Op Amps: Multiple Feedback Bandpass Filter

In this video we examine a basic bandpass filter using the multiple feedback template.
References: Operational Amplifiers and Linear Integrated Circuits: Theory and Application; Chapter 11, section 7.
My free texts and lab manuals are available for download at my college web site www.mvcc.edu/jfiore and at my personal site www.dissidents.com
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If you like my texts and videos, and would like to help defray the costs of making and maintaining them, consider making a modest donation at www.buymeacoffee.com/Professo...

[@CrackHead_Official]

I read your book "Operational Amplifiers & Linear Integrated Circuits: Theory and Application". I was truly eye opening. Thanks.

[@mavamQ]

This is very easy! This is the first time I have ever understood how to calculate an active filter. Hint: Using Paint, draw the schematic of the Filter type you are interested in. Copy and past it so you have four schematics down the page, (leave a little extra space between schematic 2 and 3). Label schematic 1 with the math formulas. Q, 2q, etc. On the right side add (Q=) so you can pencil in the numbers for FΘ / BW and solve for Q . On schematic 2, enter the numbers for the math you do on the symbols in schematic 1. In the room you left between 2 and 3, write the Wo=1rad/per sec and The Wo=2pi x your FΘ and solve. On schematic 3, write in the values you get when you do the F Scale. And finally, on schematic 4 you do the Z Scale and enter the values found on that schematic. Label the top with (type Filter) and save it as a picture. Then when you want to make a filter, print it, fill it out, you have an easy path for your math work and you have a record. I like this so much, I ordered the book. I'm eager to get it.

[@realchristopher4334]

Whoa, the mathematics is truly crazy! I, however, finally understood what the heck 'bandwidth' is and how to choose a center frequency. It is because operational amplifier band pass filter calculators online are either convoluted or not enough data teaching people how to actually use it. Maybe my case is a bit different. I go for the lowest and highest frequencies, rather than center frequency. Easier on my tiny brain cells.

[@ElectronicswithProfessorFiore]

Upper and lower limits are useful for wide bandwidth circuits, like say, an audio amplifier. For narrow bandidth circuits, a center frequency and Q are more useful and more common.

[@realchristopher4334]

@@ElectronicswithProfessorFiore Ah! No wonder whenever I used the calculator online, the Q value keeps on giving me (or showing me) unnaturally narrow bandwidth. I thought those calculators are trolls or I was missing too much. Many calculators do not offer many variables such as that common highest and lowest frequencies, and than calculate the center one. Which I thought it is how it was calculated this way. It, however, appears that it is not that simple by adding those two frequencies and divided by 2 as in the conventional way. Something to do with relative (log or decibel) number rather than absolute number. But then I have this little app called Iowa Hills OpAmp Filter Designer. It does have all the variables. Extremely powerful but sadly it is not being upgraded anymore. Nothing to improve because the mathematics is already there and considered standard, and no body complains, perhaps.

[@MasterMindmars]

Very good. The program is "Tina-Ti". Which version do you used? I've been simulated it in Multisim and works perfect. Thank you.

[@ElectronicswithProfessorFiore]

There is only one version of TINA-TI. It's the free version 8 of TINA (which is now up to version 12).

[@JonDeth]

I have another question. I am deeply out of practice and haven't done any significant math for engineering for about a decade(I did go to college though), and am taking on a couple simple projects for myself to regain competence and confidence. I was told I can build a multitude of bandpass filters all tied in parallel to a single node as the output to summarize the input signal it's divided into. Is this accurate? It seems it would be and I can use a pot on the feedback loop to adjust gain for each filter.

[@ElectronicswithProfessorFiore]

Whether or not the summation was perfect would depend on how the response curves overlap. That, in turn, depends on the chosen center frequencies and associated Q of each section. If not done correctly, the summation could have "holes" or "peaks" in the response.

[@JonDeth]

This also got me thinking, will matching our resistor value to the reactance of the capacitor at the defined corner frequency improve filter stability and quality? I know it's also a voltage divider and this would seem a potential criteria in some schemes.

[@ElectronicswithProfessorFiore]

The interaction between R and Xc is "baked into" the computations. There is nothing extra to do.

[@JonDeth]

@@ElectronicswithProfessorFiore ah I think I get it! The computation results in the voltage division across the 2 being virtually equal at the cornered frequency because we're plugging in known values in the first place?

[@ElectronicswithProfessorFiore]

@@JonDeth Pretty much.

