For getting balanced

*output we use this circuit. In this circuit two source is present, so the superposition theory is applied to get the output. In this circuit both the inverting and non-inverting terminal is working.It rejects the common-mode voltages, so it is very useful in noisy environments.***differential**
A differential input and differential output amplifier using two identical

__. It is most commonly used as a preamplifier and driving__*Op amp**arrangement. The differential input and output are inphase or the same polarity provided***push-pull***V*and_{in }=V_{x}– V_{y}*V*

_{o }=V_{ox}– V_{oy }
When we want to find out the 1

^{st}op-amps output*V*, we will use the superposition theory._{OX}
When we get

*V*is active ,_{X }*V*is inactive then ,_{Y }
In non inverting terminal

*V*

_{1}= (1+*)V*

_{X}
When we get

*V*is active,_{y }*V*is inactive then,_{x }
In inverting terminal,

*V*

_{1 }= -*V*

_{y}
So

*, V*_{ox}= V_{1}+V_{1}

_{ }*=(1+*

*)V*

_{X }-*V*

_{y}
Fig: The circuit diagram of the differential input and output amplifier.

When we want to find out the 2

^{nd}op-amps output*V*,we will use superposition theory._{Oy}
When we get

*V*_{X}_{ }is active ,*V*is inactive then_{Y }
In inverting terminal ,

*V*

_{2}= -*V*

_{x}
Again, when we get

*V*is active ,_{y }*V*is inactive then_{x }
In non inverting terminal we get,

*V*

_{2}= (1+*)V*

_{y}
So,

*V*_{oy}= V_{2}+V_{2}*=(1+*

*)V*

_{y}*-*

*V*

_{x}
So the output result

*V*

_{o }=*V*

_{ox}– V_{oy}*= (1+*

*)V*

_{X }-*V*

_{y }*–[(1+*

*)V*

_{y}*-*

*V*

_{x }]*=*

*(1+*

*) ( V*

_{X }- V_{y }) +*( V*

_{X }- V_{y })*= ( V*

_{X }- V_{y })*(1+*

*)*

**Design**

To design a input and differential output amplifier,

*taking a differential output of at least 3.7**V*and the differential input*V*_{in }=10V.
We know,

*V*

_{o}= ( V_{X }- V_{y })*(1+*

*)*

Or,

*3.7 = (0.1)**(1+**)**Or,*

*37 =*

*(1+*

*)*

*Or,*

*36 =*

*Or,*

*R*

_{f }= 18 R_{1}
Let

*, R*then_{1}= 100Ώ,*R*_{f }= 1.8 KΏ .

_{ }
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