Active Low Pass Filter and Active High Pass Filter Explained
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- čas přidán 26. 08. 2017
- 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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When i start working, i will definitely donate something to this channel!
when are u going to start..??
@@rohanmarkanda9933 I so not know
@@abpdev god bless u
Ye sab kehne ki batein hai baad me sab bhul jate hai
R u alive bro?
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.
For the last two resistors, you used Hz, and not ohms.
where did u get the value of R as 3.18KHz ?? and what do u mean by get value??
he calculated using the formula @@SUBHANKARDEY-ji6xs
god bless u bro. thanks!
The little pause between words is awesome... The content is even more awesome 👍
Great work sir! Thank you for sharing knowledge with us. 😍
Thanks for the great and simple explanation!
you are the teacher explained better than nptel
Thanks. This is helping me in the revision.
as usual, great narrative, thanks
You are the best! thanks
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.
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
actually it is wrong, when the f is larger than f_c, f/f_c square will be very small and you can assume it is zero so that in low pass filters |V_out/V_in| = Av, but in high pass filters in numerator there is f/f_c term so we cannot assume v_out/ v_in = Av. In this question Av is unknown and and V_out/V_in with f and f_c values known so you can find Av then it will be equal to the 1+Rf/1 ( do not forget f= 10kHz, fc= 5kHz and C value should be assumed smt and with this you can find R, after finding Av, Rf/R1 will be equal to some number and regarding that number you can choose Rf and R1
Thank you very much. You are a genius.👍👍🙏🙏🔝🔝👌👌
For Voltage Follower case 2:31 , if I write it in the equivalent transfer function I should get Vo/Vin = A/(A+1). Shouldn't the gain A be infinity in order to obtain the Vo that is close to Vin?
Thanks!
I really appreciate your support !!
16:32 now i can understand why this capacitor is there in the subwoofer circuit
Thanks
Great work Apu!
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!!
Thank you sir you are a life saver
Great video.
AMAZING thank you
So nice!!!
great video thank you
You are doing such a nice work. The content in the video can understand easily. Could you please explain about the convolution of DTS and CTS? Thank you so much.
Thank you Chathura. Yes, I am planning to make videos on signals and system. But it will take some time. Maybe in a couple of months, you may see videos on signals and system and many other subjects.
sir please make an video on loading effect
In the output voltage expression of non inverting amp...we should consider voltage at V+ right?? Which in this case comes out to be 66.67mV and design values comes out to be r1=1k and rf=15k?
Is it correct?
Excellent video
Hi, in my lecture notes, for active high pass filter , instead of the inverting terminal having the resistors connected to the ground, it is connected to Vcc/2 and one of the supply terminals of the op amp is grounded. Can you please explain why
sir at 15:12, should you consider the total impedance of the filter to calculate the gain? so that Av = -Z/ R1, where Z is the impedance.
R = 318.3 ohm
C = 0.1 micro farad
R1 = 1k ohm
Rf = 9k ohm
Very helpful for viva
if i wanna design a hight pass filter for the adc for audio do i need to pay someone to do it or buy a din rail and tweak it please help
nice video sir....
Fc=159HZ in Active LPF. What is the output if we connect input of 10KHZ.
Great work Sir..
But there is one mistake between 13:30 to 13:37 in voice....
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
That's correct. 👍
ALL ABOUT ELECTRONICS thank you! I learnt because of u an important topic of electronics....subbed
do you have video about Frequency Response of FET?
thank you for this video series
the gain formula of active LP filter has two frequencies f and fc, fc is 1/ 2*pi*C*R, how will calculate for f?
wow.......good lesson
It's just amazing
R =3.18kohm,C=1nf,Rf=9kohm,R1=1kohm. Will it work for given requirements?
For the circuit given at 15:26
If we were about to calculate the gain for that circuit in terms of Ron do we say Zin?
Do we add the impedance of the capacitor and Rin together or does it just stay Rin?
thanks sir. but i think how you used the voltage divider rule to obtain the total input impedance should be elaborated.
..Ansah Dominic, accra ghana
.
great course
Gain is -Rf/R1 for inverting opamp.
