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Op amp trouble!

  1. Aug 3, 2011 #1
    Here is the Problem.

    http://img577.imageshack.us/img577/3769/screenshot20110803at231.png [Broken]

    Uploaded with ImageShack.us
    I know the first part where the signal goes from A to B and i know what the signal will look like at B, as i used the inverting op amp formula.

    Vo=Vin x -Rf/R1

    I understand from the image the op amp between A and B is an inverting op amp and which Resistor is which, Rf=100kOhms and R1=3kOhms.

    I worked Vout at B to be 0.4V so the signal drawn would just be the same shape as the signal above but drawn mirrored above the the 0V line and peak at 0.4V.

    The trouble is with the second part between B and C.

    I think the second op amp is a non-inverting op amp as the input from B goes into the + input of the op amp but i do not know how to calculate the output as the layout of the resistors and the use of a Capacitor are confusing me.

    Any help on this would be great as i feel i can understand it but the layout of the circuit is boggling my brain.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 3, 2011 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    I believe you are correct for the first question.

    The second stage would appear to be in a "comparator" configuration. That is, it is not operating as an opamp with negative feedback. Instead, a comparator compares the two signals at its + and - inputs, and the output drives to the rails depending on whether the input comparison gives a positive or negative result.

    So look at the voltage at the - input from the voltage divider -- what is that voltage? And which way will the comparator drive when the input voltage at "B" is above that reference voltage? Which way will it drive when it is below that reference voltage?

    Now as for the output of the comparator, it is drawn incorrectly, or at least you are not given enough information to sketch the waveform. If the comparator output can drive both up to the + rail and down to ground, and if the output impedance of the comparator is low, it will overdrive the RC at its output. If the output impedance is comparable to the impedance of the resistor, then there will be a voltage divider effect. If the output of the comparator is open-drain or open-collector, then things change....

    Do you have any more information about the comparator (like a device designator, or its output impedance)?
     
    Last edited by a moderator: May 5, 2017
  4. Aug 4, 2011 #3
    no sorry thats all the information you are given in the image.

    Can anyone else make sense of the 2nd part?
     
  5. Aug 4, 2011 #4

    berkeman

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    Staff: Mentor

    Well if that's all you are given, they are wanting you to do some independent research to figure it out. Pretty tricky of them....

    Hint -- I checked, and the 2nd component has the same pinout as the jellybean LM111 comparator. Take a look at the datasheet for the LM111 comparator to see what kind of output drive circuit it has. That will let you answer the 2nd question. Let us know what you figure out.
     
  6. Aug 5, 2011 #5
    Thanks for all your help, I know they wont be wanting me to do any independant reasearch becuase this is a simple 1st year electronics class assignment so it shouldnt be too hard im just not good at it. i could send you the paper to look at if you think im missing out something?

    Thanks
     
  7. Aug 5, 2011 #6

    berkeman

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    Just look at the datasheet. That will give you what you need to figure out the output characteristics of the comparator that is in your problem.
     
  8. Aug 5, 2011 #7

    berkeman

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    Also, have you covered comparators yet in your coursework?
     
  9. Aug 5, 2011 #8
    yes i have covered comparators but i dont really understand much of it.
     
  10. Aug 5, 2011 #9

    berkeman

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    What is the output stage of the LM111 comparator? How does it pull its output low? Can it drive its output high? How will the output stage affect the waveform at the RC circuit connected to the comparator's output?
     
  11. Aug 5, 2011 #10
    I dont think it is an LM111 comparitor cause we never learnt things as complicated as learning the names of various comparitors, just what a comparitor is and what it can be used for aswell as some simple calculations, nothing like LM111 or anything. and as for pulling the output low and driving high i dont understand at all. Im rubbish at this stuff. sorry.
     
  12. Aug 5, 2011 #11

    berkeman

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    Staff: Mentor

    It may not specifically be an LM111, but I am pretty sure they mean for you to assume that the comparator has an output stage similar to an LM111. There is no other practical way to solve the problem, IMO.

    Please read the datasheet of the LM111, with an eye toward what the output characteristics are. Your answer is in that.
     
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