The accordion amplifier also yields unity when working into an optimal load. The series connection does double the effective rp of the output tubes, but since the output transformer's impedance is also doubled, the output impedance remains the same. Wait a minute: doesn't the top tube function as a cathode follower, which would greatly reduce the output impedance? The answer is that even if the top triode did function as a cathode follower, which it need not, the bottom triode's rp would still spoil the chance of a lower output impedance. Depending how the drive circuit is arranged, we can configure both output tubes as either cathode followers or grounded-cathode amplifiers. But as drawn, the accordion amplifier uses both triodes as grounded-cathode amplifiers, as the top triode's input is not referenced to ground, but to its cathode. Thus a voltage pulse applied to the secondary would reflect back to the primary, forcing the bottom triode's plate down and the top triode's cathode up. And since the top tube's input signal is referenced to its cathode, its grid will see the same positive pulse. In other words, the voltage pulse across the primary sees only the two output tubes' own internal resistance as a load. Thus, as configured, the rp is doubled. Interestingly enough, the mu is also effectively doubled while the gm remains the same, as the same change in current results that would have resulted with only one output tube. Why does the mu increase? The answer lies in the series arrangement. The amplification factor of the two tubes in series must be added together, as the load impedance sees twice the gain that it would with only one output tube. Mathematically this makes sense, as the constant gm against the doubled rp equals twice the mu. The formula to memorize is; mu = rp · gm Compare this to the parallel configuration, wherein the gm is doubled, but the rp is halved, which equals the same mu, as mu = rp/2 · 2gm
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