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_- \$
Best Answer
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.