Want to know more about this op-amp configuration

Question

When I see op-amp inverting and non-inverting amp configurations, I see the one on the left. (only inverting circuit is shown).

This op-amp circuit has a gain equal to \$ {-R2}\over{R1} \$.

However, if I do the math I can see that the configuration to the right also works (notice the inverting and non-inverting pins are swapped.) as an inverting ampifier, with gain equal to the same \$ {-R2}\over{R1} \$.

PS: I have checked this in a simulator and found that the circuits are indeed the same and the one on the right works exactly like the one on the left.

Thus my question is why do we use the Left circuit and not the one to the right, if they behave the same?

My derivation is here:

Here is the simulator:

Answer 1

You are somewhat correct. When you analyze both circuits, you come up with the same transfer function:

$$ H=-\dfrac{R_2}{R_1}$$

The problem with the positive feedback one is, if the output is perturbed the slightest, it will drive the output towards one of the rails (+Vcc or -Vcc). In the negative feedback configuration, the feedback network always tries to steer the output towards an equilibrium point.

What you've done is what's called a 'static analysis' of the opamp circuit. But the dynamics of it, won't allow you to use the positive feedback version in the linear region, it will saturate.

A model of the opamp that includes dynamics looks like:

This will show you how the negative and positive feedback affect the behavior of the circuit. Now consider the following:

All we have here is an opamp with both negative and positive feedback and the dynamic model will show what happens.

If you go through the math and find the differential equation regarding the output voltage (\$v_o\$) you find that:

$$ \dfrac{dv_0}{dt}+\dfrac{v_o}{T}=0$$

Where \$T=\dfrac{RC}{A(\gamma^--\gamma^+)}\$

And \$\gamma^-=\dfrac{R_3}{R_3+R_4}\$ (Negative feedback network),

\$\gamma^+=\dfrac{R_1}{R_1+R_2}\$ (Positive feedback network)

You can find all the details of how things were derived here, this is an MIT paper.

So back to the equation, if:

$$\gamma^->\gamma^+ $$ This means the net feedback is negative and therefore \$T\$ is positive, this results in

$$v_o = Ke^{-\dfrac{t}{T}} $$ And this is an stable solution!

if: $$\gamma^-<\gamma^+ $$ then the net feedback is positive and the solution to the DE becomes: $$v_o = Ke^{\dfrac{t}{T}} $$ This is an unstable solution since the output voltage will grow unbounded.

Keep in mind that we don't have anything driving the input. This just shows how perturbations will drive the output voltage to either an equilibrium point or to saturation. The constant K would come from initial conditions.

Answer 2

The problem is that you are assuming that you can achieve stable conditions with positive feedback. You can't. The transfer function is non-linear. Your circuit is a non-inverting Schmitt trigger.

With the response time of the op-amp \$ V_P \$ will always lag \$ V_O \$, the error will magnify and the positive feedback will drive the amplifier to one rail or the other.

There's more detail and maths on page 11 of Operational Amplifier Circuits Comparators and Positive Feedback by Chaniotakis and Cory. 6.071 Spring 2006.

See also How are positive and negative feedback of opamps so different? How to analyse a circuit where both are present?.

Answer 3

It's difficult to deduce that the positive feedback configuration is unstable, particularly if we ignore the dynamics of the op amp and just work with gains. So, let's include the dynamics in their simplest form - a 1st order lag - by letting the TF of the op amp be: $$\small \frac{V_O}{V_P}=\frac{K}{1+\large \tau s}\:\:\:\:\:\:...\:(1) $$

where \$\small \tau\$ is a small valued time constant, and \$\small K\$ is the (large) op amp gain.

Also, for simplicity, but without detracting unduly from rigour, let \$\small R_1=R_2\$.

With zero current into the non-inverting terminal, \$\small R_1\$ and \$\small R_2\$ form a potential divider, and we can write \$\small V_P\$ as: $$\small V_P=\frac{V_O+V_S}{2} $$

Substituting for \$\small V_P\$ in equation (1), and re-arranging:

$$\small V_O=\frac{K}{1+\large \tau s}\times \frac{V_O+V_S}{2}$$

Solving for the overall TF of the configuration:

$$ \small \frac{V_O}{V_S}=\frac{K/2}{(1-K/2)+\large \tau s}$$

\$\small K\$ is the op amp gain, hence \$\small K/2\$ is large compared with 1, and the TF reduces to:

$$ \small \frac{V_O}{V_S}=\frac{K/2\large \tau}{-K/2\large \tau+ s}$$

which has a right-half pole, \$\small s=K/2\large \tau\$, and is, therefore, unstable.

Answer 4

Consider the circuit on the left: This is a standard inverting op-amp circuit with gain \$ - \frac{R_2}{R_1} \$. Lets assume for now this was not quite true and the output was too high. As a result the inverting input is higher than it should be and the output falls. If the output falls too far then the output is lower than it should be and the output rises. We have negative feedback and any error is self correcting. This is an inverting op-amp and works.

Now consider the circuit on the right: Arguing that both inputs are the same we get gain \$ - \frac{R_2}{R_1} \$ as before. However, lets assume for now this was not quite true and the output was too high. As a result the inverting input is higher than it should be and the output rises. This is the opposite of what we want: we have positive feedback and this leads to saturation.

The circuit on the right does not work as an inverting amplifier.