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Regarding the second page of this document, in the first example it calculates the DC CMR due to line imbalance Rs and an explanation is given in the first paragraph below (ignore the second example):

enter image description here

I don’t understand where the numbers 0.9990 and 0.0009990 come from.

I redrew the circuit model to simplify it and named the nodes as follows:

enter image description here

Can’t we show the path taken by the differential current flow direction using the nodes order such as a, b, c, d, e and the common mode currents direction as a, b, c, m, k and e, d, c, m, k?

If true, the differential voltage gain and common mode voltage gain seems not as given in the document. For instance, the differential gain to me seems like 20Meg/(20Meg+10k) but the document shows it as 0.999. I'm confused about how to calculate both differential and common mode gains mathematically here. How should they be calculated separately to find the CMR?

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Think about the voltages at nodes \$b\$ and \$d\$, and the voltage divider formed by \$R_s\$ and \$R_{in+}\$. For simplicity, it's assumed that the amplifier gain is 1. Noticed that the gain does not affect the CMRR calculation because it shows up in the numerator and denominator.

Differential mode:

no noise => \$V_e=V_d=0\$; \$V_{in}=V_a\$

\$V_{out}=V_b-V_d=V_{in}\frac{10MΩ}{10MΩ+10kΩ}-0\$

\$\frac{V_{out}}{V_{in}}=\frac{10MΩ}{10MΩ+10kΩ}=0.999\$

Common mode:

no signal => \$V_a=V_e=V_d=V_{in}\$

\$V_{out}=V_b-V_d=V_{in}\frac{10MΩ}{10MΩ+10kΩ}-V_{in}=V_{in}(\frac{10MΩ}{10MΩ+10kΩ}-1)=-V_{in}\frac{10kΩ}{10MΩ+10kΩ}=-0.000999\$

For CMRR purposes, we don't really care about the sign of the gains, so we get:

\${CMRR}=20log(0.999/0.000999)=60dB\$

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  • \$\begingroup\$ Why not the equation is 20M/(20M+10k) for the DM?? \$\endgroup\$
    – cm64
    Sep 4, 2019 at 16:31
  • \$\begingroup\$ That would have been the case if not for the fact that the signal source has the bottom side connected to ground, what effectively shorts \$R_{in-}\$, leaving \$R_{in+}\$ as the only load for the signal source. \$\endgroup\$
    – joribama
    Sep 5, 2019 at 1:23
  • \$\begingroup\$ This happens because the signal source is single-ended and not balanced as someone would expect to see in front of a differential amplifier. \$\endgroup\$
    – joribama
    Sep 5, 2019 at 1:59

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