Summing Amplifier with non-ideal op-amp

John Woodgate · Mar 9, 2005

I read in sci.electronics.design that Spehro Pefhany <speffSNIP@interlog
DOTyou.knowwhat> wrote (in <j77u21dsvph99tuhtli5oov46fi5si3cja@4ax.com&gt

about 'Summing Amplifier with non-ideal op-amp', on Wed, 9 Mar 2005:

Assume all the N summing inputs save one and the output of the op-amp
are grounded (the op-amp is doing nothing). With a voltage Vin applied
to the remaining summing input, the voltage at the inverting op-amp
input is Vin* 1/(N+1).

I think we are discussing different configurations. I presume a feedback
resistor R, and N individual series input resistors, also R. The output
is definitely NOT grounded; it has voltage V appearing on it. The
voltage at the inverting input is zero for an ideal op-amp and is V/Ao
for an open loop gain of Ao.

--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Spehro Pefhany · Mar 9, 2005

On Wed, 09 Mar 2005 09:22:51 -0700, the renowned Jim Thompson
<thegreatone@example.com> wrote:

On Wed, 09 Mar 2005 11:20:33 -0500, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

On Wed, 9 Mar 2005 15:34:21 +0000, the renowned John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Spehro Pefhany <speffSNIP@interlog
DOTyou.knowwhat> wrote (in <n12u21hdrsh1h7hkl2bgqqipu0kob0v6ri@4ax.com&gt
about 'Summing Amplifier with non-ideal op-amp', on Wed, 9 Mar 2005:

The open-loop signal at the input is reduced by a factor of 1/(N+1)

Sorry, where does 1/(N+1) come from? This 'open loop signal' is due to
the op-amp open loop gain not being infinite?

Assume all the N summing inputs save one and the output of the op-amp
are grounded (the op-amp is doing nothing). With a voltage Vin applied
to the remaining summing input, the voltage at the inverting op-amp
input is Vin* 1/(N+1). That effectively degrades the Vos, GBW, DC gain
characteristics of the op-amp by a factor of ~1/(N+1). Or to look at
it from another direction, you have a passive ~1/N attenuator followed
by a non-ideal amplifier with gain ~-N (N>>1).

Best regards,
Spehro Pefhany

Eh? Does this "summing" amplifier not include a feedback resistor?

When you close the loop, yes.

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Yes, -Rf/Rin, with an ideal input. Or -1 for a gain-of-one summing
amplifier from a given input to the output.

But shunt that virtual ground to ground through a relatively low-value
resistor and you degrade the performance. That's effectively what we
have here.

The unused inputs have no effect on the gain... HOWEVER, they do add
_offset_voltage_ terms.

...Jim Thompson

I think they do affect the non-ideal gain (and GBW) as measured from a
single input to the output with the other inputs at 0V.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Dave · Mar 9, 2005

Jim Thompson Mar 9, 8:22 am wrote:

The unused inputs have no effect on the gain... HOWEVER, they do
add _offset_voltage_ terms.

So the lower the parallel equivalent to ground (unused inputs) the
higher the offset voltage... but the noise increases or decreases?

Jim Thompson · Mar 9, 2005

On Wed, 09 Mar 2005 11:51:19 -0500, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Wed, 09 Mar 2005 09:22:51 -0700, the renowned Jim Thompson
thegreatone@example.com> wrote:

On Wed, 09 Mar 2005 11:20:33 -0500, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

On Wed, 9 Mar 2005 15:34:21 +0000, the renowned John Woodgate
jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Spehro Pefhany <speffSNIP@interlog
DOTyou.knowwhat> wrote (in <n12u21hdrsh1h7hkl2bgqqipu0kob0v6ri@4ax.com&gt
about 'Summing Amplifier with non-ideal op-amp', on Wed, 9 Mar 2005:

The open-loop signal at the input is reduced by a factor of 1/(N+1)

Sorry, where does 1/(N+1) come from? This 'open loop signal' is due to
the op-amp open loop gain not being infinite?

Assume all the N summing inputs save one and the output of the op-amp
are grounded (the op-amp is doing nothing). With a voltage Vin applied
to the remaining summing input, the voltage at the inverting op-amp
input is Vin* 1/(N+1). That effectively degrades the Vos, GBW, DC gain
characteristics of the op-amp by a factor of ~1/(N+1). Or to look at
it from another direction, you have a passive ~1/N attenuator followed
by a non-ideal amplifier with gain ~-N (N>>1).

Best regards,
Spehro Pefhany

Eh? Does this "summing" amplifier not include a feedback resistor?

When you close the loop, yes.

