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Summing integrator circuit?

Discussion in 'Electronic Design' started by Len Thomas, Mar 11, 2005.

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  1. Len Thomas

    Len Thomas Guest

    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?

  2. 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).
    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.)
    Unless you are running out of loop gain, the
    inputs act independently and the integrator
    gain is -1/(s*(R*C)) for each input R.
    The phrase "unity gain" in reference to an
    integrator is confusing to me. What response
    do you want?
    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.
  3. Active8

    Active8 Guest

    On Fri, 11 Mar 2005 16:18:17 -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

    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.
  4. 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.
  5. Active8

    Active8 Guest

    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.
  6. [Severe cuts applied to get to the essence here.]
    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
    Calling an RC high-pass filter a differentiator is much like
    calling an RC low-pass filter an integrator. So I have to
    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.
  7. Active8

    Active8 Guest

    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? :)
  8. Fred Bloggs

    Fred Bloggs Guest

    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
  9. 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.
    Guess what, Fred. Integrators generally have either
    a reset mechanism or are used in a feedback loop
    that obviates the "problem" you describe.
    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.
  10. Fred Bloggs

    Fred Bloggs Guest

    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.
    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?
    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...
  11. [Fetid stream cut for space and relevance.][Fetid stream cut for space and relevance.]
    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.]
  12. Fred Bloggs

    Fred Bloggs Guest

    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.
  13. I read in that Larry Brasfield <donotspam_larry_b
    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
    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.
  14. lemonjuice

    lemonjuice Guest

    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.
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