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RE: Owww!

Posted by cpotl on June 14, 2021 at 03:08:39:
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.