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Yes, -Rf/Rin, with an ideal input. Or -1 for a gain-of-one summing
amplifier from a given input to the output.

But shunt that virtual ground to ground through a relatively low-value
resistor and you degrade the performance. That's effectively what we
have here.

The unused inputs have no effect on the gain... HOWEVER, they do add
_offset_voltage_ terms.

...Jim Thompson

I think they do affect the non-ideal gain (and GBW) as measured from a
single input to the output with the other inputs at 0V.

Best regards,
Spehro Pefhany

480)460-2350 | |
| E-mail Address at Website Fax

John Woodgate · Mar 9, 2005

I read in sci.electronics.design that John Popelish <jpopelish@rica.net>
wrote (in <422F2B5B.3888B45F@rica.net&gt

about 'Summing Amplifier with
non-ideal op-amp', on Wed, 9 Mar 2005:

Jim Thompson wrote:

Eh? Does this "summing" amplifier not include a feedback resistor?

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Make that nearly a virtual ground. It is only an actual virtual ground
with an infinite gain amplifier. So, when there is finite gain, there is
some small voltage (output voltage over open loop gain) at the - input
that drives the output, and that small voltage also drives some of the
input current through the other input resistors. When the effective
parallel resistance of all the other input resistors is so low that it
soaks up half of the total input current (from the one input in
question), there will be only half of the input current to go through
the feedback resistor, and so the gain will have fallen to half.

The resistance looking into the virtual ground is R/Ao. Since Ao is sort
of big, you need an awful lot of grounded inputs to get to a shunt
resistance equal to that! OK, if you are pushing a 741 op-amp to 20 kHz,
there is a problem, but who does that these days .....? (;-)
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Jim Thompson · Mar 9, 2005

On Wed, 09 Mar 2005 13:20:01 -0500, John Popelish <jpopelish@rica.net>
wrote:

John Woodgate wrote:

I read in sci.electronics.design that John Popelish <jpopelish@rica.net
wrote (in <422F2B5B.3888B45F@rica.net&gt about 'Summing Amplifier with
non-ideal op-amp', on Wed, 9 Mar 2005:
Jim Thompson wrote:

Eh? Does this "summing" amplifier not include a feedback resistor?

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Make that nearly a virtual ground. It is only an actual virtual ground
with an infinite gain amplifier. So, when there is finite gain, there is
some small voltage (output voltage over open loop gain) at the - input
that drives the output, and that small voltage also drives some of the
input current through the other input resistors. When the effective
parallel resistance of all the other input resistors is so low that it
soaks up half of the total input current (from the one input in
question), there will be only half of the input current to go through
the feedback resistor, and so the gain will have fallen to half.

The resistance looking into the virtual ground is R/Ao. Since Ao is sort
of big, you need an awful lot of grounded inputs to get to a shunt
resistance equal to that! OK, if you are pushing a 741 op-amp to 20 kHz,
there is a problem, but who does that these days .....? (;-)

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of how
many inputs you can have (actually how low a shunt resistance you can
tolerate) must include the gain bandwidth required, as well as the
opamp in question.

See.......

Newsgroups: alt.binaries.schematics.electronic
Subject: Re: SED: Summing Amplifier with non-ideal op-amp -
GBWwithMultipleInputs.pdf
Message-ID: <umeu21pb89n19dl7sdkp1rupc0s9o2oq5m@4ax.com>

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice

480)460-2350 | |
| E-mail Address at Website Fax

John Popelish · Mar 9, 2005

John Woodgate wrote:

I read in sci.electronics.design that John Popelish <jpopelish@rica.net
wrote (in <422F2B5B.3888B45F@rica.net&gt about 'Summing Amplifier with
non-ideal op-amp', on Wed, 9 Mar 2005:
Jim Thompson wrote:

Eh? Does this "summing" amplifier not include a feedback resistor?

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Make that nearly a virtual ground. It is only an actual virtual ground
with an infinite gain amplifier. So, when there is finite gain, there is
some small voltage (output voltage over open loop gain) at the - input
that drives the output, and that small voltage also drives some of the
input current through the other input resistors. When the effective
parallel resistance of all the other input resistors is so low that it
soaks up half of the total input current (from the one input in
question), there will be only half of the input current to go through
the feedback resistor, and so the gain will have fallen to half.

The resistance looking into the virtual ground is R/Ao. Since Ao is sort
of big, you need an awful lot of grounded inputs to get to a shunt
resistance equal to that! OK, if you are pushing a 741 op-amp to 20 kHz,
there is a problem, but who does that these days .....? (;-)

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of how
many inputs you can have (actually how low a shunt resistance you can
tolerate) must include the gain bandwidth required, as well as the
opamp in question.

