SaM - CMRR • Design Rules for Resistance Values of an Instrumentational Amplifier

We start with the CMRR (Stage II), more specifically with $A_{1_{II}}$ and $A_{2_{II}}$ and we consider a non ideal amplfiier, specifically $A\ne \infty$, so we can write:$\begin{array}{l}{A}_{1_{II}}=-\frac{R_2}{R_1}\frac{\beta A}{1+\beta A}\\ A_{2_{II}}=\frac{R_4}{R_3+R_4}\frac{A}{1+\beta A}\end{array}$Where:$\beta =\frac{R_1}{R_1+R_2}$We use the two following design rules:

1. $A_1=-A_2$ (So that we obtain an infinite CMRR)
2. $\frac{R_4}{R_3}=\frac{R_2}{R_1}=K$ (like for a balanced bridge)

So using the (1.) design rule, we can simplify $\frac{A}{1+\beta A}$ from both terms, and we get: $\frac{R_2}{R_1}=\frac{R_4}{R_3+R_4}\frac{R_2+R_1}{R_1}$ Using the (2.) design rule: $K=\frac{K}{1+K}(K+1)$And this is true $\forall k\in \mathbb{R}$. So if we ensure that $\frac{R_4}{R_3}=\frac{R_2}{R_1}$ we will have an infinite CMRR, even if the amplifier is not ideal.

---

Memory Card

---