SE, the gain of either section is ~mu/2.
-This is the CCS keeping the current sum of both sections constant. The CCS forces a cathode signal that cuts off the non-inverting section by half the amount the inverting section *WOULD have wanted to conduct( reducing its mu by 1/2 ), IFF it were a stand-alone grounded cathode stage.

The '~' is so I can ignore the math showing output voltage based on resistive loading. The load magnitude of that resistance will determine how far away from mu it actually is, and so far nobody is worrying about, or failing to understand that effect...LOL

Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Current Change, posted on June 15, 2021 at 05:21:17
cpotl
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"In the case of Crowhurst, however, the statement regarding the necessity of I1 and I2 being unequal isn't referring to AC current flows. It refers only to the instantaneous, simultaneous value of the two currents at a single point in time, a reference to the various static values along the see-saw diagram."

Yes, I agree with this. The currents I1 and I2 are the instantaneous values of the *changes* in the anode currents from their quiescent values. Essentially, one treats the problem in terms of linearised perturbations around the quiescent state. By this means the problem is reduced to a set of N linear equations that can be solved for all the N unknowns, in terms of one given quantity (the input voltage Vg1 on the grid of tube 1). As I said before, this set of N equations gives a well-posed system, in the sense that there are exactly the right number of equations to allow all the N unknowns to be solved for uniquely.

Luckily in this problem there are no capacitors or inductors that play any important role in the essential basic discussion, and so the whole thing can just be viewed in terms of static "snapshots." This must be what Crowhurst is doing. I've not been through all his equations yet, but I take it he has written down the set of linearised equations, and then solved them. As soon as I get the chance, I'm going to go through this exercise myself, to familiarise myself with all the details.

To come back to the equation we have been discussing, (I1 - I2) Rk = (Delta Vk), this is, of course, just one out of the complete set of equations that together form the "well-posed system" I was speaking of. I shall assume for now that we are discussing the case where Rk is finite; that is to say we are discussing an old-style LTP where the current through Rk is not going to be constant.

(I1 - I2) Rk = (Delta Vk) is an equation that must certainly hold, but it does not, by itself, allow one to solve for all the unknowns of course. I think it can be a bit misleading to take this single equation in isolation and say "if I1 and I2 were equal then (Delta Vk) would be zero." Yes, this is a true statement, but it is based on a counter-factual assumption. (Again, I emphasise that for now I am discussing the case where Rk is finite, not the Rk --> infinity CCS limit.) The full well-posed system of equations does not admit I1 = I2 as a solution (when the input voltage Vg1 on grid 1 is non-zero). So supposing that I1 = I2 is a little bit like supposing that 2 + 2 = 5; one can end up getting misleading or incorrect conclusions by supposing counter-factuals to be true!

The way I would say it is the following: If we write down the complete set of N well-posed equations we can solve for all N unknown quantities in terms of the input voltage Vg1 on grid 1. In particular, it will be the case that in this solution, I1 and I2 are unequal, and this difference is related to (Delta Vk) by that equation (I1 - I2) Rk = (Delta Vk). So what happens if one were to try saying "let me suppose that I1 = I2"? Well, the answer is that this would be an (N+1)'th equation and the system of equations would now be over-determined, admitting no solution at all (if Vg1 is still viewed as freely specifiable). Or, to put it equivalently, this (N+1)'th equation would mean that one could now solve also for Vg1 in addition to the other N unknowns. The solution would be Vg1=0. In other words, there would be no solution with Vg1 non-zero.

This is why I think it is a little misleading when Crowhurst (or Rozenblit) says something like "in order to get any drive at all for V2 it is necessary to have unequal values of I1 and I2...". The way it is phrased, it makes it sound almost as if this is an instruction to the builder, saying "you had better make sure that you build this circuit so that I1 and I2 are unequal." Whereas in fact, a better way to say it would be "when you solve the system of equations, you will find that I1 and I2 are unequal."

Now, if we pass to the CCS limit, by sending Rk to infinity (it always being understood that the negative supply voltage Vneg on the bottom end of Rk is adjusted appropriately so as to keep the quiescent current through Rk fixed), then the equation (I1 - I2) Rk = (Delta Vk) continues to hold as one of the system of equations. And now, when you solve the full set of N well-posed equations it will turn out that (Delta Vk) comes out to be some particular finite and non-zero result (as a function of the input voltage Vg1), and hence by taking (I1 - I2) Rk = (Delta Vk), dividing it by Rk, and then sending Rk to infinity, we see that I1 - I2 goes to zero. That is, I1 = I2 in the CCS limit.

By the way, in your final paragraph when you discuss examples of values for I1 and I2, you should keep in mind the sign conventions Crowhurst is using. He is defining I1 as the *increase* in current through tube 1 from its quiescent value, whereas he is defining I2 as the *decrease* in current through tube 2 from its quiescent value. (This is why that equation reads (I1 - I2) Rk = (Delta Vk) rather than (I1 + I2) Rk = (Delta Vk).) In other words, I1 is positive when the current through tube 1 increases, and I2 is *positive* when the current through tube 2 decreases. Since the currents in the two tubes see-saw up and down, this means that at any given instant I1 and I2 as he has defined them are either both positive, or they are both negative. They will never be of opposite signs at any given instant.

