# Proving formula for feedback for operational amplifiers

Positive and negative feedback circuits : let the following circuit

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_- \$$

• As it is drawn, R1 and R2 form a Schmitt trigger, as it is positive feedback. This will saturate the output. Commented Mar 13 at 20:54
• hey sparky, thanks for your insight, i dont understand which R1,R2 pair are you talking about, infinite thanks :) @Sparky256 Commented Mar 13 at 20:55
• 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. Commented Mar 13 at 21:00
• 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 Commented Mar 13 at 21:02
• in general, crudest model = write it out for a voltage gain of A, and then take lim A->infinity. Commented Mar 13 at 21:51

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: -

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.

• 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 Commented Mar 13 at 21:45
• 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. Commented Mar 13 at 22:03
• Yes i did it now Commented Mar 13 at 22:08
• 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. Commented Mar 13 at 22:25
• Not tonight, it's too late for me now but, send a comment tomorrow for more help 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.

• this is by far the best answer i've seen thanks simon! 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):

Acl=+[R2/(R1+R2)]/{[R3/(R3+R4)]-[R1/(R1+R2)]}

(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)