unstable opamp ringing oscillating and such
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From: Ian Iveson on 12 Aug 2010 05:53 Phil Hobbs wrote: >>> 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.) >>> >>> 2. 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. >> 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 >> = -Vs*(Rf/Ri) >> > > "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 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
From: George Herold on 12 Aug 2010 10:19 On Aug 11, 8:44 pm, Jamie <jamie_ka1lpa_not_valid_after_ka1l...(a)charter.net> wrote: > George Herold wrote: > > On Aug 11, 12:57 pm, Tim Wescott <t...(a)seemywebsite.com> wrote: > > >>On 08/11/2010 09:45 AM, George Herold wrote: > > >>>On Aug 11, 10:49 am, panfilero<panfil...(a)gmail.com> wrote: > > >>>>can anyone in here explain why a OpAmp could oscillate or become > >>>>unstable if there are 2 poles before it reaches 0dB on a plot of gain > >>>>vs frequency? > > >>>>Here's what I've gathered thus far... there's a cap internal to the op- > >>>>amp, that cap is there for some reason and it puts the dominant pole > >>>>in the op-amp's transfer function. That pole also puts a 90deg phase > >>>>shift on the output of the opamp... > > >>>>I think the phase shift is with respect to the input signal, so now > >>>>output is 90deg out of sync with the input > > >>>>adding another pole to the transfer function would cause our gain to > >>>>drop by another 20dB/decade (or 6dB/Octave) and also another 90deg > >>>>shift on our phase... so now we're 180deg out of phase with respect to > >>>>our input... > > >>>>but why would this be bad? > > >>>>Is there a way to explain this where it makes intuitive sense? or is > >>>>this something that can only be seen coming out of the math in a non- > >>>>intuitive kind of way? > > >>>>much thanks! > > >>>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. > > >>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.. > > >>-- > > >>Tim Wescott > >>Wescott Design Serviceshttp://www.wescottdesign.com > > >>Do you need to implement control loops in software? > >>"Applied Control Theory for Embedded Systems" was written for you. > >>See details athttp://www.wescottdesign.com/actfes/actfes.html > > > 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. > > 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. 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 = 1 at the frequency where, 1/(2*pi*f*C) = R) George H.
From: Paul E. Schoen on 13 Aug 2010 16:24 "Ian Iveson" <IanIveson.home(a)blueyonder.co.uk> wrote in message news:2xP8o.104498$LU.81628(a)hurricane... > > 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. 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: http://cds.linear.com/docs/Application%20Note/an25fa.pdf 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 : http://cds.linear.com/docs/Application%20Note/an47fa.pdf Paul
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