resistor R2, rather than the sum of resistors R1 and R2. How do we find the values for resistors R1 and R2? The first step is to arbitrarily define the power supply voltage (Vb) and the idle current (Iq). The next step is to find the value of the cathode resistor (Rk), which will equal the sum of both resistors R1 and R2.
    Rk =  R1 + R2
Given RL, Vb and Iq,
    Rk = Vb/2Iq - rp 
                  mu +1.
Next we must find the value of either R1 or R2 and by subtracting one of these resistor values from Rk find the value of the other resistor. And since R1 is added to limit the maximum positive current equal twice the idle current (2Iq) and to make the inverse of the top triode's effective Gm equal the value of R2, resistor R1 must be first determined:
    R1 = Vb/2Iq - RL - rp 
                    mu +1,
thus,
    R2 =  Rk - R1
The peak current into the load can be no greater than twice the idle current and still remain in strict Class A operation. If the ratio between resistors R1 and R2 is set correctly, the maximum peak current the top triode will pull is equal to twice the idle current:
     Ipk = Vb / 2(rp + (mu +1 )Rl + RL) = 2Iq.
     In case some readers imagine it would be cool to design a Class AB SRPP circuit, consider this :when the bottom tube completely stops conducting current, it can no longer control the current flowing through the load impedance. Here is an analogy that might make this point clear: an SRPP circuit is like a driver-training car with two steering wheels and two drivers. Each is able to steer the car simultaneously. The problem is one driver is blind, but as long as the sighted driver tells the blind driver when to turn left or right, all goes well. Obviously, if the sighted driver decides to let go of wheel and stop talking to the blind driver things do not go so well. To control, the bottom tube must conduct.

    Now to give these formulae a test run, let's use a 6DJ8 with B+ of 200 volts and an idle current of 10 mA. Let's also continue to use a load impedance of zero ohms to simplify the math. First we find Rk:
     Rk = Vb/2Iq - rp 
                     mu + 1,
substituting variables
     Rk = 200/2x0.01 - 3300 
                          33 + 1
     Rk = 197.
Next we find R1,
    R1 = Vb/2Iq - RL - rp 
                     mu +1,
     R1 = 200 /(2x0.01) - 0 - rp 
                         33 + 1
     R1 = 50
thus,
     R2 =  Rk - R1
     R3 =  197 - 50
     R2 =  147
Next we find the peak current Ipk,
     Ipk = Vb / 2(rp + (mu +1 )Rl + RL)
     Ipk = 200 / 2(3300 + (33 + 1)50 + 0)
     Ipk = 0.02 A,
and as
      2Iq = 0.02 A
our result is on the mark. In this example, as the load impedance was set to zero, the effective transconductance of the top triode is given by:
      Gm´ = mu / [rp + (mu + 1)R1],
substituting variables
      Gm´ = 33 / [3300 + (33 + 1)50]
      Gm´ = 6.6 mA/v.
and
      151.5 = 1 / 0.0066,
which is close to the 147 value of resistor R2. In other words, we have met both goals: a peak positive current swing of 2Iq (20 mA) and the top tube's effective Gm equal to the inverse of resistor R2. Consequently, this circuit will linearly work into a dead short, a severe test. 
    But will this new circuit variation work with an actual load impedance? Let's test it with a 300 ohm load and an idle current of 7 mA.