The 'classic' non-inverting amplifier configuration uses negative feedback. What happens when the + and - terminals are switched? Based on my pencil-and-paper analysis and a Circuitlab simulation, it seems that the result is the same. Is this correct? Can we make a non-inverting amplifier with positive feedback?

Derivation for the positive feedback non-inverting amplifier: I assume that input impedance is infinite, output impedance is zero, and the output voltage is given by

$$ V_{out} = A(V^{+} - V^{-}) $$

Now using the voltage divider equation,

$$ V^{+} = V_{out} \frac{R_b}{R_a+R_b} $$

and for the negative terminal, we have

$$ V^{-} = V_{in} $$

Substituting into the equation for the output voltage,

$$ V_{out} = A(V^{+} - V^{-}) = AV_{out} \frac{R_b}{R_a + R_b} - AV_{in} $$

$$ V_{out}(1 - A \frac{R_b}{R_a + R_b}) = -AV_{in} $$

$$ V_{out} = \frac{-AV_{in}}{1 - A \frac{R_b}{R_a + R_b}}$$

If the input frequency is relatively low, and both resistors are of a similar order of magnitude, then

$$ A \frac{R_b}{R_a + R_b} >> 1$$

$$ V_{out} \approx \frac{-AV_{in}}{ - A \frac{R_b}{R_a + R_b}} = V_{in} \frac{Ra + R_b}{R_b} $$

Which is the formula commonly given for the output of the classic non-inverting amplifier. So is there any difference between these two circuits? If so, where does my analysis fail? What is a more accurate way of analyzing the positive feedback circuit presented below?

Non-inverting Amplifiers

EDIT: As suggested by Jasen, the positive feedback circuit I posted is a Schmitt Trigger. The Wikipedia article on Schmitt Triggers explains it fairly well. It turns out that they key to analyzing it is to solve for the voltages at the input terminals and then treat the Op-Amp as a comparator. That is, find equations for V+ and V-, and then ask the question "When does V+ > V- hold?"

  • \$\begingroup\$ You need to take into account the dynamics of the op-amp in order to do a stability analysis. See this pdf for an example \$\endgroup\$ Commented Nov 8, 2015 at 6:04
  • \$\begingroup\$ \$V_+\$ is equal to \$V_-\$ when there is a lot of gain in the op-amp and when there is negative feedback through passive components. the output continually adjusts itself to whatever value it needs to be for the voltage difference across \$V_+\$ and \$V_-\$ to be virtually zero. \$\endgroup\$ Commented Nov 8, 2015 at 6:39

2 Answers 2


If you do that the amplifier will amplifiy its own output and the ouput will go high or low and stick there until a sufficiently low or high input is presented, which will make it go the other way (low or high).

In essence you will have created a schmitt trigger.

What the formulas don't show is that the solution for this amplifier configuration describes the metastable point,

Another limitation of those equations is that they don't represent the limited output voltage range of real amplifiers.

  • 1
    \$\begingroup\$ An opamp with positive feedback cannot be used as an amplifier because the quiescent point is at one of the suppl rails - and not within the linear amplification range. You have created a Schmitt trigger (not a Schmitt "buffer") or a bistable multivibrator. \$\endgroup\$
    – LvW
    Commented Nov 8, 2015 at 14:50

1-Note that positive feedback is used (yes) ... to increase gain towards quasi infinity. Remenber the old receiver with super-reaction ? But must be used with some caution. 2- Your calculation is made (quasy steady-state, level 0) with approximations which are incomplete (making this calculus is VERY difficult). Some electronic simulators will see nothing but fire. It is essential to take into account the frequency response as well as the transient response of these components (and "variable" non linear parameters). 3- I suggest you changing simulator for getting best results. I use microcap12 from Spectrumsoftware (free). Making simulations further.


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