From: George Herold on 11 Aug 2010 12:45 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. George H.
From: Tim Wescott on 11 Aug 2010 12:57 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 Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Phil Hobbs on 11 Aug 2010 13:30 Tim Wescott 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. > Just make it a superregen. ;) Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
From: GregS on 11 Aug 2010 14:22 In article <4C62DE51.7010700(a)electrooptical.net>, Phil Hobbs <pcdhSpamMeSenseless(a)electrooptical.net> wrote: >Tim Wescott 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. >> > >Just make it a superregen. ;) CMOS inverters. geg
From: panfilero on 11 Aug 2010 15:05 On Aug 11, 10:43 am, Phil Hobbs <pcdhSpamMeSensel...(a)electrooptical.net> 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. > > Cheers > > Phil > > -- > Dr Philip C D Hobbs > Principal > ElectroOptical Innovations > 55 Orchard Rd > Briarcliff Manor NY 10510 > 845-480-2058 > hobbs at electrooptical dot nethttp://electrooptical.net 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)
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