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In Reply to: RE: Current Change posted by Triode_Kingdom on June 14, 2021 at 15:32:51
"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.
Follow Ups:
"(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.
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