SaM - CMRR • Calculation of the CMRR for the First Stage of an Instrumentational Amplifier

Look at this structure: We can define $V_2^{\prime }-V_1^{\prime }$ and $V_2^{\prime }+V_1^{\prime }$ based on $V_1$ and $V_2$, so: $(V_2^{\prime }-V_1^{\prime })=\frac{(2R_B+R_G)}{R_G}(V_2-V_1)$ But we also have that: $V_2^{\prime }+V_1^{\prime }=V_2+V_1$

NOTE:IMPORTANTE $R_B\simeq 50\text{k}\text{ohm}$

We have calulated the CMRR for the next stage, already and we found:${\text{CMRR}}_{II}=\frac{1}{2}\frac{\frac{R_4}{R_3+R_4}\frac{1}{\beta }+\frac{R_2}{R_1}}{\frac{R_4}{R_3+R_4}\frac{1}{\beta }-\frac{R_2}{R_1}}$Or:${\text{CMRR}}_{II}=\frac{1}{2}+\frac{R_2(R_3+R_4)}{R_1{R}_4-R_2{R}_3}$ If we add what we just said:

• $(V_2^{\prime }-V_1^{\prime })=\frac{(2R_B+R_G)}{R_G}(V_2-V_1)$
• $(V_2^{\prime }+V_1^{\prime })=1\cdot (V_2+V_1)$

So we can say:

& \text{CMRR}_{TOT} = \\[5px] & = \frac{1}{\frac{(2R_B + R_G)}{R_G}}\cdot \text{CMRR}_{II} \\[5px] & = \frac{R_G}{2R_B+R_G}\cdot \text{CMRR}_{II}\end{matrix}$$