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

Here’s the first stage, just remembering these formulas:

$V_{0} = V_{C} A_{C} + V_{d} A_{d}$
$V_{d} = V_{2} - V_{1}$ & $V_{C} = \frac{V _{2} + V _{1}}{2}$
$A_{d} = \frac{A _{2} - A _{1}}{2}$ & $A_{C} = A_{2} + A_{1}$

And the structure of the Stage II:

We first calculate $A_{1}$ for $V_{1}^{'}$ : $V_{0} ∣_{V_{2}^{'} = 0} = - \frac{R _{2}}{R _{1}} \frac{β A _{d}}{1 + β A _{d}} V_{1}^{'}$

Then $A_{2}$ for $V_{2}^{'}$ : $V_{0} ∣_{V_{1}^{'} = 0} = \frac{R _{4}}{R _{3} + R _{4}} \frac{1}{β} \frac{β A _{d}}{1 + β A _{d}} V_{2}^{'}$ So: $A_{1} = - \frac{R _{2}}{R _{1}} \frac{β A _{d}}{1 + β A _{d}}$ $A_{2} = \frac{R _{4}}{R _{3} + R _{4}} \frac{1}{β} \frac{β A _{d}}{1 + β A _{d}}$ Finally: $CMRR_{II} = \frac{1}{2} \frac{\frac{R _{4}}{R _{3} + R _{4}} \frac{1}{β} + \frac{R _{2}}{R _{1}}}{\frac{R _{4}}{R _{3} + R _{4}} \frac{1}{β} - \frac{R _{2}}{R _{1}}}$ Remember: $β = \frac{R _{1}}{R _{1} + R _{2}}$

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

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