Yes bro, -Rf/R1 is for inverting amplifier....
Thanks bro
Very heavy Indian accent but the content is great, thank you for your sharing
indian accents are beautiful.
Sir if I obtain the value of C as 3.18 nF assuming R = 10 k ohms , is it correct ???
Yes, it is correct for the given cut-off frequency. Theoretically your calculation is correct. But practically suppose if you design this filter, then I doubt you will get 3.18 nF capacitor. So, it is good practice to choose C first and calculate R accordingly, because it is easy to adjust R using Trim POT (variable resistor). Your answer is correct but just keep this little thing in mind for the future designs. Good luck.
@@ALLABOUTELECTRONICS sir, i would like to ask how do we decide the value of C?
I choose c= 0.01uF, then i got R=3.18kohm from the law of cut off frequency (fc), Here input is 100mv is equal to 0.1v, So gain is Av=Vo/Vin = 1/0.1 =10, So we have to make the gain 10. this is non inverting op-amp so we know Av= (1+Rf/R1)= 10 if i choose Rf= 90kohm, R1=10kohm.
will it right?
thanks
In active low pass rc filter in high frequency it act as voltage follower, then it can pass high frequency also,how could it would be?
I have that same doubt!
Sir gain =1+Rf(/R1)=1+(4×10 Power 3 /
You could have given values also directly simply 5 as the answer...
Sir, at 10:33 when a supply voltage of 1V is there, you write the output will be 5V but sir wouldn't the gain of the rc filter be multiplied to it to give the final output?
Yes, it's true. The gain of the RC filter will also get multiplied. But as the cut-off frequencies 159 Hz, so, at 1Hz, there will be hardly any attenuation. And Vo is approximately equal to 5V.
I hope it will clear your doubt.
wow wonderful explanation.. but please reply that what is difference between in active low pass filter in non inverting configuration and inverting configuration.. ? please
The only difference in the output phase. In the inverting configuration, there will be an additional 180-degree phase shift in the output.
Correct me if I’m wrong, but I think you made a mistake at 10:19; the capacitance C = 0.1uF which is 0.1 x 10^-6, however when calculating the cut-off frequency, you used 10^-7 as your capacitance value - the answer should be 1.59KHz
0.1 × 10^-6 = 10^-7. Isn't it ??
That's why for simplicity, I have taken 10^-7 during calculation. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you for your response, I appreciate you helping to clear this confusion. 0.1 x 10^-6 is the same as 1 x 10^-7, therefore it would be 10^-8, wouldn’t it?
can you please elaborate why are we avoiding the use of buffer in this circuit at the input
We are assuming that the input is applied to the circuit/filter using the ideal frequency source. But if the input is also coming through another circuitry then it is better to use the buffer at the input as well.
Please check at 7: 25
+ALL ABOUT ELECTRONICS fine thnx...
The numerical done for finding the cutoff frequency, in 10:40, why aren't we taking the Rf value in the cutoff Frequency, instead we took R value... Please explain...
In the example, the op-amp resistors Rf and R1provides the gain. In fact, if you find the frequency response (Vo/Vin) then you can see that, (1+Rf/R1) is the gain.
I hope it will clear your doubt. If you still have any doubt, let me know here.
Hey, i like your video, but there have a small mistake at 3:14 ,u said 1 divided by rf +r1, for non inverting op amp, hopefully the subtitles is correct .
Yes Chan, you are right, that should be 1+ Rf/R1
Can you make another video just non inverting low pass filter?
What is the value of the w(omega) variable. What is its value? Is it always constant? Particularly at 5:49.
w (omega) is the angular frequency. Its value depends on the input frequency(f) to the filter, and w = 2*pi*f.
c=47 uF, R=.677 ohm, rf/r1=21.36/1
Assuming c=1uf
Gain Af=10
R=31ohm
Rf=340ohm
Sir, please clarify my doubt at 15:13 , in this circuit diagram, will both the resistors decide the cut-off frequency?
It will depend on R1. Rf will decide the gain.
@@ALLABOUTELECTRONICS Thank you, sir❤
in non inverting input signal are applied with the negative terminal of opamp....and gain are =-rf/ri
It is exactly the opposite of what you said.
inverting through positive terminal
1.@8:54 as for high frequencies too we are getting the output then,how is that a low pass filter?
