Thus, the only way the Vg of the top triode can equal Vg of the bottom triode is if the plate resistor equals the inverse of the transconductance of the triodes being used. (The test to put any tube circuit equation through is to try the equation with a 6AS7 and then with a 12AX7 to check the equation for  absurdities.)
  What happens if we chose to start with infinite ohms as a load instead of 0 ohms. The answer is the same, the optimal value for the plate resistor is the reciprocal of the Gm of the triodes used, or what is the same quantity, rp/mu.

   The math can become quite thick here, but if we think abstractly enough, it will not be too difficult to follow. We know that if the top triode sees a +1 volt pulse at its grid, its cathode will follow to some degree less than +1 volt. Whatever this outcome may be, we will refer to it as "Vg." Now Vg/rp equals the increase current (Ip) flow through the entire circuit, as all components are in current series with each other. Ip times the plate resistor (Ra) equals the voltage pulse that the bottom triode sees, which times the Gm of the bottom triode will equal Ip, if the right value of Ra has been chosen. Thus,
   VgRa/rpmu/rp = Vg/rp,
which when we solve for Ra equals:
   muVgRa/rp²  = Vg/rp
   muRa/rp = 1
   muRa = rp
   Ra = rp/mu.
Okay, what if we choose a load impedance somewhere between zero and infinity, say, 10k. Same result, Ra = rp/Gm. In this case, the load impedance is in parallel with the rp of the bottom triode. So Vg/(rp||R_L) equals the increase current (Ip) flow through the top triode and IpRa equals the pulse voltage to the bottom triode.  In this case, like the one with a shorted output, we have true Class A output current swing capability, so as the bottom tube approaches cutoff, the top tube's current conduction will near twice its idle value. And, of course, vice versa for negative input voltage swings. Thus,
   VgRa/(rp||R_L)mu/rp = Vg/(rp||R_L),
which when we solve for Ra equals:
   VgRa/(rp||R_L)mu/rp  = Vg/(rp||R_L)
   VgmuRa/rp = Vg
   muRa/rp = 1
   muRa = rp
   Ra = rp/mu.

Optimization and Zo
   We can use the stock, long, complex equation for output impedance for the White Cathode Follower or we can realize that we have

White Cathode Follower  with an infinite impedance output load

    Now let us step back and look at what is happening with this circuit in broad terms. Without an external load the rp of the bottom triode will define the sole load impedance for the top triode, remember we had defined an infinitely high impedance load. Since the gain of this circuit is less than unity, the cathode voltage will slightly lag the grid's and this gap is the change in the grid-to-cathode voltage that will prompt a change in current flow through both the top triode and  plate resistor, which in turn will give rise to a change in voltage across that resistor, which will then be relayed to the bottom triode's grid. We need to ensure that that bottom tube receives an identical grid-to-cathode voltage signal as the top tube.

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