--
John Popelish

Larry Brasfield · Mar 9, 2005

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in
message news:9xqKalCRSzLCFwf1@jmwa.demon.co.uk...

I read in sci.electronics.design that John Popelish <jpopelish@rica.net
wrote (in <422F2B5B.3888B45F@rica.net&gt about 'Summing Amplifier with
non-ideal op-amp', on Wed, 9 Mar 2005:
Jim Thompson wrote:

Eh? Does this "summing" amplifier not include a feedback resistor?

If there IS a feedback resistor, the OpAmp maintains a virtual ground
at the inverting input, so the gain is simply Rf/Rin.

Make that nearly a virtual ground. It is only an actual virtual ground
with an infinite gain amplifier. So, when there is finite gain, there is
some small voltage (output voltage over open loop gain) at the - input
that drives the output, and that small voltage also drives some of the
input current through the other input resistors. When the effective
parallel resistance of all the other input resistors is so low that it
soaks up half of the total input current (from the one input in
question), there will be only half of the input current to go through
the feedback resistor, and so the gain will have fallen to half.

The resistance looking into the virtual ground is R/Ao. Since Ao is sort
of big, you need an awful lot of grounded inputs to get to a shunt
resistance equal to that! OK, if you are pushing a 741 op-amp to 20 kHz,
there is a problem, but who does that these days .....? (;-)

I can remember designing a number of op-amp circuits
where loop gain was a concern, some fairly recently.
Of course, the frequency at which loop gain became
a concern was higher than 20 kHz.

In any configuration where loop gain is needed to get
accuracy (which includes driving down the natural,
open-loop distortion of the amplifier), and where
adequate loop gain is not a foregone conclusion, if you
add inputs then you have likely degraded loop gain and
need to reevaluate its adequacy. This can happen in
real situations; you should not defer that kind of analysis
for (nominally) unity gain DC summers with hundreds of
thousands of inputs.

In an earlier post on this subthread, you wrote: "The
voltage at the inverting input is zero for an ideal op-amp
and is V/Ao for an open loop gain of Ao." Very true,
but if you look at the named subject, and consider the
formula for closed-loop gain (which I posted earlier):
Avcl = G / (1 + GH) == 1 / (1/G + H)
you should be able to see what happens as H (which
designates the gain of the feedback network) is made
so small that it approaches 1/G. The closed loop gain
is no longer well approximated as 1/H, which is the
result that obtains with an ideal op-amp.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Spehro Pefhany · Mar 9, 2005

On Wed, 09 Mar 2005 10:26:09 -0700, the renowned Jim Thompson
<thegreatone@example.com> wrote:

So the "Rin" are very small?

Time to write some equations and impress the students ;-)

...Jim Thompson

Simplified case, unity gain inverting summing amplifier, consider
finite gain only.

Op-amp: Vo = -AoVin

|\|
+----------|+\
| ___ | >----+----o Vo
Vx o-----|-|___|-+ -|-/ |
| _R_ | |/| |
+-|___|-+ |
| _R_ | ___ |
+-|___|-+---|___|--+
| R |
|
| etc. N inputs
| ___ |
+-|___|-+
| R
|
|
===
GND

Vin = (Vo/R + Vx/R) * R/(N+1)

Vo = - AoVin => Vin = -Vo/A

-Vo/Ao = Vo/(N+1) + Vx/(N+1)

-Vo (1/Ao + 1/(N+1)) = Vx/(N+1)

Vo/Vx = -Ao/((N+1) + Ao) (closed loop gain)
~= -1 for Ao >> N+1

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

Fred Bartoli · Mar 9, 2005

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> a écrit dans le message de
news:HQIUy1BbcyLCFwLe@jmwa.demon.co.uk...

I read in sci.electronics.design that Spehro Pefhany <speffSNIP@interlog
DOTyou.knowwhat> wrote (in <j77u21dsvph99tuhtli5oov46fi5si3cja@4ax.com&gt
about 'Summing Amplifier with non-ideal op-amp', on Wed, 9 Mar 2005:

Assume all the N summing inputs save one and the output of the op-amp
are grounded (the op-amp is doing nothing). With a voltage Vin applied
to the remaining summing input, the voltage at the inverting op-amp
input is Vin* 1/(N+1).

I think we are discussing different configurations. I presume a feedback
resistor R, and N individual series input resistors, also R. The output
is definitely NOT grounded; it has voltage V appearing on it. The
voltage at the inverting input is zero for an ideal op-amp and is V/Ao
for an open loop gain of Ao.

