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## Re: [Amps] 807 transmitter

 To: "The Bowers" , Re: [Amps] 807 transmitter "Tom W8JI" Mon, 5 Feb 2007 18:54:18 -0500
 ```> Has anyone built "K5DH's 807 CW transmitter" (Google) ? > I've built just the > amp section but can't get it to load. I'm driving it with > a sig gen at 3.9 > KC. > > According to the on line calculators, L3 is at 10.3 uh, > but should be 18.5 > uh., for 75 meters. Dana, There is a VERY wide range of tank Q's that will work just fine. People go way overboard fussing around to get a certain Q that is really just a wild approximation anyway and often wrong. The minimum Q you can use and have the network behave like a Pi is the square root of the impedance ratio plus a tiny bit. This would be a phase shift of about 90 degrees. The highest Q is really limited only by the components and how narrow you want the tuning. If the tank components are reasonable quality you won't notice much difference in performance over a VERY wide range of Q's, so don't get caught up in the "Q-steria". Now I'm assuming you are running deep class C and you meant 3.9 MHz and not kHz :-) and your generator has enough output to drive the tube (a few watts anyway if it is running directly into the grid circuit) and anode voltage is about 750 volts. If so, the plate voltage would swing about 500 volts RMS. That would be about 5500 ohms plate loadline with 50 watts. Let's say you want a pi that would match 20 ohms as a minimum impedance. The tank Q would have to be a minimum of sqrt 5500/20 = 16.5 plus a little. Let's say 17 would be a good number. You can see from this how the idea all tanks need a Q of 12 falls apart. The plate C would be 125pF, peak voltage just over 700. The inductor would be 13.3uH and current would be 1.6 amperes RMS. The loading cap 500pF Then when you have a 50 ohm load (or higher) all you need to do is close up the loading cap and re-dip the plate. Loading C would be 1155pF and plate C 135pF and Q would be 18.2. With 75 ohms the settings would be loading 1050pF, tuning 138pf, and L would still be 13.3 uH. Now the loading moves in a normal direction because Q is significantly higher than the minimum needed (phase shift 150 degrees). If you use the values I just gave with a tank inductance of 13.3 uH and capacitors of 150pF and 1300pF you could match any pure resistance between 20 and a hundred ohms or more. The reason the load cap works "backwards" at low Z is we are running right on the edge of minimum Q. Phase shift is only around 100 degrees. So you see the fellow who wrote the article is probably closer than the on-line calculators. The tank inductor would have to be 13.3 uH or LESS to have enough Q to match 20 ohms to a few hundred ohms. he probably had a little more Q than necessary, assuming his cap values were correct (I didn't read it). Remember the rule. You need a tank Q of at least the square root of the largest impedance ratio you expect to match plus a small safety factor. That would set the maximum L value at the highest frequency for any band. using less L than that would be safe, and more L means it would not tune under the impedance extremes. 73 Tom _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/mailman/listinfo/amps ```
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