Summing integrator circuit?

[Len Thomas]

I know what a summing amp and integrator are as separate circuits, but
can someone please explain the design criteria for a "summing
integrator". IOW one op amp doing both tasks.

The integrators I have used in the past have a feedback resistor in
the MegOhms and cap in parallel. If used with 10K series for each of
the several summed inputs, this would affect the gain and seemingly
limit options in this regard.

The frequencies involved are 10-40Hz, subaudio and I am looking for
unity gain. Input is 6V.

Any suggestions please?

Len

 

[Active8]

On Fri, 11 Mar 2005 16:18:17 -0800, Larry Brasfield wrote:

<snip>

Use the above gain formula, picking a common
C value (since there is but one feedback C) and
weight your R values according to the relative
gains need for each input.

Yeah, with one additional consideration - the application. Going too
far with the RC time constants (or range thereof) might cause a
prob. The 1/(2.pi.R.C) cutoff freq and the input signal need to be
taken into consideration. IOW, it won't be an integrator if the
signal freq falls within the passband of the corresponding low-pass
filter.

In case the OP hasn't done so:

National Semiconductor AN-31 might help a bit.

I don't know where I got it, but it looks like a lecture note from

Electronics II Theory. It's BASICOPA.pdf -- It's got a few things on
integrators. Try google.
--
Best Regards,
Mike

 

[Larry Brasfield]

"Active8" <reply2group@ndbbm.net> wrote in message
news:d1mnfagu5jwf$.dlg@ID-222894.news.individual.net...

On Fri, 11 Mar 2005 16:18:17 -0800, Larry Brasfield wrote:

snip

Use the above gain formula, picking a common
C value (since there is but one feedback C) and
weight your R values according to the relative
gains need for each input.

Yeah, with one additional consideration - the application. Going too
far with the RC time constants (or range thereof) might cause a
prob. The 1/(2.pi.R.C) cutoff freq and the input signal need to be
taken into consideration. IOW, it won't be an integrator if the
signal freq falls within the passband of the corresponding low-pass
filter.

Sorry, but that is nonsense. While it is true that the
app note you reference below shows this equation
fc = 1 / (2 pi R1 C1)
(where R1 and C1 are the principle compoents)
for the integrator, there is no such cutoff frequency.

That frequency is where the gain reaches 1, but
the gain versus frequency is a simple -20 dB/decade
out to the unity gain crossover of the op-amp, as it
should be for an integrator (up to that real "cutoff",
dictated by the op-amp, not the R and C values).

If you doubt this, try running a simple-to-setup
simulation rather than quoting some authority of
unknown provenance. (You should know or
become aware that app notes are notoriously
unreliable as sources of engineering expertise.)

There are, of course, practical limits on the R and C
values that can be used, but they relate to the desired
gain relative to the op-amp open loop gain, and the
impedance level at which the circuit operates relative
to leakage currents, stray capacitance, and the output
impedance of the amplifier.

In case the OP hasn't done so:

National Semiconductor AN-31 might help a bit.

It says less than what you cut when quoting my post.

I don't know where I got it, but it looks like a lecture note from

Electronics II Theory. It's BASICOPA.pdf -- It's got a few things on
integrators. Try google.

Basic op-amp theory is quite sufficient to cover
how well an op-amp integrator will work.

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

 

[Active8]

On Sat, 12 Mar 2005 11:42:12 -0800, Larry Brasfield wrote:

Yeah, with one additional consideration - the application. Going too
far with the RC time constants (or range thereof) might cause a
prob. The 1/(2.pi.R.C) cutoff freq and the input signal need to be
taken into consideration. IOW, it won't be an integrator if the
signal freq falls within the passband of the corresponding low-pass
filter.

Sorry, but that is nonsense. While it is true that the
app note you reference below shows this equation
fc = 1 / (2 pi R1 C1)

I didn't get that from the app note. It's the cutoff freq for a
single pole lowpass filter. I dug up the appnote for the OP.

(where R1 and C1 are the principle compoents)
for the integrator, there is no such cutoff frequency.


That frequency is where the gain reaches 1, but

It's also the freq where the reactance equals the resistance, which
is the cutoff freq of a one pole RC filter. Notice the equation you
gave is similar to the eq for a simple passive RC integrator/LPF ?
Yeah, they called it an integrator in the Army - when that was it's
intended purpose. Good enough for gov't work? You don't need to
point out that the charging current will not be constant as the cap
charges up. They used RC passive "differentiators" too. Square in,
sharp pulses out.

