Bill wrote: >Push-Push Jfet amplifier with parallel inputs and a Toroid output >transformer, no secondary along with a simple filter using a 10 MHz >series resonate crystal connected to one drain and an adjustable >capacitor connected to the other would work fine. You connect the >other ends of the two together and a loading resistor to >ground. The capacitor is used to neutralize or null out the shunt >capacitance of the crystal so that a capacitive path for the other >frequencies , 5, 15, 20, etc is eliminated. I concur with what Bruce said regarding crystal filters (or any narrow bandpass filter) at the output frequency. More fundamentally, I'm not sure I understand your description of the circuit. You say it is a pair of FETs with parallel input and a transformer (autoformer) output. To me, that suggests the circuit pictured below (one feeds the sources in parallel, the other feeds the gates in parallel -- it doesn't make any difference in how the circuit operates). The usual push-push doubler feeds the FETs differentially, and takes the common-mode output. The diagrammed circuit reverses this -- it feeds the FETs in parallel (common-mode) and takes the 10MHz output differentially. As drawn, the circuit would have essentially no output at the input frequency or any of its harmonics (only that due to the mismatch between the FETs). The only signals it would amplify are uncorrelated signals -- i.e., the FETs' input noise voltages. A quick simulation confirmed no significant output at the input frequency or its harmonics for matched FETs. Simulating mismatched FETs produced a 5MHz signal rich in harmonics, but at a very low level and with no suppression of the 5MHz and its odd harmonics. I assume I misinterpreted your description and that you had a different circuit in mind, or that if you did have this circuit in mind I'm missing something about its operation. Can you please describe again what you had in mind, and how it generates 10MHz? Best regards, Charles