No, you're discussing the same configuration, a summing inverter with N
equal gain inputs.
From the POV of _one_ input, the equivalent schematics reduces to:

R1 R
___ ___
Vin -|___|------+-----|___|-.
| |
| |
| |\ |
+-----|-\ |
| | >--+--Vout
| .-|+/
.-. | |/
R1/(N-1) | | |
| | ===
'-' GND
|
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

Obviously the gain is (about) -R/R1, but the noise gain, i.e. the loop gain,
reduces from G(w)R1/(R1+R) for the one input case to G(w) R1/(R1+N R) for
the N input case.

Suppose you use an OP77 (700kHz GBW product) you want gain=-2, have 5 inputs
and need 1kHz BW, then your loop gain is only 700/11 = 63. Not that much,
and easily further reduced as asked higher gain or with input numbers.

One cute trick if you have an opamp to spare, or you add another one because
you want improved performance, is to add a negative resistor between the
minus input and GND.

R1 R
___ ___
Vin -|___|------+-----|___|-.
| |
| |
___ | |\ |
.-|___|---+------------+-----|-\ |
| | | | >--+--Vout
| /| | | .-|+/
| /+|---' .-. | |/
+--< | R1/(N-1) | | |
| \-|---. | | ===
| \| | '-' GND
| ___ | |
'-|___|---+ |
| ===
.-. GND
| |
| |
'-'
|
|
===
GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

You'll have to pay attention to some stability pb if you want to push the
limit, but if you want a reasonable loop gain (say maybe 1/3 to 1/4 of the
main opamp gain) this is easy to use.

--
Thanks,
Fred.

Larry Brasfield · Mar 9, 2005

"Fred Bartoli" <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo>
wrote in message news:422f591f$0$16355$626a14ce@news.free.fr...

..., but the noise gain, i.e. the loop gain,

Those two concepts should not be confused. The
term "noise gain" refers to the closed loop gain for
a signal arising between the amplifier inputs, while
the term "loop gain" refers to the gain around the
loop consisting of the forward path from amplifier
input to output and the feedback path from that
same output to that same input.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Fred Bloggs · Mar 10, 2005

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:422EC4E8.9000108@nospam.com...

Larry Brasfield wrote:

(Responding to the question: "If an OP-77 connected as simple
inverting summing op-amp stage has N inputs, how could you
estimate the practical limit for N ?")

In addition to the issues mentioned by Mr. Grise and Mr. Shoppa,
as N approaches the open loop gain of the op-amp divided by
the hoped-for realized gain from each input, the loop gain will
approach something near 1, leading to limited accuracy of the
actual closed loop gain and increased distortion.

Oh yeah- people are running that check all the time in their
calculations. Why don't you review OpAmps 101 before you post your
irrelevant pedantic drivel........

You amaze me, Fred. You show signs of some
intelligence but seem to have never applied it to
understanding your own limitations.

That is a response to be expected from a pretentious little punk
pseudo-intellectual like you. We know you are the PUNK born again xtian
maggot from Washington state who has been plaguing the NG with his
superficial trash for some months now.

Do you deny that loop gain imposes a practical
limit on how many inputs a summing op-amp
circuit can have? If so, you are the one that
needs remedial training in op-amp theory.

If you are merely suggesting that the limit I have
stated is never applicable, why have you not
given some hint as to what the real limit on N is?
My bet is that you cannot.

You're quite unimpressive to look at your posting history. You spout a
bunch of pseudo-theoretical elementary garbage, typical of a punk
know-nothing 2nd rate EET student or wannabe, and soon evaporate once
the thread becomes focussed. You certainly skulked out of the encoder
divider thread without doing much of anything other than putting on airs
about being an experienced motion controls designer- which we know now
is a despicable fiction. You're a cowardly little hypocrite born-again
xtian maggot and a hot air bag.

Fred Bloggs · Mar 10, 2005

[..snip Woodgate drivel...]

For that topology, the feedback gain is:
(1/N) / (1 + 1/N) == 1 / (N + 1)
Clearly, as N approaches the open loop
gain of the op-amp, the loop gain will be
approaching 1. This means that the gain
accuracy will be reduced relative to the
usual case where loop gains exceed one
by quite a bit.

What a bunch of pseudo-intellectual garbage- and look at that damned
complex math- and who the hell waits until loop gain gets near one to
conclude they have an "accuracy" problem? What a pile of horse manure!

Consider the formula for
closed-loop gain:

Consider? Consider?! You really are losing it. Who the hell are you to
tell anyone to consider that pathetic, boring, and irrelevant pedantry-
what a damned joke- stinking moron regurgitating trash from his textbook.

