
resistor, Rk: Rak = Rk = ( Vb / 2Iq)  rp mu +1 The value of Rl seems to be missing from this equation, but it is implied in the calculation of the idle current Iq. But we could not find Iq until we found Rak. Reducing both formulae to one formula gives us our answer: Rak = Rk = rp + 2Rl mu 1 While this formula gains much by virtue of its simplicity, it hides from the tube circuit designer some key information, such as the value of the idle current, and by extension, the tube dissipation and the peak current into the load impedance. Still, it is intriguing to see Vb and Iq both drop out of the mix, which shows just how decisive the triode's mu and rp are in this circuit's functioning. This formula promises to give us the value of Rak that yields the biggest, cleanest voltage swing into the load resistance. Let's give it a test. If we set the load impedance to zero ohms, the formula reduces further to Rak = rp / (mu  1), which would yield 103 ohms as the correct value for a 6DJ8, as 103 = 3300 / (33  1). To verify this result, let's plug this value into the following formula: Gain = muR / (rp + R), Which is the formula for the gain of a Grounded Cathode amplifier, Gain = 33 x 103 / (3300 + 103), 0.9988 = 3399 / 3403. The result easily rounds to 1, which means that a 1 volt pulse at the bottom tube's grid results in an increase in current draw sufficient to develop a 1 volt pulse at the top tube's grid. Substituting 200 ohms for the value of Rak shows an asymmetrical drive voltage: Gain = 33 x 200 / (3300 + 200), 1.88 = 6600 / 3500, as does substituting 50 ohms for Rak: Gain = 33 x 50 / (3300 + 50),

