Now to finally answer your question about how to best drive a MOSFET. It is often said that MOSFETs function more like tubes than as transistors. Do they? Not as much as I would like. One reason usually given for such a view is that the transistor is primarily current driven, while the MOSFET, like the vacuum tube, is primarily voltage driven. The MOSFET's extremely high input impedance would seem to imply that is insensitive to the driving impedance. And it is at DC and low frequencies. But as the frequency increases, likewise the problems. High power MOSFETs carry the price of having a fairly stiff amount of capacitance. (In fact, the capacitance is not constant, but increases with the MOSFET is turned off.) In a source follower amplifier, the gate-to-drain capacitance is the most worrisome, as the source's following hauls the gate-to-source capacitance with it (bootstrapping of sorts). In the grounded source amplifier, the gate-to-source capacitance is fully realized, but as the gate sees only small voltage swings, this capacitance is not to taxing, but the gate-to-drain capacitance becomes Miller Effect amplified, which is taxing. So is a low output impedance all that is needed to overcome the MOSFET's high input capacitance?

No. Or rather, in itself it is not enough. For example, a 12AX7 based cathode follower has an output of about 600 ohms, yet it cannot drive more than a trivial amount of capacitance. Capacitance must be charged and charging requires current. Here is a thought experiment: carry a high voltage oil capacitor by holding one terminal in your hand as you scuff your leather soles across the nylon carpet. Now touch the door knob with the other capacitor terminal. Have you captured the full thousands of volts your scuffing generated in the capacitor? No, only a few millivolts. (If you had retained the full voltage potential, we would not have an energy crises in California today.) Scuffing your feet just does not generate that much current. So just how much current is needed to drive the MOSFET to full power? Slew rate times capacitance equals current:

       I = Slewrate x C.

Slew rate can be found from:

       Slewrate = 2piFVpk / 1,000,000,

where F equals the highest frequency you wish to reproduce and Vpk is the highest peak voltage swing. For example, 30 volts peak at 30 kHz requires a slew rate of 5.65 volts per microsecond. This slew rate against, let's say, 1200 pF equals 7.5 mA. Just to be safe, doubling this amount would not hurt. What is the total input capacitance of the MOSFET output stage this amplifier uses? There actually three poles to the MOSFET grounded-source amplifier and the output stage should be evaluated in a high-end SPICE program. But a guess would be that effectively there might be as much as 4000 pF of capacitance, which must be driven to maybe ±2 volt peaks that would require only 1.5 mA of current. Using a fudge factor of ten to one, 15 mA would be a good idle current of the driving stage, as I am sure that something closer to 5 mA would be the true minimum.