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ADS42LB69: Purpose of Digital Gain

Sean Ferguson
Prodigy 80 points
Part Number: ADS42LB69

Hi,

Looking at the manual for the ADS42LB69, section 8.3.1 on digital gain, I'd like to better understand how this feature is implemented. From the manual:

"The device includes gain settings that can be used to obtain improved SFDR performance (compared to no gain)...SFDR improvement is achieved at the expense of SNR; for a 1-dB increase in digital gain, SNR degrades approximately between 0.5 dB and 1 dB (refer to Figure 15 and Figure 16). Therefore, gain can be used as a trade-off between SFDR and SNR. Note that the default gain after reset is 0 dB with a 2.0-VPP full-scale voltage"

Why would increasing digital gain after sampling occurs have any impact on the SNR or SFDR? Is digital gain applied uniformly to all samples or some function of input amplitude? If want to change my input range from 2 Vpp to 1 Vpp I would expect an (analog) amplifier to boost my 1 Vpp signal to 2 Vpp prior to digitization in order to make use of the full 16 bits of the converter. If I apply 6 dB of digital gain, which restricts my input voltage range to 1 Vpp, am I not losing my MSB and hence reducing my theoretical SNR by 6 dB?

Thanks!

-Sean

  • 0
    TI__Guru* 91947 points

    Sean,

    You are correct. Digital gain is only to rescale smaller full scale signals back to 0dBFS.

    6dB lower full scale basically result in 6dB worse SNR.

    Regards,

    Jim

  • 0 in reply to jim s
    Prodigy 80 points

    Thanks, Jim. The reduction in SNR by using a 1Vpp with 6dB digital gain compared to 2Vpp with 0dB digital gain makes sense. However, I'm unclear how SFDR is improved. Figure 15 shows that the SFDR improves with digital gain. At the top of the page it's stated that the signals are -1dBFS. The only way this makes sense to me is if the amplitude of the input signal is changing as a function of gain since the FS input range also changes as a function of gain according to Table 3. For example, with a digital gain of 0, FS= 2.0vpp so -1dBFS is 1.7825Vpp, resulting in some SFDR and SNR. If I now apply a digital gain of 6dB, my input full scale is now 1Vpp so -1dBFS reduces to 0.891Vpp. I suppose we would get a better SFDR since we would be less impacted by nonlinearity but then our SNR goes down due to quantization noise. Is this what was inferred by tradeoff between SNR and SFDR with digital gain? If so, then it's really just the tradeoff between linearity and quantization noise, right?

    Finally, what is the point of the digital gain feature? Is to compensate for loss in the acquisition path?

    Thanks for your help.

    -Sean

  • 0 in reply to Sean Ferguson
    TI__Guru* 91947 points

    Digital gain.docxSean,

    See attached.

    Regards,

    Jim

  • 0 in reply to jim s
    Prodigy 80 points

    Perfect, thanks!