CMRR equations with resistor error for two-OP instrum amp and three-OP instrum amp

[allennlowaton]

Hello EDA fellows,

I would like to ask help from you regarding the CMRR equations of the two-OP IA and three-OP IA.
Shown below are the diagrams and the pertinent equations.



My question is how to derive the CMRR,min equations?

Attached also is the paper I referred to.

Thank you.

 

[LvW]

.................
My question is how to derive the CMRR,min equations?

CMRR is defined as the ratio between the differential gain and the common mode gain.
Both gain values can be calculated from the circuit diagram (assumption opamp ideal).
However, you must not apply ideal matching for the resistors (like Rx=Ry); instead use different indices for all parts.
At the end you can check if for ideal matching conditions the common mode gain is zero (CMRR infinite).
The ratio of both gains gives you an expression that shows how resistor tolerances influence tth CMRR value.
If you like, you can combine this CMRR value, which results from tolerances only, with the finite CMRR of the opamp.
For this purpose, the inverse values are addded: 1/CMRR=1/CMRR(opamp) + 1/CMRR (tolerance)

---------- Post added at 11:02 ---------- Previous post was at 10:45 ----------

Some additional hints:

From the common mode gain expression one can see which resistor - for worst case calculations - should have a positive or negative tolerance.
Then set: Rx=Rx,nom*(1+rx) with rx=dRx/Rx,nom (here d means DELTA, not a differential quotient). The value rx can be positive or negative.
Then you can apply the matching conditions like Rx,nom=Ry,nom.
As a result you have an expression for the common mode gain that contains only the tolerance factors r.