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How to use ADA4932-1 with CM on the input

I am using the ADA4932 as a SE to DE ADC driver.  The design notes and reference designs show several examples, but they all have input signals centered on 0V.  Are there some design guidelines for dealing with input signals that have a common mode voltage?  I have an input signal riding at 2.5V.  Output CM is 2V.   Thanks.

Thread Notes

  • Hi Harry,

    I did do that.  I see that the output has offsets shifted by 0.632V (up on Voutp, down on Voutn) with the Vn input termination to GND.  If I use the 'auto input tracking', it says to use 1.298V instead for Vn termination.  I'm wondering if there are design references to show how input CM levels are factored in.

    George

  • OK.  That does make some intuitive sense.  I'm using the last example from the data sheet here.  The 50 Ohm source is terminated with an impedance of 50 Ohms.  If I assume that the input Vcm is also divided by two as is the signal, I would set the Vn input termination voltage level to 1.25V.  It gets me close, but there is still a small 24mV offset in output Vcm.  It is not intuitively obvious (well, to me anyways)  how to trim that offset to zero. 

  • FormerMember
    0 FormerMember
on May 13, 2015 11:08 PM

George,

  Download the DiffAmp calculator tool:

ADI-DiffAmpCalc | Analog Devices

Harry

  • George,

      Think of it this way.  You want to take the difference between 2.5V and your signal.



    Harry

  • OK, I think I understand now.  I also came across AN-1026. Calculating the input impedance was straightforward, however it was your earlier comment about taking the diference between the signal and input CM that got me thinking. The voltage level that the input termination to Vn input is simply the Thevenin equivalent CM voltage at the Vin+ input.  In my case, with 50 Ohm source and 53.6 Ohm input termination and a 2.5V CM level, I get 1.2934V, which brings the output to the desired CM with little to no offset.

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