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From: panfilero on 11 Aug 2010 10:49
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!
From: Tim Williams on 11 Aug 2010 11:10
"panfilero" <panfilero(a)gmail.com> wrote in message
news:9f2536b2-028b-4005-bcbc-dc1df9c25eac(a)w30g2000yqw.googlegroups.com...
> 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?

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

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms


From: Phil Hobbs on 11 Aug 2010 11:43
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!

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.)

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 net
http://electrooptical.net
From: Tim Wescott on 11 Aug 2010 11:46
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 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: John Larkin on 11 Aug 2010 12:12
On Wed, 11 Aug 2010 10:10:42 -0500, "Tim Williams"
<tmoranwms(a)charter.net> wrote:

>"panfilero" <panfilero(a)gmail.com> wrote in message
>news:9f2536b2-028b-4005-bcbc-dc1df9c25eac(a)w30g2000yqw.googlegroups.com...
>> 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?
>
>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

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

John

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