I have an analog 27 MHz transceiver which i am planning to convert to 10 m. The radio is an AM / SSB unit with a single conversion superhet receiver with an IF of 7.8 MHz using a crystal filter. This radio does not have any faults due to component failure, and I'm trying to align the receiver and have a problem understanding one aspect of the alignment.
Normally with other radios of this type, alignment of the IF strip generally involves connecting a signal generator to the antenna socket and then peaking coils and transformers for maximum signal level.
However when i do that with this model radio (i have 4 and they all do the same thing), the local oscillator used to inject carrier into the product detector used for SSB receive seems to appear as a signal on the S meter. I have to adjust the first transformer in the IP strip off to the side of the point where i see maximum signal generator level on the meter and then it seems to work properly, however with reduced receiver sensitivity.
Other radios I have aligned don't show this behavior and you just set the alignments in the IF strip for maximum signal and you never see the local BFO appear as a signal on the S meter.
Can anyone explain to me what is going on when I am aligning this radio ?
In the schematic below, the 7.8 MHz IF strip is Q201, Q202, Q203, and you can see the IF output from the IF transformer T203 at the top middle going to T401 and then to the SSB product detector Q405 and Q406 at the lower right. The product detector has an input from a 7.8 MHz local oscillator which you can't see in the picture at the junction of Q405 and Q406 emitters. The 7.8 MHz crystal filter FL-1 is at the top left. The separate detector for AM are the diodes at the top right. The audio outputs from the 2 detectors are connected to R419 which connects to the audio amp input.