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Opamp Circuit with Feedback Problem

Discussion in 'Electronic Basics' started by Guest, Dec 19, 2003.

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  1. Guest

    Guest

    8
    0
    Nov 7, 2014
    Hello,
    In the Electronics Club of our university, we are weekly solving
    several problems at our meetings. But this time, noone could solve
    this problem.

    http://it.isikun.edu.tr/tango/electronics/circuit.jpg


    If anyone is able to solve this, we would be grateful because many
    friends of mine are on the job but still no solutions available.
    (although there are speculations about the solution).

    Thank you very much,
    Electronics Club
     
  2. Ban

    Ban Guest

    || Hello,
    || In the Electronics Club of our university, we are weekly solving
    || several problems at our meetings. But this time, noone could solve
    || this problem.
    ||
    || http://it.isikun.edu.tr/tango/electronics/circuit.jpg
    ||
    ||
    || If anyone is able to solve this, we would be grateful because many
    || friends of mine are on the job but still no solutions available.
    || (although there are speculations about the solution).
    ||
    || Thank you very much,
    || Electronics Club

    it is very simple:
    Vo= Vdc + A*V1 + B*V2
    Vdc is 10V, thats the voltage on the non inverting input. The DC-gain is 1
    A = ((10k||20k)/(10k||20k)+77k)*(100k/10K) = 0.797
    B = -(100k/10k) = -10
    assuming we are at at high frequency, the RC-time constants you should be
    able to insert yourself, I'm not gonna do all the work.
    Hey what kind of university is that where you are not even able to solve
    this simple task?
     
    Last edited by a moderator: Dec 11, 2014
  3. Don Pearce

    Don Pearce Guest

    Almost right, but not quite(a bracket needs moving, and the last term
    is wrong):

    A = ((10k||20k)/(10k||20k+77k))*((100k+10k)/10K) = 0.876

    d

    _____________________________

    http://www.pearce.uk.com
     
    Last edited by a moderator: Dec 11, 2014
  4. Ban

    Ban Guest

    ||||| Hello,
    ||||| In the Electronics Club of our university, we are weekly solving
    ||||| several problems at our meetings. But this time, noone could solve
    ||||| this problem.
    |||||
    ||||| http://it.isikun.edu.tr/tango/electronics/circuit.jpg
    |||||
    |||||
    ||||| If anyone is able to solve this, we would be grateful because many
    ||||| friends of mine are on the job but still no solutions available.
    ||||| (although there are speculations about the solution).
    |||||
    ||||| Thank you very much,
    ||||| Electronics Club
    |||
    ||| it is very simple:
    ||| Vo= Vdc + A*V1 + B*V2
    ||| Vdc is 10V, thats the voltage on the non inverting input. The
    ||| DC-gain is 1 A = ((10k||20k)/(10k||20k)+77k)*(100k/10K) = 0.797
    ||| B = -(100k/10k) = -10
    ||| assuming we are at at high frequency, the RC-time constants you
    ||| should be able to insert yourself, I'm not gonna do all the work.
    ||| Hey what kind of university is that where you are not even able to
    ||| solve this simple task?
    ||
    || Almost right, but not quite(a bracket needs moving, and the last term
    || is wrong):
    ||
    || A = ((10k||20k)/(10k||20k+77k))*((100k+10k)/10K) = 0.876
    ||
    || d
    ||

    THX for the correction Don. I was too fast!
    A = (10k||20k)/(10k||20k+77k)*(1+100k/10K) = 0.876

    ciao Ban
    Bordighera, Italy
    electronic hardware designer
     
    Last edited by a moderator: Dec 11, 2014
  5. Genome

    Genome Guest

    20K and 10K off 30V puts the non-inverting (plus bit) at 10V. That's Mr Ohms
    law. V = IR. 30 volts across 20K plus 10K iz 30 volts across 30K which gives
    a whole 1mA of current. 1mA of current through a 10K resistor gives 10 of
    the volts.

