Positive and negative feedback circuits : let the following circuit enter image description here

I want to prove that this system is stable (meaning the op-amp isnt getting saturated) if and only if \$ k_+ = \frac{R_{1+}}{R_{1+}+R_{2+}} < k_- = \frac{R_{1-}}{R_{1-}+R_{2-}} \$ Here is what i tried to do, since \$ V_s = A_0 \varepsilon , \varepsilon = V_+ - V_- = k_+ V_s - k_- V_s = V_s(k_+ - k_-)\$ but i dont know how can i relate the saturation to the difference between the two k-factors

Response for Andy For a standard inverting amplifier, we have \$i_- = -i_s \iff \frac{V_-}{R_1} = \frac{V_s}{R_2} \iff V_s = \frac{-R_2}{R_1}V_- \$

  • \$\begingroup\$ As it is drawn, R1 and R2 form a Schmitt trigger, as it is positive feedback. This will saturate the output. \$\endgroup\$
    – Sparky256
    Commented Mar 13 at 20:54
  • \$\begingroup\$ hey sparky, thanks for your insight, i dont understand which R1,R2 pair are you talking about, infinite thanks :) @Sparky256 \$\endgroup\$ Commented Mar 13 at 20:55
  • \$\begingroup\$ The ones connected to the + input. Omit them if you want normal op-amp behavior. The - input sets the gain based on the ratio of R2/R1. \$\endgroup\$
    – Sparky256
    Commented Mar 13 at 21:00
  • 1
    \$\begingroup\$ ah yes i understand that i should omit the resistances at the (+) terminal, but the problem asks when the presence of both feedbacks, the k+ should be inferior to the k-, and it asks for a proof. @Sparky256 \$\endgroup\$ Commented Mar 13 at 21:02
  • \$\begingroup\$ in general, crudest model = write it out for a voltage gain of A, and then take lim A->infinity. \$\endgroup\$
    – Pete W
    Commented Mar 13 at 21:51

3 Answers 3


I want to prove that this system is stable

If you were deriving the gain equation for a standard inverting op-amp you would recognize that the inverting input is ideally at the same potential as the non-inverting input and, that is still true for your circuit up until the point at which positive feedback exceeds negative feedback.

So, assume that inverting input voltage equals the non-inverting input voltage and all will come clear.

EDIT (as requested by OP and alluded to above but simpler)

Consider the non-inverting op-amp and my amendment on the right: -

enter image description here

Notice that \$V_P\$ is now derived from \$V_{OUT}\$: -

$$V_{OUT} = V_P\left(\dfrac{R1+R2}{R2}\right)\hspace{0.5cm} = \hspace{0.5cm}V_{OUT}\left(\dfrac{R4}{R3+R4}\right)\left(\dfrac{R1+R2}{R2}\right)$$

This is manipulated to: -

$$V_{OUT}\left[1 - \left(\dfrac{R4}{R3+R4}\right)\left(\dfrac{R1+R2}{R2}\right)\right] = 0$$

So, \$V_{OUT}\$ will remain stably at 0 volts unless R1, R2, R3 and R4 form a composite value (within the square brackets above) that equal unity (at which \$V_{OUT}\$ becomes indeterminate). To show this more clearly, we can make the simplification that \$R3+R4 = R1+R2\$. Then we are left with R4 and R2 as independent values: -

$$V_{OUT} \left[1 - \dfrac{R4}{R2}\right] = 0$$

So, if R4 is the same value as R2 we have equal values of positive and negative feedback and the result for \$V_{OUT}\$ is indeterminate. If R4 is less than R2 we have the stable case of \$V_{OUT}\$ remaining at 0 volts.

If R4 is slightly less than R2 (and we have a real circuit with noise and offsets), we can see that \$V_{OUT}\$ will start to acquire significantly more noise and offsets.

