For getting balanced differential 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.
A differential input and differential output amplifier using two identical Op amp. It is most commonly used as a preamplifier and driving push-pull arrangement. The differential input and output are inphase or the same polarity provided Vin =Vx – Vy and
Vo =Vox – Voy
When we want to find out the 1st op-amps output VOX , we will use the superposition theory.
When we get VX is active , VY is inactive then ,
In non inverting terminal
V1= (1+ )VX
When we get Vy is active, Vx is inactive then,
In inverting terminal,
V1 = - Vy
So, Vox = V1+V1
=(1+ )VX - Vy
Fig: The circuit diagram of the differential input and output amplifier.
When we want to find out the 2nd op-amps output VOy ,we will use superposition theory.
When we get VX is active , VY is inactive then
In inverting terminal ,
V2= - Vx
Again, when we get Vy is active , Vx is inactive then
In non inverting terminal we get,
V2= (1+ )Vy
So, Voy = V2+V2
=(1+ )Vy - Vx
So the output result
Vo = Vox – Voy
= (1+ )VX - Vy –[(1+ )Vy - Vx ]
= (1+ ) ( VX - Vy ) +( VX - Vy )
= ( VX - Vy ) (1+ )
To design a input and differential output amplifier, taking a differential output of at least 3.7V and the differential input Vin =10V.
Vo = ( VX - Vy ) (1+ )
Or, 3.7 = (0.1) (1+ )
Or, 37 = (1+ )
Or, 36 =
Or, Rf = 18 R1
Let, R1 = 100Ώ, then Rf = 1.8 KΏ .