2.and how loading effect is reduced by using capacitor in the feedback circuit? Please tell sir🙏
In the filter, what matters is the difference between the passband and stopband. In case of this LPF, let's say, at DC the gain is 100. (or 40dB). And the gain at high frequency is close to unity. (or 0dB).
So, the difference between the passband and the stopband is 40dB. That means the filter attenuates the high frequencies by 40 dB. In any practical filter, you will never get complete rejection in the stopband. That means the output will never be zero in the stopband. That's why when you see the filter characteristics then as a designer one used to see, how much attenuation in dB is provided by that filter.
I hope, it will clear your doubt.
At 9:20 you did not explain where that formula comes from. Since the least value we are getting for Vout = Vin how does it relate to the cutoff frequency that drops to 1/sqrt(2) * Vin is not clear how you got that formula just changing R to Rf.
can anybody explane to me, how we find the H(s) of the inverted active low pass filter? thank you.
Only we have multiply the gain of inverted opamp to previous formulae
Jst try writing the equation for a general opamp circuit using opamp and take parallel Rf and C equivalent to Z and then later simplify or you can do it bytaking Laplace transform also.... If futher explanation needed then do ...comment
in active low pass filter at 9:07, if frequency is high gain is 1. then it is still giving amplitude equal to input then how did it become low pass. If should have reduced the output. please correct me
Here, we are amplifying the input signal. So, at lower frequencies the amplification will be more, and at higher frequencies, the amplification will be less. And very high frequencies, the output is the same as input. That means higher frequencies are not getting amplified at all. For example, if your input is coming from a microphone (where the signal level is mV or even less than that, then higher frequencies will not get amplified and will not be able to be heard in the speaker). So, still, this circuit works as a low-pass filter with amplification. I hope, it will clear your doubt.
Values are C=10 micro F , R1=1kohm, R= 1k ohm , Rf= 9.1k ohm
To prevent change of cutoff frequency in low pass filter ,you set the capacitor in parallel with feedback resistor, but you do not do anything like that for the active high pass filter. My question is If a circuit is attached at the high pass filter terminals, will that change the cutoff frequency or not? Or, would we need to add a buffer for the active high pass filters? Please let me know.
It can be connected directly. Because the op-amp itself will act as an buffer. And in fact, that is the advantage of the active filter.
@@ALLABOUTELECTRONICS thanks for replying back. The OP amp acts like a buffer for the load, right? But I am talking about at the vin end. For the low pass filter, you say that at the vin end if there was another circuit connected, the cutoff frequency of the low pass filter might change. That's why you put the capacitor parallel to feedback resistor. However how do you prevent this if a circuit with impedance and voltage were to be connected at the vin end of the high pass filter? For the high pass filter, you did not change the configurations. Please let me know. I would really appreciate it.
If the input to the high pass filter is also coming through the op-amp, then there won't be any issue. For example, if the input to the active high pass filter is the output of the active low pass filter, then it won;t affect the overall response.
But if is coming through some other source, its good to use the buffer before applying it to the input.
I hope it will clear your doubt.
At 9.00 in both case(w=0,w=infinite) we are getting output,then how it is a low pass filter
At low frequencies, the circuit provides the gain of 1+Rf/R1, while at high frequencies, it acts a buffer. So, it is not amplifying the high frequencies. So, in a way, it is low pass filter. Because as the frequency increases the gain provided by the circuit will reduce. I hope, it will clear your doubt.
* At 9:05, if gain is 1 then, how is it eliminating that frequency? * Isn't it simply allowing the high frequency to pass with the unity gain and if so, then how is it behaving as a filter?
What I mean to say here is, it amplifies the low-frequency signals and it passes the high-frequency signal as it is.
And if you plot the frequency in dB then for the low frequency it provides gain and at the high frequency, it does not provide any gain.
This is the case in many communication system, where the signal and the unwanted frequency signals are very low (in microvolt) and you want to boost the desired signal with the gain.
I hope it will clear your doubt.
