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How do Sallen key bandpass filters really work?

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Megamox

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Hi all,

Does anyone know the actual mechanics of how a Sallen Key Band pass achieves its unique response? The Sallen key design looks like:
**broken link removed**
As you can see it comes quoted with a formula to calculate its centre frequency, where R(eq) is the equivalent resistance R1||R2 and both capacitors are considered to have the same value. Everything I've read suggests that it's combination of low and high pass filters within it acting like a band pass. So I redrew the circuit below as (1) and then transformed the input side to its thevenin equivalent (2).
**broken link removed**
Now looking at the newly redrawn circuit (2) I do indeed begin to see a low pass filter (made up of R(equivalent) and C1) and a high pass filter made up of C2 and R3. So since both filters have their own individual cut off frequencies, I can work out their geometric mean (ie the centre frequency of a band pass filter made from two such filters) and compare it to the centre frequency formula given with the Sallen Key Band pass filter in the first diagram. I did the calculation and they're exactly the same (Shown on the right - both capacitors considered the same here too).

This is where I'm stuck as to how the circuit performs. I can guess that at really high and really low frequencies, no signal appears at the inverting input because it's blocked by at least one of the filters and so the op amp keeps the bottom rail grounded. Therefore the output of the filter is 0v for exceptionally high and low signals. But there's a particular centre frequency where the gain becomes -R3/2R1 and I'm not sure how this happens.

I even thought this could be described as some sort of oscillator where at a particular frequency the output of the op amp reinforces the input signal and hence you get a Q type response. However I don't know if this is correct as I cannot see how you'd get an additional 180 deg phase shift by going through the passive filter network back to the input which would help to reinforce the incoming signal. The inverting arrangement already provides the other 180 deg shift needed.

So I'm a bit stuck, maybe all of this is in the wrong direction, maybe I'm missing something, but I'd really like an idea of how the circuit works, especially around the centre frequency where the whole circuit seems to turn into just an inverting amplifier of gain R3/2R1 (It's as if R2, C1 and C2 either cancel out or disappear from the circuit!).

Any help or directions for further reading would be appreciated!

Thanks,
Megamox
 
You DO NOT have a Sallen and Key bandpass filter that uses a non-inverting opamp.
Your filter is called a Multiple Feedback bandpass filter that uses an inverting opamp.
Look in Google for an article describing how it works.
 

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Thanks, you're right of course, this topology is a multi-feedback filter. I've checked online and I've been unable to find a coherent explanation of how the circuit performs the way that it does. The mathematical analysis shows clearly it behaves like a band pass filter but I don't see how the circuit aligns itself to do this. Is there anybody who wants to give it a shot and try and explain how the circuit gain at the centre frequency is conveniently just the result of R3/2R1 with no other component dependency. I've added a diagram below which explains my confusion, somewhere there's a flaw in my logic but I can't spot it!

**broken link removed**

Thanks,
Megamox
 
A rough explanation is that, at the center frequency, the high pass effect of C1 is cancelled by the low pass effect of C2. Thus the gain is then equal to the resistor values in the circuit.
 
I think you're right. After all the op amp is supposed to be standing in for an inductor so we can get a sort of 'resonance' happening. In a typical series LCR circuit, when resonance occurs the complex imedances cancel and you're left with an entirely real impedance. It looks like the same thing is happening here. The only way that i can imagine these two identical capacitive impedances to cancel out is to have a phase shift difference of 180 deg between the current going through each of them. Have no idea why R2 is not included either, it's effect must be cancelled out somewhere else.
 
R2 is included. Look at your original post. R(eq) is the parallel (not series) combination of R1 and R2, or 1/2 of R1 if R1=R2.

Edit: Regarding the cancellation of C1 and C2 signals, note that the signal to C2 from the op amp output is 180° out of phase with the input.
 