---/\/\/\/---+--- Vo/Vi = 1/(1 + RCs)
|
|
---
---
|

Another difference is, Vo/Vi above will be 1/2 at F_c

the gain versus frequency is a simple -20 dB/decade

A one pole LP filter.

out to the unity gain crossover of the op-amp, as it
should be for an integrator (up to that real "cutoff",
dictated by the op-amp, not the R and C values).

If you doubt this, try running a simple-to-setup
simulation rather than quoting some authority of
unknown provenance.

I'm quoting no one.

(You should know or
become aware that app notes are notoriously
unreliable as sources of engineering expertise.)

I'll keep it in mind. I trust (unless it really sounds or feels
wrong) but usually verify. And you're right. My own feeling was
wrong. I was thinking (always have) that if a square wave is in the
passband, it'd slow down the rise time, and the thing would charge
up to the rail (or as close as it can get, of course) until the
square wave goes negative - provided it stayed at that voltage long
enough to hit the rail. But [duh!] I know that a constant charging
current gives a cap a linear voltage ramp. That should have been all
and I threw a monkey wrench at it.

There are, of course, practical limits on the R and C
values that can be used, but they relate to the desired
gain relative to the op-amp open loop gain, and the
impedance level at which the circuit operates relative
to leakage currents, stray capacitance, and the output
impedance of the amplifier.

In case the OP hasn't done so:

National Semiconductor AN-31 might help a bit.

It says less than what you cut when quoting my post.

There was nothing to be gained by quoting text irrelevant to my
reply. As I said, I checked the mfg and AN # for the OP. First you
ream me for not cutting enough chaff from my code, now you're
complaining about me snipping posts

I don't know where I got it, but it looks like a lecture note from

Electronics II Theory. It's BASICOPA.pdf -- It's got a few things on
integrators. Try google.

Basic op-amp theory is quite sufficient to cover
how well an op-amp integrator will work.

What do you think BASICOPA stands for? I just looked at it again. It
looks ok.
--
Best Regards,
Mike

 

[Larry Brasfield]

[Severe cuts applied to get to the essence here.]
"Active8" <reply2group@ndbbm.net> wrote in message
news:1fsh9mu87cw03$.dlg@ID-222894.news.individual.net...

On Sat, 12 Mar 2005 11:42:12 -0800, Larry Brasfield wrote:
[Active8 once wrote:]
The 1/(2.pi.R.C) cutoff freq and the input signal need to be
taken into consideration. IOW, it won't be an integrator if the
signal freq falls within the passband of the corresponding low-pass
filter.

Sorry, but that is nonsense. While it is true that the
app note you reference below shows this equation
fc = 1 / (2 pi R1 C1)

I didn't get that from the app note. It's the cutoff freq for a
single pole lowpass filter. I dug up the appnote for the OP.

Ok, I assumed you got it from the app note you quoted
because it seemed too long a coincidence that the same
misconception would appear independently.

(where R1 and C1 are the principle compoents)
for the integrator, there is no such cutoff frequency.

That frequency is where the gain reaches 1, but

It's also the freq where the reactance equals the resistance, which
is the cutoff freq of a one pole RC filter. Notice the equation you
gave is similar to the eq for a simple passive RC integrator/LPF ?

Similar, perhaps, in the appearance of a similar
term, (2 pi R C), but not similar enough to turn a
low-pass filter configuration into an integrator.

Yeah, they called it an integrator in the Army - when that was it's
intended purpose. Good enough for gov't work? You don't need to
point out that the charging current will not be constant as the cap
charges up.

That is is an important aspect of the difference between
a low pass filter and an integrator. In the passband of
an RC LPF, the cap does more or less charge up so
that there is little loss across the R and, more importantly,
the output magnitude asymptotically approaches the
input magnitude. That never happens in an integrator
except, in an op-amp realization, at the absurdly low
frequency where the amplifier operates nearly open
loop.

They used RC passive "differentiators" too. Square in,
sharp pulses out.

