My analog design book says the following:

"As a signal \$ S_\epsilon \$ propagates around the loop and returns to the summer \$ \Sigma \$ as \$ S_f \$, it experiences a frequency-dependent phase shift, which we shall denote as \$ph\ T(jf) \$. If this shift reaches \$ -180^\circ \$, then feedback turns from negative to positive."

Why is feedback positive only when the phase shift is \$-180^\circ\$? If we consider the loop gain \$T(jf) = \beta a(jf)\$ then surely the feedback is negative for \$-90^\circ < ph\ a(jf) < 90^\circ\$, where \$a(jf)\$ has a positive real part, and positive for \$ -270^\circ < ph\ a(jf) < -90^\circ\$, where \$a(jf)\$ has a negative real part (or equivalently \$ 90^\circ < ph\ a(jf) < 270^\circ \$).

I would imagine the necessary condition for oscillation, instead of being \$ a(jf_{-180^\circ})\beta \geq -1\$, should be \$|a(jf)| \cdot \sin{(90 + ph\ a(jf))} \cdot \beta \geq -1\$.

Thoughts on this?

Is it only -180 degrees because any other phase shift, once you take into account the summer, would give something that's not a multiple of \$2\pi\$ and would change for every pass through the loop? Like comparing it to waves how it might be "constructive interference" but it wouldn't meet resonance criteria?

Is this a case of people not being precise with language? Resonance, i.e. oscillation, for \$ a(jf_{-180^\circ})\beta \geq -1\$ but positive feedback in general for \$ -270^\circ < ph\ a(jf) < -90^\circ\$?

  • \$\begingroup\$ If you flip polarity, you are not changing the fundamental signal. \$\endgroup\$ Jun 1, 2023 at 17:33
  • \$\begingroup\$ The feedback signal is subtracted from the input signal. The result then feeds into the error amplifier. Changing polarity of the feedback signal will absolutely change the output signal. Not sure what you're getting at. \$\endgroup\$
    – Walls55
    Jun 1, 2023 at 17:39
  • \$\begingroup\$ Barkhausen states α A = 1 (Attenuation × Gain) for oscillation to occur. Fundamentally, it matters not what the signal is (sinewave, square, etc.) and with positive (0°) or negative (180°) feedback. You can do it in steps of 90° (RC circuits), but the goal is a stable oscillation. \$\endgroup\$ Jun 1, 2023 at 17:53
  • \$\begingroup\$ No - that is not correct. Barkhausens necessary conditions applies to one single sinusoidal frequency only! \$\endgroup\$
    – LvW
    Jun 2, 2023 at 9:46

1 Answer 1


Many other questions have been asked about the Barkhausen Stability Criterion as it is quite misunderstood.

  1. For starters, it is not even a stability criterion, as it does not determine stability, and it does not guarantee a system is unstable, or that it will oscillate.
  2. If a linear circuit meets the Barkhausen "Stability" Criterion for some frequencies it means that it might oscillate at those frequencies. But it does not guarantee that it will do so.
  3. Some books and instructors try to make the description more wordy and intuitive but end up just creating confusion. Look at the criterion and we see the actual statement and not the wordy explanation, which I adapted to use a negative feedback as your book does. It goes like this, for frequencies \$f\$ such that
  • \$ |a(jf)\beta| = 1 \$

  • \$ \text{phase}\left[ a(jf)\beta\right] \equiv 180^\circ \$

    the circuit might oscillate. The notation \$\equiv 180^\circ\$ means being equal to \$180^\circ\$, or \$360^\circ+180^\circ\$, or \$2\cdot360^\circ+180^\circ\$, and other equivalent angles.

A few other resources on the whole Barkhausen Criterion saying that things loop around and phase shift makes feedback go from negative from positive is the following answer here in EESE https://electronics.stackexchange.com/a/252049/227586 and this note http://web.mit.edu/klund/www/weblatex/node4.html

  • \$\begingroup\$ Yes - the original Barkhausen criterion was (and is) a necessary criterion only. That means: When a circuit oscillates it will fulfill this criterion. However, this criterion has recently been supplemented by a third condition (in addition to the two conditions for magnitude and phase). Now, we have a sufficient condition for oscillation at w=wo: Loop gain at w=wo with (a) magnitude equal to "1" and (b) zero phase shift and (c) negative slope of the phase function [d(phi)/dw<0]. \$\endgroup\$
    – LvW
    Jun 2, 2023 at 8:03
  • \$\begingroup\$ A short comment to the second given link (edu/klund): The author (K. Lundberg) considers it necessary to write "down with Barkhausen". Possibly he does not know that Barkhausen's original criterion is only a necessary (but not yet a sufficient) condition for oscillation. \$\endgroup\$
    – LvW
    Jun 2, 2023 at 9:05
  • \$\begingroup\$ I understand Barkhausen is a necessary, and not a sufficient, condition and I'm feeling a bit better about oscillation now. However, can you comment on the definition and use of the phrase "positive feedback"? It's bothering me. It seems people use the phrase positive feedback only when you have oscillation at \$-180^\circ\$. Basically I'm wondering if it is correct to say you have positive feedback anytime the real part of \$ a(jf)\$ is negative before reaching the summer? \$\endgroup\$
    – Walls55
    Jun 2, 2023 at 17:38
  • 1
    \$\begingroup\$ By negative feedback I mean that at the summing junction the signal being fed back is subtracted from the reference signal (which some call input signal). While by positive feedback I mean that at the summing junction the signal being fed back is added from the reference signal. But some call negative feedback if the feedback stabilizes the system (for a step eventually the output is constant) while a positive feedback is something that makes the system become oscillatory or unstable. I would avoid this terminology and just say whether the system is stable or not. \$\endgroup\$
    – jDAQ
    Jun 2, 2023 at 18:06
  • 2
    \$\begingroup\$ @Walls55 An oscillator needs both: Negative feedback for DC (stable DC bias point) as well as positive feedback for one single frequency only! Therefore, we always need a frequency-selective feedback network which can fulfill both requirements. \$\endgroup\$
    – LvW
    Jun 3, 2023 at 8:45

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