MOSFET has a Gm of 2 A/V or 4 A/V, as using a Rak of 0.5 ohm and 0.25 ohm respectively would ensure symmetrical current swings into the load.

   Had we used a MOSFET with twice the Gm, then we would have had to use a Rak resistor with half the value so that the 1 amp increase in current would have yielded half the drive voltage for the top MOSFET, which would then also decrease its current draw by 1 amp.
   How do we ensure equal current swings for top and bottom triodes? We could begin by finding the peak current flow, halving that value to specify the idle current, which then let's us determine the correct value for Rak. In other words, the idle current must be set to one half the value of the peak current draw when working into the specified load resistance with a known rp and a unknown Rak value:
    Iq = Vb / 4(rp + Rload + Rak).
It is a bit circular, is it not? We can ignore the value of Rak on the first try and then add to the second try, as the answer will be close, as long as the mu is high. Or for those who prefer graphically solving tube circuit operations, the load resistance is plotted at half the B+ voltage in same way it would be in the design of a Grounded Cathode amplifier and the intersection with the zero volts grid line marks the peak output current and half this value equals the idle current Iq.

   For example, given an idle current of 2 amps and a load impedance of 10 ohms, a positive 1 volt pulse delivered to the bottom MOSFET's gate will result in a 1 amp increase in current flow through both the bottom MOSFET and resistor Rak.

   This increase in current flowing through Rak will result in a 0.5 volt increase in voltage across the resistor, which in turn will provoke a 1 amp decrease in current flow from the top MOSFET, as 0.5 times 2 equals 1. The difference between the 3 amp conduction of bottom MOSFET and the 1 amp conduction of the top MOSFET equals 2 amps of current into the load, which equals 20 volts of negative voltage swing.

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pg. 9