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I am trying to apply two-port feedback analysis to a circuit similar to the one shown below. I can't represent the forward amplifier as a two-port network because the input signal and feedback signal go to different inputs.

Is it a valid operation to refer the feedback signal to the non-inverting input of the forward amplifier to form a two-port network as shown in the second figure?

Then I can calculate the forward amplifier gain and the feedback quantity, and the summing node is clear. Referring the feedback signal to the non-inverting input can be done using equations for the ideal inverting and non-inverting gains of the forward amplifier, because the non-ideality (i.e. finite loop gain) is the same for both gains.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

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  • \$\begingroup\$ It's useless and error prone to make the circuit more complex and different than it is. Assuming opamps ideal and the circuit stable: From your first figure one can see that Vfb=(Uout) * (R7)/(R7+R6). And Uout = -(Vfb) * (R2/R1) + (Vin) * (R2+R1)/(R1). You can build a control theory style feedback loop block diagram which obeys these equations if you want. To solve Uout as a function of Vin eliminate Vfb. The stability was an assumption. Proving it true demands more and much more if the amps are non-ideal practical devices. \$\endgroup\$
    – user287001
    Aug 12 '21 at 20:40
  • \$\begingroup\$ @user287001 say I want to build a feedback loop block diagram that obeys those equations (because that is how I am used to analyzing stability), don't I have to do exactly what I did in the question to have the input and feedback signal at the summing node? I think the classic block diagram relies on every block being a two-port, with the drawn connections being between one pair of terminals, and the ground connection is implicitly between the other pair. The forward amplifier in the first schematic has 3 input terminals (ground is not drawn). \$\endgroup\$
    – DavidG25
    Aug 12 '21 at 20:57
  • \$\begingroup\$ I wrote my equations by assuming the system is stable and founds a certain Uout for given Vin. Those equations are 100% useless for stability analysis because the stability was assumed. Opamps can be modeled as a summing junction (another input is taken as minus) followed by more or less ideal amplifier which can contain a low pass filter. About your 2nd circuit: It is in accordance with my equations but it has assumed stable operation which allows you to take the Vfb into the account by adding it as gain compensated to Vin. So, for stability analysis it's as useless as my equations. \$\endgroup\$
    – user287001
    Aug 12 '21 at 22:04
  • \$\begingroup\$ @user287001 you're saying that referring VFB to the non-inverting input of the forward amplifier makes an assumption about the gain of the amplifier consisting of OA1, R1, and R2, right? I agree with you, but the stability of that amplifier can be assessed on its own, outside of the circuit (same can be said about the buffer). I am trying to get the closed loop transfer function of the entire circuit, after I have confirmed that the forward amplifier is stable on its own. I don't think my second circuit makes assumptions about the overall loop. \$\endgroup\$
    – DavidG25
    Aug 12 '21 at 22:17
  • \$\begingroup\$ Do you assume opamps have some real word slowness which reduce gain and lags the phase as the frequency grows? That assumption should be seen also in the gain compensated Vfb which is added to Vin. If the opamps are ideal, your formula is right. But proving that the whole circuit founds certain Uout for certain Vin with ideal opamps is tricky. \$\endgroup\$
    – user287001
    Aug 12 '21 at 22:37
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If all R values are equal, the use of Vin- vs Vin+ for feedback is irrelevant in this 1st circuit.

The Vin- current is relative to Vout , while the Vin+ is relative to 0V. So the gain of Vo/Vin is a huge difference due to Aol.

So the real problem is the feedback is not differential. The 2nd problem is choosing a stable gain for a differential feedback.

To make this easier to understand...

The non-inverting signal input with negative feedback causes both O.A. inputs to follow the input with a "virtual" null difference. So if all R's are equal the gain= +2 and the use of a Vout/2 buffered then matches and follows the input.

But using the Vin+ referenced to 0V using Vin- with a gain of -1 will not produce a Vfb of 0V to match the grounded reference of Vin+. Now both inputs are 0V but Vfb is still Vout/2 changing it's polarity won't help.

Using 741's here

enter image description here

The label nodes are connected.

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  • \$\begingroup\$ Sorry, I need some help understanding your answer. What do you mean by the feedback being "not differential"? \$\endgroup\$
    – DavidG25
    Aug 12 '21 at 23:09
  • \$\begingroup\$ THe difference between Vout and Vin \$\endgroup\$ Aug 12 '21 at 23:13
  • \$\begingroup\$ I have a few guesses at what your answer is saying, please correct me if I am wrong: 1) "the use of Vin- Vin+" is saying that the loop gain is independent of where the signal input is, 2) "Vin- current" suggests you did a Norton transformation at the node labeled "VFB". The feedback signal is a shunt connection and the input is a series connection, making it difficult to take the difference, or "the feedback is not differential". Point 2 is exactly what I am trying to address with the manipulation in my question. Can the feedback current be referred to the non-inverting input as a voltage? \$\endgroup\$
    – DavidG25
    Aug 13 '21 at 15:34

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