Explain this preamp circuit

[pdf64]

Does it seem strange that the old 82k and 100k anode load resistors were still used? As the current source should result in them causing an imbalance.

[vintagekiki]

If is Rt=0, does Cg2 get enough excitation to tube operate as phase inverter?

[vintagekiki]

Does it seem strange that the old 82k and 100k anode load resistors were still used? As the current source should result in them causing an imbalance.

The imbalance is relative. You forgot that HiFi has a completely different philosophy.

[Dave H]

If is Rt=0, does Cg2 get enough excitation to tube operate as phase inverter?

Cg2 grounds the right triode grid, excitation of the right triode is from the left triode via the cathodes. The tail resistor is Rb + Rt. If Rt is shorted the balance will be be very poor because the low value of Rb is a poor approximation to a CCS (Rb is a 'short' tail) but if Rb is replaced by a CCS balance will be excellent because a CCS is effectively an infinitely long tail.

[vintagekiki]

It is in the tail It's ensuring a constant tail current through R6. I think R6 is redundant. It could be replaced by a short.

Cg2 grounds the right triode grid, excitation of the right triode is from the left triode via the cathodes. The tail resistor is Rb + Rt. If Rt is shorted the balance will be be very poor because the low value of Rb is a poor approximation to a CCS (Rb is a 'short' tail) but if Rb is replaced by a CCS balance will be excellent because a CCS is effectively an infinitely long tail.

Rb is the cathode resistor.
Rt is the tail resistor.
I probably didn't understand something best. Sorry

[Helmholtz]

Does it seem strange that the old 82k and 100k anode load resistors were still used? As the current source should result in them causing an imbalance.

The schematic shows equal 100k plate resistors (?).

[Helmholtz]

Effective tail resistance is the sum of bias resistance and lower tail resistance. Regarding the tail it doesn't matter where it is tapped, as there is no current through the grid leak resistors.

Some thoughts:

The second triode is driven via its cathode. Means that there needs to be some signal voltage at the top of the total tail. (For the same reason the tail must not be bypassed).
Now if the signal currents through both triodes were exactly balanced, the tail signal would be zero, because the signal currents are out-of-phase and cancel.
Consequently some current imbalance is essential to make the circuit work. The signal current of the first triode always needs to be somewhat larger.

The higher the tail resistance (impedance) the less current imbalance is required to produce/drop the same cathode signal voltage.
So high tail impedance improves balance.(and lowers common mode gain).

[pdf64]

The schematic shows equal 100k plate resistors (?).

Haha, my mind was playing tricks on me, I would have sworn that R12 was 82k

[vintagekiki]

100k / 82k uses Marshall.

[Helmholtz]

(Referring to my post #22)

I think it works like this:

For ideal balance (with identical plate resistors) both triodes require equal (but opposite phase) grid-to-cathode signals. This happens when the cathode signal is 50% of the PI input signal.
More than 50% would increase the signal current of the second triode and thus lower the cathode signal until it's back at 50%.
Lower than 50% would have the opposite effect.
So the circuit should be self-adjusting.

Practical example:

Input signal 1V, cathode signal 0.5V.
Necessary differences between triode signal currents with total tail impedance of:
1k. 0.5mA
10k: 50ľA
1M: 0.5ľA

The current difference will reflect as gain difference at the plates.

Makes sense?