Operational amplifier

Discussion in 'Homework Help' started by regexp, Dec 28, 2010.

  1. regexp

    Thread Starter New Member

    Nov 20, 2010
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    Hello,

    I have the following operational amplifier:

    [​IMG]

    My task is to calculate the expression that yields the quotient ~~\frac{u_0}{i_s}

    Hm, i don't really know how to start, any advice?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Consider the features of an ideal opamp.

    1. Infinite open loop gain.
    2. Infinite input impedance on both input terminals
    3. Output impedance is 0 ohms.

    Also consider that an opamp's mission is to drives its output to whatever voltage it takes to make the two input terminals equal in voltage. In this opamp configuration, the positive terminal is at 0v. That means that the opamp is going to drive its output to a voltage that will force its negative terminal to be 0V.


    hgmjr
     
  3. shteii01

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    Feb 19, 2010
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    \frac{U_0}{I_s}=R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}-1)
    Can someone check me?
     
  4. Ron H

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    That's not correct.
     
  5. thatoneguy

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    Feb 19, 2009
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    Think about it as "What current will flow through R3?"

    Since R1 and R2 are voltage dividers across the current source, and R3 is the tap to the output.....
     
  6. shteii01

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    Ok. So. There is no current in R3? Then U0=R2(Is)?
     
  7. Ron H

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    No one said there is no current in R3. How did you conclude that?
     
  8. shteii01

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    I am just stumbling in the dark. When I solved it the first time, I thought R2 and R3 would act as current divider, that was shut down. Now I am out of ideas.
     
  9. Jony130

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    Feb 17, 2009
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    Here's a tip:
    You can convert the T-pad network to a PI-pad network using the Y-Δ transformation:
    http://en.wikipedia.org/wiki/Wye-delta

    Or you could just write out your node equations

    [​IMG]

    Is = ( Va - Vb) /R1 (1)

    Is = Vb/R2 + (Vb - Vout)/R3 (2)

    Va = 0V ( thanks to op amp "action" )
     
    Last edited: Dec 29, 2010
  10. Ron H

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    Are you the same person as regexp?
     
  11. hgmjr

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    Jan 28, 2005
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    Greetings Shteii01,

    Like RonH, I too am confused. Your replies are phrased in such a manner that it sounds like you are the originator of this thread.

    Can you please confirm that you "are" or "are not" regexp?

    hgmjr
     
  12. shteii01

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    Feb 19, 2010
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    This is not my thread. I am trying to solve it because I am not very good at op-amp circuits, I thought it would be good practice for me. Note that I only posted my answers, I still want regex do his own homework.
     
  13. shteii01

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    Feb 19, 2010
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    I see what you mean. I think I can do node equations on the original circuit.

    Ok, I redid the problem using node voltage method on the original circuit. I got following: \frac{U_0}{I_s}=-R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}+1)
     
    Last edited: Dec 29, 2010
  14. Jony130

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    Your equation look good for me
     
  15. Ron H

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    Yeah, that's what I got too.
     
  16. BillO

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    Nov 24, 2008
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    The minus sign looks wrong. Everything else looks right.
     
  17. hgmjr

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    I obtained the same answer as shteii01 using Millman's Theorem.

    hgmjr
     
  18. Ron H

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    The output voltage is bound to be negative, due to the direction of the current and the feedback.
     
  19. BillO

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    Nov 24, 2008
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    Hmm, suit your self, but there are valid ways to work this out so that you do not end up with negaitve ohms as your answer.
     
  20. Ron H

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    Put your money where your mouth is.:D
     
    Last edited by a moderator: Dec 30, 2010
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