I read this article on the feedback and stability concept of an op amp.

The article describes the relationship between open loop gain, closed loop gain, loop gain and some oscillation effects for the following non-inverting op amp: enter image description here Here is the bode plot I plot for the gains: enter image description here

Here is the summary of what I understood so far:

Basically the gain of the circuit G = Vout/Vin is actually dependent on the open loop gain Ao. G is approximated to Vout/Vin = 1/β for low frequencies, however the general form from the feedback equation is Vout/Vin = Ao/(1+βAo). Ao shows single pole LPF behaviour and decreases in such fashion as in the plot. Since Ao decreases with frequency and since the circuit gain Vout/Vin is directly related to Ao, one can obtain the blue plot in LTspice as Vout/Vin. And open loop gain Ao as in the green plot can be plotted by the ratio Ao = Vout/Vdiff. Another point the article mentions is that when the freq. increases Ao decreases and the circuit fights with this change and tries to keep the Vout same. Since Vout=Vdiff*Ao and when Ao decreases Vdiff increases by feedback.

My question is about interpreting the following plot and seeing the weirdness when the β*Ao=-1 in time domain. The article continues as:

Ao = Vout/Vdiff

β = Vn/Vout (Vn is the inverting input)

β*Ao = Vout/Vdiff * Vn/Vout = Vn/Vdiff

Since βAo = Vn/Vdiff, and I am looking for the frequency where βAo = -1; here is the bode plot for β*Ao = Vn/Vdiff. (I plot Vn/Vdiff in LTspice):

enter image description here

How can I interpret the above plot for βAo? I mean is there a point in this plot where there βAo=-1 (unity gain with 180 degree shift)? If not, does that mean this circuit is never unstable?

And is it possible to see this β*Ao=-1 effect on LTspice time domain?


I followed the user analogsystemsrf's suggestion and added a 10u capacitor at the output. Below the plot of β*Ao.

enter image description here

Here is the new Bode plot of Vout.

enter image description here

Below is the plot of Vout in linear scale:

enter image description here

And below I sweep the 10mV AC input upto 4kHz

enter image description here

Here are my questions:

1-) If you look at the βAo plot, 0dB and 180 degree never crosses. When 180 degree phase shift the dB is -40dB and when 0dB the phase shift is 178 degree. So it means its never βAo=-1. But still one can see some peculiarity on the plot. Why is that so? I mean for oscillation is this a must there must be a point exactly where β*Ao=-1 ?

2-) If you see my plot Vout in linear scale you notice that it makes a peak at around 1.7kHz. But when I sweep in time domain the peak I observe is at around 3.4kHz(the last plot above). Why is Bode and time domain shows this twice different way?


2 Answers 2


Here is a high-speed opamp (OPA211, 35MHz) with Rout of 25,000 ohms (edited up from specsheet 25 ohms). To cause oscillation, we add 1uF Cload. Rout+Cload are INSIDE the feedback loop, thus providing another 90 degrees phaseshift. Result? Notice the 25,000 ohm Rout drops the Fring by sqrt(1,000).

enter image description here

For more understanding, read up on Barkhausen's 2 criteria. They relate to the Nyquist plot/point crossing.

  • \$\begingroup\$ Thanks for the nice plots. So I understand that you add a 1u capacitor between point 1(opamp output in your program) and the ground. I understand the idea is to add an extra 90° phase shift. But I have two questions here: Shouldn't the βAo=-1 to have oscillation? When I plot βAo at unity when dB=0 the shift must be exactly 180°? I will try to edit my question. \$\endgroup\$
    – user16307
    Commented Apr 19, 2017 at 11:18
  • \$\begingroup\$ I edited my question, I would be glad to hear your opinion. \$\endgroup\$
    – user16307
    Commented Apr 19, 2017 at 13:18
  • \$\begingroup\$ As the amplitude of oscillation increases, the finite supply voltages constrain the gain. Instead of gain of 2.485X (for example) at exactly 180 degrees, the gain drops to 1.0000X, at which point the 2 requirements of Barkhausen are satisfied. \$\endgroup\$ Commented Apr 19, 2017 at 15:32

Op-amps are multi-stage high-gain differential amplifiers, and all stages behave like first-order low-pass filters. Combining 3+ stages, at some frequency, there is a 180° phase shift.

If, at this frequency, the gain was ≥ 1, then the simple voltage-follower circuit (with op-amp’s output directly connected tho the inverting input) would be unstable.

For this reason, most op-amps are compensated: an extra low-pass filter is added, on purpose, to ensure that the gain will be < 1 at the frequency were the phase shift is 180°. They are called “unity-gain stable“ op-amps.

With the circuit you consider, the resistors attenuate the signal by a factor 10 before feeding it back to the inverting input. If the op-amp is unity-gain stable, then, with your circuit, the total gain at the frequency were the phase-shift is 180° is < 0.1. So yes: your circuit is never unstable.

Note that there exist some op-amp that are not unity-gain stable. They are specified for a minimum gain, and may be unstable if you try to configure them with a lower gain. (Yes, as strange as it may sound, a low gain may cause instability, not a high gain.) Although they’re not really op-amps en example of such components is Microchip’s MCP6N11, with some versions unity-gain stable, and other that are only stable at gains ≥ 100.

  • \$\begingroup\$ How can I make that circuit in my question "not stable" I mean an exact phase-shift of 180° at unity gain? I want to see the unstableness effect in LTspice. \$\endgroup\$
    – user16307
    Commented Apr 18, 2017 at 21:50
  • \$\begingroup\$ You don’t need gain = 1 where phase shift is 180°, you need gain ≥ 1. To experience unstability, your best option would be to find a non-unity-gain-stable op-amp and use it as a voltage-follower. Or you have to amplify the op-amp’s output signal before feeding it back to its inverting input (with a high-enough-bandwith amplifier), but that would be quite en uncommon configuration!). \$\endgroup\$ Commented Apr 18, 2017 at 21:59
  • \$\begingroup\$ It seems like one pole system is always stable. Why is that? \$\endgroup\$
    – user16307
    Commented Apr 18, 2017 at 22:14
  • \$\begingroup\$ @user16307 Sorry, can’t help you on this one. \$\endgroup\$ Commented Apr 18, 2017 at 23:08
  • \$\begingroup\$ A 1-pole system can only generate 90 degree phase shift. The father of oscillators --- Barkhausen --- says you need N*360 degrees phaseshift from Input to Output, where N is 0/1/2/3/4... \$\endgroup\$ Commented Apr 19, 2017 at 4:24

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