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Here is the extract from the manual: enter image description here

I want to have an idea about how much voltage error occurs when applied 10V signal. And is this CMRR zero if differential ended configuration is used? How can I calculate the Vout error for 10V input? Whats the difference between LLGND and GND in the manual? Is CMRR measures due to LLGND?

Here is the manual: http://www.mccdaq.com/pdfs/manuals/PCI-DAS6034-35-36.pdf

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Common mode rejection ratio (CMRR) does not give you an absolute common mode error. What it does tell you is how much common mode signal (which is usually noise or some bias voltage) is attenuated. For example, 85dB CMRR means that signal common to both differential pins attenuated \$10^{\frac{85\text{dB}}{20}} \approx 17782\text{ times}\$. You can apply that to your common mode signal to find the common mode error. This applies to differential and single ended systems. If you're asking this question, you should be making differential measurements.

As to the difference between LLGND and GND, I'm not going to sift through that manual to find the difference between the two, but one is probably analog and the other digital. In practice you can probably connect them if they're not already connected on the card. Grounding does not effect the CMRR.

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  • \$\begingroup\$ I'm making single-ended measurement. Isn't cm voltage eliminated in differential-ended case since it subtracts the common mode voltage just like noise? So if the input is a known 10V DC in single-ended mode, cant I roughly talk about the Vout error with respect to ideal case(with infinite CMRR)? If cm voltage was not an issue I would measure Vout without any cm voltage error. Since it is not ideal there must be a way to quantify this in terms of voltage.? \$\endgroup\$ – user16307 Oct 27 '15 at 14:12
  • \$\begingroup\$ If you're making SE measurements, then CMRR is irrelevant. It only applies to the attenuation of a differential measurement common mode. For instance, connect both inputs together (so no differential input) and take them both to +10v. The worst case reading will be -85dB(10v) = +/- 0.56mV, instead of 0.000v \$\endgroup\$ – Neil_UK Oct 27 '15 at 14:38

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