For example, the intersection of the plate resistor loadline with the 0-volt gridline defines the maximum current draw that both triode and load resistor will experience. And, obviously, the least amount of current these elements can draw is 0. So by making R_l = 2r_p, we now know that the peak voltage swing positively and negatively will equal B+/3; we also know that the maximum current swing in either direction will equal the idle current, that the cathode-top voltage equals 2/3B+, and that the gain will equal 2/3 of the mu). This 1:2 ratio cleanly divides the operating rages of current and voltage swings into thirds, as shown in the graph below.

Magic Thirds and 2rp
      If we specify that the plate resistor, R_a, equals two times the plate resistance and set the idle current to V_b/6r_p, some very sweet results obtain. How?
   We must first understand some of the limitations the triode imposes. The first is that the triode's grid ceases to be high impedance once the grid becomes positive relative to the cathode, as it then becomes a diode's anode that can conduct current. So if we retain the 0-volt gridline as a boundary, then certain relationships develop. (Fixed bias will be used in this example to make the concepts clearer.)

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