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ADS5294: Does my signal need to be stabililed before the aperture delay?

Maxime Puech1
Prodigy 220 points
Part Number: ADS5294


Tool/software:

Hi,

I plan to use the ADS5294 to perform the acquisition of a high speed analog image sensor, and I need some help understanding the implication of the aperture delay specification.  

My sensor is outputing a new pixel analog value every 25 ns and I need to make one acqusition per pixel. I will therefore use a 40 MHz sampling frequency.

The rising time of the analog pixel is typically 23ns. If I set the ADC clock phase so that its rising edge (beginning of the hold phase) is at 23 ns, the sampling switch will actually be fully open 4 ns later (aperture delay specification), which give me 23 + 4 = 27 ns. As you can see, this is too long, as the analog image sensor is already outputing its next pixel at 25 ns. 

So in order to avoid that, my idea is to change the ADC clock phase so that is rising edge is at 19ns, which means the sampling switch will be fully open at 23 ns. Can I do this and simply compensate the aperture delay by anticipating it, or is there another concern I should be aware of?

If I anticipate it like I described above, the analog pixel will continue to rise while the sampling switch is opening ; can this cause an error? Which value will be sampled by the ADC?

In a few words, does my analog signal need to be stabilized before the aperture delay phase begins?

Best regards,

Maxime Puech

  • 0
    TI__Expert 7725 points

    Hi,

    Your methodology should be fine. You can say sample at 20ns and after aperture delay sampling will happen. Also in the datasheet you can see there is a spec for aperture delay variation (2.5ns) . So considering this you have to calibrate each device .

    One more point to consider , device has 550MHz input bandwidth. Your  analog pixel output bandwidth should be less than that to sample the point correctly .

    One general question :

    Why are you particular about sampling point . Your input information is preserved if you follow nyquist rule . What would be the analog pixel output bandwidth ? 

  • 0 in reply to Sachin H K
    Prodigy 220 points

    Hi Sachin,

    Thank you for your answer. 

    If I understood well, I can consider that the aperture delay is only a delay, and that it does not introduce an additional error on the value of the sampled signal. My concern was that the value of the sampling would be slighly affected by the fact that the sampling capacitor was charged through the increasing impedance of the sampling switch opening gradually during the aperture delay. 

    Now, to answer your question, the output frequency of the analog image sensor is 40 MHz. But it has 23 ns rising edges (0.1% to 99.9%), which gives an equivalent frequency of about 50 MHz ( BW(-3dB) = 1.1 / settling_time(0.1% to 99.9%) ). Therefore, if we wanted to respect the Nyquit rule, we would need to sample with a frequency bigger than 2*50 = 100 MHz.

    We do not want to to this for 2 reasons : 

    1. A bigger than 100 MHz sampling frequency would be very complex to achieve because we need 8 channels acquire 14 bits in the same time. It would require a way more expansive ADC, and also an interface and a FPGA that can deal with this troughput. 

    2. We do not care about the value of the analog pixel when it is in its rising/falling phase, and therefore we do not want to sample it. The only valuable information given by the analog pixel is when the pixel is settled after 23 ns. Because the image sensor is outputing pixels at a fixed frequency, we thought that if we correclty set the sampling point to happen after the pixel is settled, we can afford not to respect the Nyquist rule and use a 40 MHz sampling frequency.

    Do you think this approach is correct?

    You'll find below a graph showing the image sensor signal vs the ADC sampling clock (sorry about the french, but I think you'll get the idea)

    Best regards,

    Maxime Puech

  • 0 in reply to Maxime Puech1
    TI__Expert 7725 points

    You need to obey Nyquit rule if you want to reconstruct the signal back as it is . But if your analog sensor output is low frequency then you can just sample it without doing this delay calibration and get back the signal. Only high frequency content in signal will go away.

    What is the frequency of analog output if you assume the output rise time is 0 instead of 23ns?