Active Low Pass Filter and Active High Pass Filter Explained

In this video, Active Low pass Filter and Active High Pass filters have been discussed.

What is Active Filter:

The active filter is the electronic filter which is designed using the active components like Op-Amp and transistors.

So, in this video, active filters which are designed using op-amp have been discussed.

Advantages of Active filters: (Disadvantages of passive filters)

In case of passive filters, the gain will be always less than 1, or in other words, the output will be always less than the input.

Also, when multiple stages of such passive filters are cascaded then the overall output will be attenuated severely.

The cut-off frequency of such passive filters also depends on the load. Depending upon the load, the cut-off frequency will be shifted slightly.

All these problems can be overcome by using the active filters.

So, active filters not only provides the gain to the input signal, but they also act as a buffer and isolate the load from the input filter circuitry. (Particularly, when they are designed using op-amp)

In this video, active low pass and active high pass filters have been designed by using the non-inverting configuration of Op-amp. And at the end, it is shown that how to design these filters using the inverting configuration of Op-amp.

The timestamps for the different topics in the video is given below:

0:22 Advantages of Active Filters

1:38 Active Low-pass filter design using Op-Amp (Non-inverting)

9:40 Example on Active Low Pass Filter

10:42 Active High-pass filter design using Op-Amp (Non-inverting)

14:30 Active filter design using Op-amp in inverting configuration

15:32 Example on Active High Pass Filter ( Try it yourself)

This video will be helpful to all students of science and engineering in understanding and in designing the active high pass and the active low pass filters.

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Music Credit:

[abpdev]

When i start working, i will definitely donate something to this channel!

[Math.Lampron.official]

The little pause between words is awesome... The content is even more awesome 👍

[fleabag]

For those who are not able to understand the last example,
We will start off by noticing that this is a Non Inverting OPAMP therefore, Av=1+(Rf/R1)
We have already derived that Vout/Vin= Av*{(f/fc)/sqt[(f/fc)^2+1]}
It's always better to assume the value of Capacitor because it's not feasible to get capacitor of obtained value in the market. We can easily obtain Resistor value using a POT. Hence, I assumed C=10nf;
We know, Fc= 1/(2*pi*R*C); Fc= 5*10^3 (given);
Get value of R; R=3.18KHz.
Now, It is given that Vout= 1V, Vin= 100mV; Hence, Vout/Vin=10 and f=10KHz (given)
Substitute all the values in the eq:
Vout/Vin= Av*{(f/fc)/sqt[(f/fc)^2+1]}
GET VALUE OF Av; Av=11.18.
Av=1+(Rf/R1); Rf/R1= 10.18
Assume Rf=10.18KHz
Therefore, R1= 1KHz.

[TronicYaka]

Great work sir! Thank you for sharing knowledge with us. 😍

[aadisingh]

16:32 now i can understand why this capacitor is there in the subwoofer circuit
Thanks

[zarafshan]

you are the teacher explained better than nptel

[vakhariyajay]

Thank you very much. You are a genius.👍👍🙏🙏🔝🔝👌👌

[PetakyahBuckley-htiz]

What is the value of the w(omega) variable. What is its value? Is it always constant? Particularly at 5:49.

[vsilte]

Thanks for the great and simple explanation!

[lifephilic]

Thanks. This is helping me in the revision.

[midclock]

It's strange how you say divided by, it took me 3/4 of the video to get the word, but I'm not English mother tongue :)

Thanks for sharing this theory!!

[shivg]

DESIGN FOR HIGH PASS FILTER ;
Let C=0.1uF ; (Easily available and compatible for the practical circuitry)
Given : Gain, A=10.

Formulas; a) Gain, A=1+(Rf/R1)
b) Cutoff Frequency, fc=1/(2*pi*R*C)
We get,
C=0.1uF
Rf=9kHz
R1=1kHz
R=318.3 Ohms= 318 Ohms

[nguyenangtuandung]

Very heavy Indian accent but the content is great, thank you for your sharing

[sheetalmadi]

2.and how loading effect is reduced by using capacitor in the feedback circuit? Please tell sir🙏

[piyushgupta]

Sir if I obtain the value of C as 3.18 nF assuming R = 10 k ohms, is it correct ???

[sanjuhazra]

R = 318.3 ohm
C = 0.1 micro farad
R1 = 1k ohm
Rf = 9k ohm

[kazemzahedi]

Hi, thanks for your great video. What is the difference between non inverting op amp with no feedback resistor with the one with feedback resistor. Are thy both stable and reliable?

[Jono.]

That last problem was a fantastic problem - thank you for suggesting it.
I looked up capacitor values on google and used the first one I found -> C = 5.6pF
Using this value I get a resistor in the filter to be R = 5.68E6 Ohm.
This produces a voltage at the op-amp of .0894V.
Finally, I chose Rin to be 1000 Ohm. This gives me a Rf of 10, 185 Ohm.
I believe I am correct and this really gave me a good understanding of designing these circuits.

[divyeshcvora]

Great work Sir..
But there is one mistake between 13:30 to 13:37 in voice....

[Saikumar-kblf]

Ur awesome video made me to get following values of the problem given by you:
Rf = 10235.9 ohms, R1 = 1000 ohms, R = 9645.754126 ohms, C = 3.3nF
please judge my answer & provide your feedback if it's wrong