Easy Analog Circuit: low pass RC circuit

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.

Wednesday, 30 July 2014

Peeping into AC1 class room and low pass filter

Analog Circuit I is included in B Tech Syllabus under M G University for Electronics & Communication Engineering(EC 010305) and Applied Electronics & Instrumentation Engineering(A010 305).

Hope this blog will show an easy way for Analog Circuit I for u S3ans.

Its better now we think about module 1 only.

Module I includes study of RC Circuits,Rectifiers,Voltage regulators and clipping & clamping circuits. RC circuits and rectifiers are familiar in your first year(OR a nightmare, no problem i will help you to give a twist.Together we can lit a torch into that darkness ).

You have to ~~study~~ chew the following things -yaa,"The Syllabus"
('The' is for showing respect - not for comparing it's size with that of 'The Sun' and 'The Earth").

Section I

1. RC circuit:
   Response of low pass and high pass RC circuits to –
       sine wave, step, pulse and square wave.
        Properties of such output waves - Tilt and Rise time
Differentiator, Integrator
2. Rectifiers:
Analysis of Rectifiers – Half wave, Full wave, Bridge
Analysis of Filters – L, C, LC, π
3. Clipping and Clamping circuit
4. Voltage Regulators :
5. Zener voltage regulator
Transistor series (with feedback)
Short circuit and foldback protection
Shunt voltage regulators
6. Small signal diode model for low and high frequencies

RC Circuit

NEED some Refreshments

1. Resistor (R)- If there is a resistor in a circuit, there is a voltage (voltage drop) across it. That voltage drop is in proportion to the value of resistance. Certainly it is in direct proportion because V = RI.

2. Capacitor(C) – Easily allows AC (high frequency) but opposes DC (low frequency). In other words, Capacitor offers low resistance to AC and high resistance to DC. This resistance is represented as Xc.

Just for pushing the above fact into brain, watch this:

3. Voltage division

     V = $10v\frac{{20\Omega }}{{20\Omega + 10\Omega }}$    = 6.67v

      The voltage divided across 20Ω and 10Ω in 2:1.
       Voltage across 10Ω = 10-6.67 = 3.33v

First note: A simple RC circuit can be act as a passive filter ( don’t afraid, here passive means ‘no transistors’ and filter means it filtered out some portion from input and fed the output side with the rest). It may be either a low pass filter or a high pass filter depending upon the positions of R and C relative to each other.

Low pass RC Circuit:

We all are equipped with a low pass filter in our body. Guess which one is that organ...Our ears- rejects all the frequencies above 20K Hz. SHHH....accurately it is a band pass (20Hz to 20 KHz). Great scientists always adopt some assumptions, then why we can’t???. We also can take as 20Hz is near to zero. So 0Hz to 20 KHz and low pass.

Let’s Start !!!

Consider a first order RC circuit. \\first order = one pair of R and C \\

Circuit diagram:

Working: How it act as a low pass filter?

The reactance of capacitor is Xc = $\frac{1}{{2\pi fc}}$ \\ reactance = resistance of a capacitor \\

f – frequency of input AC (In India frequency of AC is ?? 50Hz) c - capacitance of the capacitor C (eg: 10 µF, 50 µF etc)

Equation reveals that Xc is inversely proportional to frequency f. Now just think that we have options to change the frequency of input AC. \\ Its very easy with an equipment called Signal generator which is in your college lab \\

At first we give a low frequency input signal, look into equ, what happens to Xc? Xc become very large. \\ Just imagine the case when ‘f’ is nearly zero, Xc is as large as infinity. Then you compare Xc and R.R has a fixed and finite value. So Xc >> R\\ .

Then consider voltages across Xc and R. Which one is greater? Yaa, that across Xc. Therefore voltage drop across the capacitor is very large compared to voltage drop across the resistor. In a similar way, at high frequency, Xc << R. So voltage drop across R is greater than that across Xc.

Summarising : for low frequency AC or for DC Vc >> VR larger output voltage
for high frequency AC Vc << VR smaller output voltage

\\ say, we are changing the frequency of input AC from 10Hz to 1,00,000Hz (but keeping its amplitude fixed eg:10v). For low frequency range, output voltage is nearly 10v(may vary depends on filter type). For higher frequency range, output will be much less than 10v.//

From the above statements it is clear that this circuit produces a high voltage for low frequency inputs and it attenuate the output voltage for high frequency input signals. That why it has the name ‘low pass filter’.

Equation for output voltage

Vout = Vc = $\frac{{VinXc}}{{\sqrt {{R^2} + X{c^2}} }}$ Vin indicates amplitude of input signal

Frequency response of first order Low pass filter: