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I want to make a bandpass filter like this

From here

Im making a bandpass filter such that I want to make band pass range 40 to 400Hz such that at 40 and 400Hz value is -3dB. what centre freq and what values I should take for res,cap and etc. things in image. these are formulas that I got by derivation and are fully correct

I have tried everything from my side even AI. still not getting desired result. what values I should take for A0, Q, W0, C, Ra, Rb, Rc

As per my work...from one more formula f0=underroot(fl.fh)...where fl and fh are 40 and 400 Hz.
I got centre freq as 126.5Hz(795 rad/sec). and bandwidth as fh - fl = 400 - 40 = 360 Hz and Q=f0/BW=126.5/360 = 0.35..dont know if this is in right direction as Im not getting right values. What values I should take and what approach to follow. Please give simple understandable answer. Don't give PhD level or difficult answers

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  • \$\begingroup\$ When the fractional bandwidth exceeds about 0.8 it is better to implement using a high pass and low pass. Your case is about 2.8. \$\endgroup\$ Commented Jan 3 at 18:01
  • \$\begingroup\$ Please add a source link for the 1st diagram (required by site rules.) || Strong advice: You are free to present your questions as you wish, but people will tend to downvote or ignore questions phrased in informal /sloppy ways. eg "Im" no apostrophe (twice) , "what values ... " uncapitalised .|| "I don't know" <-- dont know. || NB I'm not seeking to be pedantic but rather to convey things that people really do care about. || FYI - you may know this: Lines ending in a carriage return with one space at the end DO NOT give a line break. Two spaces and a carriage return gives a new line. \$\endgroup\$ Commented Jan 4 at 12:57
  • \$\begingroup\$ @RussellMcMahon ok noted. This is diagram link : electronicsforu.com/electronics-projects/hardware-diy/… \$\endgroup\$ Commented Jan 4 at 13:06
  • \$\begingroup\$ Again fyi: Free hand drawings are acceptable as long as tidy and readable. Yours are reasonably OK although tending towards a little marginal in places. (But tidier tha anything I hand draw unless I take vast effort :-) ). Again, my reason for mentioning this is not to be cranky but to avoid bad reactions from others. You have +1 and -1 votes. The upvote is mine. I do not know why the downvotae but maybe for the above reasons. Or not. \$\endgroup\$ Commented Jan 4 at 13:06

3 Answers 3

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You could just implement a simple high-pass filter followed by a low-pass filter with a voltage buffer in-between: -

enter image description here

  • At 40 Hz, the cut-off is 3.1 dB
  • At 400 Hz the cut-off is 3.0 dB
  • Light blue shows resistor values using same capacitance

Because the two filter sections are independent you can alter them at will without one section affecting the other. One thing to note though is that the output impedance is 3900 Ω at low frequencies and, falls towards zero when you get past 400 Hz. This might be an issue to you but, you can always add another op-amp buffer at the end of the signal train. Op-amps are cheap and don't take up much room.

can you make the desired filter for multiple feedback bp image also I gave because thats part of my question I have to show

Of course if you were really interested in understanding the MFB band-pass filter then that's a slightly different exercise: -

enter image description here

You have to fiddle around with it a bit more but, it's doable as an equivalent MFB band-pass filter. And yes, the centre frequency will always be (for both) the geometric mean of upper and lower cut-off frequencies.

The cut-off frequencies and 3 dB points are identical to the earlier filter design so that should tell you something about how things really work. Yes, you can use a pile of math but, nothings beats insight.

Regarding the strange requirement to keep both capacitors the same value, this cannot be achieved unless you are prepared to significantly lose pass band gain from near unity to maybe up to 20 dB lower. Now, this may not be a problem so here's some help from Okawa Denshi's website:-

enter image description here

Take note of the restrictive formula in the grey area on the right that I have put inside a red box. So, if K is unity inverting gain at the mid-frequencies and, you want a Q value of circa 0.3 (required to achieve the 40 Hz to 400 Hz response), the capacitor ratio needs to be about 10:1 (as per what I used in my circuit above).

If you are going to use equal value capacitors, the ratio of gain magnitude (|K|) to \$Q^2\$ has to be less than 2. This means that Q has to be equal to or greater than \$\frac{1}{\sqrt2}\$ (0.7071) for a gain magnitude of unity. But, you won't get anything like the required bandwidth (40 Hz to 400 Hz) with a Q of 0.7071; Q needs to be about 0.3 to get the required bandwidth (with unity gain magnitude).

It's far more practical to use capacitor values of 10:1 for such low-Q applications or, just use the circuit in the upper diagram I posted. You could use a C1 value equal to C2 (100 nF) if you made R1 into 39 kΩ but, it seems you are fixated on using the MFB band-pass filter maybe?

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You have chosen the well-known classical bandpass configuration in MFB (multi-feedback) topology. Design formulas for all relevant bandpass parameters can be found in all books and contributions dealing with active filters.

Here are some hints:

  • Set Ra=R; Rb=xR; Rc=yR and two equal C`s.

  • Then from the transfer function:

    wo=(1/RaC){*SQRT[(1+x)/xy]}; and Q=SQRT[y(1+x)/4x] ; and midband gain Ao=y/2.

Now you can freely select two suitable capacitors C and then fix the values for the filter parameters Ao, wo and Q and solve for the components (start with Ao).

