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?

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.

  • \$\begingroup\$ Why can't I use the linear gain if the system is unstable? As far as I know, LTI systems can be unstable, but that doesn't mean they don't have a transfer function. Or maybe I am mixing up things here? \$\endgroup\$ – Tendero Nov 12 '16 at 19:22
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    \$\begingroup\$ A system that is unstable has no fixed DC operating point where the transfer characteristic (from input to output) could be linearized. But this is the precondition for defining small-signal parameters like "gain", "input resistance",.... \$\endgroup\$ – LvW Nov 13 '16 at 10:28

It all comes down to how you interpret the gain equation.

In an amplifier, things do not happen instantaneously. Extremely quickly - yes - but there is always a small time delay, \$\delta t\$, (or a lag) before the input is operated on to produce the output.

To take account of this, write the gain equation as: $$y(1-af)=ax$$

where \$x\$ and \$y\$ are the input and output voltages, respectively.

Now, in the passage through the amplifier, \$y\times af\$ and \$a \times x\$ are subject to the delay, \$\delta t\$, and the equation may therefore be written: $$y_n-(af\times y_{n-1})=a\times x_{n-1}$$ or $$y_n=(a\times x_{n-1})\:+\:(af\times y_{n-1})$$

where the \$n\$ subscript means the current value of time, and \$(n-1)\$ means the previous value of time, \$\delta t\$ earlier.

If, now, you take an input of 1 volt, \$a=10,\: f=0.5, \: af=5\$, and calculate the \$y\$ values as time proceeds, you get: $$y_1=10$$ $$y_2=10+50=60$$ $$y_3=10+300=310$$ $$y_4=10+1550=1560\:\: ...$$ which is unstable.

If, however, you take an input of 1 volt, \$a=10,\: f=0.01, \: af=0.1\$, and calculate the \$y\$ values as time proceeds, you get: $$y_1=10$$ $$y_2=10+1=11$$ $$y_3=10+1.1=11.1$$ $$y_4=10+1.11=11.11\:\:...$$

which is stable

Clearly, we can let \$\delta t\$ be as small as we wish; it just means that the difference equation is executed more rapidly and the output reaches it's final state, be it finite or infinite, more quickly. In practice, the amplifier's characteristics dictate the speed of response.

  • \$\begingroup\$ Shouldn't y1=0, since x[n-1]=0 and ax=10*0? Otherwise there would have to be initial conditions specified. \$\endgroup\$ – a concerned citizen Nov 13 '16 at 7:06
  • \$\begingroup\$ Not if x[0]=1, y[0]=0 \$\endgroup\$ – Chu Nov 13 '16 at 13:46

You almost gave yourself the answer. As \$A=-2.5\$ violates the prerequisite of a positive feedback, your example is not - so called - well defined. The formula is only valid, if the fed back output is positive.

Just play your example by going some rounds through your loop. You'll see, that the output gets bigger and bigger.

  • \$\begingroup\$ Why do you say that? As I see it, that value of A would mean that whatever I put in the input will apear at the output inverted and multiplied by a scalar. That doesn't seem unstable to me. What am I thinking wrong? \$\endgroup\$ – Tendero Nov 11 '16 at 21:37
  • \$\begingroup\$ The negative value of A only gives proof, that the formula is not suitable for your case. Try and feed the loop of your example with a value of 1. After the gain stage, the output is 10. Feeding it back adds 5 to the first 1. So the new input for the gain stage is 6, makes 60 at the output. ... And so on, increasing the numbers rapidly. Nothing negative, just out of scope of the closed-loop gain formula. \$\endgroup\$ – DPF Nov 11 '16 at 21:43
  • \$\begingroup\$ I see that now. But why does that imply that the system is unstable? The fact that I can't use a formula doesn't mean that the output will diverge, does it? \$\endgroup\$ – Tendero Nov 11 '16 at 21:48
  • \$\begingroup\$ Mmm... diverging for DC = frequency of 0 Hz ? \$\endgroup\$ – frr Nov 12 '16 at 9:08
  • \$\begingroup\$ I've seen the Nyquist's stability criterion formulated in various ways. For the lay man who still has a problem with the Laplace domain (like me :-) possibly the following explanation is understandable: observed in the frequency domain, the system becomes unstable if you can find any frequency where open-loop phase shift is an integer multiple of 360 degrees (including 0 degrees) and open-loop gain greater than 1. This amounts to the definition of an "effectively positive feedback" (at DC or at some non-zero frequency). \$\endgroup\$ – frr Nov 12 '16 at 9:17

There are different ways to think of stability. The example of frequency stability you use is one. A more general definition of stability is if a system is perturbed, does it come back to the starting point. The classic example of this is a ball bearing in the bottom of a bowl. This is a stable system. If the bowl is given a little shake, the ball will move but return to the bottom of the bowl. Now if the bowl was turned over, the ball could be balanced right at the top of the bowl. Say the bowl is a is a half dome with no flat. Any little perturbation will cause the bearing to roll away. It is not stable.

