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Question about a schematic

Discussion in 'Electronic Basics' started by Guest, Aug 5, 2003.

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  1. Guest

    Guest Guest

    I was wondering if someone knows how the following circuit works:
    http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF
    I understand most of it, but have a few questions.
    Firstly, I don't understand why we need the first diode
    (the one which is connected between C and GND)
    Then, why do we not use the 3 pins of the variable resistor, surely
    we are meant to trim the voltage by "taking" only part of it, so for
    this I would have though the two "ends" of the pot should be
    connected to the diode and GND, the the output should be the
    middle pin rather than the top pin ?
    Tia
     
  2. A capacitor cannot pass DC current. If the first diode was missing,
    there would be a brief current out, and then the cap would charge up
    with enough voltage to keep the other diode reverse biased over the
    whole cycle. The first diode keeps the cap output side from going
    more than one diode drop below ground during the negative half cycle,
    so that once the wave starts to swing positive, it starts from that
    voltage. This diode capacitor combination is called a voltage
    doubler, because it produces and output that swings from zero to the
    whole peak to peak voltage of the wave (twice the center voltage to
    peak difference).
    I might have used the pot as a voltage divider on the output, and
    added a second capacitor across the while resistor, to give a variable
    DC output.

    Their way, the resistor and output capacitor form a voltage divider
    with a variable time constant that is discharging the cap, pulse by
    pulse.

    The best method depends on what you want.
     
  3. Guest

    Guest Guest

    A capacitor cannot pass DC current. If the first diode was missing,
    1.
    Regarding this voltage doubler, can we say that if our power supply is
    -5V to +5V than the output can be from 0V to 10V ?

    2.
    Can we say that the first diode is responsible for out output to start from
    0V rather than from 0.6V
    I forgot to mention this but I think you understood anyway, this circuit
    is supposed to give a voltage output which is proportional to the sound
    level. Given that, and to phrase is in your terms, what do we want ?
     
  4. It would if the opamp could produce a full supply output swing. But a
    real amplifier can seldom do this. The output stage of a 741 op amp
    can only approach within about 3 volts of the supply voltage, so you
    will be lucky to get a +- 2 volt swing from it when it is powered by
    +- 5. In this case, it is powered by a single 9 volt battery, so its
    output can swing from about +3 to +6 volts. The output cap and diode
    pair reference this swing from about -.6 to +2.4. Long after the 741
    was designed, opamps have been made that have input and output stages
    that work to one or both supply rails. The very common LM324 (quad)
    and LM358 (dual) have inputs and outputs that function all the way to
    the negative supply rail, and to within 1.5 volts of the positive
    supply rail. So these can provide 7.5 volts of swing from a 9 volt
    battery. There are cmos amplifiers that work with the full supply
    voltage at both inputs and outputs.
    You could if it were a perfect diode (zero volts forward drop, no
    reverse current).
    The method shown gives a meter reading (which is the average of the
    sound waveform) that gives a fast response to changing signals, but
    doesn't show much imformationb about brief loud peaks. The extra
    capacitor and pot approach I mentioned, acts more like a peak hold
    that will register the loudest peak and fade it out slowly (depending
    on the output RC time constant). Neither of these is very accurate
    because of the diode voltage drops.
     
  5. Guest

    Guest Guest

    Take a look at: Can't understand how this works, but thanks.
    Nice link, lots of usefull circuits, no explanations as to how they work though.

    What about this, eliminate the C and the first D, leave 2nd D and then
    add a big C (electrolotic) from D to GND. Basically like in the very
    basic power supplies which have one diode and C. Will be get a DC
    voltage which is proportional to the sound level ?
     
  6. Each end ot the transformer winding puts out a sine wave referenced to
    the ground on the center tap of the winding. So both ends of the
    winding do the same thing, but on opposite halves of the cycle.
    Taking the top half...
    The first series cap is just like the one on your opamp. The first
    diode keeps the negative half cycle from going below ground, by making
    the right side of the cap charge up positive with respect ot the left
    side, so that when the wave turns around and starts positive, it
    starts from ground, instead of a negative voltage. The second diode
    dumps the peak positive voltage from this pushed up wave into the
    right side of the first cap along the middle of the circuit (the one
    that is connected to ground). That full peak to peak positive voltage
    is then used as the reference point that the third diode uses as its
    clamp voltage for the second capacitor in the top row, which passes
    the same AC wave that went through the left one. but now the wave is
    forced to swing from the positive voltage from the first center row
    cap to the full peak to peak voltage more positive that that, etc.
    etc.

    The top and bottom rows both dump charge into and draw it from the
    middle row of caps on alternate half cycles.
    That would work, using only the positive half of the sound wave. But
    it will still waste the first diode drop before any positive signal
    comes out. The capacitor will charge up on the biggest peaks and
    discharge very slowly (set by what resistance you connect across it).
     
  7. Once you have the basic black box concept of how an opamp works, most
    of these are pretty obvious. Do you have a conversational
    understanding of the concept of an opamp?
     
  8. Guest

    Guest Guest

    Yes, you are most likely right about that, it is so trivial they
    omit explanations. I simply like explanations if only to
    confirm what I think. Yes, the basic principle of the op-amp
    with a ve- feedback is that the minus input follows the plus
    input.
     
