Today I've had a real long look at the datasheet, trying to figure out the lowest sawtooth frequency possible given a certain clock frequency CLK. This information is critical to see if I need two separate IC's or if I can do with a single AD9106.
From the four lines on page 27 (Rev. A), I assume that the sawtooth generator runs at CLK and you can divide the clock up to 64x by means of SAW_STEPx field, thus each sawtooth step would take 64/CLK seconds. To calculate the sawtooth frequency, I guess I would need to know the number of steps, i.e. the number of bits of the sawtooth waveform counter?
Let's say that CLK = 180MHz, the sawtooth generator has 12 bits (i.e. 4096 steps) and the SAW_STEP register is 63, the lowest sawtooth frequency would be (180MHz/63)/4096 = 697.5Hz, correct? Can anyone confirm the number of steps for a full sawtooth waveform?
I do realise that one can lower the frequency by choosing a lower CLK frequency, the question comes down to 'what is the biggest difference in frequency between the DDS sine wave and sawtooth waveform?'
Thanks in advance!
