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From: panfilero on 11 Aug 2010 15:36
On Aug 11, 10:46 am, Tim Wescott <t...(a)seemywebsite.com> wrote:
> On 08/11/2010 07:49 AM, panfilero 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?
>
> 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 <--, just made bigger or smaller.
> If it's been made bigger, then chances are that the op-amp will oscillate*.
>
> Note that the 180 degree phase shift can be misleading -- how can
> reversing the sign of the signal make things bad?  The reason is because
> you're designing for the op-amp to be connected with negative feedback
> (i.e. the output of the op-amp feeding back to the - input).  The fact
> that you're feeding into the negative input adds it's own 180 degrees of
> phase shift, so 180 in the input plus 180 in the amp ads up to no phase
> shift at all.
>
> This article probably has more math and less intuition than you'd like
> -- but it has way more hand-waving and less math than a professor would
> like, so that's OK.  At any rate, toward the end it discusses Bode plot
> design for stability.  It does so in the context of discrete-time
> systems, but the rules for design are pretty much the same.  So you may
> find it helpful.
>
> http://www.wescottdesign.com/articles/zTransform/z-transforms.html
>
> * Not always, but the exceptions -- while fortuitous -- have to be made
> to happen.
>
> --
>
> 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

Thank you it's starting to make more sense now
From: Phil Hobbs on 11 Aug 2010 15:56
panfilero wrote:
> 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)
>

"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
--
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: George Herold on 11 Aug 2010 18:02
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.
From: Jamie on 11 Aug 2010 20:38
panfilero 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!
Miller?


From: Jamie on 11 Aug 2010 20:44
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.


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