Re-doing HF amplifier output calculations

The critical component is the output transformer, this has to match the push-pull output MOSFETs to the Ro = 50R load.

Load needed

The transistors need a load of RL = (Vcc)^2/Pout. and the transformer ratio to match this has to be n = ns/np = sqrt(Ro/RL). Calculating these things give:

Rl = (13.8)^2/10 = 19R

and

n = sqrt(50/19) = 1.6 So a transformer with a turns ratio of 3:5 will be good.

Saturation

But we also have to make sure the core never runs into saturation, to do this we need its primary reactance to be x4 RL, or 76R. The reactance of the primary is given by 2pifL. Calculating we get:

2pi x 7.1 x 10^6 x L = 76, or L = 76/(2pi x 7.1) = 1.7uH minimum.

With a type 61 core, of the chosen Binocular ferrite BN-61-202, with an Al = 385 we have L = n^2 x Al/1000 uH, so for 3t on the primary we get

L = 9 x 385/1000 = 3.2uH which is sufficient.

Here's a repeat of the final proposed circuit:



As you can see the MOSFETs Vcc is fed by a couple of RFCs, and their is the 3:5 transformer on the output.