I designed an amplifier using an instrumentation amplifier and an operational amplifier. Here, I used a 2 Hz high pass and 100 Hz multi feedback low pass filter, but I could not get the efficiency I expected from my filters. What I need is for the pass band between 2 Hz and 100 Hz to have a constant gain, unfortunately, I can only have the constant gain I want between 15 Hz and 30 Hz. I tried to visually express the problem I was experiencing in the third image.

I would like your solutions and suggestions on where I should focus to improve the response of the filters.

AC sweep analysis


  • 2
    \$\begingroup\$ How constant do you need it? It is impossible for it to be perfectly flat. \$\endgroup\$ Oct 10, 2023 at 22:19
  • \$\begingroup\$ @atif As Kevin mentions, "how flat" between 2 Hz and 100 Hz do you want? You don't get to have a lot of flatness without moving your skirts wider or else without greatly increasing the filter order (which is another set of problems all by itself.) Also, you are well within the 1/f noise area for opamps. Which suggests chopping of some means. \$\endgroup\$ Oct 11, 2023 at 7:29
  • \$\begingroup\$ @atıfkoçak If you want to increase the fractional bandwidth over-which the gain is fairly constant you must widen the skirts. The skirts are the part where the gain rapidly declines. But this requires either high-order filters (to keep the things tight) or else the acceptance that you have to accept some frequencies at a level that may be annoying. The order of the filter factors in here. \$\endgroup\$ Oct 11, 2023 at 9:24
  • \$\begingroup\$ Why did you reduce clarity of the question by removing the schematic and some description? \$\endgroup\$ Oct 12, 2023 at 21:22

1 Answer 1


You get to control two things for each 2nd order filter: its shape and its corner frequency. That's it.

You also get to choose between: (1) designing a single higher order bandpass filter that is represented by one or more 2nd order bandpass filters combined and designed as a unit; or else, (2) break it up into a combination of one or more 2nd order low-pass and one or more 2nd order high-pass filters.

But as a rule, you pick #2 for anything anywhere even close to the span you are talking about. So in your case, you do in fact make the choice to combine low-pass and high-pass to make the band-pass. Your circuit illustrates this choice. So we are agreed on that point.

You've also indicated that each of these are 2nd order. Well into their stop-bands this means \$-20\:\text{dB}\$ per decade. Filter shape doesn't influence that, sufficiently far beyond the corner frequency. But filter shape has everything to do with what happens when nearing the corner frequency:

enter image description here

The above shows the exact same 2nd order low-pass filter with its gain set to 1 and the corner frequency set to \$f_c=200\:\text{Hz}\$. I've also shown a line to draw your attention at \$70\:\text{Hz}\$.

Some things to also notice.

The skirt is the steeply slanted line on the right. Note here that the skirt isn't altered by the filter shape. All 2nd order shapes have the same skirt in the stop band. So, if you need something steeper than that, then you need to increase the order of the filter. But this means that all you need to do is specify the magnitude difference at \$10\times\$ the corner frequency to completely determine the filter order. In this case, it is \$-40\:\text{dB}\$ at \$2\:\text{kHz}\$. So this is a 2nd order low-pass filter.

This is important for you to grasp. You need to understand how these things behave so you understand how to communicate your needs to others in a way that they can also follow and then help you.

The shape impacts what goes on as frequencies approach (from either side) the corner frequency. The green line has \$Q=2.5\$ or \$\zeta=0.2\$ (damping factor.) Note that this probably is not what you want. The others, in descent on that picture are \$Q=1\$, \$Q=0.71\$, \$Q=0.5\$, and \$Q=0.42\$.

This is also important for you to grasp. I've added a \$70\:\text{Hz}\$ line that digs into those shape lines. You should be able to see that there is still some variation at \$70\:\text{Hz}\$ despite the fact that this filter's corner frequency is at \$200\:\text{Hz}\$! Things only get flat at about \$\frac1{10}\times\$ the corner frequency! You need to understand this.

