unstable opamp ringing oscillating and such

Because 180 degree phase shift means the op-amp sees a falling signal, so it produces a rising signal. Which gets phase shifted to a falling signal. So the op-amp produces a rising signal. Which gets phase shifted...

In other words, it turns negative feedback into positive.

Two pole systems are "interesting" to compensate (example: forward converter switching supply, which has an LC output filter). You can use dominant-pole compensation, but this sucks, because transient response, to both line and load changes, is very slow. The solution is pole-zero (or lead/lag) compensation, which quadruples the number of components to adjust (instead of one capacitor, you have at least two capacitors AND two resistors). The resulting system is at least third order, which means you might have overshoot that's uncontrolled relative to other parameters, like rise time. The waveforms can be interesting.

Tim

The math is pretty intuitive actually, if you get the basic concept.

Do you understand how negative feedback works?

1. Can you explain in a sentence or two how an ideal op amp voltage follower works, i.e. why its output follows its noninverting input closely, even though the amp itself may have a gain of a million? (No frequency response funnies for now, just the basics.)

1. If you have an inverting amp whose feedback network has zero poles (ideal case again), why does the output equal -R_FB/R_IN?

If you can get that far, all you need for the rest is to understand what phase shift is.

Cheers

Phil

The 180 degree phase shift means that what appears on the output of the op-amp is exactly out of phase with the input. The "input", in this case, is the difference between the + and - inputs of the op-amp. So if you take that signal and feed it back to the - input, then what comes out of the op-amp is --> that signal

Most switchers only work because of the ESR of the output caps.

John

Something that you might find fun (and informative) is to try and make an opamp into an oscillator. There is a "phase shift oscillator" that uses a bunch of series RC sections. There is also an oscillator that uses two all-pass opamp sections in series.

George H.

Not to mention several oscillators where you start by trying to make a new and unique amplifier topology, only to find out that its not new, and there's a reason that it's never suggested as an amplifier topology.

Just make it a superregen. ;)

Cheers

Phil Hobbs

CMOS inverters.

geg

ok, let me take a crack at these

1. In a voltage follower the output is tied back into the inverting input, and then you apply your input voltage at the non-inverting input. Because the op-amp will try and keep both its inputs equal, the output will force the inverting input its tied to to be equal to the non-inverting input, therefore the output and the non-inverting input are the same.
2. The output equals the voltage drop across the feedback resistor. This is because the op-amp is keeping its inputs at zero (I'm assuming the non-inverting input is grounded) the voltage drop across the resistor is the product of the current going through it (Vs/Ri) and the feedback resistance value Rf. Therefore Vout =3D -Vs*(Rf/Ri)

Thank you it's starting to make more sense now

"The op amp will try" is just the sort of vagueness that'll get you into trouble. You might say instead, "If the noninverting input is initially epsilon volts different from the inverting output, the amplified error is -A_VOL times epsilon. The output will drive the inverting amplifier negative until V_out = V_in+ and then stop." That's negative feedback.

If you wire it up the other way, with the output to the NONinverting input, the op amp will actively try to make the inputs NOT be equal--any initial error will be multiplied indefinitely until the output hits the rail. This is of course positive feedback.

In the second case, the output only equals the voltage across the feedback resistor if feedback is forcing the two input voltages to be equal. Again, wire it up backwards and it's a Schmitt trigger (positive feedback) and not a nice linear inverting amplifier.

Cheers

Phil Hobbs

Well, I mostly just use opamps. Still oscillations or at least gain peaking is a constant problem.

Say you've got a high input impedance opamp circuit. Would you rather have the capacitance on the inverting or non-inverting input?

The inverting input gives you gain peaking, but if you tame, it seems you can get a bit more bandwidth.

George H.

Miller?

I always thought most OP-AMPs were designed with some Miller effects in it to help prevent oscillation? Of course, if you have one that is designed for a much higher freq than what you are subjecting it too? I guess you could run into problems. Hence one of the reasons comparators don't fall into the AMP family..

Have a goof day.

Er...

The real problem here appears to be that "panfilero" hasn't managed so far to get time into his equations.

