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let's consider a generic differentiator circuit, i.e. a circuit whose transfer function is \$\text{constant}\cdot j\omega\$. For instance, you may consider this circuit with Op-Amp:

enter image description here

I was reading this paragraph about it ("Pulse And Digital Circuits", Prakash Rao):

enter image description here

Which is the precise reason for its unstability? In general, I know that a linear system is stable if and only if its transfer function has only poles with negative real part. In this case I do not see any poles, and so?

Moreover, I have seen that usually negative feedback (as that present in this circuit) improves the stability: is, despite this fact, this circuit unstable?

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  • \$\begingroup\$ ... instability, qualitatively, means it can go into uncontrolled oscillation. \$\endgroup\$ – vicatcu Jan 1 at 20:08
  • \$\begingroup\$ ... under ostensibly stable input conditions \$\endgroup\$ – vicatcu Jan 1 at 20:20
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At high frequencies, it becomes more prone to noise. When noised is introduced to an amplifier, well... you can imagine what an amplifier will do with noise. If you get into control theory, you'll find that a differentiator has an infinite gain margin with 90 degrees of phase margin. The gain margin won't make the differentiator unstable; the phase margin will. The oscillation goes rampant at high frequencies.

You can also prove instability comparing this system to a Linear Time-Invarient (LTI) system. There's a concept called: Bounded-Input, Bounded Output (BIBO) stability and when you do the math on that, you'll find that the system is also unstable, which is a more accurate way to determine if the system is stable.

The physics behind this unstable behavior may be another topic but think about the nonlinearity of the capacitor.

There are ways to combat this and STMicrosystems has a good application note to one of their amplifiers on that. It is a good read if you are curious about how to compensate for this unstable system.

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With the ideal differentiator, there is a phase lag in the negative feedback network. Combined with the phase shift in a real opamp this can result in making the overall phase shift around the loop reach 360 deg and cause oscillation. Even if it doesn't cause oscillation it can reduce the phase margin at high frequencies.

To understand this imagine that Vin is grounded. The network R and C will give a phase shift that is asymptotic to 90 deg at high frequency.

A typical opamp is designed with dominant pole compensation to provide stability - this adds another 90 deg of phase shift at high frequencies. In addition, there are usually one or two additional poles internal to the opamp that can give more phase shift.

Note from the diagram that although the phase shift is ~90 deg over much of the operating range it goes beyond that above ~100kHz while the gain is still >1. This can cause instability when combined with the phase-shift of the negative feedback network. (note: the curve that is horizontal for much of the graph is the phase shift, the sloping line is the open-loop gain).

Typical phase shift and gain vs Frequency of opamp

OPA227 Datasheet

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  • \$\begingroup\$ how one can differentiate between which curve is phase and which curve is gain? \$\endgroup\$ – muyustan Jan 1 at 21:12
  • \$\begingroup\$ The one that slopes down from 20Hz is the gain. The phase is constant at 90 deg for much of the graph. \$\endgroup\$ – Kevin White Jan 1 at 21:49
  • \$\begingroup\$ I mean, the ones with enough experience on these topics will of course understand which one is which one, but for a datasheet, I think it should've been more explatory. \$\endgroup\$ – muyustan Jan 1 at 22:00
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    \$\begingroup\$ @muyustan The one with the G beside it is the gain and the one with the Phi beside it is the phase. \$\endgroup\$ – Spehro Pefhany Jan 2 at 19:04
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    \$\begingroup\$ @Chu - I changed the diagram for one that has a single decade per major division. As Sphero points out this one does show which curve is which. \$\endgroup\$ – Kevin White Jan 2 at 19:07

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