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Help on Butterworth bandpass filter design

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nego

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Hi all, i am designing a Butterworth Bandpass filter to pass frequencies between 100Hz and 10kHz with a gain of 2dB using LM741. The attached figure is my design and also bode plot of my circuit in NI multisim. However, my design seems incorrect and my bode plot is totally wrong. Can anyone please help? Thank you.
 

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You need to swap the values of R3 and R4. If you set R3=10K, R4=20K, it works fine.

gain of 2dB
No, gain is *2. That's 6dB

BTW, it would be better to put the U3 inverting gain stage in front of the filter stages, instead of after them. That will give constant input impedance vs frequency.
 

You need to swap the values of R3 and R4. If you set R3=10K, R4=20K, it works fine.


No, gain is *2. That's 6dB

BTW, it would be better to put the U3 inverting gain stage in front of the filter stages, instead of after them. That will give constant input impedance vs frequency.

Thanks for your reply. filter1.png this is the bode plot i get after modified the circuit based on what you suggested. However, it is not correct..Ain't it suppose to be like filter2.gif ?? Furthermore, may i know how to get the bandpass filter with gain of 2dB(as it is requested in the question).
 

My simulation of your circuit with the corrected values gave the correct response, with 6dB gain. If you want 2dB, that's a gain of *1.259 so in your circuit, R6 must be = 1.259 * R5.

I don't know why your simulation only shows about 4.2dB gain. The response shape looks like it might be correct, but you're only showing it between 200Hz and 3KHz. When you run the simulation, set the frequency range to 10Hz - 100KHz. That will give you a good picture.
 
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    nego

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My simulation of your circuit with the corrected values gave the correct response, with 6dB gain. If you want 2dB, that's a gain of *1.259 so in your circuit, R6 must be = 1.259 * R5.

I don't know why your simulation only shows about 4.2dB gain. The response shape looks like it might be correct, but you're only showing it between 200Hz and 3KHz. When you run the simulation, set the frequency range to 10Hz - 100KHz. That will give you a good picture.

Hi, i have tried changing the dispaly range for my circuit. Here is my new circuit.simulation diagram.png. However, i still unable to get the desired response of standard bandpass filter. I am really confusing now. Here is my bode plot bode plot.png. Please have a look and tell me what's wrong with my circuit. Thanks so much.
 

You forgot to swap the values of R3 and R4. Set R3=10K, R4=20K.
 

You forgot to swap the values of R3 and R4. Set R3=10K, R4=20K.

Hi, i have changed the swap the R#3&4. This is the bode plot i obtained.2.png1.png. May i know what modification should i do in order to get a proper response of a typical butterworth bandpasss filter? Thank you :)
 

You're only showing the response between 200Hz and 3KHz again. As I wrote before: "When you run the simulation, set the frequency range to 10Hz - 100KHz. That will give you a good picture".

You almost got it right in post 5. There you showed 100Hz - 100KHz.
 

You're only showing the response between 200Hz and 3KHz again. As I wrote before: "When you run the simulation, set the frequency range to 10Hz - 100KHz. That will give you a good picture".

You almost got it right in post 5. There you showed 100Hz - 100KHz.

I get what you mean actually..actually that response can't be set. It come out automatically when i simulate. How to say? Actually there is nothing to be displayed on the left hand side..that range in post#5 is also not set by me..even if i set the range 100-1kHz for the bode plot, the minimum frequency display is always around 300 Hz for my circuit.

- - - Updated - - -

You're only showing the response between 200Hz and 3KHz again. As I wrote before: "When you run the simulation, set the frequency range to 10Hz - 100KHz. That will give you a good picture".

You almost got it right in post 5. There you showed 100Hz - 100KHz.

sorry cause i misunderstanding your meaning. I make some alteration and finally i get what i want. Thanks alot.
 

Oh, good! :grin:
What does it look like now?
 

Are you going to actually build this circuit, or is it only a simulation for learning purposes?

If you are actuallly building the circuit, I would suggest to change the LM741 opamp for something more modern.
The LM741 has a terribly slow slew rate specification, and I doubt that it will be able to pass a 10Khz signal (if it is higher than a few hundred millivolts) without distortion.
 
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Are you going to actually build this circuit, or is it only a simulation for learning purposes?

If you are actuallly building the circuit, I would suggest to change the LM741 opamp for something more modern.
The LM741 has a terribly slow slew rate specification, and I doubt that it will be able to pass a 10Khz signal (if it is higher than a few hundred millivolts) without distortion.

Yup. I am going to build the circuit using LM741 since it is requested in the assignment question. However,may i know what is the other opamp that you would suggest to replace the LM741?
 

"Yup. I am going to build the circuit...."

There are literally thousands of different opamps. Texas Instruments, all by itself has 1382 different devices.
Additional questions to properly select a device are required:

Where in the world can you purchase components? Not everything is available everywhere.
Are you on a tight budget, or can you afford high performance (read = expensive) devices?
Are you confortable with SMT devices?
Are there any other circuit restrictions? (i.e. single vs dual power supply, audio quality with low THD, low noise/offset).

Without having the answers, I can recommend the TL074. It has enough slew rate for your bandwidth; It is old enough to be low cost and widely available; it is available in DIP and SMT packages and has decent THD and noise specs. And requires dual power supply, like you show in your schematic. Lastly, it comes 4 opamps per package, so with a single IC you can realize your project.

Now, I'm pretty sure other posters will suggest their favorite devices.
 
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With minimum specified slewrate of 0.3 V/µs, LM741 can still output 3.5Vrms of undistorted 1o kHz sine. Not mind blowing but probably sufficient for this application.
 
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    tpetar

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    nego

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FvM;
you are absolutely correct..... full power bandwidth is

f = SR/(2*pi*G*Apeak)

However, as I asked in the questionaire, we require to know exactly what his requirements are, to suggest an optimal device. For instance....If he is going to use this filter in an audio application, it would be best to use a low noise, low THD opamp.
 

Yes, of course. 100 to 10 k doesn't sound like high end audio, however.
 

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