In this example, all resistors are of equal value, but they are so only because each element sees an equal voltage differential. If we had wanted the triode to see 200 volts of potential, then both the plate and cathode resistor would have been set to 100k and 200k for the second cathode resistor. The value of the bypass capacitor is found by C = 1/(2¶RF) or by dividing 159155 by R and by C. In this example, with a cutoff frequency of 10 Hz, at the very least, 0.1 µF and 1 µF would be better.

    Actually, the output stage is  loaded by both the reactance of capacitive load and the resistance of plate resistors. This is not ideal. The simple plate resistor loading of the output stage suffers because of the passive nature of the plate resistors: the resistor must be high in value to work and that value must remain fixed. Conversely, an active load would, by virtue of its varying conductance, offer more aggressive voltage swings into the capacitance presented by the electrostatic headphones.
  To help better understand the liability imposed by the plate resistors imagine the situation wherein one triode has been completely driven into cutoff and the other triode is conducting to its maximum. At low frequencies, one the stator will see almost the full +300 volts, while the other will see -250 volts. At high frequencies, one the stator will see only a portion of +300 volts, say +200 volts, as charging of headphone's capacitance will require much more current than was required at low frequencies, which will define a greater voltage drop across the plate resistor, 100 volts, in this example. But the other stator will still see the same -250 volts.

Alternate input to the electrostatic amplifier

  Since the phase splitter's cathode voltage is only a volt greater than ground, the connection to the Differential amplifier only needs one  input capacitor (from the plate output), which results a savings of one capacitor, which considering the price of oil capacitors (up to $200 and beyond) is worth considering.

Balancing The Output Stage
   The output stage's balance really needs no improvement in terms of the balance between output signals from each plate. Given a balanced input signal, the output will be in balance. But within one output, a better balance is possible between positive and negative voltage swings. In the original circuit, the stators bridge the plates of the output stage and each stator is terminated by an equal impedance at idle. But as the output voltage swings from one extreme to the other, particularly as the frequency increases, an imbalance develops.

One triode is turned off and the other conducts fully

   A better scenario would be an equal voltage spread regardless of frequency. Replacing the plate resistor with active devices, i.e. more triodes would help to restore balance at the extreme voltage swings from the output stage even at 20 kHz.

pg. 3

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