Op-amp: finding out how it works

Question

With both feedbacks, why do I use something like "stronger negative feedback" or "stronger positive" feedback?

I've read that when I supply a negative potential to the negative input, the output must be positive, but it was when the positive input was grounded.

But when it is not grounded, then V+ is unknown, and V- is also unknown (before V+ was grounded so V- was known). So why can't V- have a positive potential when I have a negative potential on Vin which is connected to the negative input? I don't know how it works.

The output is positive if I supply a negative voltage to the negative input, but V- is equal to 0 V, so looking at this: Aol(V+ - V-). So why can't V- be positive when I have a negative source connected to a negative input ?

My professor told me that Vout can be anything, even if I supply a negative potential on V-.

I tried to use this equation: Vout = Aol(V+ - V-). But I don't understand it when I tried to understand both feedbacks.

Also (second question), why do we say that when we have only negative feedback and we supply a negative potential to V-, then the output must be positive? V- is = 0 V because of V+ which is also grounded (I'm talking here about a different simple op-amp with only negative feedback). V- is 0 V so it is positive nor negative, so why does the output act like that? And why, with both feedbacks, we look at it as if V- is positive, even though it has a negative source connected to it, then positive feedback is "stronger", so everything goes into saturation.

Calculation for V+ and V-:

And a different version:

The last one is wrong even though I have calculated V- and V+. The answer I got was "because the positive feedback is stronger". I don't get it. Also I was told that Vin is connected to V+, so the output must be positive. So why can't V+ be negative?

Answer 1

why do I use something like "stronger negative feedback" or "stronger positive" feedback?

I don't know why you "use" something like that. It sounds like nonsense, and is not helpful.

I tried to use this equation: $$V_{OUT}=A_{OL}\left(V_+-V_-\right).$$

Great! That's all you need.

Suppose we have both inputs driven with some voltages:

^{simulate this circuit – Schematic created using CircuitLab}

Both feedback circuits are simple voltage dividers, thus $$ V_-=V_1 + (V_{OUT}-V_1) \cdot Z_1/(Z_1+Z_2),\\ V_+=V_2 + (V_{OUT}-V_2) \cdot Z_3/(Z_3+Z_4). $$

To simplify, let's substitute \$g_n=Z_1/(Z_1+Z_2)\$ and \$g_p=Z_3/(Z_3+Z_4)\$ into the above:

$$ V_-=V_1\cdot(1-g_n) + V_{OUT}\cdot g_n,\\ V_+=V_2\cdot(1-g_p) + V_{OUT}\cdot g_p. $$

Now we can substitute these voltages back into the original equation:

$$\begin{aligned} \frac{V_{OUT}}{A_{OL}} &= V_2(1-g_p) - V_1(1-g_n) + V_{OUT}(g_p-g_n) \\ V_{OUT}\left( \frac{1}{A_{OL}} + g_n - g_p \right) &= V_2(1-g_p) - V_1(1-g_n) \\ V_{OUT} &= \frac{1}{1/A_{OL} + g_n - g_p} \left[ V_2(1-g_p) - V_1(1-g_n) \right] \\ \end{aligned}$$

Now we need to take the limit to get the DC output value for an ideal op-amp with "infinite" DC open loop gain:

$$\begin{aligned} \lim_{A_{OL}\to \infty} V_{OUT} &= \frac{V_2(1-g_p) - V_1(1-g_n)}{g_n - g_p} \\ &= V_2 \frac{1-g_p}{g_n - g_p} - V_1\frac{1-g_n}{g_n - g_p}. \end{aligned}$$

Note how when \$g_n \approx g_p\$, the gain tends to infinity.

Answer 2

Using NFB and PFB

It's really nice that you are writing so much in response to others. You are trying at something. I'm not sure what it is, but I'm going to pick up a few of your bread-crumbs and avoid writing what others have already written to see if this kind of different approach gets there for you.

I'm going to look at this as negative and positive feedbacks as part of a feedback system. Maybe that's what you've been grasping at. If not, I'll just add another answer to the heap of answers that have failed you. But at least I tried.

Let's dig into it:

^{simulate this circuit – Schematic created using CircuitLab}

On the left is my schematic. The topology is the same as in Kuba's answer and is a generalization of your schematics (neither \$v_{\text{a}}\$ nor \$v_{\text{b}}\$ is grounded, but instead left to be assigned any arbitrary voltage value.)

On the right is a feedback system that is equivalent. I've not yet specified the values of each gain block (except for the \$A_{_\text{OL}}\$ block, which we agree is some value specifying the open loop gain of the opamp.)

(I've chosen very different naming conventions for the voltage nodes, intentionally. I want to avoid any confusion with my writing, your writing, and the writing of others.)

It's not difficult to see that \$B_1=\frac{R_2}{R_1+R_2}\$, \$B_2=\frac{R_1}{R_1+R_2}\$, \$B_3=\frac{R_4}{R_3+R_4}\$, and \$B_4=\frac{R_3}{R_3+R_4}\$. (Please take a moment to see why each of those is correct.)