Sir I obtained r=31.8ohm by assuming c=1uf, now how to find R1 and Rf ?????
Bro, for a butterworth filter Gain is 1.586, ie 1+(Rf/R1) = 1.586 =》Rf/R1 = 0.586, Rf =0.586R1, let R1= 1K, Rf = 586 ohms
I chose Rf = 9 kΩ, R1 = 1k Ω, R = 3.18 kΩ, C = 10 nF
That's correct design👍
Shouldn’t we be taking into account Vout/Vin = Av * (f/fc)/sqrt(1+(f/fc)^2)? In the example we are supplying a 10 kHz signal and obtain a ratio of Vout/Vin = 10. Plugging in f/fc = 2 to the above equation, I determined Av = 11.18 so I chose a 10.18 k for Rf and 1k for R1. I chose the same R and capacitance values as above.
Yes, that's right Christian. We also need to take into account the Vout/Vin = Av * (f/fc)/sqrt(1+(f/fc)^2).
Your calculated values are correct.
How did you find the values for R and C
Please can you tell me how did you find the value of R ?
At 0：13 A/2^1/2=3 db. At that point on Y axis
1)If my input is coming from another circuit, how the impedance of the that circuit will effect the cut-off frequency of normal active LPF design.?
2)How Xc=0 will make it unity gain amplifier ?
1) in the first case, I am referring to the output impedance of that circuit from where the signal is given to the filter.
2) When Xc= 0, Rf will get short-circuited. So, the output will get short-circuited with the inverting input terminal.
Hence, it will act as, unity gain amplifier.
I hope it will clear your doubts.
@@ALLABOUTELECTRONICS but as there is short circuit there will be no current flowing in the circuit also invertting terminal is grounded due to which Vin will become 0 so V0 will be zero
In Active low pass filter , to calculate cut of frequency , 4K ohm is not feedback resistance . why you used 10k is feedback resistance ?
Because the filter is formed by 10 k resistor and 0.1 uF capacitor. The other resistors (4k and 1k resistors) provides the gain to the filter. I hope, it will clear your doubt.
I use Rf = 1k ohms & R1 = 1 k ohms and C = 0.1 micro F.
So R becomes 3.183 k ohms..
is this right?
I think there is mismatch in gain of circuit by using your values.. (1+Rf/R)=(1+1/1)=2 i.e Av is 2 but as per question output is 1V and input is 100mV.. So gain should be.. 10..... Hence your value.. Rf/R must be 10-1=9 ...Do mention if I am wrong
If Rf & R1 be 3k & 330 ohm respectively ... The gain Av will fulfill the value as 10 ... To determine the capacitor in this circuit, if we consider the value of R= 15 k ohm.. We will get the value of capacitor as 2.12 nF ... Practically , during my lab work.. I have found all these type of resistors
the value of C is 10 nF because the gain is 10 (Vout/ Vin)
👍👍
Sir I don't came to know how to choose these values.please sir make another video to clear my queries
Hello Anjali, you can refer my video....
Superb
You should have derived phase shift too.
Sir, what is cutoff frequency in active high pass filter with inverting input? You have talked about the non inverting input cutoff frequency only.
The cut-off frequency will remain the same.
@@ALLABOUTELECTRONICS ok sir
@@ALLABOUTELECTRONICS
You mean 1/2piRfC?, asking because in non inverting case(active high pass filter) there is a resistor R but in inverting there is no R and in inverting case fc = 1/2piRC.
@@sumankumar_phy0987 At 15:19, the cut-off frequency is 1/2Pi*R1*C.
@@ALLABOUTELECTRONICS Okay sir. Thanks for your time!
I choose R1=1k ohms so i got Rf as 9k ohms, C=15.9F,R=2k ohms
feed R = 9k ,R=1k and filter parameters R=6.33kC=0.01micro farad
Please check the values again buddy.
Try this values, R = 3.18 kΩ, C = 10 nF, Rf= 10.18 k and R1= 1k
@@ALLABOUTELECTRONICS i think the guy is right because gain in (1+Rf/R1) not (-Rf/R1) so if we take your values gain will be (1+10.18/1)=11.18 (a total gain of 10 is required according to the problem)
You keep on
Thanks, but what is the transfer function useful for?
