Will the halving of the power supply noise trick from the circuit work here? Yes, if we carefully choose the value of capacitor C1, then the amount of noise at the top of C1 will match that at the top of 33µF capacitor and the balancing of the power supply noise will be assured. These two capacitors cannot share the same value, as they do not see the same shunt impedance. The first stage represents a much lower impedance than the split-load phase splitter. If the value of resistor Rk is correctly chosen, then the total effective impedance of this first stage must equal twice the value of resistor Ra. Whereas, the split-load phase splitter's effective impedance equals rp + (mu + 2)Ra, a truly huge number. In short, capacitor C1's value will have to be somewhat greater than 33µF.

Split-Load and Class-AB
    What if we are not willing to pay the price of high current draw and heat generation that Class-A demands? The alternative is what 99.9% of power amplifiers run in: Class-AB operation. (In fact, many (most?) advertised Class-A amplifiers actually run in Class-AB. If the amplifier's output stage does not at least draw half of the peak output current swing, it is not Class-A.) Class-AB provides a compromise between Class-A and Class-B. It is not as distorting as Class-B and not as inefficient as Class-A operation.

     Eliminating power supply noise is still possible within a Class-AB amplifier; it just requires a little more thought. The circuit below is a case in point. A bipolar power supply is used and the output is DC coupled. The bottom output device must see the same power supply noise at its input as it does at its connection to the -B connection.

    When the connection is made at the cathode side, the previous circuit works perfectly, as both the first stage and the phase splitter share a common B+ connection. In other words,  in this circuit it does not matter that the output devices are providing gain.

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