Push Pull bypass cap - 1 Rk

[lowell]

I just had a thought last night. If you have a push pull circuit that is sharing Rk there may be no point in using a Ck. Being that tubes on both sides of the OT are sharing Rk, there will be no negative internal feedback right because the signal on the cathodes are phase cancelling? Correct me if I'm wrong here.

[loudthud]

That idea works in theory when the drive to each side is balanced and the power tubes are matched and perfectly linear until... you overdrive one side, the other side is cut off and Rk degenerates the overdriven side. 25 or 50uF really doesn't bypass the typical 250 ohm resistor at low frequencies. Try 1000uF to see what real bypass is like. Almost like fixed bias.

[lowell]

That idea works in theory when the drive to each side is balanced and the power tubes are matched and perfectly linear

Ok, got it

until... you overdrive one side, the other side is cut off and Rk degenerates the overdriven side. 25 or 50uF really doesn't bypass the typical 250 ohm resistor at low frequencies.

What do you mean by "degenerates?" Also, using Fc=1/2*RC - You get about 13hz as the cutoff point. Seems quite low to me, but still don't see it making an audible difference in my initial post's case. If the power tubes are overdriven I'm not sure how that affects the cathode. Can you explain further?

Try 1000uF to see what real bypass is like. Almost like fixed bias.

Fc=1/2*RC=.64hz, wow that's low. Is it like fixed bias because the cathode's so clipped that the cap charges up?

[pdf64]

While both tubes are conducting for the whole of the signal waveform, then assuming a well balanced pair of signals to their grids, the signals at the cathodes will cancel out. Then there will be no local negative feedback (ie degeneration of the plate signal by the cathode signal).
When the signal level increases such that the tubes enter cut off for part of the waveform, then current from the cut off tube won't appear at the shared cathode, so the signal from the other tube will appear at the cathode and local negative feedback (for that part of the waveform) will occur. This will reduce/distort the plate signal and likely reduce maximum power output.
With regard to the bypass cap value, it's capacitive reactance has to be low compared to the tube's cathode resistance, as well as the cathode resistor - a tube's cathode resistance is very low, which is why cathode followers act as impedance buffers.
How do you work out the value of a tube's cathode impedance? It's quite difficult, but a general rule of thumb is that it tends to be about the same as that tube's normal cathode resistor. As the equivilant circuit puts the cathode resistance in parallel with the cathode resistor, use Rk/2 in the equation.
However, it gets more complicated when cathode resistors are shared, in which case it might be more practical to temporarily split the cathodes (to stop cancellation muddying the water), put a smallish bypass caps in place and measure what the -3dB freq is, in order to derive the total equivilant R value for the individual tube. Then use that R/2 value to calculate a suitable cap value for a shared cathode resistor.

[lowell]

pdf64,
hmmm that's interesting. Cathode internal resistance is in parallel w/ Rk and is about equal to it as well. Good to know!

[pdf64]

See
Designing Common-Cathode Amplifiers
That's based on small signal triodes, I don't know how much of it is applicable for power pentodes, but assume that the general principles apply.