I am much too late to said that but obviously, the schematic Bobby Dipole show cannot be used like that with 6CZ5, like smoking-amp said, 355V on the grid #2 is too high, according to Sylvania, the maximum is 285V ... Also, the grid leak resistors should be changed for 100K and the bypass capacitors for 0.22uF but the load of the 12AU7 will then be bigger and the amplifier response different. The fixed bias most also be adjusted for the new circuit.

But there is no mention of the primary plate to plate impedance of the output transformer and this is very important to know. Fortunately, it is very easy to find out what the transformer impedance ratio is by applying a high voltage to the primary (120V is OK) and measuring the voltage at the secondary without any load, assuming the right speaker impedance, usually 4, 8 or 15 ohms.

The impedance ratio is the square of the voltage ratio, this mean if you apply 120V on the primary ends where the plates are plugged and you get 3.92V at the 8 ohms tap on the secondary, the voltage ratio is 120/3.92 = 30.61:1 then the square of this ratio is 937:1 and 8 x 937 = 7496 ohms or just about 7500 ohms, perfect for a push-pull fixed bias amplifier using two 6CZ5 ... giving 21.5 watts with only 1% THD (according to Sylvania).

Notice something important about push-pull transformers, the impedance "view" by each output tube is only one fourth of the impedance plate to plate and not half because each tube use the half of the primary winding. In the case of a 7500 to 8 ohms transformer, the ratio of the voltage from half the primary winding to the secondary will be only 30.61/2 = 15.305:1, the square is 234.243:1 then the impedance ratio for each tube is about 234.243 x 8 = 1874 ohms ... one fourth of 7500 ohms ...

I explain that for novice tube amplifier designers, it is important to know for drawing correctly the AC load line and calculate the output power. Doing it for a single end amplifier is very easy but for a push-pull amplifier, it is much more complex. Another thing many people don't understand about load line is the operating point by which pass the load line, representing the position of the zero signal voltage and current, is the supply voltage less the DC drop in the output transformer primary. If the supply is 355V like in the schematic show by Bobby, and there is only 350V reaching the plate, that mean the drop in the winding is 5V, if the current is 23ma per tube, the DC resistance of half the winding should be about 5/0.023 = 217 ohms ...

Many will wonder how come the voltage between the cathode and plate of the tube can reach almost twice the supply voltage, at the lower end of the load line ? There is no magic, the ohm law said E=RI (voltage = resistance x current), in a tube amplifier, a small voltage variation on the grid produce a big current variation between the plate and the cathode. The peaks of this current vary from much under the operating point current to much lower, the impedance of the speaker is fixed at a medium frequency then is reflected impedance at the primary is also fixed. Like I said before, if the impedance for each tube is 1874 ohms and

**for example** the maximum peak current reach 151ma alternatively for each tube, 1874 x 0.151 = 283V, then the peak to peak current is 302ma and the resulting peak to peak voltage is 566V. Because the supply voltage at the plate is 350V, the voltage between the plate and the cathode vary from 350 - 283 = 67V to 350 + 283 = 633V.

Those peak voltages and currents in each tube depend on the operating class and

**are just an example** to illustrate what I try to explain and really apply only to class B because in class AB1 and AB2, both tube conduct part of the time and in class A all the time ...The formula to calculate the RMS output power is peak to peak voltage x peak to peak current / 8 or 0.302 x 566 / 8 = 21.4 watts RMS, this is the power you can reach with a class AB1 push-pull circuit using two 6CZ5, "less the power loss in the transformer" because the winding resistance and the magnetic loss ... The perfect transformer don't exist !

I hope this will help some new designers because it took me many years and Aspirin to understand how a output transformer is working, specially in a push-pull circuit, most electronic books are not really clear about that.