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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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