by AB2EZ » Wed Jul 27, 2005 5:06 pm
Bill
Perhaps what you are thinking of, when you think about the Q of the output circuit, is something entirely different from what I am thinking of. Therefore, perhaps that is why you come up with a different result.
I am thinking only about how good a job the tuned output circuit does in blocking the 2nd harmonic of the output voltage waveform (i.e., the voltage waveform that appears across the secondary of the output transformer) from reaching the 50 ohm load that I am assuming (for this calculation) is connected at the antenna port.
The output circuit (as I define it for my purpose) consists of: the inductor, the tuning capacitor and the load represented by the antenna ... all in series. For this purpose, I am ignoring the loading capacitor. The load represented by the antenna (for the purpose of this calculation) is 50 ohms. The inductor and the capacitor in Steve's design each have a reactance (one positive and one negative) of 150 ohms at resonance. Thus, as I am defining it here, for this purpose... the Q of the output circuit is 3.
Here is some text from Steve's Web site description of the operation of the 6-FET Class E 75 meter transmitter
"The approximate reactance of the input inductor and the series capacitor are 150 ohms, or about 11 or 12 times the 12.8 ohm load. Now, getting a value for the inductor: Xl = 6.28 x F x L therefore L = Xl/(6.28 x F). Putting in the values we obtained, and assuming a frequency of 3.9mHz (75 meters): L = 150 / (6.28 * 3,900,000) which works out to 6.125uH.
The series capacitor value may be figured in the same way: Xc = 1/(6.28 x F x C), or C = 1/(6.28 * F * Xc). Using our 150 ohm reactance: C = 1/(6.28 x 3,900,000 x 150) = 272pF. "
While Steve talks about a 12.8 ohm load (in the above)... he is referring to the effective output impedance of the 6-FET amplifier ... as it appears at the output of the 1:2 (step up) output transformer. When I speak of "the Q" of the output circuit... what I have in mind is the following question: how much less current will flow through the tuned output circuit (and into the 50 ohm load that is presented by the antenna) at the second harmonic frequency v. the fundamental frequency... for a given amount of voltage across the input to that tuned output circuit. At the fundamental frequency (where the circuit is tuned to resonance) the net load is ~50 ohms. At the second harmonic frequency the load is ~ j150 ohms x (2 - 0.5) + 50 ohms = [jQ (1.5) + 1][50 ohms]... because the impedance of the inductor is 3x the impedance of the antenna load (Q=3) at the fundamental frequency... and (of course) as we move up to the second harmonic frequency, the impedance of the inductor is twice as much as it is at the fundamental frequency. Likewise, the impedance of the capacitor drops from 150 ohms to 75 ohms as we move to twice the freqeuncy. All of this, is based on the assumption that the rig is looking into a constant 50 ohm antenna system. Naturally, when one takes into account the characteristics of the antenna/antenna tuner combination... the whole situation gets more complicated to analyze.
Based on the above (and again, using the simple case of a 50 ohm antenna load that is independent of frequency)...
The tuned circuit attenuates the 2nd harmonic (relative to the fundamental) of the voltage on the secondary of the output transformer by
10 log [(1+ j1.5Q)(1-j1.5Q] dB = 10 log [ 1 + 2.25Q**2]
where Q**2 means QxQ
By using Q =6 instead of Q =3, I increase the attenuation of the 2nd harmonic (relative to the fundamental) from 13.3 dB to 19.3 dB... i.e., a 6 dB increase in second harmonic suppression.
When I look at the actual r.f. output spectrum of my new 1-FET transmitter on a scope that can perform math functions (like Fourier transform), driving a 50 ohm dummy load...
it is 28 DB down from the fundamental.
When I look at the "class E" waveform on the drain of the FET... using the same scope... the secoond harmonic as about 5.5 dB down from the fundamental.
When I look at the class E waveform on the other side (i.e., the secondary side) of the output transformer... the second harmonic is about 5 dB down from the fundamental. This means the the transformer is adding just a bit of distortion... probably due to core saturation effects. As you would expect, the waveform on the secondary of the output transformer is just about twice as big as the waveform on the primary... and looks nearly identical.
Thus, the tuned circuit appears to be producing about 23 dB of selectivity (28 - 5).
Going back to the fomula above... the measurement indicates that the Q of the output tuned circuit is about 9.3. This is (of course) larger than the value of 6 that I calculate based on a 50 ohm load... but the actual load is less than 50 ohms because of the loading capacitor that is in parallel with it. The loading capacitance I am using is about 850 pF... which corresponds to -j48 ohms... so the effect of this on the behavior of the output tuned circuit is to increase the selectivity.
Stu
Stu
Bernardsville NJ