Avcl = G / (1 + GH) == 1 / (1/G + H)
Usually, we neglect the 1/G term because
it is small relative to H. But as N becomes
comparable to the open-loop gain (G), that
simplifying assumption becomes invalid.

What a bunch of retarded crap- the moron is so damned dumb he/she/it
thinks a working engineer with 60+ years of practice needs that low
level minutia explained to him. What an unbelievable little wimp punk
born-again xtian scumball.

Fred Bloggs · Mar 10, 2005

Larry Brasfield wrote:

I can remember designing a number of op-amp circuits...

Ahh- you did some homework on your own...whew-i-m-p-r-e-s-s-i-v-e.

[ snip boring pedantic drivel...]

>

Fred Bloggs · Mar 10, 2005

Larry Brasfield wrote:
[...snip usual pretentious Brasfield- fake name trash...]

I note that you have well developed name-calling skills,
but are unable to answer the challenge I posed. I take
this to mean that your "review OpAmps 101" suggestion
is nothing but empty pretense.

You can take this as my response to your other two
content-free posts.

You should talk about "content-free" posts, you little punk, superficial
engineer wannabe. Hey!- Where's your "tricky" current source, maggot?
Don't bother playing your trump card about having nothing to prove,
because so far you're just a skulking little punk hot air bag.

Dave · Mar 10, 2005

John Popelish on Mar 9, 10:20 am wrote:
John Woodgate wrote:

The resistance looking into the virtual ground is R/Ao. Since
Ao is sort of big, you need an awful lot of grounded inputs to
get to a shunt resistance equal to that! OK, if you are pushing
a 741 op-amp to 20 kHz, there is a problem, but who does that
these days .....? (;-)

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of
how many inputs you can have (actually how low a shunt resistance
you can tolerate) must include the gain bandwidth required, as well
as the opamp in question.

So could this ever be considered a legitimate technique for gain /
slew-rate adjustment? Or is it just too stupid?

Jim Thompson · Mar 10, 2005

On 10 Mar 2005 08:53:37 -0800, "Dave" <galt_57@hotmail.com> wrote:

John Popelish on Mar 9, 10:20 am wrote:
John Woodgate wrote:
The resistance looking into the virtual ground is R/Ao. Since
Ao is sort of big, you need an awful lot of grounded inputs to
get to a shunt resistance equal to that! OK, if you are pushing
a 741 op-amp to 20 kHz, there is a problem, but who does that
these days .....? (;-)

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of
how many inputs you can have (actually how low a shunt resistance
you can tolerate) must include the gain bandwidth required, as well
as the opamp in question.

So could this ever be considered a legitimate technique for gain /
slew-rate adjustment? Or is it just too stupid?

480)460-2350 | |
| E-mail Address at Website Fax

Jim Thompson · Mar 10, 2005

On Thu, 10 Mar 2005 12:31:52 -0500, John Popelish <jpopelish@rica.net>
wrote:

Dave wrote:

John Popelish on Mar 9, 10:20 am wrote:

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of
how many inputs you can have (actually how low a shunt resistance
you can tolerate) must include the gain bandwidth required, as well
as the opamp in question.

So could this ever be considered a legitimate technique for gain /
slew-rate adjustment? Or is it just too stupid?

Putting shunt elements across the two inputs is a sometimes used
method to improve/alter the stability of an opamp, by changing its
effective open loop gain and phase response. I have seen it used to
turn a comparator into an opamp. It doesn't affect the slew rate,
which doesn't involve the input gain stage.

480)460-2350 | |
| E-mail Address at Website Fax

John Popelish · Mar 10, 2005

Dave wrote:

John Popelish on Mar 9, 10:20 am wrote:

The point is that with any opamp, the effective bandwidth starts
dropping when you shunt the input to ground. So the question of
how many inputs you can have (actually how low a shunt resistance
you can tolerate) must include the gain bandwidth required, as well
as the opamp in question.

So could this ever be considered a legitimate technique for gain /
slew-rate adjustment? Or is it just too stupid?

Putting shunt elements across the two inputs is a sometimes used
method to improve/alter the stability of an opamp, by changing its
effective open loop gain and phase response. I have seen it used to
turn a comparator into an opamp. It doesn't affect the slew rate,
which doesn't involve the input gain stage.

--
John Popelish

Mar 10, 2005

Jim Thompson <thegreatone@example.com> wrote:

Which reminds me... sometimes bypassing the summing node to ground can
improve system noise problems.

...Jim Thompson

When you "improve a problem" do you make it better or worse?

Jim

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Summing Amplifier with non-ideal op-amp

John Woodgate

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