"robert casey" wrote

Would an easy fix for this be to insert a resistor in the
path between the cathode of the phase splitter and the grid
of the next tube?

You know better. Is this a game I don't understand?

With equal loads, a concertina has equal output impedance top and
bottom. With unequal loads (due to different input impedance of two
halves of following stage, either because of increasing dominance of
unbalanced miller at hf or when one valve is driven into +ve grid,
or at even higher frequencies to unequal parasitics in the
concertina itself), the impedance of the cathode differs from that
of the anode.

The Williamson escapes nearly all of this by the buffer stage
between concertina and output stage. What remains is not worth
bothering about, particularly if the driver valves are not perfectly
matched.

If a concertina is used for driving the output stage directly, then
some series resistance can be added to the cathode output, but it is
a trade-off. You get slight imbalance for most of the time, but less
gross imbalance at the extremes. Vanderveen doesn't, Jones does, for
example. Best not to drive with a concertina.

The key to all this is to view the concertina as a single unit, with
feedback, and not as separate CF and AF. Current lost at the anode
is also lost to the cathode, and vice-versa.

cheers, Ian