  • \$\begingroup\$ ah hello andy, i just wish if you could mathematically express what you want me to do to attain the point where i see that the positive feedback exceed the negative feedback, thanks \$\endgroup\$ Commented Mar 13 at 21:45
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    \$\begingroup\$ Can you do this for a standard inverting amplifier? If so, please add it to your question as a recognizable new section at the end. Then I'll help you take it to the full circuit in your question. \$\endgroup\$
    – Andy aka
    Commented Mar 13 at 22:03
  • \$\begingroup\$ Yes i did it now \$\endgroup\$ Commented Mar 13 at 22:08
  • \$\begingroup\$ Not quite enough; you need the formula to include the voltage at the non-inverting input (call it v+) as an independent signal to the system. Then you replace that with the potential divider from the output to finish the job off. Sorry for not being explicit enough. \$\endgroup\$
    – Andy aka
    Commented Mar 13 at 22:25
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    \$\begingroup\$ Not tonight, it's too late for me now but, send a comment tomorrow for more help \$\endgroup\$
    – Andy aka
    Commented Mar 13 at 23:16

Here are some labels to help:


simulate this circuit – Schematic created using CircuitLab

The goal is to determine whether the arrangement is stable (negative feedback dominates) or unstable (positive feedback dominates). Formally, we consider if output potential \$V_Z\$ at Z were to be perturbed, how that would change \$V_P\$ and \$V_Q\$, and how those changes would further influence the output.

We assume that the output \$V_Z\$ is related to open loop gain \$A\$ and inputs \$V_P\$ and \$V_Q\$ as follows:

$$ V_Z = A(V_P-V_Q) $$

From that we ascertain that if \$V_P-V_Q\$ were to increase (change towards \$+\infty\$), then \$V_Z\$ also increases (moves towards \$+\infty\$). Conversely, if \$V_P-V_Q\$ decreases (changes towards\$-\infty\$), then \$V_Z\$ will also decrease (change towards \$-\infty\$).

The direction of perturbation of \$V_Z\$ is important, as is the resulting direction of change in \$V_P-V_Q\$. If an increase of \$V_Z\$ would result in \$V_P-V_Q\$ decreasing, that would effectively tend to cancel the initial perturbation, since the output would be changing in a direction opposing that perturbation. This would consititute negative feedback, which tends to induce stability.

By contrast, if a perturbation of output \$V_Z\$ causes input \$V_P-V_Q\$ to change in such a way that would push \$V_Z\$ further in the same direction, then that is positive feedback, a condition which will result in the output swinging wildly off to some extreme.

To determine which of those two possible conditions prevails, we need to hold potentials \$V_A\$ and \$V_B\$ steady, at some arbitrary values, and evaluate how \$V_P-V_Q\$ changes as a function of some change in \$V_Z\$. To simplify the algebra, I'll set \$V_A=0\$ and \$V_B=0\$:


simulate this circuit

The op-amp's inverting and non-inverting inputs (P and Q) don't draw any current, so the op-amp itself can be considered a voltage source producing \$V_Z\$, but may otherwise be disregarded. The resistors therefore form simple potential dividers between Z and ground.

$$ \begin{aligned} V_P &= V_Z\frac{R_1}{R_1 + R_2} \\ \\ V_Q &= V_Z\frac{R_3}{R_3 + R_4} \\ \\ V_P - V_Q &= V_Z\left( \frac{R_1}{R_1 + R_2} - \frac{R_3}{R_3 + R_4} \right) \\ \\ \end{aligned} $$

For negative feedback, stability, you need to show that a change in \$V_Z\$ causes a change in \$V_P-V_Q\$ in the opposite direction. In other words, you must show that \$V_Z\$ and \$V_P-V_Q\$ have opposite signs, which I think you can handle from here.

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    \$\begingroup\$ this is by far the best answer i've seen thanks simon! \$\endgroup\$ Commented Mar 14 at 21:24

Knowledge Seeker was asking: "I just wish if you could mathematically express what you want me to do to attain the point where i see that the positive feedback exceed the negative feedback"

The answer is simple- just write down the closed-loop gain for Ve2=0 (your first diagram) using the feedback factor (two portions) Hfb=-R1/(R1+R2)+R3/(R3+R4). (As you see, I have used another notation for the positive part of Hfb).

With input damping (forward factor) Hfw=-R2/(R1+R2) the closed-loop gain is

Acl=(Hfw * Aol)/(1 - Hfb * Aol) and

Acl=-Hfw/Hfb for Ao approaching infinity.

Therefore (after inserting Hfw and Hfb):


(For R3=0 the above equation reduces to the known gain of an inverting opamp Acl=-R2/R1).

As you can see, when the positive feedback portion R3/(R3+R4) in the denominator exceeds the negative portion, the resulting gain Acl would be positive. However, this is a contradiction to the expected negative closed-loop gain Acl. Therfore, we always require R3/(R3+R4)<R1/(R1+R2)


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