One Leg Ground Connection
   In the case where ground is referenced to one output cathode, the signal to each tube must have a balanced grid-to-cathode amplitude. This means that relative to the ground, the signal to the tube T2 with the grounded cathode will be much smaller than the signal to the tube T1 whose cathode moves with the output signal. Making this adjustment is, once again, easy with an interstage coupling transformer and only a little more difficult with resistors and capacitors. The circuit to the right shows a simple amplifier suitable for small power output, say a headphone amplifier, as the single gain stage's gain is meager considering the loss of gain from the output stage.

expanding voltage across the plate resistor.
    This is how this phase splitter splits the phase of the input signal into a balanced output. In the absence of the introduction of the resistor in series with the cathode resistor and the capacitor that attaches to the output, this circuit would yield the best balance of all the phase splitters. But when the resistor and capacitor are added to this circuit, the output swing will be very much skewed in favor of the plate. Of course, the real balance we seek is at the output tube's grid-to-cathode voltages and this modification ensures it.
  The phase splitter's cathode does not know that these components have been added and it just follows the grid. As the output tube T2 receives all its input signal from the phase splitter's cathode, this signal defines its grid-to-cathode voltage. The phase splitter's plate resistor, on the other hand, is sensitive to the addition of these parts.
  When the first stage's output swings negatively, the voltage the cathode, which causes the output tube T2 to conduct less, while the other output tube T1 conducts more from its connection to the plate of the phase splitter, which is swinging positively. The amplifier's output is then pulled positively by the T1's cathode. This swing is relayed through the capacitor C, which will force even less voltage across the phase splitter's cathode resistor. Less voltage equals less current; and less current through this resistor will further decrease the voltage across the plate resistor, which means the voltage swing from the plate will exceed that from the cathode to the degree that the output of the amplifier falls behind the cathode of the phase splitter. Balance is thus restored, as each output tube see the same grid-to-cathode voltage. 
  Let us imagine two scenarios. The first is that the output stage yields a perfect unity gain, i.e. the sum of what is present at the output tube's grids is present at the output:
      Vout = Vg1-Vg2.

   Starting at the grid of the first triode, the signal enters here and is inverted and amplified at the plate. This signal then cascades into a Split Load phase splitter whose balanced outputs drive the output tubes. As in the previous design example, a feedback loop can be tied from output to the first tube's cathode.   
   Now in greater detail, the first stage constitutes a Grounded Cathode amplifier. The signal introduced at the grid is amplified at the plate, but phase inverted. Signal introduced at the triode's cathode is also amplified at the plate, but without phase inversion. If the same signal is simultaneously present on both the grid and cathode, the output will diminish to nearly zero.
    When the first stage's output swings negatively, the phase splitter's cathode will follow, swinging negatively. As the voltage across the cathode resistor decreases, so too the current through this resistor will decrease and as the phase splitter's plate resistor is in the only current path from cathode to B+, it will experience the same amount of current and, consequently, the same amount of voltage across its leads as the cathode resistor. But as the top lead of the plate resistor is effectively grounded at the power supply connection, when the cathode swings closer to ground (thus decreasing the voltage across the cathode resistor), the plate will be pulled up by the collapsing voltage across the plate resistor.
    When the first stage's output swings positively, the phase splitter's cathode will still follow swinging positively. As the voltage across the cathode resistor increases, so too the current through this resistor will increase and the phase splitter's plate resistor will experience the same amount of voltage across its leads as the cathode resistor. So when the cathode swings further from ground, thus increasing the voltage across the cathode resistor, the plate will be pushed down by the

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