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In Reply to: RE: Rosenblit Also Got This Wrong posted by Triode_Kingdom on June 13, 2021 at 15:30:56
Well there must be something wrong with my LtSpice since it confirms Brosky and Rozenblit.
AM
Follow Ups:
"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.
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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Buy Chinese. Bury freedom.
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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Buy Chinese. Bury freedom.
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.
"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.
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.
"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.
and cathode (v2) Gm = (mu +1)/Rp
Edits: 06/13/21
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
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.
okay CCS.
still doesnt change the fact that gain is slightly higher with grounded grid.
No?
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.
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.
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.
page 137 might help.
.
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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Buy Chinese. Bury freedom.
Well, THAT made my brain hurt! Excellent text, thanks for the link!
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Buy Chinese. Bury freedom.
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.
Edits: 06/14/21
"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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Buy Chinese. Bury freedom.
"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.
"(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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Buy Chinese. Bury freedom.
" "(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.
Edits: 06/14/21
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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Buy Chinese. Bury freedom.
Edits: 06/14/21
"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.
Edits: 06/15/21 06/15/21
"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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Buy Chinese. Bury freedom.
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.
Edits: 06/15/21
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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Buy Chinese. Bury freedom.
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.
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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Buy Chinese. Bury freedom.
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.
Edits: 06/14/21
just for giggles...-
separate the two tube halves, using a CCS for each one and couple the cathodes together with a cap.
"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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Buy Chinese. Bury freedom.
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.
What are we looking for?
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Buy Chinese. Bury freedom.
balance or lack thereof. cap should be ~8uf for Fc decade below 20hz, right?
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.
my random number generator chose 5. so 5uf.
'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.
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