gain - Positive feedback and unstability

In Wikipedia's positive feedback article it is stated that given the closed loop gain $$A=\frac{a}{1-af}$$ the system is unstable if \$af>1\$.

I don't really get this. If \$a=10\$ and \$f=0.5\$ (just to give a very simple example), I just see that \$af>1\$ but \$A=-2.5\$, which is not infinite. So what is really happening here?

I know that a system is unstable if the transfer function (i.e. the gain in Laplace domain) has poles in the right-half complex plane. But here, \$A\$ would be a constant so I don't see why unstability would occur.

This question arised when I was trying to analyze a Schmitt trigger using feedback. Quantitavely, I see why the output will go to saturation voltages. I just don't see it mathematically. Suppose that the Op-Amp was ideal (so it has infinite gain and it doesn't depend on frequency). Then why would, mathematically, anything diverge in this circuit, if \$A=\frac{-1}{f}\$ which is a finite value? That's the question that led me to thinking about positive feedback and unstability in general.

To sum up:

• Why is positive feedback often related to unstability?



• Why does \$af>1\$ imply that a system is unstable if using positive feedback?

Answer

Simple answer: The open-loop gain of a=10 indicates a non-inverting (positive) amplifier. However, after applying feedback with af>1 the formula gives a resulting gain A which is NEGATIVE. Did you expect such a result?

For af<1 the gain A is - as expected - still positive; and for af=1 it goes (theoretically) to infinite values (stability limit). That means: For af>1 the amplifier is already "beyond" the stability limit. Hence, you were not allowed to apply the (linear) gain formula for af>1.