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In this case the phase splitter's cathode resistor sees exactly twice the voltage swing that the grid sees. This also means that the phase splitter's plate will also see twice the voltage swing the grid sees. Thus, tube T1's grid-to-cathode voltage will equal input voltage minus its output voltage, which would equal 2Vg-Vg or Vg. And T2's grid-to-cathode voltage would equal Vg, as its cathode is locked at ground potential.
Now, the second scenario: imagine that the speaker or headphone is replaced with a thick wire, which will reduce the output stage's gain to zero. In this case, the phase splitter's cathode and plate output voltages do not differ in amplitude, only in phase. This is the result of effectively grounding the bottom lead of the phase splitter' cathode resistor through capacitor C. Therefore, since each output tube's grid would see the same amplitude input signal and each output tube's cathode is locked to ground potential, the grid-to-cathode voltages of the output tubes must balance.
An infinite spread of potential output stage gain exists between 0 and 1, yet in all cases, the grid-to-cathode voltages of the output tubes balance. At the center of these two extremes (Gain = 1/2), an example is shown in the schematics to the right.
The output impedance of this amplifier is not very low, in spite of what the Cathode Follower like appearance of the output stage. In fact, in the absence of a global feedback loop, it is equal to only rp/2 and this is true only over the Class A portion of the output signal. How can this be? The addition of the capacitor from output to the phase splitter not only balances the grid voltages, it cancels the Cathode Follower functioning of the output stage. In fact, the output stage in this circuit is actually made up of two Grounded Cathode amplifiers working in parallel.
Replacing the speaker with a 10k resistor helps to make this point clear. Starting with T2, a positive input voltage will cause the tube to conduct more current. This increase of current is entirely in series with the 10k resistor, which causes a large voltage to develop across its leads (the same amount as would be the output of a Grounded Cathode amplifier with two of the same tubes in parallel and a 10k plate resistor). But as even the triode with the lowest rp is still several times higher in impedance than a 32 ohm headphone, let alone a 4 ohm speaker, the gain
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