There's a general rule that's good to know and keep close to your heart. All important variational details happen between \$\frac1{10}\times\$ and \$10\times\$ the design point. Outside of those boundaries, you only need to know how things start and how things end.

So in this case of the 2nd order low-pass filter, there is the starting point which has a flat gain of 1. That's all you need to know about that. There is the ending point where the gain falls off at the rate of \$-20\:\text{dB}\$ per decade. That's all you need to know about that. And finally there is everything else that happens between \$\frac1{10}\times\$ and \$10\times\$ of the design point's \$200\:\text{Hz}\$ corner frequency. And here, those details are described entirely by the shape you choose.

Everything I just wrote and presented about the low-pass filter also applies to the high-pass. The only difference between them is that you flip the diagram around, left over to right, on the vertical axis formed by the corner frequency.

When you compose a wide-band band-pass filter (which is what you want), you do that by combining a low-pass and a high-pass filter. Just as you show in your circuit. So again, we agree. But now you also have enough terminology to be able to specify where you want the two corners located and also to carefully examine the above image to select your desired shape. You can, if you want, select a different shape for each (though many tend to keep the shapes the same.)

So. For 2nd order high-pass and low-pass, you only get to say two things: the corner frequency and the shape. That's it. You don't get to do more.

(If you want much more complex band-pass filters then the order goes up fast and the design details are, of course, more complicated.)

It's not up to us to tell you how much variation you can accept in the pass-band and therefore also dictating to you what shape corners you can accept. That's entirely up to you. And you also won't have much control over frequencies that are not of interest to you when you are limited to combining two 2nd order filters. What you can do is specify the two corner frequencies and the shape you want near those corner frequencies.

Finally, just to be crystal clear, you never get to specify a filter that looks like this:

\$\quad\quad\quad\$enter image description here

Not in this universe, anyway. You'll need to go find a very different one for that.

So, if you want to play around with what is achievable then here's an LTspice deck:

.param flo={.1} fhi={10k} f0={sqrt(10*70)}
.param fractional={3} q0={1}
V1 Vin 0 0 AC 1
E1 lowpass 0 Vin 0 laplace=1/((s/wh)**2+(s/wh)/q0+1)
E2 highpass 0 Vin 0 laplace=(s/wl)**2/((s/wl)**2+(s/wl)/q0+1)
E3 bandpass 0 Vin 0 laplace=(s/wl)**2/((s/wl)**2+(s/wl)/q0+1)/((s/wh)**2+(s/wh)/q0+1)
.param fl={f0*(sqrt(fractional**2+4)-fractional)/2}
.param fh={f0**2/fl} wl={2*pi*fl} wh={2*pi*fh}
.ac dec 2000 {flo} {fhi}

Just open up a new schematic, use the S key to call up the Spice command deck dialog, copy the above text and paste it into the dialog box, hit OK, and place it on the schematic. Once done, right-click on the schematic and select Run. You will get a blank plot. On the blank plot, right-click and select Add Traces and pick V(bandpass) to plot that up.

You will want to edit the fractional value. It can be as small as zero or some larger value. It sets the fractional bandwidth. Play with it and you will see what it does.

And you may want to edit the f0 value (I've set it to the center between \$10\:\text{Hz}\$ and \$70\:\text{Hz}\$.) So you may want to just leave this one alone, for now.

Finally, you likely will want to adjust q0. I've set it such that you likely will want to make it a little smaller. But this is the shape factor. So have fun with it, too.

You don't get to set the gain in the band-pass. But once you have nailed down what you really want to use for the fractional bandwidth and q, that can always be quickly worked out later. Focus on the fractional bandwidth and shape, for now.

  • \$\begingroup\$ Not to mention, the transient response of a brick-wall filter would terrifically distort (that is, in terms of the waveforms visually) whatever EEG signals one expects to see! \$\endgroup\$ Oct 12, 2023 at 21:24

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