Type of Op Amp

Discussion in 'Homework Help' started by tquiva, Dec 22, 2010.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
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    [​IMG]

    I identified the basic op amps in the figure, but don't know what type of op amp is the one circled in green. Could someone please help me finish this problem?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    You may have mislabeled some components. The two components circled in green are resistors. The triangle is the op amp. It is probably an ideal one, so it has no real-world counterpart.
     
  3. tquiva

    Thread Starter Member

    Oct 19, 2010
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    Yes, it's an ideal op amp.
    But how would I go about solving the one circled in green according to these equations?

    [​IMG]
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I got Vo/Vi=½.
     
  5. tquiva

    Thread Starter Member

    Oct 19, 2010
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    How did you calculate 1/2?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The resistors circled in green form a simple voltage divider. So the voltage presented to the +ve op-amp terminal V+, is related to Vi by V+=(1/3)*Vi. You can assume this is the case because an ideal op-amp would have zero bias current into the input terminals.

    From the +ve terminal to the op-amp output terminal the circuit is a non-inverting stage with the output voltage given by (1+4/2)*V+.

    Between the op-amp output terminal and the output point (node Vo) there is another voltage divider with divider factor 5/10=0.5

    This may help you to find the overall gain Vo/Vi.
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You have three circuits there. V1 goes into voltage divider, result is V1'. V1' goes into non-inverting amplifier, result is Vo'. Vo' goes into voltage divider, result Vo. Do the math. I get ½.
     
  8. tquiva

    Thread Starter Member

    Oct 19, 2010
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    I'm still lost on how 1/2 is the answer:
    For the first voltage divider on the left, I obtained:

    V1 = (1/3) Vi

    Then for the voltage output of the non-inverting op-amp, I got:
    V2=(1+ (4/2) ) * V1 = (3)*(1/3) Vi=Vi

    Then I obtained Vo with the following voltage division:
    Vo = [ 5 / (5+5) ] *V1 = 5/10 * Vi = (1/2) Vi

    So I divided Vo/V1 = [ (1/2)Vi] / [(1/3)Vi] = 3/2

    What did I do wrong?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's not clear what your terms Vi and V1 indicate.

    In any case I think you really want the relationship Vo/Vi where Vi is the left most input node.

    You already have

    Vo = [ 5 / (5+5) ] *V1 = 5/10 * Vi = (1/2) Vi


    So Vo/Vi=1/2
     
    tquiva likes this.
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Somebody double check me please.
     
  11. t_n_k

    AAC Fanatic!

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    Looks fine.:)
     
  12. haykp

    New Member

    Oct 7, 2010
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    I want to give some general notices/ideas about op-amp without going in deep details in equations.
    So op-amp always tries to bring the voltage difference in its inputs to 0:
    prov:
    As ideal opamp has infinite gain, hence out =A(inpls - inmns), so as A is infinite and out is constant => (inpls - inmns) should equal to 0 => inpls = inmns
    Based on this principle you can understand all the above brought schemes.

    BTW: GOOD Question!!
     
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