Let's take into account a "*Basic Feedback System*" (hereinafter BFS) block diagram first:

^{simulate this circuit – Schematic created using CircuitLab}

We can write:

\$ V_{OUT}=A \cdot (V_{IN}+ \beta V_{OUT}) \$

Therefore the BFS overall gain:

$$
G= \frac{V_{OUT}}{V_{IN}}=\frac{A}{1- \beta A} \> \> \> \> (=\frac{1}{\frac{1}{A}- \beta})
$$

if ( \$ 1- \beta A \$ ) → 0 , then G → \$ \infty \> \> \$ (the system becomes unstable)

so, for the stability of such a system it is required: \$ \> \> \beta A ≠ 1 \$

It shows that system stability depends on the \$ \beta \$A product - the *open loop gain* (see the *Nyquist stability criterion* for instance for more details).

(For an ideal OpAmp with A → \$ \infty \> \> \$:
\$ \> \> \> G= -\frac{1}{\beta}) \$

Now let's analyze those two cases in question: (starting with case 1; an inverting amplifier)

**A)**

^{simulate this circuit}

\$ v_+ =0 \$

\$ V_{OUT}=A \cdot (v_+ - v_-)=-A \cdot v_- \$

=> \$ v_- = - \frac{V_{OUT}}{A} \$

\$ ( i_1 = ) \$
\$ \frac{V_{IN}-v_-}{R_2} \$ = \$ \frac{v_--V_{OUT}}{R_F} \$ \$ (=i_2) \$

then:

\$ \frac{V_{IN}}{R_2}=v_- \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

Substituting now the above expression for \$ v_- \$, we obtain:

\$ \frac{V_{IN}}{R_2}=- \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

and the overall gain is as follows:

$$
G= \frac{V_{OUT}}{V_{IN}}= \frac{(-1)}{ \frac{1}{A}(1+ \frac{R_2}{R_F})+ \frac{R_2}{R_F}} \> \> \> \> \> (1)
$$

(Note that the denominator of this expression never can be 0! ; presuming A and both \$ R_2 \$ and \$ R_F \$ being positive, of course)

if A → \$ \infty \$ :

\$ G=- \frac{R_F}{R_2} \$

Comparing it now with the BFS:

\$ A'=-A \frac{R_F}{R_F+R_2} \$

\$ \beta = \frac{R_2}{R_F} \$

(here A' stands for /is analogical to/ the A in BFS)

Then:

\$ \beta A'=-A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=-A \frac{R_2}{R_F+R_2}<0 \$ always (provided A>0, of course)

=> always* stable ( \$ \beta A' \$ ≠ 1)

*For "real" OpAmps this may not apply - under certain conditions (the phase angle between \$ V_{OUT} \$ and \$ (v_+ - v_-) \$ changes with rising frequency)

Continuing with the case 3 (positive feedback):

**B)**

^{simulate this circuit}

\$ v_- =0 \$

\$ V_{OUT}=A \cdot (v_+ - v_-)=A \cdot v_+ \$

=> \$ v_+ = \frac{V_{OUT}}{A} \$

\$ (i_1=) \frac{V_{IN}-v_+}{R_2}= \frac{v_+-V_{OUT}}{R_F} (=i_2) \$

=> \$ \frac{V_{IN}}{R_2}=v_+ \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

Substituting now the above expression for \$ v_+ \$, we obtain:

\$ \frac{V_{IN}}{R_2}= \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

and the overall gain is as follows:

$$
G= \frac{V_{OUT}}{V_{IN}}= \frac{1}{ \frac{1}{A}(1+ \frac{R_2}{R_F})- \frac{R_2}{R_F}} \> \> \> \> \> (2)
$$

(Note that the denominator in this case can be 0!)

if A → \$ \infty \$ :

\$ G=- \frac{R_F}{R_2} \$

Now, the limit values of the overall gain G (when A is approaching \$ \infty \$ ) are the same in both the cases A) and B):

$$
G=-\frac{R_F}{R_2}
$$

So it looks like it is the same at first sight...

**BUT!**

Comparing now the current case with the BFS:

\$ A'=A \frac{R_F}{R_F+R_2} \$

\$ \beta = \frac{R_2}{R_F} \$

(here A' again stands for /is analogical to/ the A in BFS)

\$ \beta A'=A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=A \frac{R_2}{R_F+R_2}>0 \$,

so, if \$ \frac{R_F}{R_2}=(A-1) \$ then G → \$ \infty \$ => unstable!