Your example of a comparator is an example of an unstable system that is useful. It is useful because it is limited by the realities of the physical world. It stops at the voltage rails. So a comparator is like the bowl if we said it could control whether the bowl is tilted to the right or the left. So not being stable is not the same as not being useful.

If you look at a voltage regulator, it is stable. If you increase the load the voltage will try to stay the same. If you remove the additional load, the regulator will return to where it started.

So if you put any little input into your a=10 f=0.5 system the algebra says the gain is -2.5. But if you put 1V in you are not going to get -2.5 Volts out. You are going to get a ball bearing rolling around on the floor, unstable.

  • \$\begingroup\$ input = 1, other input of summing junction = -5/4, output of summing junction -1/4, output of system = -5/2. It works, so why is it unstable? \$\endgroup\$ – Adil Malik Nov 11 '16 at 22:46
  • \$\begingroup\$ Your numbers are just the ball bearing at the top of the upside down bowl. If you shake it, you'll never put it there again. \$\endgroup\$ – owg60 Nov 12 '16 at 0:18
  • \$\begingroup\$ Indeed, so that proves the point that af>1 does not guarantee or imply certain instability in anyway, it just shows it might be possible under certain circumstances. \$\endgroup\$ – Adil Malik Nov 12 '16 at 0:22
  • \$\begingroup\$ So if you balance a broom on the palm of your hand for a second you call that a stable system? \$\endgroup\$ – owg60 Nov 12 '16 at 0:30
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    \$\begingroup\$ No i agree that in reality this might be an unstable system, but as the OP said he doesnt see how the math he presented proves this notion, and as it stands it doesnt prove it. I like the bearing in a bowl analogy! \$\endgroup\$ – Adil Malik Nov 12 '16 at 0:33

The circuit simulator that I commonly use (PSPICE) shows results for AC Analysis with no distinction between positive and negative feedback - circuits are strictly linearized around a bias point. Results are valid for the positive-feedback case, at least for that brief span of time before a circuit runs off the rails.
Circuits with positive feedback yielding hysteresis are non-linear and fail to give hysterical results in SPICE's AC analysis.
SPICE Transient Analysis does give a proper result for all positive-feedback circuits, including bias point shift as non-linearities come into play. But in this case, your simple feedback equation blows up - it is strictly limited to linear circuits, just as SPICE's AC analysis.
SPICE's AC analysis is useful in seeing the operating region around the point where your equation's denominator goes to zero. For example, regenerative radio-frequency amplifiers (where +ve feedback is applied to an inductor/capacitor resonator) can be examined with AC analysis. Circuit Q does become infinite at the critical transition point, and is less-than-infinite for too little feedback, as well as too much feedback. But look at results carefully - AC analysis often displays magnitudes (by default) and it is often not clear if you have an amplifier or an oscillator.

  • \$\begingroup\$ I can confirm that AC analyses in SPICE-based simulators show "normal" gain properties even for positive feedback - as far as the magnitude is concerned. However, the "un-normal" phase behaviour (rising phase for falling gain) is an indication for instability. \$\endgroup\$ – LvW Nov 12 '16 at 14:56

The best way to understand the differences between negative feedback and positive feedback is study the case when feed-forward block is dynamic system then extend the results to ideal linear block, let suppose that the transfer function of feedforward system is:

  • The Feedforward block has a single pole at \$s=-p_1\$:

$$ A(s)=\dfrac{a_0}{1+\dfrac{s}{p_1}} $$

  • When we use the negative feedback, the closed loop transfer function is: $$ T(s)=\dfrac{A(s)}{1+\beta A(s)}=\dfrac{a_0}{1+\dfrac{s}{p_1}+a_0\beta} $$ the pole is shift from \$s=-p_1\$ to \$s=-(1+a_0\beta)p_1\$ as the pole in LHS, the system is stable. the Ideal linear block is the same, but with \$p_1\rightarrow \infty\$, whatever the value of \$p_1\$ the system is unconditionally stable as the value of p_1,\beta and a_0 >0.
  • In the case of positive feedback, the closed loop transfer function is $$ T(s)=\dfrac{A(s)}{1-\beta A(s)}=\dfrac{a_0}{1+\dfrac{s}{p_1}-a_0\beta} $$

this system in this case is not stable if new pole is in the RHS. $$s=-(1-a_0\beta)p_1$$ the stability of the closed loop system dictate that the \$(1-a_0\beta)>0\$ which means: $$a_0\beta<1$$ from these results we can understand the Nyquist criteria of stability.

In conclusion, the ideal linear block is a linear system with pole locate in the negative infinity, and the stability of the closed loop system depend on the sign of \$1-a_o\beta\$.


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