Measuring CMRR in single-supply simulation

[jmw]

In Analog Device's app note MT-042, they describe this circuit and formula for measuring CMRR:



First off, is this formula right? When I solve the circuit, I get:

\[v_+ = \frac{R_2}{R_1 + R_2}v_{in}\]
\[v_- = \frac{R_2}{R_1 + R_2}v_{in} +\frac{R_1}{R_1 + R_2}v_{out}\]
\[v_{out} = A_d(v_+ - v_-) + \frac{1}{2}A_{cm}(v_+ + v_-) = \frac{A_{cm} v_{in} \frac{R_2}{R_1+R_2}}{1 + \frac{R_1}{R_1 + R_2}(A_d - \frac{1}{2}A_{cm})}\]

Under the assumption the denominator is large, the "1" drops out, and if \(A_d \gg A_{cm}\), then \(A_d - \frac{1}{2}A_{cm} \approx A_d\), so the result is

\[v_{out} = \frac{v_{in}}{A_d/A_{cm}}\frac{R_2}{R_1} = \frac{v_{in}}{\mathrm{CMRR}}\frac{R_2}{R_1}\]

Did I make a mistake in simplifying? They didn't show their work in the formula for the app note... Next, in doing simulations, what is right setup in a single-supply configuration? Should the input reference the virtual ground, like this:



Or be AC-coupled to ground?

[Marco]

I'm not going to run the formulas, but if R2 over R1 is large you can ignore the 1 in their formula as well.

Also they meant to show the naive circuit couldn't work ... 1 ppm matching dividers don't exist outside of SPICE.

[David Hess]

It is sufficient to trim the resistance ratios but the more practical version uses another resistor between the inverting and non-inverting input to raise the noise gain.

https://www.changpuak.ch/electronics/downloads/bob_pease_lab_notes_2005.pdf