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WaveVision's (5) ENOB calculation

Florian Busch
Prodigy 150 points
Other Parts Discussed in Thread: ADC16DV160HFEB

Hi,

I have a simple question regarding WaveVision's ENOB calculation or better:

Is the ENOB increasing with increased Vpp or decreasing ?

We have the ADC16DV160HFEB eval board and we are playing around with different clocking strategies. For that we measure the ENOBs at different Vpp values of the input signal for a given clocking circuit.

We see the following ENOB(Vpp) curve (BLUE) when applying different Vpp values for given clocking circuit.

 

The IEEE spec says [1] (equation 71) that the ENOBs depend on Vpp and the amount looks right (see ORANGE curve, which is the relevant part of equation 71). But the spec also says, that the value is reduced by the amount, so we would expect a curve like the YELLOW one, approaching the highest ENOBs (~12.5) for maximal Vpp (2V or 2.4V).

If we now modify the „squared“ oscillator clock source to a more sinusoid waveform, the theory says, that we get less ENOBs. WaveVision 5 shows instead, that the ENOBs increase for a given Vpp.

So my assumption is, that WaveVison does a false ENOB calculation (I know I shouldn't do this). This results into the fact, that the customer is optimizing in the wrong direction and against the theory and mathematics ;-)

There is also a paper [2] from Walt Kester, who might now be working for TI, which says that the ENOBs increase with increasing Vpp (although he has a typo in the relevant section of equation 2). This is not what we see in WaveVision.

Can you shine some light into this ?

Looking forward to you answer,

Best Regards, Florian

PPS: I had posted a question a few days ago about the general dependency of ENOB and Vpp. I started a new post now about this - what I think – more Wavevision related problem.

[1] IEEE Standard for Terminology and Test Methods for Analog-to-Digital Converters, IEEE Standard 1241-2010

[2] http://www.analog.com/media/en/training-seminars/tutorials/MT-003.pdf

  • 0
    Prodigy 150 points

    Unfortunately, the beautiful diagram disappeared when posed, so here are the numbers in a table format. I hope you can connect it to the question in the original post.

    Increasing Vpp: 0,1 0,3 0,5 0,7 0,9 1,1 1,3 1,5 1,7 1,9 2,1
    Oscillator (Diff), Wavevision numbers, Decreasing ENOB over increasing Vpp (blue): 12,28 11,38 11,15 11 10,91 9,65 9,56 9,38 9,09 8,88 8,75
    log2(1*Vpp/2.4V), IEEE spec (orange): 4,58 3,00 2,26 1,78 1,42 1,13 0,88 0,68 0,50 0,34 0,19
    expected curve, increasing ENOB over increasing Vpp (yellow): 8,09 9,67 10,41 10,89 11,25 11,54 11,79 11,99 12,17 12,33 12,48


    I there a way to delete a post ? Is there a way to add graphics / pdfs ?

  • 0 in reply to Florian Busch
    TI__Expert 4340 points

    Florian

    In WaveVision, the ENOB is calculated as (SINAD-1.76)/6.02 where the SINAD is in units of [dBFS]. Due to the fact that reference signal is the full scale range and not the input sinusoid amplitude, SINAD (and ENOB) tends to get larger as the input signal amplitude decreases due to the overall reduction in noise and distortion while the reference amplitude stays constant.

    Equation 70 of the IEEE standard you reference shows that ENOB is not dependent on the input amplitude by definition, rather it depends on the amount of noise and distortion in the spectrum. In reality, the NAD value changes with the type and power of the input signal and influences the ENOB with a trend as I have described above. Equation 71 links SINAD to ENOB but the dependance of the SINAD (in units of dBc) term and -log2(2G*A/FSR) term on the input signal amplitude cancel each other out. This description is consistent with Kester's. Kester does comment that "If the signal level is reduced, the value of SINAD decreases, and the ENOB decreases," but this is in reference to using Eq.1 for reduced amplitude which necessitates Eq.2. His Eq.1 assumes that SINAD is in units of [dBc] which is different from my equation above where SINAD is in units of [dBFS]. His Eq.2 is equivalent to my equation.

    You have mentioned that changing the clock source waveform type influences the ENOB. This is expected. In general, you can achieve better performance (better ENOB) with the following clock properies:

    - Fast edge rate

    - Low in-close phase noise. Must choose a good PLL for the clock source

    - Low broadband phase noise. Must choose a good clock source or filter the clock source to remove the noise.

    Regards, Josh

  • 0 in reply to Joshua Carnes
    Prodigy 150 points

    Josh,

    first of all, thank you for your thoughtful answer.

    ENOB (Vpp): [dBFS] vs [dBc]

    Okay, I guess I understand now. It leaves the question, how to convert it.

    Optimization:

    In both cases (dBFS and dBc) you would try to increase the ENOBs. So for a constant Vpp (of the input sinusoid)  a clocking circuit with the higher ENOB is preferable. The problem is, that we see different results (in WaveVision) as to what we would expect (in theory), which made us questioning the WaveVision results – sorry for that.

    So for example, an RLC circuit after the squared PLL output generates a perfectly looking sinusoid clock. Here the ENOBs increase by 1-2 at (Vpp=1V) compared to the circuit without the RLC but a squared clock. This is not necessarily in-sync with the theory, because you basically have a slow edge rate. Anyway, I will summarize my findings in a new more detailed question and a new blog and I would be delighted, if you can share some light into this as well. Thanks a lot so far.

    Regards, Florian