How does an active filter transfer function translate to a block diagram?

Question

I am having troubles trying to translate when finding a transfer function of an active filter is it in a 'Close loop' or 'Open loop' state?

When finding the transfer function of these active op-amps how come we don't include the characteristics of the op amp itself?

For example a Op-Amp has an Op-loop gain I would imagine you need to include that in your transfer function otherwise you could use any op amp/disregard any op amp?

Example: Sallen-Key Butterworth Filter of 2nd Order

$$ H(s) = \frac{1}{C_1 C_2 R_1 R_2 s^2 + C_2 R_1 s + C_2 R_2 s + 1}$$

Its confusing as the OP-amp has a feedback of 1.

Closed Loop:

Open Loop:

Answer 1

I am having troubles trying to translate when finding a transfer function of an active filter is it in a 'Close loop' or 'Open loop' state?

Of course, the filter function is available in a closed-loop condition only. It is the frequency-variable feedback network which gives the desired filter characteristic.

When finding the transfer function of these active op-amps how come we don't include the characteristics of the op amp itself?

It is common practice to assume idealized opamps only (infinite gain, no frequency dependent gain). Of course, this simplification causes errors in the transfer characteristic - however, this is acceptable as long as the operating frequency range is limited to a region where other uncertainties (parts tolerances) are dominating. Of course, this means that opamp based filters are used not in the upper MHz range (where opamp non-idealities play a remarkable role).

Please note, that there are many, many different lowpass circuit alternatives which ALL would have the same transfer function in practice (reality) - under the assumption of IDEAL opamps. The difference between these alternatives can be revealed only for REAL opamps. In this context, it is important if the opamp is used as a fixed-gain positive amplifier (Sallen-Key) or as a high-gain amplifier (multi-feedback) or as an integrator (state-variable structures). All these alternatives have different sensitivities against opamp non-idealities.

For example a Op-Amp has an Op-loop gain I would imagine you need to include that in your transfer function otherwise you could use any op amp/disregard any op amp?

Of course, theoretically we could include the open-loop gain of the amplifier into the overall transfer function....but for which purpose? To slightly improve the accuracy of the function ? The price for it would be a very complicated transfer function which is very hard to use for practical realisations.

Remark: There are studies which show how the frequency-dependent open-loop gain of an opamp can be exploited - together with a pure resistive feedback network without external capacitors - for realizing active filters ("R-filter). However, as a severe drawback, it is necessary to know the exact open-loop gain opamp characteristic, which has very large tolerances. Buth these filters have no practical relevance....

Comment 1: Of course, it is possible to create a block diagram in the classical form with a feedback block (and with a summing junction at the input) and a block with the opamps open-loop gain - but for which purpose?

Comment 2: Actually, as an advantage of block diagram visualization, it is very easy to see HOW the S&K lowpass works: There is a passive lowpass (Hforward) with a bad quality factor (pole-Q) Qp<0.5.

However, in the "critical" frequency region (exactly at the pole frequency) the bandpass in the feedback path enhances the amplitude of the lowpass because the bandpass has the same pole frequency (zero phase shift, positive feedback effect). This effect produces a larger Q value (Example: Qp=0.7071 for a Butterworth response).

EDIT: Block diagram: Both passive transfer functions (lowpass, bandpasss) are derived from the original circuit. Vp is the signal voltage at the positive (non-inverting) opamp input terminal.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

@LvW Answer is just perfect.

But just for fun, I think you can accomplish your task. From this base schematic:

^{simulate this circuit – Schematic created using CircuitLab}

You can model the amplifier as a input impedance and controlled voltage source in series with an output impedance.

^{simulate this circuit}

Of source we can rearrange the circuit to become more easy to visualize its meshes and nodes:

^{simulate this circuit}

Now, I have used mesh analysis (tough I tried with node and didn't like were it was going) to obtain the filter transfer function (after many more hours than I thought it would be necessary and many possible mistakes):

\$H(s)= \frac{A_{ol}(s)Z_{i}(s)+Z_o(s)[1+s C_1 (R_2+Z_i(s))]}{A_{ol}Z_i(s)[s^2C_1C_2 R_1 R_2+s C_2(R_1+R_2)+1] + s^2C_1C_2 \frac{R_1R_2Z_o(s)Z_i(s)}{R_1//R_2//Z_o(s)//Z_i(s)}+ s[ C_2(R_1+R_2)(Z_o(s)+Z_i(s)) + C_1((Z_i(s)+R_1)(Z_o+R_2)] + (R_1+R_2) +Z_o(s) + Z_i(s)}\$

Curiously, but not a proof per se, if you rearrange terms to only have \$1/Z_i\$ and \$1/A\$ format and take the limit of \$Z_0 \rightarrow 0\$, \$Z_i \rightarrow \infty\$ and \$A_{ol} \rightarrow \infty\$ (only the terms with \$Z_i\$ \$A\$ are nonzero) it will become the given original and simpler transfer function.

\$H(s)= \frac{A_{ol}(s)Z_{i}(s)}{s^2C_1C_2 R_1 R_2 A_{ol}Z_i(s)+ s C_2 (R_1+R_2) Z_i(s)A_{ol}(s) + Z_i(s)A_{ol}(s)} = \frac{1}{s^2C_1C_2 R_1 R_2 + s C_2 (R_1+R_2) + 1}\$

I think this may be partially useful just to show that there isn't a simple block transfer multiplication, but it will take into account very relevant limitations of the amplifier. Any additional limitation will imply that linear operators and transfer functions can't properly describe it.

Since this took me a lot more hours than I expected I am planning to write another one or a edit to this one with a very different approach exploiting a similar idea that @LvW diagrams leads to, unfortunately yesterday the diagram was not uploaded yet.

Edit 1 : Following @LvW suggestion, we can model the amplifier as \$A_{ol}(s) = G\frac{\omega_0^2}{s^2+2 \xi \omega_0s +\omega_0^2}\$ and \$Z_i(s)=R_i\$ and \$Z_o(s)=R_o\$. Curiously, after doing this and finding some 2 algebraic manipulation errors, its not hard to separate the amplifier gain characteristic indeed, as corrected above. The new transfer function with the second order filter is now:

\$H(s)= \frac{G R_{i}\omega_0^2+(s^2+2\xi\omega_0+\omega_0^2)R_o[1+s C_1 (R_2+R_i)]}{G\omega_0^2R_i[s^2C_1C_2 R_1 R_2+s C_2(R_1+R_2)+1] + (s^2+2\xi\omega_0+\omega_0^2)[s^2C_1C_2 \frac{R_1R_2R_oR_i}{R_1//R_2//R_o//R_i}+ s[ C_2(R_1+R_2)(R_o+R_i) + C_1(R_i+R_1)(R_o+R_2)] + (R_1+R_2) +Z_o(s) + Z_i(s)]}\$

Which is truly an horrific expression.

Answer 3

When finding the transfer function of these active op-amps how come we don't include the characteristics of the op amp itself?

Generally you want to operate at frequency/gain underneath the gain bandwidth product. That way the op-amp dynamics don't come into play.