Monday, 11 August 2014

RC Integrator



How Low pass RC filter work as an Integrator ?- It is explained here.

Low pass RC circuit is explained in detail here - Low pass RC filter , plz visit. Then move to next course - but have same ingredients -that's Integrators . Oops ...Integration. Is this our AC1 or Maths??? ....

Arranging the table for serving...

In electronics also there are Integrators and differentiators because of the following reasons. Many circuits need signals other than sine wave, like square wave, cosine wave, triangular wave, pulses etc. But the basic signal available as input to any device is our Alternating Current, which is sine wave. Square wave can be produced easily either from sine wave or from a dc voltage with the help of some simple circuits. By integrating and differentiating these two waves(square and sine), it is  possible to produce other waves also. In order to realize integration and differentiation, we are   using integrator and differentiator circuits.

Integrator

An integrator produces an output which is proportional to integral of its input.




Circuit diagram:



AC – AC source that provides the input - Vin
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.



Low pass RC circuit itself is acting as an integrator provided it obey one condition. Lets check that condition.


Equation for output voltage    Vout = Vin\frac{{1/CS}}{{R + 1/CS}}      
                                       1/CS –reactance offered by C in terms of Laplace transform

                \frac{{Vout}}{{Vin}} = \frac{{1/CS}}{{R + 1/CS}} = \frac{1}{{1 + RCS}}

Laplace transform s=jw and w=2πf (w-omega)

                   \frac{{Vout}}{{Vin}} = \frac{1}{{1 + RCjw}}
if we are adopting a condition as wRC >> 1, then , in denominator 1+jwRC is nearly equal to jwRC. Then equation become

                  \frac{{Vout}}{{Vin}} = \frac{1}{{jwRC}} = \frac{1}{{RCS}}

Re arranging,

                  Vout = \frac{1}{{RC}}\frac{1}{S}Vin       Here 1/RC is taken as a constant


Taking inverse Laplace transform,

                 Vout = \frac{1}{{RC}}\int {Vindt}

//Those who are not familiar with Laplace-- while converting integral into Laplace 1/s is adding with the equation. During inverse Laplace, 1/s indicates an integration.//


Above equation shows that Vout is proportional to integral of input voltage. So if we have a low pass RC circuit which obeys the condition wRC >> 1, it will act as integrator.


//The product RC is known as time constant(T). With the value of RC there will be changes in output waveform.//


In order to achieve a good integration, the following conditions must be satisfied:
  • The time constant RC of the circuit should be very large as compared to the time period of the input signal.
  • The value of R should be 10 or more times larger than Xc.


Output for different input waves:


1. For Step input:










 2. For square wave input


Three output waveforms are shown. These are with increase in values of time constant ‘T’.

Applications:
Use to perform integration in certain electronic circuits.
Use to convert square wave into triangular wave.

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