  9. Guest

    Guest Guest

    What is this beast? My guess is its an audio noise detector for intrusion
    Thanks, I sort of get it now, I mean with the capacitor and 2 diode's.
    Firstly, I assume that the first diode can have any meaning only because
    the op-amp has a ve- voltage and because it's o/p can go below GND.
    Then, when the signal is -ve, the capacitor charges through D1. Then,
    when the voltage becomes +ve, the capacitor conducts, but because
    in the +ve cycle the current goes through R, it does not change it's
    charge and what we get is the o/p voltage plus the volateg from the
    -ve cycle which is still on the capacitor, and this is the doubling effect.
    Is this about right ?

    Regarding as to the purpose of this circuit, the answer is that it is some
    exercise on the net to demonstrate how sound level decreases with
    distance, and I think they conclude there that is follows the inverse
    square law relationship. You can play with the link I provided by
    subtracting the directories one by one, this will reveal all their files,
    eventaully you can reach this
    http://hop.concord.org/s1/mess/s1ma2dgrm.html#anchor884816
    The reason I adopted this circuit is that I want do build a proper
    sound level meter, so I though why not use this one (I found it
    from yahoo). Do you think it will be OK ? The only other option
    I believe is to use a proper audio amplifier ( preferabley in a discrete
    component form) instead of an op amp, but I am not sure if we need
    this.
     
  10. I don't quite know what you mean by, "the circuit gets stuck". But if
    R is infinity, then the charge that gets pumped into C by the first
    diode is trapped there, and the output voltage just goes up starting
    at zero, and back down during one cycle.

    I think of the two diodes as having the following function. The cap
    is like a bucket that is raised and lowered by the opamp output
    voltage. The first diode represents the bucket going below the
    surface of water, and the water spills into the bucket. The second
    diode is like a tip mechanism that dumps the bucket little by little
    as it is raised out of the water (emptying it through the resistor).
    This is a very rough analogy, but gives the general sense of how the
    doubler works.
     
  11. -----------------
    These guys are guessing.

    The output stage is NOT a voltage doubler, though there is a slight
    resemblance, this is merely a half-wave rectifier. The diode to ground
    simply allows cap recharge.

    The op-amp is merely an inverting amp and the gain is -1000. The 10K
    to ground is merely the parallel of 10K and 10M for optimal common
    mode biasing. The upper input is (-) and the lower is (+) (inverting
    and non-inverting).

    The input device looks like an electret mic which is why there is a
    bias resistor from +9V and why there is a cap to isolate the offset
    and pass only AC.

    This is just a sound meter that relies on a -dB setting on the
    voltmeter.
    -Steve
     
  12. Guest

    Guest Guest

  13. Guest

    Guest Guest

  14. I spent a few minutes modelling this circuit using spice.

    The really odd thing about this circuit is the biasing seems wrong; the
    circuit runs without a -V rail, but the V+ input is grounded. Seems like the
    V+ rail should be pulled up to Vcc/2.

    In addition, there doesn't seem to be any DC path for current to ground,
    except through the input of the opamp. I'm not sure why, but this seems to
    push the output to float up to about 2V even when the V- input is > the V+
    input. I think it depends on the leakage of the V- input.

    If the input amplitude is big enough, it just manages to push the V- input
    below the V+ (which is at ground + some small amount) and generates a
    positive spike on the output, because of the huge gain of the amp
    configuration. The length of the spike is determined by how long the input
    pushes V- below V+, and thus the 'duty cycle' is determined by the amplitude
    of the wave.

    I'm not sure why the diodes and variable resistor are there. They don't seem
    to do anything important. It looks like somebody was trying to design a
    charge pump. If so, it doesn't work as advertised. It actually attenuates
    the output at 1k, which is probably an important frequency for a sound
    detector.

    The thing that bugs me about this circuit is the lack of proper biasing. I
    think the duty cycle will depend on the leakage of the input, which isn't a
    predictable parameter. It may be that for a given opamp, its always the
    same, and so tuning the circuit will only have to be done once. However, it
    may also make the output temperature sensitive.

    Seems like a simple integrator would be a better solution, assuming that
    this is a sound level meter...

    Regards,
    Bob Monsen
     
  15. Are we looking at the same circuit?
    http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF

    I missed it the first glance, too, but there is a -9v supply in pin 4.
    The + input is grounded, and the 10 meg feedback resistor pulls the -
    output to ground also. This is probably too much resistance for the
    bias current or a 741, but there is a path. At least the + input
    should also have a 10 meg resistor to ground, instead of 10k to
    balance the voltage drops caused by the input bias currents.
    It is just a linear amplifier with a gain of 1000.
    A charge pump is exactly what it is. Otherwise known as a voltage
    doubling rectifier.
    Certainly, when the pot is set to zero resistance it not only
    attenuates the output but puts an AC short circuit to ground on the
    opamp output. Not a particularly elegant design.
    This is just not right.
    An integrator does not measure average level but just low pass filters
    a signal. I think an ideal rectifier followed by a peak hold circuit
    might have been more useful, but this thing will provide a DC volt
    meter reading that is roughly proportional to sound level (except for
    the diode losses).
     
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