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I'm sorry, I mean how come R2 isn't included in the formula for the Gain. I've simulated a 3khz bandpass filter circuit and got some readings. The output is shown in red as compared to the input (1v Input), always shown in black. The output is (and this surprised me too), not 180 deg out of phase which is usually the case with an inverting amplifier. The output seems to be leading the input by 90 deg. Interestingly too, I've captured the waveform at 'x' which I've only magnified to show the phase, but from its miniscule value, I'd have to say it's acting like a virtual earth. If this is the case, maybe this is why, together with the fact that no current flows into an op amp input, that the whole combination of R2 (268 ohm) and C1 (0.01uF) can be ignored. If this is true the entire feedback path will consist of C2 and R3. At the right frequency when the reactance of C2 = R3, the feedback impedance drops to R3/2 and the gain works out to R3/2R1. This does give the answer we want but I believe this is how you calculate inverting amplifier gain... and our output doesnt seem to conform to a 180deg phase shifted output expected from such a calculation..!

Megamox

**broken link removed**
 
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I'm sorry, I mean how come R2 isn't included in the formula for the Gain. .......................
But it is. As I stated in my previous post, R(eq) in the gain formula is the value of R1 in parallel with R2. Your figure showing 2R1 for the input resistor in the simplified equivalent circuit is incorrect.
 
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You're right, R(eq) is definitely the value of R1 in parallel with R2. For this type of band pass filter though, the gain is quoted (at centre frequency) as R3/2R1 which has no mention of R2 in it, nor R(eq). The equivalent diagram I've drawn was not really meant to show 2R1 as the equivalent to R1||R2 but to show the final gain of the whole filter at the centre frequency in terms of how it's calculated for an inverting amplifier ie Rf/Ri. Perhaps I'm misunderstanding you, but let me know if I'm missing your point.

I've taken some more readings to have a look at the phase difference between x and vout as the frequency sweeps past the centre frequency (3khz). From what I can see you've got a phase difference of 90 deg at frequencies mucher higher or lower than the centre frequency, and a phase difference of 180 deg at the centre frequency. So perhaps then, qualitatively, you've got a one pole filter acting in the lower part of the frequency (the low pass), you've got a one pole filter acting in the higher part of the frequency spectrum (the high pass), and you've got a double pole acting at the centre frequency (both of them).

Megamox
**broken link removed**
 
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Hi Megamox,

here is another explanation from the system point of view:

From system theory it is known that there is a passive circuit that is able to produce a conjugate complex pair of ZEROE`S: Bridged-T network.
Therefore, put such a network into the feedback path of an opamp and you are able to produce a conjugate-complex POLE pair.
In your case the T consists of C1, C2 and grounded R=R1||R2 . This T is bridged by R3.
More than that, it is known that you can use each grounded node to couple in an input signal without touching the pole distribution (denominator of the transfer function).
 
But it is. As I stated in my previous post, R(eq) in the gain formula is the value of R1 in parallel with R2. Your figure showing 2R1 for the input resistor in the simplified equivalent circuit is incorrect.

Sorry crutschow, that`s not correct. The resistor R2 does NOT appear in the gain expression.
The transfer function shows that the gain is:

Ao(w=wo)=R3/[R1(1+C2/C1)].

This expression approaches

Ao=R3/2R1

for equal capacitors C1=C2.

(Comment: Because R2 appears in the numerator as well as in the denominator as R1||R2 it cancels out.)
 
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Megamox, if you like here is another interesting interpretation of your circuit:

* The input impedance of the whole circuit at the common node of R1 and R2 (without these resistors) resembles a "frequency-dependent negative resistor" element (known as FDNR) with parallel capacitive losses.
Such an FDNR element always can be forced to come into resonance at one single frequency with a "normal" resistor (here provided by R1||R2). The mentioned losses determine the quality factor Q of the circuit.

* This approach is in full compliance with the interpretation of an multi-feedback lowpass (other element allocation if compared with the bandpass).
In this case, the input impedance at the same node is identical to a lossy inductor which - together with a grounded capacitor (C2 replacec R2 in the bandpass) - forms a tank circuit that exhibits a resonance effect.
 
I've been unable to find a coherent explanation of how the circuit performs the way that it does. The mathematical analysis shows clearly it behaves like a band pass filter but I don't see how the circuit aligns itself to do this.

