I've been working on the crystal filter of the Barbados Barebones Superhet receiver. This was designed by Doug DeMaw in 1982. This one was built by Dale Parfitt W4OP and then repeatedly modified by me. It is now on 17 meters with a crystal-switchable VXO. Earlier I had made a very crude attempt to broaden the filter from its original very narrow CW configuration. This week I did this again, but this time I actually characterized the crystals and used Wes's LDA and GPLA software (from EMRFD) to design the filter. I played with the capacitor values and finally got the 3 kc bandwidth I wanted, but I'm having trouble getting rid of the ripple. I know this is dependent on the impedances at the two ends. The programs say I need 2000 ohms. I'm kind of puzzled about how Doug DeMaw did this with his original design. For his crystals and his 250 Hz (!) bandwidth he said he needed 450 ohms. He used 4.7:1 turns ratio transformers at either end and said that by putting 10k resistors across these transformers he got the needed impedance. I can see how this would work looking into the gate of the 40673 IF amp, but looking back at the drain of the 40673 mixer, I'm not so sure that that would yield 10k. (See schematic below.) But who am I to doubt Doug? So I assumed he was correct about the 10K and I re-wound the transformers with a 2:1 turns ratio, thinking that would get me closer to the needed 2k. But the ripple is still there. I guess I could use a return loss bridge at this point... I don't know whether this is worth messing with anymore. The receiver sounds nice. The 3kHz bandwidth gives it a nice sound, and the ripple doesn't seem to be noticeable That FAR circuits board is tightly packed and difficult to work with. So, should I leave good enough alone, or should I proceed with fanatical ripple eradication. Any advice? BTW: Why is it that receivers always seem to sound better when opened up (as above) on the workbench?
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