    Mr 10uF capacitor blocks the DC path to Mr V1 so Mr 77K does not enter into
    the DC party.

    Mr Op-amp sticks his output so that his inputs (plus and minus bits) sit at
    the same volts. Mr other 10uF capacitor blocks the DC yellow brick road to
    Mr V2 so Mr Op-amp can stick his minus bit the same as his plus bit without
    driving current through his 100K feedback resistor bit.

    No current times 100K is no volts. So Mr Op-amp sticks his output at 10 of
    the volts.

    If Mr V1 goes for a wobble then his 10uF friend couples his wobble to Mr 77K
    resistor.

    If Bill is the 20K resistor and Ben is the 10K resistor then they find their
    way to Ground and become parallel with a value of (Bill x Ben)/(Bill + Ben).
    That's because Mr Bill might be connected to 30V..... but 30V is a voltage
    source connected to ground. Voltage sources have no Bill or Ben.

    Try it like this.

    Call Martha a volt and let her wiggle her ass across Bill. The sexual
    excitement, current, for Bill is Martha/Bill. Now we are probably talking a
    bit of face sitting or getting into oral sex here.......

    Since Martha is a woman she can multitask and wiggle her volt over Ben at
    the same time. That gives a Martha/Ben.

    The total sexual excitement is Martha/Bill + Martha/Ben. Cross pollinate and
    you get.... Martha x Ben/Bill x Ben + Martha x Bill/ Bill x Ben which
    makes....

    (Martha x Ben + Martha x Bill)/(Bill x Ben) is the current

    Extract a Martha and

    Martha x (Ben + Bill)/(Bill x Ben) is the current......

    Obviously Martha has a strapon and Bill and Ben are begging for it.

    I think I've lost the plot here....

    DNA
     
    Last edited by a moderator: Dec 11, 2014
  6. Isik University, Istanbul. The palimpsest upon which he drew the schematic
    is interesting - take the image, mirror and invert it to better read. Best
    of luck to them!
     
  7. I read in sci.electronics.design that Walter Harley
    AIUI, they are much better at electrical power and BIG THICK cables.
     
  8. That has to be the most creative and least informative method I've ever seen
    for circuit analysis. Though I will admit to almost falling off my chair
    laughing.
    Think so?

    --Russell
     
  9. Dan Dunphy

    Dan Dunphy Guest

    Nobody picked up on the source impedance mismatch with the feedback
    resistor. This will create a large offset on any op amp with any input
    current. Yes, dcout is 10V with ideal op amp.

    Colorado Springs, CO
    My advice may be worth what you paid for it.
     
    Last edited by a moderator: Dec 11, 2014
  10. Guest

    Guest

    8
    0
    Nov 7, 2014
    Hello,
    "Walter Harley" <> wrote in message news:<[email protected]>...
    > "Ban" <> wrote in message
    > news:I1CEb.13991$...
    > > Hey what kind of university is that where you are not even able to solve
    > > this simple task?

    >
    > Isik University, Istanbul. The palimpsest upon which he drew the schematic
    > is interesting - take the image, mirror and invert it to better read. Best
    > of luck to them!


    In fact, we are not the university Electronics club, I know nothing
    about Electronics. I'm an IT student, as you may notice from the
    e-mail address.
    I made up this Electronics club thingy because noone will answer such
    a simple question, because it is obviously a homework for
    non-electronics engineering classes.

    The reverse page is a Requirement Analysis Document of a software
    project, which is an assignment of a student. It's nothing serious,
    but yeah, the English and the goals are funny :)

    Sorry for the lies..

    Thanks for the answers though, I have understood it.
     
  11. Ross Mac

    Ross Mac Guest

    Very Funny!....and creative too!
     
    Last edited by a moderator: Dec 11, 2014
  12. Jim Thompson

    Jim Thompson Guest

    See OpAmpCktWithFeedbackProblem.pdf on the S.E.D/Schematics page of my
    website.

    ...Jim Thompson
     
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