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Would someone mind explaining to me how the BFO frequency circuit works?
After all these years sometimes I get to forget!
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You have a carrier frequency which you send to a mixer then mix it with some audio then you end up with one central carrier and two side bands of audio content. The carrier is a waste of your power so you take it out and so also the other side band energy which is a mirror. So your xtal filter does the job and selects one side band. When this signal is received by you then you need to re-insert the carrier (BFO) back to make the incoming signal intelligible in audio. And if you vary your BFO frequency the incoming signal may sound tinny or bassy depending where your BFO is! Hope that helps! --- Raj vu2zap At 23/02/2017, you wrote:
I know a bit about electronics but not enough apparently! |
Q10 is the xtal oscillator with L & C in series, to pull the xtal down a bit to suit LSB.
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If you can generate a second frequency from the Si then you can remove Q10 and use Q11 to buffer it to the mixer. In the meanwhile I'll stick to my analog VFO! 73 Raj At 23/02/2017, you wrote:
I was wondering more about what the purpose of Q10 and Q11 serve. |
It's all working at about 11.998500mhz. ? Q10 is the oscillator and has some high impedance nodes on the left hand side, poking at it around the base or L5 may either disturb the oscillator or show low level signals that your instrument has trouble with. ?Q11 is an emitter follower to buffer output from the oscillator and drive a strong signal into the transformer without sucking too much power out of the oscillator. ?R104 ensures that even Q11 doesn't load down the oscillator, R104 is apparently getting stuffed with 10k these days instead of 2.2k. ?The little trimmer cap at C103 is not getting stuffed, it was found that by selecting crystals with all the same freq the BFO freq was just about right without any need for adjustment. On Thu, Feb 23, 2017 at 06:36 am, <KD2BBW@...> wrote:
Also when I measure at L5 I was getting ~12MHz on my counter and when I measure at base of Q10 I was getting ~6Mhz. ? |
Baruch Atta
Are you asking about the BFO in the transmitter circuit? The BFO in the transmitter circuit hetrodynes the 4.9-5.0 MHz single sideband signal with a 12.0 (approx) MHz BFO signal, giving 12+5=17 and 12-5=7 MHz output.? The BitX uses the 7 MHz signal which is amplified and transmitted.? See the March 2017 QST for a nice article on the original use of this method in the antique radios section.? On Thu, Feb 23, 2017 at 9:02 AM, <KD2BBW@...> wrote:
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Correct me if I'm wrong ... didn't you meant to say "The BFO in the transmitter circuit hetrodynes the audio baseband signal with a 12.0 (approx) MHz BFO signal, giving 12 MHz output DSB. ?Then it goes through crystal filter to get rid of the other side band, and then VFO heterodynes 4.8-5 with filtered 12 MHz output,?giving 12+5=17 and 12-5=7 MHz output" Just got my copy of QST yesterday, didn't have time to read yet. :) Cheers, Goran VE6GPO |
Q10 is the BFO oscillator and Q11 is the buffer between the oscillator and the load.
The buffer interfaces the oscillator and the load so that the oscillator is not "pulled" by the load. Being a crystal oscillator, Q10 produces a complete AM signal, i.e., a carrier and two sidebands, USB and LSB. This oscillator (The BFO) is tuned to the IF frequency. The load is a mixer. Its function is to suppress the carrier and inject audio modulation on transmit and to convert the IF by demodulating (recovering) the audio signal on receive. It is very like the main mixer but acts on only one frequency (the IF) instead of many (RF) The nominal IF frequency is that of the crystal -- 12 MHz. In actuality, it is slightly lower because we want to isolateand use the LSB. If we wanted USB, it would be several KHz higher. The crystal filter receives the IF with both sidebands (Remember, only the carrier was suppressed) and removes the USB(which has no information), in this case, passing on the LSB to the final mixer. How it does this is achieved by tuning the IF to the exact frequency needed to reflect the LSB side of the filter. The filter is sort of neutral; it will also pass the USB if the IF is tuned properly, but is very selective in that it will reject the unselected one because it is so narrow in its "passband". What this means to a DDS is that one must generate 2 if signals, one for USB and one for LSB, and select which one to send through the filter depending on the mode one wants. Since a DDS is the source, the IF oscillator (Q10) can be eliminated. Q11 may or may not be necessary; it depends on how well the DDS drives the mixer. The mixer might have to be tuned a bit as well. For example, let's say that the center frequency of your filter is 12,000,300 Hz. It doesn't have to be, and may be considerably different in a given rig. Every rig will be a little different from the others. Then the IF necessary to carry the LSB would be about 2KHz below that and the IF for the USB would be about 2 KHz above that. What it actually is will require experimentation and measurement. That is your challenge as a builder. john AD5YE |
Thank you, Raj, for the perfect high-level answer! I'm tweaking my Arduino code to generate an LSB signal and to "correct" the "crystal error" I thought I was measuring. ?Not surprising in hindsight that pulling the frequency down 3K makes the signal that escapes much smaller. ?:) Still learning...a LOT! ?(...and confirming that the crystal filter works... ?Hi Hi) 73 Andy |