Answer by periblepsis for Designing a Sallen-Key Band Pass Filter

Let me expand your diagram:

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^{simulate this circuit– Schematic created using CircuitLab}

I don't know if this is what your notes used. But it's what came to mind when reading the equations you provided. If I'm wrong about the above, you need to let me know.

Meantime, I'll just call this circuit \$Z\$ and write:

Zomega0 = omega0Zc2 = c2Zra = raZrb = rbZgamma = Zomega0 * Zc2Zc4 = (1/2) * Zc2Zr1 = 2 / ZgammaZr3 = 2 / (3 * Zgamma)Zr4 = 4 / ZgammaZeqv1 = Eq( v1/Zr1 + v1/Zr3 + v1*s*Zc2, v2*s*Zc2 + vout/Zr3 + vin/Zr1 )  # KCL v1Zeqv2 = Eq( v2*s*Zc2 + v2*s*Zc4 + v2/Zr4, v1*s*Zc2 )                     # KCL v2Zeqv3 = Eq( v3/Zra + v3/Zrb, vout/Zra )                                  # KCL v3Zeqvo = Eq( vout/Zr3 + vout/Zra, v1/Zr3 + v3/Zra + iout )                # KCL voutZeqop = Eq( v2, v3 )                                                     # opamp V- = V+Zans = solve( [ Zeqv1, Zeqv2, Zeqv3, Zeqvo, Zeqop ], [ v1, v2, v3, vout, iout ] )tf2( simplify( Zans[vout]/vin ) ){omega: omega0, zeta: (-3*ra/2 + 7*rb/4)/rb, P: [{A: (-2*ra - 2*rb)/(6*ra - 7*rb), N: 1}]}

The N=1 above just means it is, in fact, a bandpass circuit. That matches expectations. Good.

\$\omega_{_0}\$ comes out per the original assumption. That matches expectations. Also good.

But the gain, A above, isn't what you show. Instead, it is \$A=K_{_0}=\frac{R_{\small{A}}+R_{\small{B}}}{3.5\, R_{\small{B}}\:-\:3\,R_{\small{A}}}\$. So that's a problem.

In addition, \$Q=\frac1{2\,\zeta}=\frac{R_{\small{B}}}{3.5\, R_{\small{B}}\:-\:3\,R_{\small{A}}}\$, from the above calculations. And this is nothing like the expression shown in your notes.

If I assume your equations remain correct, then I have to assume that my schematic is wrong and you need to tell me what it really is.

If the schematic I provided is correct, then I can't get your results. And that's a different problem. Perhaps I made a mistake in my work above. Or perhaps there's another issue.

Just as a quick test of my KCL development above, let's just assume that the (+) input is \$V_{\small{IN}}\$, without all the rest. I'd like to check the gain for the simpler case:

expand(    solve( [        Eq( v3/ra + v3/rb, vout/ra ),     # KCL v3        Eq( vout/ra, v3/ra + iout ),      # KCL vout        Eq( v3, vin ) ],                  # opamp V- = V+    [ vout, iout, v3 ])[vout]/vin )       # TF calculationra/rb + 1

Okay. That's as expected.

Finally, let's ignore my earlier schematic. Instead, I'll just assume that I have no idea how \$K\$ is achieved, but that it exists as shown in your diagram. That's how Sallen & Key exhibited their own work. So no problem.

Then there's no node \$v_3\$ anymore and we can lose that equation and I need to modify the output equation:

tf2(simplify(solve( [ Zeqv1, Zeqv2, Eq( vout, K*v2 ) ], [ v1, v2, vout ] )[vout]/vin)){omega: omega0, zeta: 13/4 - 3*K/2, P: [{A: -2*K/(6*K - 13), N: 1}]}

In this case, it is true that \$Q=\frac{1}{6.5\,-\,3\,K_{_0}}\$. So now there's a match on that score. And \$\omega_{_0}\$ is still as expected. Good. The filter is a bandpass. Good. The closed loop gain is a function of \$K\$. But without \$R_{\small{A}}\$ and \$R_{\small{B}}\$ present, it's not necessarily a conflict.

Sallen & Key do document something similar:

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But note that it is slightly different, too.

I'll leave things there.