Summing integrator circuit?

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

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.

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

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.

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

[Active8 once wrote:]

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.

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.

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.

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.

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

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]

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.

Likewise,

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

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.

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

A one pole LP filter.

I'm quoting no one.

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 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 ;)

What do you think BASICOPA stands for? I just looked at it again. It looks ok.

--
Best Regards,
Mike

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

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

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

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.

Ditto. Remember the super polite Gophers cartoon? :)

--
Best Regards,
Mike

Brasfield, he is a sick little engineering impersonator

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.

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

and this is used to bound the DC output error due

the amplifier would soon saturate one way or the

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

maintain Vos at its (-) terminal equal to that of

C will be required to supply this current

the amplifier will not support this ramp for

(-) input bias current- which yields a similar

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.

[Fetid stream cut for space and relevance.] [Fetid stream cut for space and relevance.]

-20dB/decade slope. No one is talking about an ideal

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.

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.

Brasfield, he is a sick little engineering impersonator

than elementary working knowledge of electronics.

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

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

and this is used to bound the DC output error due

the amplifier would soon saturate one way or the

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?

maintain Vos at its (-) terminal equal to that of

C will be required to supply this current

the amplifier will not support this ramp for

(-) input bias current- which yields a similar

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.

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.

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

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

>

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.

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

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.

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'Summing integrator circuit?', on Sun, 13 Mar 2005:

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.

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.

--
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

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.