Hi megamox, I have to excuse myself for writing again a comment - but, as you probably have seen, I am interested in explaining the function of circuits rather than calculating the transfer function only.
Here comes another explanation:

*In one of your first postings (equivalent circuit arrangement) you have calculated the center frequency of the passive part only.
This circuitry is one of the classical passive RC bandpass filters (another similar form is the WIEN network).
The ouput of this 4-element passive RC bandpass is connected to the inverting input of the opamp.

* On the other hand - the gain of each opamp with feedback is calculated as -Hf/Hr (assuming infinite open-loop gain).
In this expression Hf is the forward function Hf=Vn/Vin with Vout=0 (Vn: voltage at the inv. input) and Hr is the return (feedback) function Hr=Vn/Vout with Vin=0.
This follows directly from the superposition theorem.

* Now - the above mentioned passive RC bandpass is identical to Hf, and it is easy to show that Hr is a "weak" band stop function with unity gain at very low and very high frequencies. At the center frequency the "gain" of this band stop circuit is, of course, below unity.

* When you now look at the ratio Hf/Hr you see that the resulting function is a bandpass (determined by Hf) that is "boosted" around the center frequency (caused by the band stop function, which causes a denominator smaller than unity). This results in an active bandpass that has a Q value (due to the band stop effect in the denominator) that can be much larger than for the passive RC band pass only (which is Q=0.5 at maximum).

*Perhaps this additional sight helps to explain the function of the circuit.
 
Hello there,


I calculate the same as you guys for the center frequency gain:
Vout(w0)=Ein*R3*C1/((C2+C1)*R1)

and again with C1=C2 we have:
Vout(w0)=Ein*R3/(2*R1)

and so the gain is:
A(w0)=R3/(2*R1)


This seems to suggest that the whole amplifier from input to output is just a gain stage with zero phase shift.
There is a catch here however, and that is the gain for amps like these is often specified without regard for the phase shift between input and output. If however we were to specify the phase shift then we would have to include that too, and we would find out that this amp turns into an inverting amplifier because the phase shift is 180 degrees from input to output at the center frequency.

Numerically, the phase shift is calculated:

D=(R1+R2-w^2*C1*C2*R1*R2*R3)^2+(w*C2*R1*R2+w*C1*R1*R2)^2
N1=w*C1*R2*R3*(R1+R2-w^2*C1*C2*R1*R2*R3)
N2=-w*C1*R2*R3*(w*C2*R1*R2+w*C1*R1*R2)
PhaseShift TH=-atan2(N1/D,N2/D)

When evaluated for w=w0 we get:
TH=-atan2(0,-(C1*R3)/((C2+C1)*R1))

and since the imaginary part is zero, for any real component values this evaluates to:
TH=-pi

and this is 180 degrees.


If perchance you see a phase shift that does not equal 180 degrees then it could be due to the op amp model being used. At 3kHz there might be some phase shift, so perhaps a better test frequency would be 100Hz or use a higher bandwidth op amp. A theoretically perfect op amp will show 180 degrees phase shift at the center frequency which makes it an inverting amplifier with gain at the center frequency.

For reference:
w0=sqrt((R2+R1)/(C1*C2*R1*R2*R3))
 
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Thanks Winterstone and Mr Al, really appreciate your comments! Glad the gain works out correctly, I was worried there might have been a mistake in the calculations. The idea of superposition was one I was thinking of using except as far as I can remember I've never used it on a dependent source, such as Vout which is why I was wary about its application here. I'll definitely do some reading up on it, perhaps I've got that really wrong. I've got to say though, the response transfer functions I've simulated based on your post look spot on. You can see how this filter achieves gain and Q from just essentially the strength and characteristic of the reverse band stop.
**broken link removed**
In the simulation above I'm using component values from the 3khz band pass filter with R1||R2 = 242 ohm and R3 = 106k.