Firstly, "Vout = -Vs*(Rf/Ri)" describes a destination, and yet the paragraph that leads up to that equation describes a process. Processes take time...it takes Vout a while to get there, during which it may do stuff that's very important to understanding how "unstable opamp ringing oscillating and such" happens.

Panfilero...how would you include time in your equation, such that it then describes the process of arrival as well as the destination?

Second, and related, is that Vs may be a function rather than a value. Is it true in algebra that a function can be substituted, willy nilly, for a term? AFAIK, only if it is entirely independent, in that the function must not itself contain or imply any of the terms already in the equation. It may appear that, if Vs = Asin(t) for example, that "Vout = -Asin(t)*(Rf/Ri)" would be OK. However, following from my first point above, "t" should already be in the equation, so sin(t) isn't independent, so the substitution is problematic.

Control systems theory addresses that problem. You can't avoid the maths if you want an analytical solution. No-one made it hard on purpose: it's as simple as it can be under difficult circumstances.

If you want to know what it feels like, get drunk and try walking in a straight line. DC input (straight line), sinusoidal output (more or less)...you're an oscillator because your slow response time puts a delay in your feedback. Now get not quite so drunk, and you'll sway a bit at first then settle. If you put a corner in the line you'll sway again after negotiating it. That's ringing. Try walking a sine wave (sine input) and you'll get the frequency right if it's low enough, but there'll be a phase error. If the sine wave is the same frequency as your swaying, you might resonate and get dizzy. If the frequency is much higher than your swaying, you'll ignore it and do the same as you did with a straight line, near enough.

It's not so much about the destination, the steady state solution; but rather the process of getting there, the transient response. It is no good at all trying to grasp any of it unless you've got time in your equations.

Ian

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Sure, a nice smooth roll-off of the gain as the frequency increases. But if you hang a capacitor on the inverting input, then along with the feed back resistor you have a gain that increases as the frequency increases. (With gain =3D 1 at the frequency where, 1/(2*pi*f*C) =3D R)

George H.

Yup, that's it. The op-amp is drunk. Probably alcohol absorbed during the flux removal process :)

Seriously, though I think the OP's confusion stems from the 90 degree phase shift of the op-amp which is really only present when it is operating at maximum gain or open loop. This can be simulated with an ideal op-amp and a small capacitor to the input. But when sufficient negative feedback is supplied, which provides a gain in a reasonable range of 1 to 1000 or so, the capacitance has little effect. At much higher frequencies, the internal capacitor is more significant, but the op-amp gain at that point may be low enough that it does not cause instability. And it's also helpful to add an external capacitor from the output to the inverting input so that fast output excursions are quickly sensed by the input and limited to very small amounts. That is my "intuitive" explanation.

There are other reasons for oscillation and ringing, which can be due to insufficient power supply bypass capacitors and inductance of components and PCB tracks, especially for high frequency circuits.

This app note, although for switching supplies, has some good information about ringing and compensation:

This is even more to the point about high speed amplifiers. There's even an appendix by D.L. Klipstein, who may be the same as Don who is active here :

Paul

OK. It's intuitive to you because you are thinking in dynamic terms, whereas the OP is trying to characterise dynamic behaviour with static equations. The static feedback equation stops being the whole story as soon as there is delay...such as arising from your cap...in the loop.

Sure. Other forms of feedback with delay. I was trying to address the importance of delay directly. Although slightly tongue-in-cheek, the drunken man example is good because there is a feedback loop with variable delay, depending on how much he drinks, and the same consequent characteristic behaviour as your op-amp. What's more, you can actually *be* a drunken man (or woman...I suppose I shouldn't make assumptions) so you can appreciate the situation from all angles.

Also, once it is realised that the behaviour of feedback control systems is universal, there are many examples outside electronics that lend themselves to intuition more directly. You can see and feel mechanical or human systems, whereas electricity is fundamentally mysterious, which doesn't help.

OK. I've only barely mastered audio amplifiers using valves, more or less. Feedback in switching supplies adds a level of abstraction in the PWM, and they work at shockingly high frequencies and currents that make experiments fraught with anxiety and hard to monitor. Scary stuff for a casual amateur like me.

Interesting, thanks.

Ian