Calling an RC high-pass filter a differentiator is much like
calling an RC low-pass filter an integrator. So I have to
hand it to you for consistency. <g>

---/\/\/\/---+--- Vo/Vi = 1/(1 + RCs)
|
|
---
---
|

Another difference is, Vo/Vi above will be 1/2 at F_c

Strictly speaking, (and going along with considering
only the magnitude), it is 1/sqrt(2) at Fc for that circuit.

the gain versus frequency is a simple -20 dB/decade

A one pole LP filter.

But there is no passband! The pole is at 0.
There is simply no pole at 1/(2.pi.R.C) as you
have claimed, which is what got me into this.

[big snip]

It says less than what you cut when quoting my post.

There was nothing to be gained by quoting text irrelevant to my
reply. As I said, I checked the mfg and AN # for the OP. First you
ream me for not cutting enough chaff from my code, now you're
complaining about me snipping posts

Well, if I was complaining it was only about this
short and almost simple statement that vanished:
[T]he integrator gain is -1/(s*(R*C)) for each input R.
That really says it all. There is no cutoff frequency
because there is no passband with a single pole at 0.

And I'm sorry if you felt reamed. I often try to
persuade programmers that they can often solve
their own problems by trying to reduce their code
to a minimal problematic version.

Best Regards,
Mike
Likewise,

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

 

[Active8]

On Sat, 12 Mar 2005 22:02:08 -0800, Larry Brasfield wrote:

"Active8" <reply2group@ndbbm.net> wrote in message
snip

They used RC passive "differentiators" too. Square in,
sharp pulses out.

Calling an RC high-pass filter a differentiator is much like
calling an RC low-pass filter an integrator. So I have to
hand it to you for consistency. <g

Now *that* was the Army's way, not mine. It did the job, so I went
along with it.

---/\/\/\/---+--- Vo/Vi = 1/(1 + RCs)
|
|
---
---
|

Another difference is, Vo/Vi above will be 1/2 at F_c

Strictly speaking, (and going along with considering
only the magnitude), it is 1/sqrt(2) at Fc for that circuit.

Am I overlooking the fact that the reactive component is 90 deg out
of phase. Must be. That would necessitate the sqrt.

the gain versus frequency is a simple -20 dB/decade

A one pole LP filter.

But there is no passband! The pole is at 0.

Again I overlooked the obvious. I'd need another R in the feedback
path to get an LPF.

There is simply no pole at 1/(2.pi.R.C) as you
have claimed, which is what got me into this.

[big snip]
little snip

And I'm sorry if you felt reamed. I often try to
persuade programmers that they can often solve
their own problems by trying to reduce their code
to a minimal problematic version.

Maybe not reamed, but ISTR it read more like someone yelling. But I
saw where I could have reduced it more and put it all in one file.

Best Regards,
Mike
Likewise,
Ditto. Remember the super polite Gophers cartoon?

--
Best Regards,
Mike

 

[Fred Bloggs]

Larry Brasfield wrote:

"Len Thomas" <lenthomas102@aol.com> wrote in message
news:b5a431l9qqec0dfchdplhlrrct8r9djqtl@4ax.com...

I know what a summing amp and integrator are as separate circuits, but
can someone please explain the design criteria for a "summing
integrator". IOW one op amp doing both tasks.


The summing could be of input currents into a
virtual ground, with a feedback capacitor balancing
the summed current and converting it into a voltage
output that would be -1/C times the integral of the
input currents. If you want a non-inverting integrator,
it gets a little trickier (or a lot if it has to be accurate).


The integrators I have used in the past have a feedback resistor in
the MegOhms and cap in parallel.


With a resistor across the integrating capacitor, it
cannot be considered an integrator except as a
crude approximation. Integrators do not need
that resistor. (I have had somebody tell me they
put one in to limit the gain for stability reasons. It
took me 15 minutes to educate that out of him.)

To the OP: Don't listen to this pseudo-intellectual fake and fraud,
Larry Brasfield, he is a sick little engineering impersonator who gets
excited regurgitating a bunch of specious bs to fool people with less
than elementary working knowledge of electronics. Plans are in the works
to rid the NG of the pest.
You are quite right about the integrator requiring a resistor in shunt
with C, and this is used to bound the DC output error due to DC input
bias currents and offset voltage of the op amp. Without a shunt R, the
amplifier would soon saturate one way or the other and become useless.
For example, the amplifier (OA) will force current through the feedback
RC to maintain Vos at its (-) terminal equal to that of the (+)
terminal- this will be a current of Vos/Rin so that without a shunt R, C
will be required to supply this current indefinitely with a voltage
buildup of Vc(t)=(Vos/(Rin*C))*t. Anyone can see the amplifier will not
support this ramp for indefinite time- running out of headroom. The
second DC error source is the (-) input bias current- which yields a
similar Vc(t)=(Ib/C)*t error- eventually forcing the amplifier into
saturation.