From the transfer function, you can find the frequency response and the stability of any circuit.
@@ALLABOUTELECTRONICS Thanks, so I can have a frequency of response of 1Khz even though my input frequency is 50Khz or something like that really?.
Thanks a lot. You are the best teaching.
@@ernestofavio6735 If the input signal has frequencies from 0 to 50kHz, then the output can have frequency of only 1 kHz. But if the input has only discrete frequency 50kHz then at the output we can't have 1 kHz frequency. ( I am talking from filter perspective).
The input should have that particular frequency, which we want to recover through the filter.
R=10k , C=3.183k ,Rf= 1k , R1=98.23 , Av=11.180
According to transfer function of low pass filter
If input is 1V, having freq of cutoff freq with gain of 5 then output should be 5/√2V ,fc not 5V
Reply me if I'm correct!
Thank you for ur great lessons!!
Yes, correct.
at 9:15, why you say when Xc =0 ( infinity freq), Vout = Vin is a low pass filter? isn't low pass filter has zero gain( Vout =0) at high frequency?
Well, it is not necessary to be a zero gain. It should have certain pass band and certain stop band and in the pass band it should provide sufficiently higher gain than stopband. ( e.g 40 dB or 50dB)
@@ALLABOUTELECTRONICS i see. thank you for the quick reply!!! i really like your channel!! thank you for providing all the instructions
C=3.18nF ,R =10kohm
Great video bro.However,how have you got both the values of Rf and R1??
I hope you are talking about the last example given for the exercise. If so then the transfer function of the high pass filter will be, Vout/Vin = Av * (f/fc)/sqrt(1+(f/fc)^2). Now here the operating frequency is 10 KHz and the cut-off frequency is 5 KHz. So f/fc will be equal to 2. Now, at 10 KHz, we require the gain of 10. So, Vo/Vin should be equal to 10. So, if you put all these values and simplify it then you will get the value of Av around 11.2. Or the ratio of Rf and R1 should be equal to 10.2. You can choose R1= 1K and Rf as 10.2 K to design the filter.
I hope it will clear your doubt.
ALL ABOUT ELECTRONICS thank you very much
@@ALLABOUTELECTRONICS bhai ek doubt h pls solve...
Bhai Vin = 100 mV & Vo = 1V he... So isse hum assume kar skte he k Gain around 10 times hua so by default Av= 10 hona chahiye.
So Av = 11.2 kyu aa rha he??
Because with gain the transfer function of the filter would be [(w/wc) / sqrt ( 1 + (w /wc)^2) ] ( 1 + Rf/R1)
And this entire terms should be equal to 10. If you put w = 10 kHz and wc = 5kHz then the first term would be 2/sqrt (5) x (1+Rf/R1) and this entire term should be equal to 10.
So, from this, we can say that the gain of the op-amp should be equal to 11.2
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yeah thanx bro.. Now I get it. Really helpful
Sir, at the combined response at 14:24 the amplitude of a curve shoulde be same .
The combined response is the shaded region. (14:29). The individual gain response (the op-amp and the high-pass filter) can have different amplitude. But the shaded region is the overall combined resposne. I hope, it will clear your doubt. If you still have any doubt, let me know here.
@@ALLABOUTELECTRONICS thank you sir you're my stress reliever and the reason of my confidence, we all are very lucky to have teacher like you Tku again sir 🙏🙏🙏 .
Sir, At higher frequencies , low pass filter must reject or attenuates the input signal But why op amp acting as voltage follower passing the high frequency input as output?????????????????
Would you please mention the timestamp where you are referring to in the video ?
@@ALLABOUTELECTRONICS 08:47 sir
@@Devil-dn2ew At high frequencies, the filter provides unity gain while at low frequencies it provides the gain of (1 + Rf/R1). So, basically if you see the frequency resposne of the filter then at low frequencies, there is 20dB or 40 dB gain while at high frequency there is a 0dB gain. Let's say yor input signal to the filter is 100 uV. And the gain provided by the filter is 100. So low frequencies will see the gain of 100, while the high frequencies will see the gain of 1. That means the detector or the circuit after the low pass filter would only be able to detect the low frequency.