[@reedreamer9518]

Excellent lecture. I was able to complete my design calculations with only one question: what if I'm using a single source opamp like an NE5532, but your example shows an opamp with a pos/neg source?

[@ElectronicswithProfessorFiore]

By "source", I assume you mean DC supply. The NE5532 does use a dual polarity supply (nice op amp, BTW).

[@reedreamer9518]

Okay, with an NE5532 (using a single 12V source) I used your design modified by placing a low-pass filter on the input, I got the result I wanted: A band-pass filter that rejects below 300 Hz (hum and tone squelch sounds) and also rejects above ~900 (radio static and QRM). Perfect for hearing the human voice clearly over ham radio.

[@ElectronicswithProfessorFiore]

@@reedreamer9518 Glad that worked for you. FYI, your filter is similar to the bandwidth limits set by standard land line telephones (they go a bit further on the top end by an octave or so; fairly steep slopes, though).

[@jonashuesenielsen9513]

i love your video. I am making the same bandpass filter as you show in the vid, but with other frequency ranges. i have downloaded tina ti bit it wont give me the correct graph. i have calculated everything exactly as u have but it wont work?. i also tryed using your values but that wont work either. when i open the ac analasys i get a lot of warnings that i havent connecten my wires and that some parts are floting but i have wired it all up? i hope u can help i am struggeling hard :-)

[@ElectronicswithProfessorFiore]

It sounds like you have components that only appear to be connected but aren't. This can happen if you simply move a component next to another. I suggest that you grab each component and move it. If the wires don't stay connected to the other components, it wasn't wired correctly. Not having the ground tied in is another common problem.

[@TheRealMonnie]

Another problem could be you are using a simulation model that doesn't match the symbol. The model might want additional pins connected that aren't on your symbol (for example, an enable pin). LTspice has been giving me a hard time with that for op-amp spice files from TI, so maybe Tina has the same problem.

[@Will-ob5ok]

Can you tune/sweep the band by replacing a resistor with a potentiometer?

[@ElectronicswithProfessorFiore]

Yes, theoretically, but it's not practical as the two pots (rheostats) have different values and the tracking would be a pain. For a tunable filter, you're better off using the state variable filter topology. In fact, there is an example at the end of chapter 11 in my op amp text that shows how to make a voltage controlled filter (VCF) using OTAs. You could just slide in a dual-gang pot where the OTAs are. Check out this video: kzbin.info/www/bejne/m6Suo6Wge6-sq80

[@simonyoungglostog]

Good brain day. I understood that. Re. Capacitors... Are there values that have better tolerances than others? I.e. 0.1uF is more likely to be within 5% than 10pF? Thank you for another great lecture.

[@ElectronicswithProfessorFiore]

It's less about size of capacitance than it is about the construction and dielectric. For example, aluminum electrolytic caps are not particularly accurate (but have good volumetric efficiency and are inexpensive). Meanwhile, polypropylene is very well behaved and accurate, but they tend to be physically large and expensive. For filters, you're better off with decent quality film caps like polyester (or better, polypropylene). Some small (

[@simonyoungglostog]

@@ElectronicswithProfessorFiore Thank you.

[@user-bb6jy8vy9p]

what if non inverting terminal is connected to 1.5V supply, does that make any difference while calculating gain or frequency ? If connected this filter in cascade with coupling capacitors is my gain or frequency affected? how to investigate that ? where can i find derivation for this gain and Cut off frequency ?

[@ElectronicswithProfessorFiore]

Why would you connect a DC voltage to the non-inverting input? Are you trying to run this with a single polarity supply?

[@user-mw2zo7dh7u]

Great video, but i was wondering how I can add gain to this circuit?

[@ElectronicswithProfessorFiore]

There is a form of this filter that has gain, but it is tied to the Q. For independent gain control, you'd be best off adding a gain stage (like a non-inverting amplifier).

[@F13TUN]

Can we use a narrow filter diagram for a bandwidth of 200 hz .

[@ElectronicswithProfessorFiore]

Not sure what you're asking here. Can you use this circuit to create a filter with a 200 Hz bandwidth? Yes, assuming the center frequency isn't too high (that is, a circuit with a very large Q).

[@cougar1861]

What's the number (1st, 2nd, 3rd etc) of the latest edition of your book?

[@ElectronicswithProfessorFiore]

Op Amps & LIC is in its third edition. The first and second editions were produced by commercial publishers (West and Cengage). I got the rights back from Cengage, updated it and produced the third edition as an OER a couple of years ago. Unlike the commercial publishers, I frequently issue updates. The most current as of today is V 3.2.6, May 7, 2021. (First digit is the edition, second is a more modest revision, third is for minor changes like fixing typos.)

