18W bile field output class A power amplifier

This is a Class A power amplifier with an output power of 18W. The final stage adopts a current series negative feedback circuit (the bypass capacitors at both ends of the self-supplied bias resistor of the output stage are cancelled), and the electrical performance is excellent. The circuit is shown in the figure.
　　The input stage is a dual triode 6N3, and the final stage is a 6P1 and IRF450 common output. Dual channels only need three electron tubes. Both 6N3 and 6P1 are peanut tubes, and the market price of the field tube IRF450 is also very low. The
　　amplification factor of 6N3 is μ=35, and the mutual conductance gm=5.9mA/V. With common cathode amplification, 6P1 outputs about 4W, IRF450 outputs 14W, and the total output is 18W. IRF450 can be replaced by other field effect power tubes, but its reverse withstand voltage must be greater than 500V and the power consumption must be greater than 100W. In this example, when the working current of IRF450 is 113mA, the gate-source voltage drop is 3.5V, so R5 is set to 68Ω (note that the data of different field tubes will be different and should be measured). If the value of R6 is changed, the output power of the machine can be changed. However, if the working current of IRF450 is too large, it is easy to cause self-excitation.
　　The output transformer needs to be made by yourself, with the cross-sectional area of ​​the core S = 10cm2, the number of primary and secondary turns is 14:1, the impedance ratio is 1560Ω:8Ω, the primary enameled wire ∮ = 0.29mm, 2100 turns, the secondary wire ∮ = 1.00mm, 150 turns, and good results can be achieved without segmented winding. Note that the iron sheets should not be crossed, and the iron clamps or iron strips that fix the core cannot be used anymore and should be redone with aluminum plates. A layer of kraft paper is placed on the three cylinders of the core as a gap layer to ensure that the core has no magnetic saturation, so that the transformer distortion can be minimized. The cross section of the inductor L core S = 9cm2, the enameled wire ∮ = 0.33mm, just wind it full, IRF450 needs a larger heat sink to facilitate heat dissipation.
　　This machine has no large loop negative feedback, good transient response, and a simple circuit. It should be noted that the working current of 6P1 should be 50mA. What needs to be adjusted is the resistance value of R6 to determine the final output power.