 

[Larry Brasfield]

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:42342C42.7010704@nospam.com...

Larry Brasfield wrote:
"Len Thomas" <lenthomas102@aol.com> wrote in message
news:b5a431l9qqec0dfchdplhlrrct8r9djqtl@4ax.com...

I know what a summing amp and integrator are as separate circuits, but
can someone please explain the design criteria for a "summing
integrator". IOW one op amp doing both tasks.


The summing could be of input currents into a
virtual ground, with a feedback capacitor balancing
the summed current and converting it into a voltage
output that would be -1/C times the integral of the
input currents. If you want a non-inverting integrator,
it gets a little trickier (or a lot if it has to be accurate).


The integrators I have used in the past have a feedback resistor in
the MegOhms and cap in parallel.


With a resistor across the integrating capacitor, it
cannot be considered an integrator except as a
crude approximation. Integrators do not need
that resistor. (I have had somebody tell me they
put one in to limit the gain for stability reasons. It
took me 15 minutes to educate that out of him.)

To the OP: Don't listen to this pseudo-intellectual fake and fraud, Larry Brasfield, he is a sick little engineering impersonator
who gets excited regurgitating a bunch of specious bs to fool people with less than elementary working knowledge of electronics.

I would urge the OP to evaluate posts on their merits
and ignore the stream of ad hominum that Fred spews.

Plans are in the works to rid the NG of the pest.

I'll look forward to that effort. What a laugh.

You are quite right about the integrator requiring a resistor in shunt with C, and this is used to bound the DC output error due
to DC input bias currents and offset voltage of the op amp. Without a shunt R, the amplifier would soon saturate one way or the
other and become useless.

Guess what, Fred. Integrators generally have either
a reset mechanism or are used in a feedback loop
that obviates the "problem" you describe.

For example, the amplifier (OA) will force current through the feedback RC to maintain Vos at its (-) terminal equal to that of
the (+) terminal- this will be a current of Vos/Rin so that without a shunt R, C will be required to supply this current
indefinitely with a voltage buildup of Vc(t)=(Vos/(Rin*C))*t. Anyone can see the amplifier will not support this ramp for
indefinite time- running out of headroom. The second DC error source is the (-) input bias current- which yields a similar
Vc(t)=(Ib/C)*t error- eventually forcing the amplifier into saturation.

By the time you "solve" this problem with a shunt
feedback resistor, you have built a low-pass filter
rather than an integrator. This means the output
will not reflect the integral of the input for signals
anywhere near the LPF passband.

What is comical about using that resistor to keep
input error from saturating the sort-of-integrator is
that you end up with a large error in the integrator
initial condition unless you have a reset mechanism.
And if you have that, you do not need the resistor.

Bottom line is that the resistor is misguided in most
situations where an integrator is the required function.

Of course, if what you really want is a simple low
pass filter, you may well use that topology.

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

 

[Fred Bloggs]

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:42342C42.7010704@nospam.com...

Larry Brasfield wrote:

"Len Thomas" <lenthomas102@aol.com> wrote in message
news:b5a431l9qqec0dfchdplhlrrct8r9djqtl@4ax.com...


I know what a summing amp and integrator are as separate circuits, but
can someone please explain the design criteria for a "summing
integrator". IOW one op amp doing both tasks.


The summing could be of input currents into a
virtual ground, with a feedback capacitor balancing
the summed current and converting it into a voltage
output that would be -1/C times the integral of the
input currents. If you want a non-inverting integrator,
it gets a little trickier (or a lot if it has to be accurate).



The integrators I have used in the past have a feedback resistor in
the MegOhms and cap in parallel.


With a resistor across the integrating capacitor, it
cannot be considered an integrator except as a
crude approximation. Integrators do not need
that resistor. (I have had somebody tell me they
put one in to limit the gain for stability reasons. It
took me 15 minutes to educate that out of him.)