[@cougar1861]

@@ElectronicswithProfessorFiore Thanks for your reply, your work/channel and your generosity.

[@TheRealMonnie]

How do you choose an op-amp? What sort of Gain Bandwidth and Slew Rate is required based on the BW and stop-band attenuation, etc.?

[@ElectronicswithProfessorFiore]

Slew rate and GBW tend to go hand-in-hand. The higher the passband, the higher SR and GBW will need to be. Look at it this way, in order for the filter to work, there has to be sufficient loop gain in the passband (i.e., gain over and above what you're looking for). For example, if the filter resonant frequency is 10 kHz and has a peak gain of 20, that would imply a required GBW of 200 kHz by itself, however, you'll still need loop gain over and above this. If we're looking at 20 dB of loop gain (probably at least that much), this points to a GBW of at least 2 MHz. SR can be computed based on the resonant frequency and the maximum output voltage swing (treat f0 as fmax for minimum acceptable SR-see section on slew rate in the text or associated video for details).

[@TheRealMonnie]

@@ElectronicswithProfessorFiore Thanks for the reply Professor! I watched several of your videos over the last few days, including the GBW and Slew Rate videos. Very well done, and much appreciated! I went to Lawrence Tech University in Michigan and had a great electronics professor, but it was >10 years ago so I'm admittedly rusty here. The part I'm not sure about is where do you measure the gain? Since this is a multiple feedback network, what constitutes input and what constitutes output? Should I just simulate it with a super fast op-amp and measure the output of the op-amp over the input signal (left side of your 15.92K resistor)? (see @15:57 for the time stamp showing your final circuit). Or do I look at the gain of each feedback separately to calculate it using superposition? In your example, it seems like a gain of 31.83K/100nF comes to 100 V/V at 500 Hz. And then 100nF/(15.92K//324.7) coming to 10. So the required GBW is the higher of the two, or 500 Hz * a gain of 100, which is 50kHz. Is this accurate?

[@TheRealMonnie]

I just went through and found the transfer function using Maple, and the gain is basically 0dB at 500 Hz, which makes sense as the output is unity at 500 Hz by design... But that's not the true gain of the op-amp - we know it's much higher... I just don't know how to calculate that.

[@ElectronicswithProfessorFiore]

@@TheRealMonnie The left side of that resistor is the generator signal. This circuit uses a voltage divider up front to compensate for Av>1, You'd like to determine the true gain of the filter, which would be from the right side of said resistor. My free text discusses both this filter and the simpler version without the divider. Check out page 472 which discusses funity. That will also apply to the circuit in the video. Ultimately, funity > 20*fo*Q^2 (which is 10*fo*true_Av, as I mentioned earlier).

[@descargarpeliculas3495]

How can I design a band pass filter using 2 op amp in cascade?

[@ElectronicswithProfessorFiore]

If you are referring to a HP/LP pair, that would be appropriate for a wideband filter (where f2>>f1). Set the HP to f1, and set the LP to f2. For narrow band applications, the LP/HP pair is not advisable.

[@eleazarmfonobong4941]

You didn't explain clearly how this setup act as a bandpass filter. Because both caps are rated equally, meaning they respond to frequency equally. For a high frequency, the system is by default a high pass filter. What or how is the feedback cap making it a bandpass? Since at high frequency, the caps short, making the feedback a unity gain, but for low frequencies, the first cap blocks the signals since it's a high pass filter.

[@ElectronicswithProfessorFiore]

OK, how's this: the circuit contains a lag network (low pass) and a lead network (high pass). Obviously, that combo can give you bandpass. Now, it may not be obvious how you get both a lead and a lag out of this. In fact, with a quick look, it might appear like there are two lead networks (which appears to be your situation). Sure enough, the lower cap (see template around 8:00) will block low frequencies. The upper cap, though, is in the Miller position (i.e., it straddles the input/output of an inverting amplifier). As such, this cap can be Millerized. The Miller equivalent would be sitting from the left end of its current position to ground. In other words, it would be in parallel with the input, and that creates a lag network (along with the resistors to its left). I will direct you to page 472 of my free text for more detail (see links in the video description, above).

[@omarsoek1]

what if a problem gives me the FC = 2900 hertz

[@ElectronicswithProfessorFiore]

Sometimes people use Fc synonymously with F0.