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question about noise detector

Discussion in 'Electronic Basics' started by fred, Mar 1, 2004.

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  1. fred

    fred Guest

    The noise detector can be found here

    I was wondering what benefits do we get by using
    two op-amps in this circuit, because gain can also
    be increased simply by increasing the feedback
    resistor, i.e. R9.

  2. Mikal Hadvik

    Mikal Hadvik Guest


    Without actually doing the math, I'm reasonably sure that the designer
    faced this problem: Op amps like the LM358 don't have high enough
    gain/bandwidth product to take this much gain in one gulp and maintain
    adequate bandwidth for the job at hand.

    Mikal Hadvik
    Decade Engineering
  3. Guest

    As the second post said, you get less bandwidth.

    This is because the ratio of the op-amp's ("open loop") gain to the
    actual ("closed loop") gain is a measure of the quality i.e. lack of
    distortion i.e. bandwidth.

    If you increase R9 then this ratio is reduced, the output is "softer"
    at the extremes because the feedback to the op-amp's negative-pin is
    less (via the bigger resistor) and the corrective-effect (that forces
    the output to follow-truly the input) is reduced.

    Taken to the limit, the op-amp with an infinite (i.e. no) feedback
    resistor will have a very distorted output because the gain might be
    10E6 in the middle but only 10E5 and reducing toward either limit. If
    the feedback is hard i.e. a short circuit then the output will be true
    right up to the limit i.e. the gain "loss" from 10E6 down to unity is
    replaced with by a fidelity increase of "true tracking" to one part in
    10E-6 i.e. one part in a million accuracy for 0.999999 of the output

    (Of course it is also a shame to waste the othe half of the LM385)

  4. fred

    fred Guest

    I have seen circuits before when the feedback ratio i.e.
    R9/R5 (or R6,R7) is 10M/10K, which gives a gain
    of 1000. This is quite resonable as 1000 is very far
    away from the op-amp limits.

    Here is another question:

    When we change the switch position from R7 to R6 to R5
    we are changing the gain by a factor of 10 every time because
    R7 = 560
    R6 = 5.6K and
    R5 = 56K
    therefore the db scale should be changing by equal amounts
    as well, i.e. 50, 70, 90, so why instead of 90 we have 85?
  5. Guest

    A 10M feedback is not a good idea because the

    ----- stray C

    extra CR delay might be enough to make it "ring" or even oscillate.
    Another "feature" would be that the input Z at the op-amp input is
    proportional to the feedback resistor (and inversly proportion to the
    open loop gain) i.e. a large FB resistor makes that node more likely
    to "pickup" or "cross-talk" Again; self oscillation-prone from

    Even if the circuit "works" on the bench, it may not in the field i.e.
    if the board gets grubby and/or condensation settles on it.


    My guess is that the reduction in bandwidth caused by the 56,000 / 560
    combination caused the 5db loss.

  6. fred

    fred Guest

    Any ideas why we need R4, loading the microphone will surely
    reduce the input voltage, and this means we will need more
    amplification, so this is strange ?
  7. Mikal Hadvik

    Mikal Hadvik Guest

    R4 presents an insignificant load to the microphone. It's there
    primarily to provide a DC bias current path for the op-amp's input
    pin. In combination with the input coupling cap, it incidentally
    limits low-frequency response to about 16Hz in the first gain stage.

    The noted 5dB gain discrepancy results, no doubt, from a lack of
    sufficient amplifier gain to achieve the ideal calculated closed-loop
    gain. Accurate 1000X boost in a single stage is simply asking too much
    from a low cost op-amp chip, even if you're willing to give up most of
    the bandwidth. The LM358 can barely achieve 60dB (1000X) *open-loop*
    gain at 1KHz!

    Mikal Hadvik
    Decade Engineering
  8. Guest

    What would happen if R4 was missing? Assume a perfect op-amp (no
    current flow into/out of the input). All is switched off: everything
    is at ground. Switch-on. The left plate of C1 will jump up to some
    voltage set by the resistances. The right plate will jump up too to
    the same voltage. This voltage may not be the optimal mid-range bias
    for the op-amp.

    In reality this voltage will "drift with the wind" dependent on
    "leakage" currents/temperature/cosmic ray inpacts etc. You *must* fix
    it deliberately at "mid-range" DC wise. R4 does this.

    R4 must be as large as necessary to avoid loading the mike. The
    mid-range voltage is set by R2/R3 and "passed without loss" through R4
    because, the current through R4 will be small (where can it go? The
    op-amp input demands very little) so the voltage drop across it will
    be very small so the voltage at either end will be just about the same
    i.e. mid-range.

