are the lateral mosfets obsolete

F7 has no stabilization of the operating point of the power transistors.
What do you think about it?

One of the advantages of latfets is their inherent thermal stability. No need for additional thermal compensation.

That allows to avoid source resistors and therefore preserve intrinsic transfer characteristic of fets which is (close to) quadratic. This arrangement is sometimes referred as quadratic push-pull class A. With good matching of P and N device, even harmonics are largely suppressed. Nelson calls F7 third harmonic amplifier. I suppose XA25 also falls in this category.

Another advantage of quadratic push-pull is much higher A class range of operation with the same quiescent current. With usual push-pull, max A class power is Rsp*(2*Iq)^2. With quadratic push-pull it is Rsp*(4*Iq)^2. That's 4x more than with usual arrangement. E.g. 25W in 8ohm can be theoretically achieved with Iq=650mA.

Big challenge with quadratic push-pull is Iq stability. And here latfets come to rescue.
 
I've done HexFETs, but mostly BJT outputs. No lateral MOS. It is very friendly of them to have a zero TC at the bias current. HexFETs are stablized just like bipolars. Calculate the temp-co, and design a Vbe multiplier to mimic that.

Ultimately, the low gm of all MOS is a deal breaker for me. I also found that FET bias starts climbing with high frequency signals. This might not be a problem when used as source followers, just a cap across a bias resistors, but I was using them common source.
 
One of the advantages of latfets is their inherent thermal stability. No need for additional thermal compensation.

That allows to avoid source resistors and therefore preserve intrinsic transfer characteristic of fets which is (close to) quadratic. This arrangement is sometimes referred as quadratic push-pull class A. With good matching of P and N device, even harmonics are largely suppressed. Nelson calls F7 third harmonic amplifier. I suppose XA25 also falls in this category.

Another advantage of quadratic push-pull is much higher A class range of operation with the same quiescent current. With usual push-pull, max A class power is Rsp*(2*Iq)^2. With quadratic push-pull it is Rsp*(4*Iq)^2. That's 4x more than with usual arrangement. E.g. 25W in 8ohm can be theoretically achieved with Iq=650mA.

Big challenge with quadratic push-pull is Iq stability. And here latfets come to rescue.

Forward transconductance for 2sk1058 is around 1A / V, for irf240 it is around 6A / V, for 2sc5200 it is around 25A / V.
Bipolar transistors give the greatest gain.
 
Gm of HEX and Planer Strip is very large and more exponential compared to being more squared like of Lats. The low Gm of MOS is primarily at low conduction, below a few hundred mA. This is avoided by using class A. In Class AB the low Gm issue is around bias level and the current crossover. There are ways to mitigate this but not simply. Mos has very high input Z at DC as it is primarily capacitive (and non-linear🙄 ). The charge needed for change of the same conduction is usually different for N vs P types. Using a lower Z driver stage such as "totem" pole drive for each output will reduce the cross-conduction at higher frequencies and at clipping. Mos typically has crazy variable capacitances that are better driven by a lower Z, IMO. On the other hand, although Hexfets (and planer stripe) require thermal compensation, the low GM at low Id does help with thermal bias stability. But of course this is a non-issue for Laterals anyway. That and the square function make them ideal for audio OPS.🙂 But the price and availability.......😱
 
I've built some laterals amps, VFA and CFA with no problem with Hitachi TO-3 2SJ50/2SK135, but Exicon version is different and more prone to high frequency oscillation, look here for some explanation: https://www.diyaudio.com/forums/solid-state/338814-lateral-cfa-120w-bsa-35.html#post6507954

Thank you for your explanations. Could you please give me a circuit diagram of the amplifier with 2sk135 / 2sj50. Both in the VFA and CFA versions. I am very curious about this solution.