AD9106 Sawtooth frequency range as function of CLK

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!

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  • Hi Larry,

    thanks for your answer. Nevertheless, there is still one critical piece of information missing to completely answer the question.

    My goal is to generate one entire ramp per pattern period, without the gap or 'silence' as can be seen in Fig. 51 (rev. A). In other words, I want the next ramp to start as soon as the previous ramp was completed. From your answer, I understand that I will have to shorten the pattern period to match the sawtooth period. To do so, I would have to set the pattern period to equal the number of steps of the sawtooth waveform. Neither from your answer, nor from the datasheet, the number of steps in one ramp is clear.

    Another way of interpreting your statements is that the sawtooth generator is in fact (a fraction of) the pattern period counter, which has 65536 counts. But then, what would happen to the sawtooth frequency when the generator is free-running, i.e. without patterns enabled?

    From another user on this forum, I've seen that the 14bit AD9102 ramps up and down from 0 to 16384.Since the AD9106 is roughly equivalent to the 12bit version of AD9102, you could also conclude that the 12bit AD9106 should ramp up and down from 0 to 2^12= 4096.

    In any case, the question comes down to 'how many counts does the sawtooth generator have?'. Can you comment on this?

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  • Hi Larry,

    thanks for your answer. Nevertheless, there is still one critical piece of information missing to completely answer the question.

    My goal is to generate one entire ramp per pattern period, without the gap or 'silence' as can be seen in Fig. 51 (rev. A). In other words, I want the next ramp to start as soon as the previous ramp was completed. From your answer, I understand that I will have to shorten the pattern period to match the sawtooth period. To do so, I would have to set the pattern period to equal the number of steps of the sawtooth waveform. Neither from your answer, nor from the datasheet, the number of steps in one ramp is clear.

    Another way of interpreting your statements is that the sawtooth generator is in fact (a fraction of) the pattern period counter, which has 65536 counts. But then, what would happen to the sawtooth frequency when the generator is free-running, i.e. without patterns enabled?

    From another user on this forum, I've seen that the 14bit AD9102 ramps up and down from 0 to 16384.Since the AD9106 is roughly equivalent to the 12bit version of AD9102, you could also conclude that the 12bit AD9106 should ramp up and down from 0 to 2^12= 4096.

    In any case, the question comes down to 'how many counts does the sawtooth generator have?'. Can you comment on this?

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