  9. fred

    fred Guest

    What would happen if R4 was missing? Assume a perfect op-amp (no
    If you think a gain of 1000 is too much for an op-amp, have a look here:

    One more question:
    I am have some problems to "accept" the value of R9=56K . This surely
    will effect the voltage divided of R2, R3. Don't you think 220K would
    have been better (adjusting of course the other resistors)?
  10. Mikal Hadvik

    Mikal Hadvik Guest

    I dunno how far you want to go with this, but the voltage drop across
    R4 *is* significant. The LM358 spec allows 250nA worst-case bias
    current, which works out to 25mV DC across R4. With a gain of 100X,
    that's 2.5V of DC offset (relative to half of the B1 voltage) at the
    output! And we forgot the worst-case input voltage (7mV) and current
    (50nA) offset specs! The 2nd stage gain of 25X (ignoring its offset,
    too) flings you out to 60V, and you're lost in the realm of Mordor.
    This circuit clearly suffers from poor design, even if the prototype
    was lucky enough to function as intended.

    Of course, there's an input resistor on the other side that develops a
    compensating input offset voltage, but that resistance is never equal
    in value to R4. At max gain, it's almost a short compared to R4. Good
    design would make it always equal to R4, with circuit gain controlled
    some other way. Other possible solutions: AC couple the 2nd stage, or
    substitute an FET-input op amp.

    Mikal Hadvik
    Decade Engineering
  11. Guest

    What I think is just an opinion. It means nothing. You have to do the
    calculations to *know* the answer (+ - limits) e.g. as Mikal has
    kindly done above.

    I found "Design with Operational Amplifiers and Analog integrated
    circuits" by Sergio Franco quite clear. I assume you already own the
    AofE, if not why not!

    It is healthy to assume that everything you see print-wise is wrong,
    more or less.

  12. fred

    fred Guest

    how did you find these values, i.e. 7mV , 50nA ?
    What about R9=56K, this will have an effect on the divider of R2/R3.
    I assume that because it's only a dc change, it will not matter much, is
    this the correct analysis? I assume R12 is similar to R4 i.e. it allows
    a dc offset to develope on C4. Next point, can we use an npn instead,
    simply moving R13 and D1 to the top and R12 to the bottom, will that
    work the same?

    I think this circuits offers quite in interesting feature or "trick" if you
    which is the voltage divider, to avoid the +-Vcc requirement of op-amps,
    but I the LM358 can work down to 0V so maybee it would have worked
    without this ?
  13. Mikal Hadvik

    Mikal Hadvik Guest

    If you think a gain of 1000 is too much for an op-amp, have a look here: I didn't mean to say that it's impossible, only that you can expect
    little bandwidth or gain accuracy. A 741 type op amp configured for
    1000X gain has a worst-case bandwidth of 437Hz, based on National's
    437KHz minimum unity-gain bandwidth spec for their LM741A. If you
    accept their "typical" BW spec, you still get only 1,500Hz BW at
    1000X! Sometimes that's okay, but it's not, in my judgement, okay for
    a sound level meter. Unless you attend rock concerts, of course, which
    can easily demolish your auditory bandwidth...
    Lower values in the feedback divider are generally better, to minimize
    effects of stray C and flatten frequency response, but the amp has
    trouble driving its own feedback network if you go too low. As long as
    the loop is closed, amplifier pin 2 tracks pin 3, so bias divider
    loading isn't a big factor. You might be able to decrese the value of
    C2 if you raise feedback divider impedance, but it's not worth the

    Mikal Hadvik
    Decade Engineering
  14. Mikal Hadvik

    Mikal Hadvik Guest

    how did you find these values, i.e. 7mV , 50nA ?
    It's true that LM358 input common mode range extends to zero volts,
    and you can squeak by with inputs that swing slightly below ground
    (i.e. microphones), but the output side must be biased well above
    ground to accommodate large output signal swings.

    BTW, another possible fix for the first-stage DC offset problem is to
    insert a capacitor in series to the wiper of SW1A. That brings your DC
    gain down to unity, while preserving AC gain. It doesn't help with
    gain accuracy or bandwidth issues.

    Mikal Hadvik
    Decade Engineering
  15. fred

    fred Guest

    how did you find these values, i.e. 7mV , 50nA ?
    could we assume that the negative
    cycle is not important and use only the positive cycle, the positive cycle
    will not be distorted and we will have a rectifing effect,
    would this change the reading as there is no load requirement ?
  16. Mikal Hadvik

    Mikal Hadvik Guest

    could we assume that the negative
    Since LM358 outputs can swing very nearly to ground, it's not
    unrealistic to attempt a half-cycle amplifier scheme. I haven't done
    this, but I would be on the lookout for saturation effects- where the
    output "sticks" to the ground rail long after it should be moving
    positive on each cycle, or even shoots positive during part of the
    (clipped) negative excursion. You'll need a scope for this test. If
    the LM358 fails, you should be able to find modern rail-to-rail I/O
    amps that behave.

    I'm not sure I understand the question about load requirement. The
    data sheet shows a very kinky Vo vs. Io plot for the output current
    sinking case, so you have to consider loading very carefully if you
    want the output to swing all the way to ground. It's probably best to
    make Q1 an NPN and rearrange the LED drive circuit as necessary, or
    eliminate Q1 and drive the LED through R13 directly from the final
    amplifier output pin, because LM358s can source a good 10mA or so from
    the positive rail. If you do this, calibration to known SPL is
    certainly necessary. Calibration seems unavoidable in any case,
    because the sensitivity spec for the microphone is not given by the

    Mikal Hadvik
    Decade Engineering
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