tweaking would be required and ensuring that the two rail noises maintain an anti-phase relation-ship to each other would prove difficult with mismatched power supply components.
   A better solution would be to use an Rk resistor with half the present value but with a negative power supply rail voltage of only half the present value as well. Playing with the transformer's taps is not the answer nor is using two transformers to build the two power supply rails.

we can have our cake and eat too. The effective impedance of a tube based current source is :
      impedance = rp + (µ + 1) Rk
If a tube with a mu of 20 and an rp of 4000 is used, the result would be 214K, a huge amount of impedance considering the 10 mA of current flowing through the current source. If the current source were replaced with a 214K resistor, then 10 mA of current would define 2140 volts across the resistor's leads.

   All that is needed is to drop half of the negative power supply voltage across a fixed voltage source, such as a zener diode. The better choice is to use two Rk resistors in series with each other and bypass one of them with a large valued capacitor. This way the noise from the negative supply rail will be transmitted by the capacitor to the other Rk resistor.

  If you think that this trick only works because tube based current sources should always be AC referenced to ground, you have not gotten it. This trick works because we have created not a true constant current source, but a very high impedance noise induced varying current source.

   The disadvantage of this arrangement lies in the need to provide a separate heater power supply for the Cathode Followers if they are DC coupled to the plates of the differential amplifier; or if the Cathode Follower's cathodes are to be kept close to ground potential, a pair of coupling capacitors to connect the plates to the Cathode Follower's grids. Still this trick will work perfectly in other differentially based circuits. Furthermore, it allows for one last trick: using a current source in place of the Rk resistor. Whoa. Didn't we conclude that using a current source would result in a PSRR of zero, that all the positive power supply rail noise would make to the output? Yes, if we build the current source the conventional way. But by referencing the current source's grid to ground rather than the negative rail,

The Phono Stage
  We will cover the last part of Yoon's request next month, but here is a sampling of the schematic we will dissect.

                                    // JRB

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