This circuit fails in one truly important aspect: in spite of the symmetry of driver parts, it does not offer an equal drive signal to both output tubes. As configured, the only case wherein the top and bottom tubes see identical drive signals is when the load being driven is 0 ohms. At any higher impedance, the top triode sees less drive voltage than the bottom triode. Probably the easier path to seeing this asymmetry is to imagine a pulse being reflected back into the output transformer's primary. The bottom triode's grid is oblivious to the negative going pulse at its plate, but the top triode's grid sees a small negative going pulse because the driver tube's rp defines a voltage divider with its plate resistor. So the positive going pulse at the top triode's cathode is not relayed in its entirety to the top triode's grid, but rather diminished, making it appear as a negative pulse to the grid. The only way to ensure a symmetrical drive signal is to use triodes with infinite rp's (FETs, MOSFETs, transistors, and pentodes come close). Or another alternative is to effectively increase the rp of the triode used.

             Cascode-like DC driver stage



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