With all that in mind, then we have:

$$\begin{align*} v_{\text{z}}&=A_{_\text{OL}}\left[v_{\text{p}}-v_{\text{m}}\right] \\\\ &=A_{_\text{OL}}\left[\left(v_{\text{a}}\,B_1+v_{\text{z}}\,B_2\right)-\left(v_{\text{b}}\,B_3+v_{\text{z}}\,B_4\right)\right] \\\\ &\therefore \\\\ v_{\text{z}}\left[1-A_{_\text{OL}}\,B_2+A_{_\text{OL}}\,B_4\right]&=A_{_\text{OL}}\left[v_{\text{a}}\,B_1-v_{\text{b}}\,B_3\right] \\\\ &\therefore \\\\ v_{\text{z}}&=A_{_\text{OL}}\frac{v_{\text{a}}\,B_1-v_{\text{b}}\,B_3}{1-A_{_\text{OL}}\,B_2+A_{_\text{OL}}\,B_4} \\\\ &=\frac{v_{\text{a}}\,B_1-v_{\text{b}}\,B_3}{\frac1{A_{_\text{OL}}}-B_2+B_4} \end{align*}$$

Now, this result is quite similar-appearing to Kuba's. And you can see, quickly, where this leads as \$A_{_\text{OL}}\to\infty\$, as well.

I just took a slightly different approach, using NFB and PFB blocks to get there.

Negative input values

You have a separate question, I think, regarding positive and negative values at the inputs (\$v_{\text{p}}\$ and \$v_{\text{m}}\$.) This is really a completely different question.

An opamp has two additional pins, one for the positive supply rail it is provided and another for the negative (relatively speaking) supply rail it is also provided. These can both be positive -- for example, it is fine to provide \$+30\:\text{V}\$ to the positive supply rail and \$+10\:\text{V}\$ to the negative supply rail. The actual values don't matter. Just the relationship that the positive rail has to be higher (more positive) than the negative supply rail.

The output of the opamp cannot be more positive than the provided positive supply rail nor more negative than the provided negative supply rail. So all these rails do is limit the range of the output.

(Usually, there are some additional 'overhead' amounts which further bracket the range of the opamp output. But there are many rail-to-rail output opamps where the output can almost reach both provided rails.)

The inputs to the opamp also need to be within this same range. And themselves may also be further limited (just like the output is.) But let's say that we provide \$+15\:\text{V}\$ to the positive supply rail of the opamp and provide \$-15\:\text{V}\$ to the negative supply rail of the opamp. And let's further add that the opamp itself has an additional limit that the inputs cannot be closer than \$2\:\text{V}\$ to the rails. Then this means that the inputs can still be anywhere from \$-13\:\text{V}\$ to \$+13\:\text{V}\$. So negative values are just fine, in this case.

Note also that the opamp is not provided access to the ground reference here. It may be the case that the provided negative rail to the opamp is ground. In this case, the opamp still works fine. But the limitations mentioned above still apply. So if we provide \$+15\:\text{V}\$ to the positive supply rail and provide \$0\:\text{V}\$ to the negative supply rail, and if the additional limit remains that the inputs cannot be closer than \$2\:\text{V}\$ to the rails, then the inputs must be between \$+2\:\text{V}\$ and \$+13\:\text{V}\$ for the opamp to behave well.

Notes

I think we already agree (you already worked this out) that \$R_2\,R_3\ne R_1\,R_4\$.

But there's another issue.

If a tiny positive change at \$v_{\text{a}}\$ (assume that \$v_{\text{b}}\$ doesn't change) causes a positive change at \$v_{\text{z}}\$, then this will feed back in a positive way and result in a runaway situation. So that cannot be allowed.

So this adds an additional constraint.

It means that \$\frac{\text{d}}{\text{d} \, v_{\text{b}}}\,v_{\text{z}}=-R_4\frac{R_1+R_2}{R_2\,R_3-R_1\,R_4}\$ must be negative. The only way that happens is if \$R_2\,R_3\gt R_1\,R_4\$. (This condition simultaneously guarantees that \$\frac{\text{d}}{\text{d} \, v_{\text{a}}}\,v_{\text{z}}=R_2\frac{R_3+R_4}{R_2\,R_3-R_1\,R_4}\$ is positive, which also needs to be true.)

In short, it's not only \$R_2\,R_3\ne R_1\,R_4\$ but the still more restrictive case where \$R_2\,R_3\gt R_1\,R_4\$. If that's not met, then \$v_{\text{z}}\$ will go to a rail and stay there (and the two inputs will almost certainly not be close to each other.)

Also, you seem to imagine that the \$V_{\left(-\right)}\$ input has to be negative for \$V_{_\text{OUT}}\$ to be positive. But that's not a true statement. What is true is that for \$V_{_\text{OUT}}\$ to be positive it must be that \$V_{\left(-\right)}\lt V_{\left(+\right)}\$. (And not by much, as \$A_{_\text{OL}}\$ is very large!)

Assume \$A_{_\text{OL}}=100\:\text{k}\$ and \$V_{\left(-\right)}=+100\:\mu\text{V}\$ and \$V_{\left(+\right)}=+110\:\mu\text{V}\$, then \$V_{_\text{OUT}}=+1\:\text{V}\$. Note that all of the voltages are positive with respect to ground.

Simulation

Suppose I decide to provide two sinusoidal inputs that are \$1\:\text{V}\$ peak-to-peak, with one of them exactly \$180^\circ\$ out of phase with the other. I also want the output to be \$10\:\text{V}\$ peak-to-peak. By using the above equations I can find that an example circuit to achieve this could be:

(In blue, above, I show the exactly calculated value for \$R_4\$.)

Now, here is what happens if I center both inputs at \$0\:\text{V}\$:

Let's now center them around \$+5\:\text{V}\$:

And now let's center them around \$-5\:\text{V}\$:

I think you can see that \$V_{\left(-\right)}\$ and \$V_{\left(+\right)}\$ can be always negative, or always positive, or sometimes one and sometimes the other... and the circuit still works fine.