OK, to sum things up:
An often suggested method to introduce de-emphasis, is putting capacitors across the B-C junctions of the PNP transistors in the 'Schoeps' circuit. This is the same as connecting the capacitors across the source and the drain resistors of the FET.
To simplify this even further, one could use a single capacitor between the drain and the source of the FET:
As Ricardo pointed out, this increases distortion. Measurements I did confirmed this.
As an alternative, it is suggested to place a capacitor across the drain of the FET only.
For low frequencies, the FET would then act as a phase splitter, but for high frequencies the FET would act as a source follower:
This works. Disadvantage is that the attenuation of high frequencies only works on one 'leg' and that the maximum attenuation can be no more than 6 dB.
In the circuit I suggest here below, the attenuation will be (more or less) equal on both 'legs'.
Also the attenuation can theoretically be infinite.
The source and drain resistor are usually 2.2 K.ohms.
The input impedance of the PNP transistors is roughly Hfe * Re = ~400 * 6800 = ca. 2.7 M.
This impedance is in parallel with the 150 K. resistor, say 140 K.
So a series resistor of 10 K (R7 and R8) between drain (or source) and the transistor would lower the output less than 1 dB.
The two 10 K resistors would even be at high frequencies a modest load for the FET.
Would this be a useable configuration, or do I overlook something?