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Posts: 1002
Location: Texas
Joined: December 6, 2009
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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.
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