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I understand that if an op-amp has a phase shift of 180\$^{\circ}\$ and a gain of >1 it will oscillate and I'm trying to get an intuitive feeling for how / why. I've read around a bit, but I don't understand why it would oscillate at a phase shift of 190\$^{\circ}\$ and not at 170\$^{\circ}\$. Any explanations or suggestions for further reading would be greatly appreciated!

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2 Answers 2

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This isn't exact.
You must have the loop phase shift of 180° (and loop gain=1). This is Barkhausen criterion. You can understand this if you draw a block diagram like this:
enter image description here

The equation is:

$$ \frac{V_{out}}{V_{in}}=\frac{A}{1+\beta A} $$

and tell you (intuitively) that if you want the loop will added to the input then the loop signal must be "negated", and then if I subtract \$-1\$ at something is like if I add \$+1\$.

You could understand better if you see at phase shift oscillator.
enter image description here

You have a inverting amplifier and a passive loop that do 180° phase shift.
Roughly the noise is the signal that start to go through the circuit. If you put this signal with 180° phase shift at inverting input of the amplifier is like you add the signal. Then, oscillation.

Deep study and better math could help more. :-)

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  • \$\begingroup\$ OK, I think I see. If you have a loop phase shift of 190° on an inverting amplifier, would you expect to see oscillation? \$\endgroup\$
    – Kathryn EB
    Jul 6, 2016 at 10:15
  • \$\begingroup\$ I'll not directly answer you but... if you add a sinusoidal signal to itself you have a bigger sinusoidal signal, but if you add a little shift, what you obtain? Intuitively, if you push always swing with the same strength and in the same time the amplitude increases, but what if you're pushing not always at the same time? Or in phase opposition? \$\endgroup\$
    – Antonio
    Jul 6, 2016 at 10:19
  • \$\begingroup\$ As in: a= sin(x), b=sin(x+2pi), c=sin(x+1.9pi) a+b gives sinusoidal signal twice as large a+c gives something that looks roughly sinusoidal but is a bit smaller in amplitude than (a+b) and slightly offset \$\endgroup\$
    – Kathryn EB
    Jul 6, 2016 at 10:29
  • \$\begingroup\$ Exact! Then the math does the rest (Barkhause criterion). Pick a good electronic book and try to understand feedbak theory. For example Sedra-Smith,or Millman-Halkias (you could find it in webarchive.org), or also docs from Electronics Producer, like this (I recommend), chapter 5 and 15. \$\endgroup\$
    – Antonio
    Jul 6, 2016 at 10:46
  • \$\begingroup\$ Kathrin - coming back to your first question (190deg): If the loop gain (excluding the "-" sign) is unity at -190 deg, we can assume that it will be >1 at -180deg. Hence, the circuit will oscillate at THIS frequency where the phase shift is 180deg. \$\endgroup\$
    – LvW
    Jul 6, 2016 at 11:39
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Kathryn - with all respect: Do you really understand why and under which conditions an operational amplifier oscillates? For example, I miss the term "feedback" in your text.

The key word for oscillation is "loop gain". The loop gain must not be larger than unity for a total phase of 360 deg. Otherwise, the circuit with feedback is not stable. The "magic" number of 180deg comes from the fact that sometimes the inverting property of the "minus" opamp input (additional 180deg) are not included in the stability criterion.

I suppose, it is not too problematic to get an "intuitive feeling" for the 360deg requirement (as mentioned above) because 360deg are identical to 0 deg.

EDIT (Example):Assume that an amplifier needs 1mV at its input for producing 100mV output. Now - if this input signal of 1 mV (without any phase shift) is derived from the output via feedback (feeedback factor 0.01) we have a system that can produce its own input signal. That is an oscillator with a loop gain of unity (100*0.01=1).

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