The maximum positive peak output signal voltage is roughly given by:
   Vp max = V(rp + Rl),
where V equals the cathode-to-plate voltage of the top tube at idle. This formula assumes that the bottom tube has ceased to conduct so that its current draw is not subtracting from that of the top triode and that the top triode only sees only zero volts on its grid-to-cathode voltage. In the last example that would not be true, as the bottom triode requires a huge negative voltage pulse at its grid to be turned off and when its conduction decreases only slightly, a huge positive pulse results at the top triode's grid, forcing its grid positive relative to its cathode. Thus, we see that the selection of the value of Rak is critical to symmetrical peak voltage swings into the load impedance.

Where to Start?
  Let us begin with the stipulation that we never want the grids of either the top triode or the bottom triode to be driven positive relative to the cathodes and that the circuit never leaves Class A operation and that Rk is bypassed. A further assumption is that the triodes used have a fairly high mu and that consequently the voltage drop across resistor Rk is not too great. If a low mu power tube such as the 6AS7 is used, then the effective B+ voltage must be used in the formulae (in actual use, the best procedure would be to use fixed bias with a low mu triode).