The *exact expressions*, (**1**) and (**2**), **substantially differ** one from another!
I suppose their difference and its consequences are clearly evident from the analysis and the resulting formulas above. Due to usually very high value of A the stable case A) with negative feedback maintains, under the feedback influence, very low voltage between the Op Amp input terminal \$ v_+ \$, which is grounded, and the "live" input terminal \$ v_- \$. The latter is therefore at very low value (close to zero), that's why it is usually called *virtual ground*. (Maybe this "maintenance effect" is what you, sdarella, mean under the "*stabilizer*", am I right?) Unlike with the unstable case B), where the positive feedback leads to either oscillations or output saturation at \$ V_{OUT\_MAX} \$ or \$ V_{OUT\_MIN} \$, depending on the input conditions (see the case C) below).

**C)**

The case (3) with positive feedback can also be used but it works as a *comparator*, with input voltage comparative levels \$ V_{IN\_LH} \$ and \$ V_{IN\_HL} \$ (i.e. input voltages at which the output voltage flips rapidly from a low level (L= \$ V_{OUT\_MIN} \$) to a high level (H= \$ V_{OUT\_MAX} \$) and vice versa, resp.). However, it is usually better to use "real" comparators made/intended right for this purpose.

^{simulate this circuit}

we can write:

\$ \frac{V_{IN}-0}{R_2}= \frac{0-V_{OUT}}{R_F} \$

=> \$ V_{IN}=-\frac{R_2}{R_F}V_{OUT} \$ , (condition: \$ v_+ =0 \$ )

Provided the saturation values of \$ V_{OUT} \$ of the Op Amp are \$ V_{OUT\_MAX} \$
and \$ V_{OUT\_MIN} \$ , we obtain the following:

for \$ V_{OUT\_MIN} (<0) \$:

$$
V_{IN\_LH}=-\frac{R_2}{R_F} V_{OUT\_MIN} (>0)
$$

and

for \$ V_{OUT\_MAX} (>0) \$:

$$
V_{IN\_HL}=-\frac{R_2}{R_F} V_{OUT\_MAX} (<0)
$$

(it's hysteresis is then \$ V_{HYST}=V_{IN\_LH}-V_{IN\_HL}=\frac{R_2}{R_F}(V_{OUT\_MAX}-V_{OUT\_MIN}) \$)

## Best Answer

This is neither an inverting nor a non-inverting op-amp configuration. This configuration looks similar to (but is not in fact) a

window comparator. See Is there a name for this kind of comparator?Normally an op-amp circuit would have a negative feedback path between the output and the inverting input, regardless of whether the signal enters the inverting or non-inverting input terminal. Without that negative feedback to provide closed-loop stability, the op-amp behaves like a comparator, greatly amplifying whatever small difference there is between the inputs, by a large but mostly uncontrolled open-loop gain factor.

Note that there are some internal design trade-offs that favor closed-loop operation versus comparator operation, so most IC manufacturers will market the chip as either an op-amp or a comparator. (Some of the early high-speed op-amps do not like having a large voltage difference between their inputs, because of internal back-to-back diode clamps...) So as a practical matter, op-amps and comparators are not really interchangeable.

Update based on comment from Floris:

Good catch! This is an unusual configuration, and I misread it at first. As drawn, the two sides of the LM393 dual comparator are wired in parallel. I'm not sure if that is intentional or a circuit error, but I think most likely one of the comparators should be flipped.

Here's what I was thinking when I read this schematic:

It's supposed to be a line-following robot, and I see two optical sensors, so probably one sensor detects when the robot is too far to the left and the other sensor detects when the robot is too far to the right. Bias is adjusted manually by trimpots R1 and R2. Then the output voltage from the left and right optical sensors goes into the LM393 comparators IC1A/IC1B, which drive the two motors through Q1 and Q2. Again, there's one motor on the left and one motor on the right, and they should be connected so that if the robot sees the line on the left sensor, then it needs to turn to the right and vice versa.

So given that expectation, I thought that one comparator was testing the left sensor against a threshold voltage and the other comparator was testing the right sensor against the same threshold. Note that there's a dotted line drawn between the "sensor" block and the "voltage" block; that misled me to see the R2/R8/R14 string as setting a bias threshold for both sensors. The way it's drawn it does resemble the window comparator pattern, except that one of the comparators got flipped over -- this violates the normal expectation that higher voltages should be towards the top of the page. So I'm left trying to figure out whether the author intended two parallel comparators driving the same signal, or intended one comparator to be connected with its inputs swapped. That would make some sense, if the comparator itself has enough hysteresis to prevent the robot from zig-zagging when it's already on the straight path.

This highlights why experienced EE's tend to insist on having a clear, properly drawn schematic -- to us, it's not just a wiring diagram, but (if it's drawn well) indicates how the signals are supposed to flow and what to expect. This is talked about in more detail in this question: Rules and guidelines for drawing good schematics