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enter image description here

I ask this because when I do the nodal analysis for the OP amp in figure 3, I end up with the same results as the inverting amplifier, even though the inverting terminal in figure 3 is grounded. So why do we put so much emphasis on which terminal we are using, since we can obtain the same results by changing the position of the ground, source and feedback around?

Are there differences between the Op amp in figure 3 and the traditional inverting amplifier from figure 2? Why do we use one but not the other?

Lastly, I cannot grasp the reason why non - inverting amplifiers in figure 2 are not stabilizers like inverting amplifiers. Perhaps I am having trouble with the concept of stabilizing the signal. Does it mean that non - inverting amplifiers will saturate the signal? Well, non - inverting amplifiers may still not saturate signals depending on feedback according to figure 2, although the amplification will be grater than 1. Does amplification greater than 1 implies amplificating any noises considerably, and that's one of the reasons why non-inverting amplifiers are considered not to be stabilizers?

Thank you very much

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  • \$\begingroup\$ What do you mean with "stabilizers"? \$\endgroup\$ – LvW Feb 8 '15 at 21:12
  • \$\begingroup\$ I have been told that inverting Op amps act as stabilizers while non inverting do not. I believe they were using the word to describe the ability of not saturating (they mentioned for example how some op amps will saturate and remain in that region). But perhaps that's not the right word to describe such a thing. \$\endgroup\$ – sdarella Feb 8 '15 at 22:08
  • \$\begingroup\$ Each opamp goes into saturation if the output amplitude reaches the limits set by the supply voltages. There is no difference between inverting and non-inv. configurations. More than hat, saturation has nothing to do with stabilization. \$\endgroup\$ – LvW Feb 9 '15 at 7:12
  • \$\begingroup\$ @sdarella, have a look at my answer to your question, I believe it helps you understand it. \$\endgroup\$ – Eric Best Mar 2 '15 at 10:24
  • \$\begingroup\$ @EricBest thanks for taking the time to write such good answer. Sorry it took me so long to see it, I had not checked back in a while. It was a nice surprise. Thanks again. \$\endgroup\$ – sdarella Apr 2 '15 at 0:37
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Let's take into account a "Basic Feedback System" (hereinafter BFS) block diagram first:

schematic

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)

schematic

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)

schematic

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.

schematic

simulate this circuit

enter image description here

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}) \$)

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  • 1
    \$\begingroup\$ Unfortunately, unreadable. For schematics, use CircuitLab. For math, use MathJax/LaTeX. Both are part of the editor for answering questions. \$\endgroup\$ – tcrosley Feb 24 '15 at 18:12
  • \$\begingroup\$ @tcrosley, I suppose you are satisfied now :). \$\endgroup\$ – Eric Best Mar 2 '15 at 1:13
  • \$\begingroup\$ @tcrosley, I was a few days offline (a place without Internet) so I was able to repair it only now. \$\endgroup\$ – Eric Best Mar 2 '15 at 1:21
  • \$\begingroup\$ Very nice. I know it's a lot of work. Retracted my downvote. \$\endgroup\$ – tcrosley Mar 2 '15 at 5:59
  • \$\begingroup\$ Wow, this is an awesome answer that deserves more attention. I wish I had seen this a year ago! \$\endgroup\$ – Greg d'Eon Mar 10 '15 at 15:30
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when I do the nodal analysis for the OP amp in figure 3, I end up with the same results as the inverting amplifier, even though the inverting terminal in figure 3 is grounded.

Your analysis is incorrect - what you have in figure 3 is a comparator with hysterisis i.e. positive feedback.

Regards your use of the term "stabilizers", this term means very little to me and I've worked with op-amps for many years now!

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  • \$\begingroup\$ I suppose then my question becomes why this kind of analysis only works for very specific configurations. \$\endgroup\$ – sdarella Feb 8 '15 at 19:42
  • \$\begingroup\$ I'm unsure of what analysis you did. \$\endgroup\$ – Andy aka Feb 9 '15 at 8:08
  • \$\begingroup\$ Nodal analysis. \$\endgroup\$ – sdarella Apr 2 '15 at 0:17
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"I ask this because when I do the nodal analysis for the OP amp in figure 3, I end up with the same results as the inverting amplifier,"

Your analysis is correct! That means: You have made no calculation error - however, you have assumed (blindly?) that the opamp would work under linear and stable conditions. And this is not the case.

For the circuit in Fig. 3 you can, indeed, find a voltage- and current distribution which gives a negative output voltage for a positive input. This is even true if you take a non-ideal (finite) opamp gain into account.

