Posts: 1002
Location: Texas
Joined: December 6, 2009
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"As you said, examining the circuit as a whole is required, and it will not show that the grounded grid side has a delivered amplification factor any different from the other half. The interaction between mu, gm, plate loads and the cathode load show the imbalance generating causes." One needs to be sure that disagreements are not merely about how the words are being used. In terms of Crowhurst's equations, if the amplification factors are called A1 and A2, and defined to be the changes in the plate voltages of tube 1 or tube 2 respectively, divided by the input voltage on the grid of tube 1, then they can be read off from his equations (92) and (94) respectively (since his Ep1 and Ep2 are defined to be the changes in plate voltages on tube 1 and tube 2 in response to putting a 1 volt input on the grid of tube 1). That is to say, A1 = Ep1 (in eqn (92)), and A2 = Ep2 (in eqn (94)). (Actually, there is a further typo, in his eqn (92); there should be a plus sign after Rp1 in the denominator of his expression.) Using the corrected eqn (96) that I had previously given, these equations then say it all. That is, all the signal voltages, namely Ep1 at plate 1, Ep2 at plate 2, and Eck at the common cathode point, are all expressed in terms of the input voltage on grid 1, the resistors RL1, RL2, Rk, and the characteristics mu1 and Rp1 for tube 1 and mu2 and Rp2 for tube 2. Crowhurst is, of course, precisely worrying about the fact that tube 1 is driving into the impedance corresponding to the paralleling of the cathode resistor Rk and the input impedance of the grounded grid amplifier that it is driving. That is the basis of his calculation. So he has taken everything properly into account. By the way, if we assume RL1 = RL2 = RL, and assume identical tubes so that Rp1 = Rp2 = Rp and mu1 = mu2 = mu, the resulting formula for A1/A2 agrees exactly with the one Ralph gave a few days ago, namely A1/A2= 1 + (RL + Rp)/((1+mu) Rk). If the tubes are taken to be different in their characteristics (unequal mu and Rp), but the two anode loads are still set equal, RL1 = RL2 = RL, then the ratio of the amplification factors is given by A1/A2 = 1 + (RL + Rp2)/((1+mu2) Rk). So the mu and plate resistance of tube 1 do not enter in the expression for the ratio of the amplification factors.
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