I'm reading this article about stability of OpAmps: Link to article.

It says: "If Aβ is less than unity at the high frequencies where phase shift reaches 180°, the high-frequency phase-shifted signals will gradually fade away instead of progressively building up into major oscillations." This is depicted in the picture below.

However, I don't understand this. Shouldn't the system also be unstable at a loop gain less than 1? The 180 degree phase shift in the loop plus the subtraction at the summing point results in positive feedback (See figure 2 below). This means that when a signal is fed into the system (at the control input) it will be attenuated by the loop gain and the phase will be shifted 180 degrees before it arrives back at the summing point. The control signal is still present at the plus input of the summing point so the two signals will be added up resulting in a greater signal than before. When the system keeps doing this the amplitude of the signal will rise to infinity and thus the system will also behave unstable.

My question: Is there an error in my analysis? Why is it that the system is stable at a loop gain < 1 and a phase shift of 180 degrees?

According to the article, if the loop gain (A*B) is smaller than 1 at 180 degree phase shift the system is stable Figure 1

enter image description here Figure 2

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    \$\begingroup\$ I think your error is thinking that \$1 + A\beta + (A\beta)^2 + (A\beta)^3 + \ldots\$ is infinity. If \$A\beta < 1\$, it is finite and equals \$1 \over 1 - A\beta\$. \$\endgroup\$ Commented Dec 30, 2018 at 19:22
  • \$\begingroup\$ Why is it finite? Let's say the input signal has amplitude 1 and Aβ = -0.5 (- for the 180 degree phase shift). Then after the first round of amplification we have 1 + 1*0.5 = 1.5. In the second round we have 1 + 1.5*0.5 = 1.75. In the third round we have 1 + 1.75*0.5. Etc.. Wouldn't that grow until infinity? \$\endgroup\$ Commented Dec 30, 2018 at 19:36
  • \$\begingroup\$ No: before the first round, the amplification is \$1 = 2 - 1 = 2 - 0.5^0\$; after the first round it is \$1.5 = 2-0.5 = 2-0.5^1\$; after the second round, it is \$1.75 = 2-0.25 = 2-0.5^2\$; …; after the \$n\$th round, it is \$2-0.5^n\$. Hence, it keeps finite. \$\endgroup\$ Commented Dec 30, 2018 at 19:45
  • \$\begingroup\$ For more about the math behind this, you may read about geometric series at Wikipedia. \$\endgroup\$ Commented Dec 30, 2018 at 19:52
  • \$\begingroup\$ Why is the amplification 2-0.5^n? \$\endgroup\$ Commented Dec 30, 2018 at 20:04

2 Answers 2


Do a thought experiment. Let the input signal be zero, and see what happens to a single cycle of sine wave, injected at the error channel. Do this for an open loop phase shift of \$\small -180^o\$, and for the three cases:
a. \$\small 0<A\beta <1\$;
b. \$\small A\beta=1\$; and
c. \$\small A\beta>1\$.
For example, take:
\$\small A\beta=0.5\$; \$\small A\beta=1\$; and \$\small A\beta=2\$.
Assume the single cycle travels around the loop discretely, as a packet.


When the gain is >1 with 180 phase shift and another 180 by negative feedback , you have a positive feedback with growing amplitude since loop gain >1 and non inverting and thus an oscillator. ( i.e. unstable or "astable" )

Where the loop gain >1 at 180 phase shift is the resonant frequency of the oscillator.

The output will be saturated which reduces the gain to 0 when saturated but by this time the stored energy ( sine voltage feedback) from the reactance that created the 180 phase shift , has enough amplitude. The DC still is negative feedback so it biases the into near 50% threshold for a square wave out.


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