OP-AMP question?

Discussion in 'Homework Help' started by error404, Apr 25, 2010.

  1. error404

    Thread Starter New Member

    Apr 25, 2010
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    [​IMG][​IMG]

    How can i find Vo voltage and Io current? This question is very complicated for me. Please help me.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    The op amp will likely go to full voltage on either power supply rail, so it depends on the component being used. Very few op amps anywhere near the power supply rails.

    You don't have a component for the op amp, you don't have power supply definitions.

    In short, as stated the question is unanswerable.
     
  3. retched

    AAC Fanatic!

    Dec 5, 2009
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    Being a homework question, it is either ideal or a 341 ;).
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    Show us how you think the solution to the problem should be set up.

    hgmjr
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    Find V- (inverting input voltage with respect to ground) and V+ (non inverting input voltage with respect to ground).

    Then equate V- and V+ (ideal op amp) and you should find Vo with respect to Vin.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Using the hint Mik3, see if you can take a stab at solving the problem and post it here.

    hgmjr
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    V+=(Vo*12k)/(12k+8k)=0.6*Vo , voltage divider rule

    current through 4k and 16k is: assume current flows from Vo to 6V

    I=(Vo-6)/(4+16)k=(Vo-6)/20k

    thus

    V-=6+I*4k=6+4k(Vo-6)/20k=6+(Vo-6)*0.2=6+0.2*Vo-1.2=0.2*Vo+4.8

    assuming V-=V+

    0.6*Vo=0.2*Vo+4.8

    thus

    0.4*Vo=4.8

    thus

    Vo=12V

    Io=12V/10k=1.2mA
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    Millman's Theorem Solution.......

    \small V_-=V_+

    \frac{\frac{6}{4K}+{\frac{V_o}{16K}}}{\frac{1}{4K}+{\frac{1}{16K}}}=\frac{\frac{V_o}{8K}}{\frac{1}{8K}+{\frac{1}{12K}}}

    \frac{6(16K)+4K(V_o)}{16K+4K}=\frac{12K(V_o)}{12K+8K}

    \frac{6(16K)+4K(V_o)}{20K}=\frac{12K(V_o)}{20K}

    \normalsize {6(16K)+4K(V_o)}={12K(V_o)}

    \normalsize {8K(V_o)}={6(16K)}

    \large {V_o}={\frac{6(16K)}{8K}}

    \large {V_o}={6(2)}

    \large {V_o}=12V

    ****************************************

    V_+=\frac{12K(12)}{12K+8K}

    V_+=\frac{12K(12)}{20K}

    V_+=7.2V

    ****************************************

    I_o=\frac{V_o-V_-}{16K}+\frac{V_o-V_+}{8K}+\frac{V_o}{10K}

    I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}

    I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}

    I_o=0.0003+0.0006+0.0012=0.0021A

    hgmjr
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    Hgmjr,

    You are right about Io. I thought Io is the current through the 10k resistor only.
     
  10. error404

    Thread Starter New Member

    Apr 25, 2010
    5
    0
    Thank you for your answers .
     
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