Integrator

Lets say I have a input voltage , where I want that , the out put should be such that ,

Means that the output is the integral of in put .

If we observe the following circuit, which have a capacitor connected to it

The output voltage is something like this

Similarly we can have one more case, where a resistor Can be introduced along with capacitor .

Now in put voltage = Vi

Out Put voltage is Vo

Voltage across Resistors is

Vr= IxR

Where I values is

Plugging in the value of I in equation, the voltage across resistor will become

We know that a voltage in any given circuit is voltage supplied , which at all instant is equal voltage absorbed by the elements. For the above circuit we can say that

We know that power dissipated by a Resistor is very large as compared to capacitor , which have a out put of . For just for the sake of easy calculation , we can make the assumption thatSo we can ignore Which gives us a new equation

Now finding

Creating a Integrator Circuit using Op Amp

For creating a op amp , we need a current derived from , and than current should pass from a capacitor , which should be somehow connected through a Op – Amp

So lets say we have a following configuration

,The current passing through the resistor will be equal to

Now as we need to make integrator, so we will be needing a capacitor . So a capacitor is connected in following configuration if a integrator purpose , as the current is in series with capacitor , so the same current will pass through it .

Answer

We know that current cannot go inside amplifier due to high resistance, so all current will pass through conductor.

Now putting values . For finding vo we apply KCl , which will give us

Differentiator

A differentiator is a circuit that can differentiate, the input signal .

, We know that capacitor have following function

Building Blocks For Differentiator

Lets se we have the following Setup , a where a resistor is connected to a inverting terminal in following configuration .

The value of vo for this will be

Now for the same , configuration lets connect a capacitor in it

We know that the V across resistor is V=I R ., If we can find the value of I , we can simply find the out put , which is equal to

We know that current across a capacitor is

Putting value of i, in equation