I don't think I've misunderstood anything in what I've said so far. For my taste, my understanding is not yet complete, because I would still like to complete the job of examining the full set of N equations for myself. I'll try to get to that as soon as I have the chance.

 

RE: Current Change, posted on June 15, 2021 at 07:29:21
Triode_Kingdom
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Very good! All correct as far as I can see! The LTP wasn't of much interest to me until recently. I've been trying to understand its mechanisms better over the last few weeks, and that's what led me to the Broskie article, and then Rosenblit. I never would have thought a circuit that appears so simple on the surface could be so difficult to fully analyze. Thanks for weighing in!





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RE: Crowhurst et al..., posted on June 15, 2021 at 08:22:43
PakProtector
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How big shall we make this cap????

:)

Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

Broskie is correct on this one., posted on June 15, 2021 at 08:34:06
Ralph
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Even with an excellent CCS the mu of the tube is rather low. The mu is the main issue here- its what is causing Broskie to make this claim.

So even with the forced coupling at the cathodes, the side that is receiving the input signal will have ever so slightly higher output. If a 12AT7 were used, this difference would be much smaller.

For this reason you usually want as much gain out of the differential circuit as possible.

Now to be clear the difference between the outputs will be slight if you have a really good CCS. Its so slight that mismatched sections might swing things the other way- but we are assuming here that the 12AU7 is in its ideal form.

I've been building this circuit for decades FWIW, using a two stage CCS tied to a B- that is the same voltage otherwise as the B+ (the circuit is in our MP-3 preamp, which is based on the first fully differential preamp, our MP-1, offered to home audio anywhere). IOW I am speaking from experience which includes making really effective CCS circuits; IMO/IME the performance of the CCS is paramount to a good differential circuit and frankly most CCS circuits I see leave performance on the table.

 

RE: Crowhurst et al..., posted on June 15, 2021 at 08:40:50
elblanco
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my random number generator chose 5. so 5uf.

 

If that is really what he said then Rosenblit Got This Wrong, posted on June 15, 2021 at 08:41:00
Ralph
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Your explanation is correct- you don't need unequal currents and they do seesaw. The reason Broskie is correct is simply that the 12AU7 isn't a perfect tube even in its ideal form- it has a fairly low mu. But if you want low gain- such as in a line stage- then you deal with it.

Improving the CCS reduces offsets, distortion, improves bandwidth and increases gain. So its really a good idea to have a good CCS. And they can be so good now that the actual offsets will likely come down to actual tube differences, which is about as good as you can expect. We don't bother with different plate values on this account- in fact we match them.

 

and pages 278-279, posted on June 15, 2021 at 08:49:23
elblanco
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.

 

RE: Current Change, posted on June 15, 2021 at 09:23:19
Triode_Kingdom
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"This is why I think it is a little misleading when Crowhurst (or Rozenblit) says something like 'in order to get any drive at all for V2 it is necessary to have unequal values of I1 and I2...'. The way it is phrased, it makes it sound almost as if this is an instruction to the builder, saying 'you had better make sure that you build this circuit so that I1 and I2 are unequal.'

That IS what Rozenblit is saying. Because immediately after he says the currents must be unequal for the circuit to work, he says "The only way to make the currents unequal is to make RL1 and RL2 unequal. The required difference is on the order of 10-20%, with RL2 always the larger of the two." I can't prove it, but this looks suspiciously like Rozenblit himself read the Crowhurst text, misunderstood it, and integrated the misconception into his own work. That's just conjecture, of course.

Crowhurst, OTOH, does understand how the circuit works, but he expresses it poorly, misleading the reader. Like you say, it appears on the surface as though it might be an instruction, but careful reading and interpretation clarifies his meaning.








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RE: Broskie is correct on this one., posted on June 15, 2021 at 09:30:13
Triode_Kingdom
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Can you quantify "ever so slightly"? The two output voltages can only be slightly different if the current flow through the two triodes is also slightly different. That won't happen if the CCS is effective. Of course, the term "effective" can be nit-picked, but I don't believe Broskie was referring to microscopic differences, based on his instruction to use different value anode resistors.






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Coupling Cap, posted on June 15, 2021 at 09:56:56
Triode_Kingdom
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"just for giggles...- separate the two tube halves, using a CCS for each one and couple the cathodes together with a cap."

There shouldn't be any difference, other than the LF rolloff of the cap. I took a quick look at this in SPICE, and it agrees.







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RE: Crowhurst et al..., posted on June 15, 2021 at 10:35:04
PakProtector
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'f'-FIX-1, and *MY random number generator took about 7 tries to hit 5( .5 actually ).

cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: Broskie is correct on this one., posted on June 15, 2021 at 11:44:33
Ralph
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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).

 

RE: Broskie is correct on this one., posted on June 15, 2021 at 12:20:32
elblanco
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its Kirchoff once CCS is involved and plate loads are equal.

