Showing posts with label differential. Show all posts
Showing posts with label differential. Show all posts
Thursday, May 26, 2011
Design a Subtractor
A basic differential amplifier can be used as a sub-tractor. We can get the difference of two input voltages in the output of op-amp as output voltage. The circuit diagram of a basic differential amplifier is drawn below.
This is a linear bilateral network. So, applying super position theorem, we can find the output voltage equation.
Let, assume that only Va is applied and Vb is short.
Wednesday, May 25, 2011
Electric Analog Computation
Electronic analog computation is such kind of electronic computation in which basic analog computing elements such as adders ,integrators ,multipliers, comparators etc are used to solve any desired equation such as differential equations etc .It is the basic concepts of analog computer.
In this a differential equation is solved by electronic analog computation. Let a differential equation be :
D2v+k1Dv+k2v-v1=0……………….(1) Where, k1 and k2 are constant terms.
In the starting I assumed that D2v is available in the form of a voltage .Then by means of an integrator I will get the voltage proportional to Dv. A second integrator gives the voltage proportional to v .Then an adder gives –( k1Dv+k2v-v1) From the equation it is equal to D2v
and hence the output of this summing amplifier is fed to the input terminal ,where I had assumed that D2v was available in the first place.
The integrator 1 has a time constant RC=1s, and hence its output at terminal 1 is –Dv .This voltage is fed to a similar integrator 2 and the voltage at terminal 2 is +v. The voltage at terminal 1 is fed to summing amplifier 1 which gain is 1 and in the output terminal 3 I get + k1Dv- v1.
where k1=(R/R1).At the end the output of terminal 2 and 3 are fed to summing amplifier 2,from where I will get D2v= - (k1Dv+k2v-v1) at terminal 4.
Fig1.1: Electronic analog computing circuit for calculating a differential equation .
The Differential Input and Differential Output
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 ]
)VX - Vy –[(1+ )Vy - Vx ] = (1+ ) ( VX - Vy ) +
( VX - Vy )
) ( VX - Vy ) +( VX - Vy ) = ( VX - Vy ) (1+
)
)Design
To design a input and differential output amplifier, taking a differential output of at least 3.7V and the differential input Vin =10V.
We know,
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 .
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