To the OP: Don't listen to this pseudo-intellectual fake and fraud, Larry Brasfield, he is a sick little engineering impersonator
who gets excited regurgitating a bunch of specious bs to fool people with less than elementary working knowledge of electronics.


I would urge the OP to evaluate posts on their merits
and ignore the stream of ad hominum that Fred spews.

Oh would you "urge" someone to do that, Laaaa...weeee, the goodie-goodie
two shoes born again xtian maggot, liar, hypocrite with subnormal IQ....

Plans are in the works to rid the NG of the pest.


I'll look forward to that effort. What a laugh.

Keep going goodie-goodie girl- every time you open your fetid mouth,
more ammunition will be found to use against you.

You are quite right about the integrator requiring a resistor in shunt with C, and this is used to bound the DC output error due
to DC input bias currents and offset voltage of the op amp. Without a shunt R, the amplifier would soon saturate one way or the
other and become useless.


Guess what, Fred. Integrators generally have either
a reset mechanism or are used in a feedback loop
that obviates the "problem" you describe.

Not in analog *audio* they aren't- you waffling little BS sh_t-head.
Trying to dodge the issue again, hypocrite little lying xtian rabble?

For example, the amplifier (OA) will force current through the feedback RC to maintain Vos at its (-) terminal equal to that of
the (+) terminal- this will be a current of Vos/Rin so that without a shunt R, C will be required to supply this current
indefinitely with a voltage buildup of Vc(t)=(Vos/(Rin*C))*t. Anyone can see the amplifier will not support this ramp for
indefinite time- running out of headroom. The second DC error source is the (-) input bias current- which yields a similar
Vc(t)=(Ib/C)*t error- eventually forcing the amplifier into saturation.


By the time you "solve" this problem with a shunt
feedback resistor, you have built a low-pass filter
rather than an integrator. This means the output
will not reflect the integral of the input for signals
anywhere near the LPF passband.

FU, moron. The OP's requirements are clear- he wants anything that puts
his signal on a -20dB/decade slope. No one is talking about an ideal
integrator. This is just your typical bs ploy to evade taking
responsibility for your ignorant bs post.

What is comical about using that resistor to keep
input error from saturating the sort-of-integrator is
that you end up with a large error in the integrator
initial condition unless you have a reset mechanism.
And if you have that, you do not need the resistor.

Nah- you're wrong about that, flake. Real engineers have been using that
technique for ages now- and somehow the world has gotten along just find
without your bs enlightenment- where you are such an pathetic narcissist
you think everyone but you is in the dark.

Bottom line is that the resistor is misguided in most
situations where an integrator is the required function.

The bottom line is that you should be bottoming as a male prostitute and
give up this charade about being otherwise useful.

Of course, if what you really want is a simple low
pass filter, you may well use that topology.

Oh really- trying to worm out of your bs pontificating...

>

 

[Larry Brasfield]

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:42346084.6060806@nospam.com...

Larry Brasfield wrote:
[Fetid stream cut for space and relevance.]
By the time you "solve" this problem with a shunt
feedback resistor, you have built a low-pass filter
rather than an integrator. This means the output
will not reflect the integral of the input for signals
anywhere near the LPF passband.
[Fetid stream cut for space and relevance.]
The OP's requirements are clear- he wants anything that puts his signal on a -20dB/decade slope. No one is talking about an ideal
integrator.

I've been following this thread from its inception,
and that was not clear to me at all. All the OP
has actually stated is "The frequencies involved are
10-40Hz, subaudio and I am looking for unity gain."
He never said anything that suggests that the LPF
corner set by a shunt feedback resistor is well
below that frequency range. And what he did say
suggests to me it was a concern: "The integrators I
have used in the past have a feedback resistor in the
MegOhms and cap in parallel. If used with 10K series
for each of the several summed inputs, this would affect
the gain and seemingly limit options in this regard."

If the OP does not care about the response below
10 Hz, one has to wonder why the input should
not be AC coupled.

How anybody can claim the OP's requirements are
clear really mystifies me. It would take mind reading
skills I have never experienced to get there.