It is only the surprising negative gain which is an indication for "something wrong". By the way - even a simulation program (ac or dc analysis) does not show any instability.

(This case can be compared with a mechanical analogon: Two balls - one is riding upon the other one. This could be - theoretically - a stable system, if there is no disturbing influence from outside).

The reason for the surprising result for Fig. 3 is the following: In your calculation, you have assumed that (a) the power was switched on long time ago, hence not producing any switch transients and (b) that there is no time delay caused by the opamp and (c) no disturbing influences. And the simulation program also assumes these ideal conditions (at least for ac and dc analyses).

However, all three conditions, (a) (b) (c), are NOT fulfilled in reality. Therefore, of course the cicuit in Fg. 3 will NOT work in reality. If you want to proof the instability of the circuit, you must do a TRAN analysis (time domain) for a real opamp model that contains a time delay. Moreover, switch on the power supplies at t=0 for producing an inrush transient effect. The output will go - as expected - immediately into saturation.

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  • \$\begingroup\$ Thank you very much for your reply. Yes, I did indeed blindly assume linear operation! I thought that was the reason why we did the analysis - to be able to find values that can yield linear behavior... as in, we choose our resistors and sources so that we can work under saturation (when desired). I think my problem lies in understanding how the assumptions you listed can affect the behavior of the different configurations (for example, the inverting amplifier seems to obey the predictions of the equations regardless of my assumptions while figure 3 does not). \$\endgroup\$ – sdarella Feb 8 '15 at 21:40
  • \$\begingroup\$ What I am interested in is the following: The questioner did consider my answer as sufficient, and somebody else has detected one error - or what was the reason for down-voting? \$\endgroup\$ – LvW Mar 2 '15 at 10:19
  • \$\begingroup\$ The standard analysis of op-amp circuits assumes the two inputs are at the same voltage. But that is only true if the circuit is stable. Circuits 1 & 2 are stable because when the output rises it forces the voltage at the inverting terminal to rise, which makes the op-amp output a lower voltage, thus correcting itself, and vice-versa for the output voltage lowering. Circuit 3 is not stable because if the output voltage rises it increases the voltage at the non-inverting terminal, making the output rise even more. Since it's not stable to begin with the normal analysis is not valid. \$\endgroup\$ – Austin Mar 2 '15 at 11:49
  • \$\begingroup\$ Circuit 3 can theoretically sit at the zero voltage condition, but as LvW said, there will always be something to push it away from that. \$\endgroup\$ – Austin Mar 2 '15 at 11:54
  • \$\begingroup\$ Austin- you are completely correct and - therefore - I agree with you. You speak about a rising signal which reveals the instability. And that`s exactly what I wanted to explain: Only a simulation which contains such a rising quantity (TRAN) allows to reveal the instability. However, an ac analysis does NOT contain such a "imbalance" and, thus, can lead to false interpretations. \$\endgroup\$ – LvW Mar 2 '15 at 12:02
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The answerer (1) is on the right track. With no signal, the circuit (3) will "saturate" at the maximum positive output voltage, depending on the source impedance. This is called: "positive feedback", which is generally not used.

Circuits (1) and (2) have negative feedback, which produces the greatest output accuracy. These are very common. Note that "ground" is arbitrary in that any point in a circuit can be called "ground". It generally depends on the ground of the input and the power circuits.

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If one uses an op amp (or almost any kind of amplifying circuit or device) without negative feedback, small deviations in amplifier behavior may cause huge deviations in output behavior. Adding negative feedback will reduce the available gain, but reducing gain to a fraction of the open-loop gain will cause the overall system behavior to be dominated by the negative-feedback components rather than the amplifier behavior.

In cases where the net feedback is positive rather than negative, circuit behavior will be unstable, and big changes in the output may be caused by arbitrarily small changes in the inputs. Sometimes that is desirable, but often times it isn't.

As an analogy, suppose one has a track with a marble on it, which one can tilt left and right. One wishes to be able to move the marble to any place on the track. If the track is perfectly level, one may move the marble by tilting the track, but it will be very hard to make it stay at any spot. If the track is concave-up, such that movement of the marble toward one end will bias the track's slope toward the other, then for any sufficiently-small amount of slope there will be a stable position for the marble. If the track is concave-down, then the marble will tend to stay at one end or the other until the track is tilted enough to make it move; if the track is tilted enough to get the marble off one end, it will be tilted enough to make the marble roll all the way toward the other.

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