The important thing is that it should be possible to establish a certain pre-emphasis curve going in to the distortion, to control the voicing of the distortion, and then apply the precise reciprocal of that curve on the output, so that the pre-emphasis applied to the distortion is cancelled afterwards, and therefore doesn't necessarily affect the final tonal balance of the output. One clear example is if the pre-emphasis is very bass-heavy. The distortion will take on a particular "sludgy" character, especially when low notes or chords are struck. But then, the post EQ can be set to reciprocally cut the same bass frequencies, so that the final tone is bright and trebly; yet, it will still have that character of breaking up heavily on low notes and chords. Of course, it's not necessary to set the post-EQ to the reciprocal curve of the pre-EQ: this is simply one tonal technique which is facilitated by having two identical EQs.
For various reasons, both technical and "emotional", I have decided to make this amp (and all of my amp projects) based 100% on tubes in the audio path. Thus, full graphic EQs or generalized parametrics become a bit expensive in terms of tube count. I have decided that the best EQ topology in terms of flexibility without excessive complexity, is three bands: fixed bass and treble, with sweepable-frequency midrange. Each band has symmetric boost and cut, around a central flat position with the knob at 12:00.
I originally sought to design this EQ using only one twin-triode tube (12AX7), but the performance compromises required were, I think, too great. Hence, I am now figuring on two tubes per EQ. Given that, I believe I can rather readily take an existing solid-state design which is known to work, i.e., the channel EQ from the Carvin MX22 series mixers, and adapt it to tubes. In this circuit, note that every opamp section has a grounded (+) input: therefore, they are simply inverting amplifiers, which are readily implemented with tube stages. I believe I can at least get into the "ballpark" by simply multiplying all resistor values by 10, and dividing the capacitor values by 10.

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