Op amp circuit help #2

Discussion in 'Homework Help' started by notoriusjt2, Dec 13, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]

    this is the first question I have received that does not state "Assume that all op amps are ideal". Should I assume that in this situation or is this all together different?
     
  2. steinar96

    Active Member

    Apr 18, 2009
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    I think you can safely assume it's ideal. Otherwise you would be given parameters which make it non-ideal such as input and output resistances, open loop gain. Etc
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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    Greetings norotiusjt2,

    What answer did you come up with? Be sure to show us your work so that we can determine any errors that you make.

    hgmjr
     
  4. notoriusjt2

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    Feb 4, 2010
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    I am coming up with answer (D) Vo = 4*Vin

    Say we had 10V @ Vin. that would mean that +5V would have to be at +/- op amp terminals.

    Since that network of resistors are all the same value (R), and based on their configuration they would basically divide the output voltage of the op amp by 4, effectively giving us 5V back at the - terminal on the op amp itself.

    This is only true if Vo = 4 * Vin... hence the answer

    am I correct in my reasoning?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    How did you arrive at that conclusion? I would think that for an ideal op-amp the voltage across the input +/- terminals would be 0V.
     
  6. notoriusjt2

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    Feb 4, 2010
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    thats correct...

    for example if you put a voltmeter across the + and - of the op amp you would get 0V.

    but if you put a voltmeter from + to ground or from - to ground you would have 5V
     
  7. Ron H

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    Apr 14, 2005
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    In post #4, you used 10V as the input. In post #6, you say that the op amp +input will be 5V. Which is it?
     
  8. notoriusjt2

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    Feb 4, 2010
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    If Vin = 10V then wouldn't the positive side be +5V and the negative side be -5V? for a total peak to peak of 10V?
     
  9. Ron H

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    The negative side is connected to ground. Ground is your reference, so is defined as zero volts.
     
  10. notoriusjt2

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    Feb 4, 2010
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    Right! so that means that the positive side of the power supply is automatically +10V?
     
  11. Ron H

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    Apr 14, 2005
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    Yes.
    Although, technically, it's not a power supply.:D
     
  12. t_n_k

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    Mar 6, 2009
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    Hi notoriusjt2,

    You seem to be confusing input voltage with the op-amp supply voltage.

    In this case the op-amp supply voltage is not stated as it is irrelevant for the purposes of solving the problem.

    If the op-amp were capable of running from either a single or dual polarity supply the outcome would be the same for this problem.
     
  13. notoriusjt2

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    Feb 4, 2010
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    I understand... I was just assigning an arbitrary value for Vin for the sake of demonstration.
     
  14. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    allright
    [​IMG]


    if
    node1 = 1V
    then
    node2 = 1V

    now that I look at it, node3 and node4 are the same point
    and node6 and node2 are also the same point

    in order for this to work correctly
    node4 would have to be 4V
    node5 would be 2V
    node6 = node2 = 1V

    no?
     
  15. Ron H

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    Apr 14, 2005
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    No. See the attachment.
     
  16. notoriusjt2

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    Feb 4, 2010
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    i see.... how did you calculate those voltages @ the resistors?
     
  17. Ron H

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    Apr 14, 2005
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    I started with these facts:

    1. The voltages at the op amp inputs are equal.
    2. The op amp input current is zero.

    Then I just used Ohm's law to find the currents in the resistors, and the voltage across each resistor. Do you see the currents that I added to the schematic?
     
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  18. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    yes I see, that all makes sense now... Thank you!
     
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