[Fetid stream cut for space and relevance.]
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Larry Brasfield wrote:

I've been following this thread from its inception,
and that was not clear to me at all. All the OP
has actually stated is "The frequencies involved are
10-40Hz, subaudio and I am looking for unity gain."
He never said anything that suggests that the LPF
corner set by a shunt feedback resistor is well
below that frequency range. And what he did say
suggests to me it was a concern: "The integrators I
have used in the past have a feedback resistor in the
MegOhms and cap in parallel. If used with 10K series
for each of the several summed inputs, this would affect
the gain and seemingly limit options in this regard."

Making excuses again? This is like the third or fourth time you "blame"
the OP for miscommunicating. Why don't you just wise up to the fact you
are a worthless p.o.s.

If the OP does not care about the response below
10 Hz, one has to wonder why the input should
not be AC coupled.

It most likely is ac-coupled- You're just now picking up on peripheral
information any knowledgeable person would understand.

How anybody can claim the OP's requirements are
clear really mystifies me. It would take mind reading
skills I have never experienced to get there.

Of course, you are a such special kind of person in your own mind, and
anything that contradicts that hallucination means something is wrong
with the external world- which you barely acknowledge given your extreme
case of narcissism.

 

[John Woodgate]

I read in sci.electronics.design that Larry Brasfield <donotspam_larry_b
rasfield@hotmail.com> wrote (in <zLZYd.11$Vg.250@news.uswest.net&gt about
'Summing integrator circuit?', on Sun, 13 Mar 2005:

I've been following this thread from its inception, and that was not
clear to me at all. All the OP has actually stated is "The frequencies
involved are 10-40Hz, subaudio and I am looking for unity gain." He
never said anything that suggests that the LPF corner set by a shunt
feedback resistor is well below that frequency range. And what he did
say suggests to me it was a concern: "The integrators I have used in
the past have a feedback resistor in the MegOhms and cap in parallel. If
used with 10K series for each of the several summed inputs, this would
affect the gain and seemingly limit options in this regard."

It seems to me that the OP really DOES want a low-pass filter, and has
been misled into describing it as an integrator. He is also confused
about the 'gain' of the circuit, however you describe it. Since the gain
is intended to vary inversely proportional to frequency, a 'gain'
specification without a corresponding frequency specification is
meaningless.

If the OP does not care about the response below 10 Hz, one has to
wonder why the input should not be AC coupled.

I don't know about AC coupling, but if I wanted an LP filter for a sub-
woofer, I'd look at a -3dB point at around 3 Hz. From 10Hz to 40 Hz (and
beyond), the frequency response would be quite adequately close to -6
dB/octave for the application, bearing in mind that the sub-woofer
itself is very unlikely to have a response flatter than +/- 1 dB.

How anybody can claim the OP's requirements are clear really mystifies
me. It would take mind reading skills I have never experienced to get
there.

--
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

 

[lemonjuice]

On Sat, 12 Mar 2005 10:35:33 +1100, Len Thomas <lenthomas102@aol.com>
wrote:

I know what a summing amp and integrator are as separate circuits, but
can someone please explain the design criteria for a "summing
integrator". IOW one op amp doing both tasks.

The integrators I have used in the past have a feedback resistor in
the MegOhms and cap in parallel. If used with 10K series for each of
the several summed inputs, this would affect the gain and seemingly
limit options in this regard.

The frequencies involved are 10-40Hz, subaudio and I am looking for
unity gain. Input is 6V.

Any suggestions please?

Len

As far as I know an integrator IS one of the many possible Low pass
filters. Compare the transfer functions or frequency response curves of
the 2 and you'll see they are identical.
A resistor put in parallel with the capacitor as someone explained
helps against saturation but it should be noted that it also limits
the integrability of the integrator as its easily shown that
Vout = 1/exp (t/R1*C) * Integral( Vin(t) /R2C) dt) + Vo. Vo is voltage
on capacitor at t=0 ... integration is done between t=0 and a defined
time. R2 is capacitor in parallel with C.

If you use the configuration you mentioned then Fourier transforming
the expression above you'll see that for a unity gain R2 = R1 and for
a cut off frequency fo at 10Hz put fo = 1/2*pi*R2*C.

BTW To get a summing integrator you have to eliminate R and place a
capacitor there. you'd have to add other capacitors to the summing
junction at the Opamp input. Normally they are switched on with FETS .
The advantage of these is mostly in ICs where IC resistors are plagued
by large tolerances , large size and so on. You also get progammability
of the transfer function determined by your FET switch time frequency.