Measuring CMRR of op-amp circuit

Question

I am trying to calculate the CMRR of following circuit. I am aware that similar circuit has been posted here before but I have read through them and they don't answer my question.

Based on my research so far, this circuit is one of the way to measure the CMRR of op-amp assuming that the resistor values for \${R}_{1a}\$, \${R}_{1b}\$, \${R}_{2a}\$,\${R}_{2b}\$ match perfectly.

In this circuit, i assume that the common mode input \${V}_{CM}\$ is \$ {V}_{IN} \frac{{R}_{2b}}{({R}_{1b} + {R}_{2b})} \$

And I learnt that the \$ CMRR = \frac{Differential-mode-Gain}{Common-mode-gain} \$. So, normally, i would find the common mode gain \$ {A}_{CM} \$ and differential mode gain \$ {A}_{DM} \$ separately.

However, i noticed, in most of the online sites, they simply measure the \$ \frac{{V}_{out}}{{V}_{in}} \$ and considered it as \$CMRR\$

My Question is: Why is the \$ \frac{{V}_{out}}{{V}_{in}} \$ of this circuit is equivalent to the \$ CMRR \$ of the circuit?

Can someone point me in the right direction?

EDIT 1:

I have added some link to the similar questions per @MarcusMuller request.

operational amplifier CMRR measurement

This question was asking what is the correct common mode input voltage. I am well are that it is \$ {V}_{IN} \frac{{R}_{2b}}{({R}_{1b} + {R}_{2b})} \$ so it doesn't help me.

Op Amp CMRR problem

This question was asking about his particular issue regarding common mode gain.

Answer 1

To put it shortly, even if you had perfect resistors (and resistors aren't perfectly matched, so this won't even remotely work), the formula as stated is wrong.

You need to vary the input voltage and observe the variation of the output voltage; the ratio of these two differences would then be the CMRR.

Again, it's not \$\text{CMRR}=\frac{V_{out}}{V_{in}}\$; it'd be \$\text{CMRR}=\frac{\Delta V_{out}}{\Delta V_{in}}\$, because that's how much a variation of common mode input gets amplified (that's literally just the definition of amplification applied to CM input).

And again, that assumes that your resistors are all perfectly matched, which they aren't. The best resistors I can buy are 0.01% accurate, which means the differential gain isn't exactly 1, but has some unknown error, and that sadly doesn't cancel, and at CMRR > 60 dB (which is pretty boring compared to what modern opamps offer), that error by far outshines our CM amplification. You'd hence need to measure that error, too, which, again, is pretty hard, because your measurement process needs to change the system much less than the CMRR, and measuring voltages does actually draw a bit of current.

Answer 2

The Op-Amp has an internal CMRR far better than any R tolerance. Yet the R tolerance error defines the output for ideal parts leaving only the Acm to have some value. Thus the CMRR of the circuit is simply \$V_o = A_{dm} * V_{dm} (=0) + A_{cm} * V_{cm} , ~~~or ~~ A_{cm}=V_o/V_{cm}\$.

Let the worst-case R tolerance error =X% for all 4 parts, thus degrades the Op Amp's CMRR = Adm/Acm = 4 * X% and thus the DM gain makes the CMRR even worse which demands precision-matched R ratio components and similarly matched wire impedance or STP cable for EMC concerns. I'll let you prove this as it was not part of your question.