The thing to look out in the filter design is how much dB difference is there between the pass band and the stop band. I hope, it will clear your doubt. If you still have any doubt then let me know here.
Thank you 😊
Got it!!
sir , at 7:00 why did u multiply Av with the 1/ root(1+(f/Fc)^2) ....i dont get it
That is the transfer function of the low pass filter. Without the filter, for DC, the gain of the op-amp will be Av. But because of the low pass filter, the gain of the entire circuit is frequency-dependent. It will change according to the frequency response of the low pass filter. That's why it is multiplied with the transfer function of the filter.
I hope it will clear your doubt.
How exactly does the buffer separate the input from the output
The input impedance of the buffer is very high and it has a very low output impedance. Meaning that the first stage won't see the impedance of the next stage. And likewise, as the next stage is seeing a very low output impedance of buffer, it will not be affected by the impedance of the first stage.
I hope it will clear your doubt.
Correct me if I am wrong, so the high input impedance basically supplies the voltage applied at the input to the opam and the low output impedance is present to make sure all the output voltage is applied to the load. So now there's very low impedance on the output side thus this impedance won't affect the next stage. Right?
Also, how does changing the position of the capacitor help?
Yes, That's correct. I didn't get your last question about tchanging the position of capacitor.
At around 8:00 mins into the video, you mention that to avoid using another buffer the capacitor should be connected in the feedback path. How does that work?
In that circuit, the input is applied at the non-inverting terminal of the op-amp. While capacitor and resistors are connected between the output and the inserting end of the op-amp. So, the input source signal and filter circuit are isolated from each other.
Me too bro
sir at around 8 min,u said changing the position of the capacitor will allow us to operate the circuit with out the use of another buffer at the input and immediately u explained it by taking the extreme values of 'w' as zero and infinity. Now my question is for both the values of 'W=0' and 'W= infinity' we are geting output. the only difference is,at 'w=infinity' the output is having gain 1 and at 'w=0' the output is having some gain. but all the frequencies are being allowed.Then how it acts like a low pass filter?
It's not always about the absolute gain in the stop band. Many times Its about the relative gain in the passband and stopband. Let's say at w= 0 if Rf/R1 is quite large (let's say 1000, considering input signal in mV) then at one extreme gain is 1000 and at w is equal to infinity, the gain is 1, which is 1000 times less. So, the circuit will amplify low-frequency signals by larger gain and high-frequency signals by smaller gain.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yes sir. Thank you so much. Soo kind of u.
bro your method of explanation is great i want to make 7 band music equalizer circuit and i want to separate the 7 different frequency of the audio signal such as low frequency filter up to 150 hz and band pass from 150 to 500 Hz and so on can you tell me an accurate formula of resistor and capacitor in those so that i can create low,high,band pass filters. will be waiting for your reply
For the first order Low pass filter, the formula is very simple f= 1/(2*Pi*R*C). You can choose let's say C=0.1 uF and find the value of R. But this filter will be a very basic filter. And it will not have a sharp cut-off, means after 150 Hz, the attenuation will not be zero. Rather amplitude will fall slowly. And it might have frequency components above 150 Hz.
So, you might need to higher order filters may be 4th or 5th order Butterworth filter. So, I recommend you to use analog devices filter wizard tool.
You can select your cut-off frequencies and gain. It will give you response and the component value and even suggest you the op-amp IC.
www.analog.com/designtools/en/filterwizard/
Using this even you can design the bandpass filter.
I hope it will help you.
ALL ABOUT ELECTRONICS thanks you so much for your idea and explanation.
ALL ABOUT ELECTRONICS I was really in need of something like this thanks you so much
Rf- 9k ohm
R1 - 1k ohm
C - 0.1 uf
R - 318 ohm
Is this correct?
Not correct, The Gain Av= 10, C = 10 nF, Rf = 3183.098 Ohm, Ri = 353.677 Ohm. R= 100K
@@Bosco12ful can you explain it how ?
What is the meaning of buffer
As its name suggests, it buffers the two stages. Basically, it isolates the two different stages from each other and in a way it avoids the loading of two stages.