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AFE58JD48: What is the minimum detectable signal (MDS) of AFE58JD48?

Tommy Tsang
Prodigy 10 points
Part Number: AFE58JD48
Other Parts Discussed in Thread: AFE5818, AFE58JD32

I am currently considering analog front end (AFE) from TI for our ultrasound imaging project.  I would like to know what the minimum detectable signal (MDS) is.

In general, in all the TI AFE offering (e.g. AFE5818, AFE58JD32, etc), which device gives the the best MDS?

Thanks.


Tommy

  • 0
    TI__Expert 4421 points

    Hi,

    Minimum signal amplitude detection depends on how much minimum SNR in your system, are you looking for. We specify the noise density performance in the datasheet which will be around ~0.9nV/rt(Hz) for maximum gain case. Now depending upon bandwidth and SNR requirement you can calculate the minimum signal amplitude. For example say you are looking for minimum SNR of say 10dB and signal bandwitdh of 10MHz (Assuming you will have post digital filter to remove out of band noise). So noise power = 0.9nV * sqrt(10M) = 2.85 uV (RMS).

    SNR = signal power/noise power =>

    10 = 20*log10(Signal RMS/Noise RMS)

    So from above equation, signal RMS will be 9 uV (RMS)

    Thanks

    Regards,

    Shabbir