Re: ubitx circuit, wiring, source code on github
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Earlier, the calibration routine put out a 10 MHz signal out the antenna (by unbalancing the diode mixer and setting the first oscillator to 10 MHz). I chucked it because it assumed that everybody had a dummy load or a 10 MHz antenna.
On the other hand, it is easier for a ham to call up someone and ask them to call CQ on a particular frequency. The way the present calibration works is like this. I call up friend to call CQ from his calibrated radio exactly at 7050 KHz. I set the dial to 7050 KHz, Then you switch on the calibration and tune until it goes dead zero beat. If only life was that simple.
First, the crystal filter has a steep skirt. We would ordinarily celebrate this. But in our case, that means, unless the signal is a 30 db over S9, you are not going to hear the zero beat, much less the swing to the other sideband.
Second issue with this is that as we recalibrate the VCO and reset the oscillators, the BFO too is corrected by an equal amount and the zero-beat will go away. To compensate for this, this is what I came up with, bear with me...
The VCO frequency is at 875 MHz. We generate any needed frequency by dividing the PLL by a fraction. So, to generate 1 Mhz, it is divided by 875. Now, suppose our reference that we are trying to align with is at 1 MHz, we need to nudge the 875 MHz VCO. If the VCO moves by 875 Hz, we can see the 1 MHz move by 1 Hz. To move the 1 MHz by 10 Hz at time, we have to move the VCO by 8750 Hz at a time. Our BFO is at 12 MHz, if we push the VCO up or down by 875 Hz, the 12 Mhz will be moved by 12 Hz. If we push the VCO around by 8750 Hz, the 12 Mhz BFO has to be corrected by 120 Hz. This is what the code does.
On Fri, Dec 8, 2017 at 1:14 AM, Jerry Gaffke via Groups.Io <jgaffke@...> wrote:
To scan the 12mhz filter passband, I'm thinking we drive CW-KEY high to unbalance the first mixer,