|
Starting with a grounded output, the circuit reduces to a long-tail phase splitter feeding two grounded plate or drain or collector followers. The drive signal balance is ensured by the equal valued plate resistors. (Actually, the second plate resistor should be larger, as even an infinitely large cathode resistor will not result in an equal gain from the two plates.)
Like the previous circuits, the output impedance is equal to the reciprocal of the transconductance of the output device, when only one output device conducts (Class-B); and half that value when both conduct (Class-A). This is shown by applying a +1 volt pulse to the output. The top output device sees this pulse as a 1 volt negative pulse at its input and the bottom output device sees the pulse as a 1 volt positive pulse at its input, as the zener-capacitor combination relays the pulse to the top of the first plate resistor, which in turn is relayed to the bottom of the resistor and from there travels to input of the bottom output device. When the output works into an 8 ohm loan, the zener-capacitor combination allows for dynamic equalization of the drive for the bottom output device.
In comparison with the split-load based circuits, this circuit's phase splitter offers gain. In other words, this last circuit can stand on its own as a complete amplifier. While some further design choices are needed, such as which triodes to use, high mu or high current, this circuit will perform well as drawn. Improving the circuit is as easy as replacing the cathode resistor with a current source, which not only improves the balance, it improves the PSRR of the amplifier. If the second triode's grid is used as a feedback port, then the current source is even more beneficial. Consider the extreme case wherein all of the output signal is fed back. In this case, the entire amplifier becomes a unity gain buffer. What comes in should come out. Thus, the first triode's grid and the second triode's grid will see the same signal in terms of phase and magnitude. Were only a cathode resistor used, the input signal would be imposed on it. For example, if the input signal consists of a +10 volt pulse, a 10k cathode resistor would see an increase of 1 mA in current conduction. This increase will have to be shared by both plate resistors, which will result in both output voltages being forced more negative than they should be. Conversely, if the input signal consists of a -10 volt pulse, the cathode resistor would see an decrease of 1 mA in current conduction.
|
