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## Q. 33.13

A differential amplifier has, $A_v = 150, A_{v(CM)} = 0.5$, and $v_{in}$ = 1 mV. If the base leads are picking up a common-mode signal of 1 mV, what is the output voltage?

## Verified Solution

The input has two components, the desired signal and a commonmode signal. Both are equal in amplitude. The desired component is amplified to get an output of

$v_{out1} = 150(1 mV) = 150 mV$

The common-mode signal is attenuated to get an output of

$v_{out2} = 0.5(1 mV) = 0.5 mV$

The total output is the sum of these two components:

$v_{out} = v_{out1} + v_{out2}$

The output contains both components, but the desired component is 300 times greater than the unwanted component.

This example shows why the diff amp is useful as the input stage of an op amp. It attenuates the common-mode signal. This is a distinct advantage over the ordinary CE amplifier, which amplifies a stray pickup signal the same way it amplifies the desired signal.