Single-ended source impedance at input of AD8476

Hello,

I'm considering using the AD8476 for converting a bipolar +/-9V single-ended signal from a Rs=50Ohm source into fully-differential signal going into a second variable gain differential amplifier, and then into a long twisted-pair cable, and finally into an ADC with differential input and Ri=800k.

On page 18 of the AD8476 datasheet it is mentioned that the source impedance should be kept below 0.1Ohm, to avoid unbalancing the resistor ratios.


Would having a Rs=50Ohm significantly degrade the CMRR?


I was thinking of not matching the input impedance of AD8476 stage since my signals are signals are low-frequency (<1MHz). Would you see an inconvenient with this? Should I still add a 50Ohm resistor at the grounded single-ended input to try to balance input resistances ?

If I want to add a filtering stage, would it be preferably to have it before or after the AD8476 stage? Maybe, if this was an active filter before the AD8476 I could try to decrease the source impedance seen by it. Any advises on topology and opamp?

Many thanks in advance,

  • 0
    •  Analog Employees 
    on May 8, 2015 9:51 PM

    Hi salparadise,

    The quick answer is yes, 50 ohms does degrade CMRR quite a bit. We trim the internal resistors to very tight tolerance to get the CMRR. The equations showing the effect of mismatched feedback factors on this differential amplifier topology have been published in a few places, including AN-1026 (http://www.analog.com/media/en/technical-documentation/application-notes/AN-1026.pdf).

    A while back, I went through the exercise of throwing those equations into a spreadsheet to see the effect of imbalance in the feedback network with this kind of differential amplifier (attached in case you want to use it as well). According to that, if all of the resistors are perfect and we add 50 ohms to one side, the CMRR comes down to about 52dB.

    For a single-ended input circuit, that might not be so bad. With respect to your SE signal, the input common-mode voltage changes in response to your signal (Vin,cm = Vin,se/2), so the finite CMRR will cause a small gain error. And perhaps the 52dB (or 58dB if you terminate) is enough in terms of noise rejection from ground bounce/interference.

    In terms of adding the balancing resistor on the other side, it tends to help the CMRR, but it won't be a perfect solution. The internal resistors are trimmed to the correct ratio, but the absolute accuracy is ±20%. For example, if you had 10k,10k on one side, and 8k, 8k on the other, you would have perfect CMRR. If you added 50 ohms to the input on each side, it would come down to 64dB. In practice it's unlikely you'll see a mismatch that big, but the idea holds true.

    I hope this helps.

    Best regards,

    Scott

    DifferentialAmplifierImbalanceCalculation.xlsx