AD8475 CMRR

Hello!


Correct me if I’m wrong. I want to estimate an error on AD8475B output caused by its finite CMRR. Let’s assume that AD8475 is powered from 5V supply (-VS=GND, +VS=5V) and is connected to an ideal voltage source in SE configuration and a gain of 0.4 (just like on datasheet page 21). Let –IN 0.4x =GND and +IN 0.4x = 10V. Taking into account 5V supply, the AD8475 CM bias (not the VOCM) is a half supply = 2.5V, so our signal has a 2.5V common mode component referred to CM bias. AD8475B CMRRmin=86dB so an output voltage error related to an input 2.5V common mode voltage will be ~ 125uV.


Am I right calculating input voltage common mode components related to AD8475 half supply bias?

  • 0
    •  Analog Employees 
    on Feb 1, 2015 6:05 PM

    Hi Konstantin,

    I think the error the CMRR contribute to the output can be computed like this way. By defintion, CMRR(dB)=20*log10(Ad/Acm) where ad is the input-output differential gain and Acm is the input to output common mode gain. In your case, Ad=0.4 and CMRR= 86dB, we get Acm= 20uV/V. With input 5V input common mode voltage change, from the average of your two input voltages, produces 100uV change in the output voltage. One good reference about CMRR is this tutorial. http://www.analog.com/static/imported-files/tutorials/MT-042.pdf

    Hopes it helps.

    Regards,

    Phil

  • Hi Phil.

    My mistake: I forgot about differential gain. So you are right: we get Acm= 20uV/V when AD8475 configured to 0.4 gain. But what about value of input CM voltage reffered to AD8475 in my example above: I think it is 2.5V not 5V cause AD8475 in example has a single supply and its midpoint is 2.5V (0V + 5V)/2 and relative to it voltages on AD8475 inputs will be: -IN 0.4x = -2.5V, +IN 0.4x = 7.5V, so Vcm_in= (-2.5V + 7.5V)/2 = 2.5V. So the main

    question: is the AD8475 bias |Vs|/2 which is common for most opamps or shifted?

    Best Regards, Konstantin.

  • 0
    •  Analog Employees 
    on Aug 2, 2018 2:54 PM
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