Voltage buffer with ideal Op-Amp (positive feedback) - Instability

Question

I know that the circuits which employ negative feedback are stable, whereas those which employ positive feedback are not stable.

How can I prove that a voltage buffer (ideal op-amp) which uses positive feedback is not stable?

I know only the dual case ( \$ A=+\infty \$ because the op-amp is ideal):

$$v_u=A \, v_{in}=A (v^+ - v^-) = A \, v_i - A \, v_u$$

$$v_u = \frac{A}{1+A} v_i = v_i$$

If I consider:

$$v_u=A \, v_{in}=A (v^+ - v^-) = A \, v_u - A \, v_i$$

$$v_u = -\frac{A}{1-A} v_i = v_i$$

The last result should be wrong.

Thank you so much for your time.

Answer 1

I know that the circuits which employ negative feedback are stable

Any feedback system is potentially unstable. What might seem like negative feedback at low frequencies can easily turn to positive feedback at higher frequencies.

whereas those which employ positive feedback are not stable.

Also, incorrect. A comparotor that uses hysteresis can be regarded as stable when one or the other threshold has been exceeded.

How can I prove that a voltage buffer (ideal op-amp) which uses positive feedback is not stable?

You can derive a stable condition for an op-amp with positive feedback but, if you introduce the slightest bit of noise as an influence, the circuit becomes a massive noise amplifier. If you then model input offset voltage and bias currents and how those change with temperature you get an unpredictable circuit.

Answer 2

Your premise is wrong with real op-amps, as has been pointed out.

To analyze the situations, start with the balanced condition and introduce a small perturbation at the input. If the resulting output change is opposite to that created by the input error you have the possibility of stability, but if it increases the error, that will change the output in the same direction, causing more input error and so on.

The two situations are a bit like a ball sitting at the bottom of a valley vs at the top of a hill. In either case the math works out, but the slightest nudge in the 2nd case leads to the ball rolling down one side or the other. Real circuits (as opposed to ideal or crudely simulated ones) have such nudges built-in in the form of offset voltages, noise, drift and so on.

Answer 3

The problem is, that while you may have found a solution to that single equation, it is no proof, that this point is stable.

You can transform the equation a little bit to make it look more familiar and comprehensible.

$$v_u = \frac{A}{A-1} v_i = v_i$$

The equation is correct. For infinite gain and the OP in linear behaviour there is only one valid solution. The output voltage is something little more than the input voltage. But that solution gives no information about the things that happen, if there are little external influences like an EMI, like a little initial offset of v_u and so on. If you put a marble on top of a hill, you may find a valid mathematical solution for all forces equalling out, but it is no proof for the marble being able to stay there.

To make things more complicated: Your first equation concerning negative feedback isn't a proof for stability either. It provides a solution, nothing more.

Answer 4

The problem is that you are assuming that this formula is stable.

$$v_u=A \, v_{in} $$

It isn't. There will be a slight time-lag between the input and output and the effect will be that the output ramps up feeding back to the input, causing a further increase in the output, which increases the input differential, etc., until you hit one of the supply rails. It might appear to work in a poor simulation using ideal amplifiers but if you introduce a little noise on one of the inputs you will see the effect.

The beauty of negative feedback is that it corrects the error. Positive feedback generally exacerbates it.