Broskie wrong on this one. But I dont wanna say totally wrong.

 

RE: Broskie is correct on this one., posted on June 15, 2021 at 12:45:21
Stephen R
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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 ;)

 

Only if the CCS is 'perfect', posted on June 15, 2021 at 13:14:27
Ralph
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Which is won't be. But you can get them pretty good these days. We made a tube CCS using a 6SN7 that was good to about 17 parts per million. These days we use a solid state CCS that is about an order of magnitude better. But I'm convinced we can do better than that, although we're well within the margin of error- so matched plate resistors work just fine :)

 

I don't think so, not if you have a good CCS, posted on June 15, 2021 at 13:19:53
Ralph
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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.

 

RE: Broskie is correct on this one., posted on June 15, 2021 at 13:32:00
cpotl
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"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.


 

And that was the point..., posted on June 15, 2021 at 13:46:21
Triode_Kingdom
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"we're well within the margin of error- so matched plate resistors work just fine ..."

But Broskie has written that it's so unbalanced - even with a CCS - as to require a workaround, and that matched resistors are not sufficient. This would be a ludicrous statement if the CCS can indeed achieve 1.7 ppm as you say, so I fail to understand why you would claim that Broskie is correct.





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RE: And that was the point..., posted on June 15, 2021 at 13:51:36
Ralph
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I fail to understand why you would claim that Broskie is correct.
Well, I'm not a fan although he comes up with things from time to time.

As I mentioned before, I think I get hung up on the nomenclature. Put another way, since no CCS will be perfect, it will therefore be unable to perfectly simulate infinite resistance connected to an infinite voltage.

It'll come close though, close enough that no-one will care if the two plate voltages aren't equal. I'm speaking from experience here- if I get them within a few volts I'm happy. Succeeding amps downstream will always help out, and if the circuit happens to be that downstream amp, it'll be getting something closer to a balanced input anyway.

At any rate its no worries :)

 

That is where the rot creeps in, posted on June 15, 2021 at 14:29:33
PakProtector
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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.

 

RE: That is where the rot creeps in, posted on June 15, 2021 at 14:50:41
cpotl
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"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?

 

RE: That is where the rot creeps in, posted on June 15, 2021 at 17:32:04
PakProtector
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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.

 

RE: That is where the rot creeps in, posted on June 15, 2021 at 18:09:40
cpotl
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"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?

 

RE: That is where the rot creeps in, posted on June 15, 2021 at 20:31:22
Triode_Kingdom
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"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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RE: That is where the rot creeps in, posted on June 16, 2021 at 01:49:52
PakProtector
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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.

 

RE: That is where the rot creeps in, posted on June 16, 2021 at 02:05:41
cpotl
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"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.

 

RE: That is where the rot creeps in, posted on June 16, 2021 at 02:25:30
Triode_Kingdom
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"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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RE: That is where the rot creeps in, posted on June 16, 2021 at 03:36:15
PakProtector
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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.

 

RE: That is where the rot creeps in, posted on June 16, 2021 at 03:39:59
PakProtector
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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.

 

RE: That is where the rot creeps in, posted on June 16, 2021 at 04:14:17
Triode_Kingdom
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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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RE: That is where the rot creeps in, posted on June 16, 2021 at 04:24:56
PakProtector
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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.

 

LTP VS Differential Amplifier, posted on June 16, 2021 at 06:59:46
Triode_Kingdom
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"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.



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Buy Chinese. Bury freedom.

 

RE: LTP VS Differential Amplifier, posted on June 16, 2021 at 07:20:24
PakProtector
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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.

 

RE: Coupling Cap, posted on June 16, 2021 at 10:11:05
elblanco
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can you spice circuit in fig. 626 and split up v1 and v2, cap couple them, adjust Rk for singles. use same RL value for each. no CCS.

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 12:51:56
Ralph
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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.

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 14:42:00
PakProtector
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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.

 

RE: Broskie is correct on this one., posted on June 16, 2021 at 14:57:00
AmadeusMozart
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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

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 14:59:51
Ralph
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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.

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 15:26:18
PakProtector
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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.

 

RE: Broskie is correct on this one., posted on June 16, 2021 at 15:30:20
Stephen R
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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.

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 15:37:19
Stephen R
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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 ;)

 

RE: What I like about the Crowhurst analysis, posted on June 16, 2021 at 16:17:46
PakProtector
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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.

 

RE: Current Change, posted on June 16, 2021 at 17:56:37
PakProtector
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Well, have fun with your PP adventures.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

RE: LTP VS Differential Amplifier, posted on June 16, 2021 at 18:11:12
cpotl
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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?

 

RE: LTP VS Differential Amplifier, posted on June 16, 2021 at 21:56:20
AmadeusMozart
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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).

 

RE: LTP VS Differential Amplifier, posted on June 17, 2021 at 02:11:00
cpotl
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"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?

 

PI Balance, posted on June 17, 2021 at 07:33:53
Triode_Kingdom
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"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.




--------------------------
Buy Chinese. Bury freedom.

 

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