This was the question I had asked myself, when I was disappointed by the results of an experiment wherein I had built a White Cathode Follower with the aforesaid tube and resistor values for driving a much more reasonable load: the Sennheiser headphones, which have an impedance of 300 ohms. After only a few millivolts, clipping occurred. My expectation was that the circuit should be able to deliver the idle current of at least 10 mA into this load, if not almost 20 mA, which would conform to classic Class A, push-pull amplifier standards. I then replaced the plate resistor with a 10k potentiometer with its center tab connected to one of its outside tabs, which allowed for easy adjustment of the plate resistor value. 
     After adjusting the potentiometer, I found the optimal value according to the trace on the oscilloscope to be 100 ohms. The lowness of the value surprised me. I then wondered what the optimal value would be for the 32 ohm load represented by the Grado headphones. Even more surprising was that the same 100 ohm plate resistor value yielded the best performance into the 32 ohms, in spite of this load being 10 times lower in value than the previous load. Moving to the other extreme, I replaced the 32 ohm resistor with a 3k resistor and retested. The 100 ohm plate resistor value once again made for the biggest and most symmetrical voltage swings. After some mathematical introspection, everything made perfect sense to me.
    For any push-pull tube amplifier to work well, there most be an almost identical signal presented to each tube. (The signals must differ in phase.) In this circuit, if the top triode sees an increase in its grid-to-cathode voltage, then the bottom triode must see an equal decrease in its grid-to-cathode voltage. How do we ensure equal drive voltages for top and bottom triodes?
    Let us start our analysis with the severest load possible, not Grado headphone, but 0 ohms, in other words, a dead short to ground via a large valued capacitor.

White Cathode Follower  with a shorted output

    The top triode now functions as a Grounded Cathode amplifier and does see the bottom triode at all.  The amount of current flowing from ground into the capacitor then into the cathode of the top triode is given by the formula:
   Ip = VgGm´,
where
   Gm´ = (mu + 1) / (Ra + rp).
Now as the bottom triode current flow is governed by the top triode's current flow into the the plate resistor, the amount of current flowing from the bottom triode's plate into the capacitor is given by the formula:
   Ip = VgGm
where Gm is the transconductance and
   Gm = mu / rp.
By rearranging the formulas for current we get Vg = Ip / Gm´ for the top triode and  Vg = Ip / Gm for the bottom triode. Obviously, the only way that the two grid voltages can match is if Gm´ = Gm. Expanding this formula out yeilds:
  (mu + 1) / (Ra + rp) = mu / rp,
which when we solve for Ra becomes
  (Ra + rp) / rp = (mu + 1) / mu
  Ra = rp(mu + 1) / mu - rp
  Ra = (rpmu) / mu + rp / mu -rp
  Ra = rp / mu
and as  rp / mu = 1 / Gm
  Ra = 1 / Gm.     

pg. 6

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