The question that does spring to mind though is that since the transfer function of the bandpass filter is unity at centre frequency, the gain is therefore dependent on the strength of the rejection of the bandstop filter. Since we know the overall gain of the entire multi-feedback filter is R3/2R1 then 1/(Reverse Band Stop Gain) should also equal R3/2R1. Essentially this means at the centre frequency the gain of the bandstop should be 2R1/R3. Since R1 is in parallel with R2 (drawn above as 242 ohm), I guess my question is how does the bandstop circuit arrange itself so that at the centre frequency the gain (Vn/Vout) of the dip is actually partially dependent on a single resistor (R1) which is actually part of a parallel pair.

Megamox
 
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.........
The question that does spring to mind though is that since the transfer function of the bandpass filter is unity at centre frequency, ......
............

Hi Megamox - nice graphics supporting my last explanation.
However, why do you think that the gain of the passive BP is unity? No voltage drop within the circuit at all?
No - that`s not the case.
This passive BP circuit has exactly the same properties as the WIEN network with a maximum "gain" of 1/3 (equal R and C).

Concerning MrAl`s contribution: Yes, the MFB topology always has inverting properties - resulting in -180 deg phase shift at w=wo.

Added later: megamox, there is another interesting circuit property:
The transfer function based on an output voltage at the first node (from the left) resembles a lowpass filter - of course with the same Q value as the bandpass function.
 
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Yes I admit, I've also been tinkering with the transfer function from various nodes to become more familiar with the topology. Had I been more familiar with it earlier, perhaps I would have spotted the mechanics of operation of the whole filter from the beginning! I just took some readings for the bandpass and bandstop filter at the centre frequency, the strange thing is that the band pass is actually producing a passive gain of 1 (the output is about 96% of the input signal). The output of the bandstop is about 1/200th of the input. Both outputs shown in red. In both cases I'm putting in a 3khz signal at 1v pk-pk. For this particular filter with these values of R1 = 26k and R3 = 106k, we should be looking at gain of R3/2R1 overall, ie about 2. From these simulations it looks like the gain function would be calculated as (Vn/Vin)/(Vn/Vout) =~ 1/(1/200) = 200 @ 3khz, a factor of 100 times bigger than what it should be. Not sure why this is. Any ideas?

Megamox

**broken link removed**
 
"the strange thing is that the band pass is actually producing a passive gain of 1 (the output is about 96% of the input signal). The output of the bandstop is about 1/200th of the input"

Megamox, something went wrong with your simulation.

For a given center frequency of 3 kHz both capacitors are C=1nF, agreed? Resistors as given by you (26k and 106k).

In this case, the bandpass gain at wo is 0.671 and the notch "gain" is 0.329.
Thus the ratio approaches a value of 2 - as expected.
 
Glad the calculation works out correctly at least! I'll tinker with the sim and see what's happening. The caps are 0.01uF, so 10nf though. At least the basic theory of the circuit as a whole makes sense now.

Megamox

Edit: Keep forgetting to add this. In case anybody wants to use this filter design and does not necessarily want to work out the component values and scale them, a really good component value calculator can be found at **broken link removed** which is free. I believe the max Q should be about 10-15.
 
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Megamox, one last comment:

From the beginning of this thread I was surprised how you could arrive at the correct center frequency because your approach using the geometric mean of the two corner frequencies was absolutely new for me.
More than that, I was sure that something must be wrong because you have isolated both parts - lowpass and highpass sections - from each other. Such an approach, obviously, is not correct.

Now - after reviewing and simulating your circuit - I have found the error you have made - and it was more or less by chance that you arrived at the correct mid frequency.

The error is in the "calculation" of the equivalent resistor Req. You are NOT allowed to use Req=R1||R2 as an input resistor.
That is simply wrong. Both resistors act in parallel for calculation of the function Hr (notch) only.

Therefore, simulating the passive parts of the circuit only (that means: Hf and Hr separately) with both resistors R1 and R2 (as given in your first posting) you arrive at the following results at 3 kHz (C=10nF):
Band pass: 10.15 mV
Band stop: 4.65 mV.

As you can see - the ratio again approaches a value of 2.
 
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