Help with this OPAMP CKT

Discussion in 'General Electronics Chat' started by Management, Oct 2, 2007.

  1. Management

    Thread Starter Active Member

    Sep 18, 2007
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    Problem I:

    Can someone help decribe what this circuit is meant to do ie. its basic operation? I am not quite sure what is going on.

    I have attached the input/output waves and the actual circuit diagram.

    The smaller signal (green) is the input (from the left) and the larger signal (yellow) is the output of the OPAMP.

    If you can just let out as much as your thoughst as possible because I might not be able to respond quickly. 13 hours time difference. Thank you in advance.

    -Andrew
     
  2. Management

    Thread Starter Active Member

    Sep 18, 2007
    306
    0
    Problem II:

    This is what I believe to be an integrator but it is not working the way it is on the board in the lab. The pictured ckt is exactly as it is on the board but is there anything I must add to correctly model this ckt so that it behaves in spice the same way it is suppose to when I use the scope in the lab.

    Attached is the input (yellow) and the output (green). In spice I always get a -4V saturation. Should I put a resister across the feedback cap?

    -Andrew
     
  3. Dragon

    Active Member

    Sep 25, 2007
    42
    0
    For your FIRST circuit; I think its meant to be some kind of an oscillator, as we have both the postive and negative feedback connections. The additional capacitor at the inverting feedback stregnthens the argument.

    However, the ocillators are known to provide a square waveform. In this case, the combination of resistors might be responsible for the sinusoid output. Further more, if you have a close look at the waveform, it reveals that for the uneven input waveform, we have a positive peak at the output when the input is staggering in its negative region and vice versa. Hence, the conlusion for an oscillator.

    The second circuit gave me a headspin. Could'nt come up with any explanations for it.
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The second circuit appears to be a low pass filter.
     
  5. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    The first circuit looks like an oscillator to me, since it uses a positive feedback network as well as a negative feedback network. But I might be wrong. What confuses me is the presence of those capacitors.
     
  6. Management

    Thread Starter Active Member

    Sep 18, 2007
    306
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    Thank you for all the responses. I will look more into this oscillator and see if I can come back with an explanation.
    Question: What will the Oscillator output voltage in this case be a function of? How would it be related to the output of a D/A converter and it's output voltage? Thank you.

    I have an additional circuit that serves the same purpose somewhere as the 2nd circuit that I posted. They do the same thing but are configured differently. At first glance I said it was an integrator, taking the average of the input signal. What you are seeing is that sine wave from the so called Oscillator feeding this circuit. At the output of the first Opamp the signal is roughly the same is the previous Problem II but shift 90 degrees. The output is similar a DC constant voltage. No matter what I do I get saturation. And for some reason I am getting an offset of approx. 3.3mV everytime at the noninverting into. I know Integrators do not like that but on the real thing there is no offset (or very small) and gives me a nice DC ouput. Why? How is this thing averaging. Is it averaging? :confused:

    BTW these are all 2kHz signals although this one may not look like it.

    Sorry if these are simple questions.

    -Andrew
     
  7. Management

    Thread Starter Active Member

    Sep 18, 2007
    306
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    In case anyone wanted to know, I now understand the first problem. It is a bandpass filter centered at 2kHz (which makes sense to me) and has 20dB of gain. Without the positive feedback it would be a bandpass filter centered at 2kHz but it is not a very good filter and has about -4 to -5 db of gain.

    If you Spice and look at the frequency response you will see. I then understand some of the circuits around it now.

    Thank you for stirring my thoughts.

    The problems I have are still Spice related with Problem II and the added similar problem. :confused:
     
  8. hon

    New Member

    Jun 22, 2006
    4
    0
    Are the capacitor values of problem II correct? a quick look at the poles and zeros of the circuit, they are dominant poles@DC and R126C112, and a dominant zero at R126C111. Assuming R125C110 provide stability at some higher frequency, the top part of the schematic plays the dominant role. It has extremely high gain at DC and decreases to some gain determined by R126 and R125 in the frequency region between the dominant zero and pole. Frequencies beyond the pole is attenuate by the pole. This type of filter (which I call an integrating LPF) is known as a lead-lag and used in control systems to set the stability of a system, for example in a phase-locked system. So C112 should be much smaller than C111. The product of C125C110 should be smaller or about the same as C126C112. If the values you showed is correct then I am not sure exactly what it is trying to accomplish.

    2Cents
     
  9. Management

    Thread Starter Active Member

    Sep 18, 2007
    306
    0
    Thank you so much for your comments and observation. Those values were indeed used and I am just trying to understand what it is doing.

    The output of this ckt is feedback to provide what I think is stability. The circuit on a whole does not work if it is not feedback. Without feedback the signal you see as the input is offset by some value. Feeding back removes that and that is why I think it is providing some stability to the system. I will read a little more into the use of a Intergrating LPF as a lead-lag. I also think that the level of the offset in the input if there was no feedback is the output DC output of this ckt if feedback is connected.

    BTW when I say is feedback I mean to somewhere way before this circuit.
     
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