Consider the difference amplifier as shown in the figure below. Assuming the op- amp to be ideal the CMRR of the circuit is __________dB.

41.63

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Solution

The correct option is A 41.63
Applying superposition at the output, we have

$V_{o} = - \frac{R_{2}}{R_{!}} . V_{1} + (\frac{R_{4}}{R_{3} + R_{4}}) . (1 + \frac{R_{2}}{R_{1}}) . V_{2}$

$= - 10 V_{1} + 10.0833 V_{2}$

Now, the differential input voltage is given as

$V_{d} = V_{2} - V_{1}$

$a n d V_{c} = \frac{V_{1} + V_{2}}{2}$

$T h u s, V_{1} = V_{c} - \frac{V_{d}}{2}$

$V_{2} = V_{c} + \frac{V_{d}}{2}$

$∴ V_{o} = (10.0833) [V_{c} + \frac{V_{d}}{2}] - 10 [V_{c} - \frac{V_{d}}{2}]$

$V_{o} = 10.042 V_{d} + 0.0833 V_{c}$

Camparing the equation from the standard result.

$i . e, V_{o} = A_{d} V_{d} + A_{c} V_{c}$

$W e g e t, A_{d} = 10.042$

$A_{c} = 0.0833$

$∴ (C M R R)_{d B} = 20 l o g_{10} [\frac{10.042}{0.0833}] = 41.63 d B$

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