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  • \$\begingroup\$ see as per your values...I wanted to make BP filter from 40 to 400hz such that at corner freq is -3dB...I chose cap = 0.1uF then Gain=1 then Q=0.35(as per formulas f0=underroot(fl.fh)...where fl and fh are 40 and 400 Hz. I got centre freq as 126.5Hz(795 rad/sec). and bandwidth as fh - fl = 400 - 40 = 360 Hz and Q=f0/BW=126.5/360 = 0.35) then I put in values and got y=2,x=1.325,Ra=12K,Rb=15.9k,Rc=24k are these values correct? if yes Im getting -9dB at 40 and 400Hz..if no whats wrong...also tell is this way of calculating Q fine? Gain I choose 1 bcs I want 0dB at centre freq \$\endgroup\$ Commented Jan 4 at 12:16
  • \$\begingroup\$ @ GM DM You have 3 equations (wo, Q, Ao) for three unknown values (R, x, y). Because you can start with selecting a suitable value vor C, it should be possible for you to solve the system. Are you able to simulate the circuit ? \$\endgroup\$ Commented Jan 4 at 14:27
  • \$\begingroup\$ Yes I stimulated on ltspice, cant attach image here. For the mentioned values, I'm getting -9dB at corner frequencies instead of -3dB. I solved according to equation. According to equations given by you... Capacitor, Gain, and Quality factor I can chose independently then find x,y and then resistors value, on what basis I should choose A,Q. Just telling chatgpt says according to these formulas I cant make corner frequencies -3dB in MFB, because this MFB filter gives you only control on Q, not on gain of corner frequencies. So I have to take cascaded HP and LP filter as andy said as one answer \$\endgroup\$ Commented Jan 5 at 6:47
  • \$\begingroup\$ As Andy aka has mentioned - the MFB topology does not allow Ao=1 and Q=0.35 at the same time. Recheck your requirements! \$\endgroup\$ Commented Jan 5 at 9:22
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Well, assuming that your transfer function is correct. We see that:

$$\mathscr{H}\left(\text{j}\omega\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{j}\omega\right)}{\displaystyle\text{V}_\text{i}\left(\text{j}\omega\right)}=\frac{\displaystyle-\frac{\omega}{\text{CR}_\text{a}}\cdot\text{j}}{\displaystyle\frac{\text{R}_\text{a}+\text{R}_\text{b}}{\text{C}^2\text{R}_\text{a}\text{R}_\text{b}\text{R}_\text{c}}-\omega^2+\frac{2\omega}{\text{CR}_\text{c}}\cdot\text{j}}\tag1$$

So, we can see:

$$\left|\mathscr{H}\left(\text{j}\omega\right)\right|=\frac{\omega}{\text{CR}_\text{a}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\left(\frac{\text{R}_\text{a}+\text{R}_\text{b}}{\text{C}^2\text{R}_\text{a}\text{R}_\text{b}\text{R}_\text{c}}-\omega^2\right)^2+\left(\frac{2\omega}{\text{CR}_\text{c}}\right)^2}}\tag2$$

Solving:

$$\frac{\displaystyle\partial\left|\mathscr{H}\left(\text{j}\hat{\omega}\right)\right|}{\displaystyle\partial\hat{\omega}}=0\space\Longrightarrow\space\hat{\omega}=\dots\tag3$$

Gives:

$$\hat{\omega}=\frac{\displaystyle1}{\displaystyle\text{C}}\sqrt{\frac{\displaystyle\text{R}_\text{a}+\text{R}_\text{b}}{\displaystyle\text{R}_\text{a}\text{R}_\text{b}\text{R}_\text{c}}}\tag4$$

And for the cut-off frequencies we get:

$$\left|\mathscr{H}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\mathscr{H}\left(\text{j}\hat{\omega}\right)\right|\space\Longrightarrow\space\omega:=\omega_\pm=\frac{\displaystyle\sqrt{1+\text{R}_\text{c}\left(\frac{1}{\text{R}_\text{a}}+\frac{1}{\text{R}_\text{b}}\right)}\pm1}{\displaystyle\text{CR}_\text{c}}\tag5$$

When we know the higher and lower cutoff frequencies, we get:

$$\text{C}=\frac{\left(\text{R}_\text{a}+\text{R}_\text{b}\right)\left(\omega_+-\omega_-\right)}{2\text{R}_\text{a}\text{R}_\text{b}\omega_+\omega_-},\space\text{R}_\text{c}=\frac{4\text{R}_\text{a}\text{R}_\text{b}\omega_+\omega_-}{\left(\text{R}_\text{a}+\text{R}_\text{b}\right)\left(\omega_+-\omega_-\right)^2}\tag6$$

and notice then that we get:

  • $$\hat{\omega}=\sqrt{\omega_+\omega_-}\tag7$$
  • $$\mathcal{B}:=\left|\omega_+-\omega_-\right|\tag8$$
  • $$\mathcal{Q}:=\frac{\hat{\omega}}{\mathcal{B}}=\frac{\sqrt{\omega_+\omega_-}}{\left|\omega_+-\omega_-\right|}\tag9$$

So, for your case we get:

  • $$\hat{\omega}=\sqrt{2\pi\cdot400\cdot2\pi\cdot40}=80\pi\sqrt{10}\approx794.767\space\text{rad/sec}\tag{10}$$
  • $$\mathcal{B}=\left|2\pi\cdot400-2\pi\cdot40\right|=720\pi\approx2261.95\space\text{rad/sec}\tag{11}$$
  • $$\mathcal{Q}=\frac{80\pi\sqrt{10}}{720\pi}=\frac{\sqrt{10}}{9}\approx0.351364\tag{12}$$
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