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Network function for an opamp circuit

  1. Mar 13, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-3-13_16-48-23.png

    2. Relevant equations


    3. The attempt at a solution
    upload_2017-3-13_16-51-46.png
    upload_2017-3-13_16-52-23.png

    I'm really unsure of what to do. I feel like Vs should not be equal to Vo. I am not sure if I have the right idea with the equations that I have written. I am confused because I don't know the value for omega. Can someone help ? Thanks.
     
  2. jcsd
  3. Mar 13, 2017 #2

    gneill

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    Staff: Mentor

    Write the node equation for node 1 (figure below) knowing that its potential is equal to that of the source voltage. Be careful with the impedance of the capacitor: What's the difference between the impedance formulas for a capacitor and an inductor?

    upload_2017-3-13_17-21-44.png

    Edit: By the way, the "network function", or "transfer function" is a function of ω, so it will remain as a variable in your result.
     
  4. Mar 13, 2017 #3
    upload_2017-3-13_19-18-28.png
    would this be the node equation, since the current going into the op amp is 0?
     
  5. Mar 13, 2017 #4

    gneill

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    You're right that the current going into the op-amp input is zero. But there are a couple of things wrong with your equation. In the third term you shouldn't have voltage in the denominator when all the other terms are divided by an impedance. The first term is written as a current flowing out of the node, so if the other terms are on the other side of the equals sign they must represent currents flowing into the node. That's not what the order of the voltages in the terms suggest.

    To avoid errors by making the process automatic, a good approach is to always write a node equation as a sum of terms that sum to zero. That is, place all the terms on one side of the equals sign and a zero on the other. Choose one direction for the assumed current direction for all the terms, either all into the node or all out of the node, then write all the terms with that one assumption. The math will then take care of the details and you don't have to try to figure out ahead of time what the "real" current directions are.
     
  6. Mar 13, 2017 #5
    upload_2017-3-13_19-53-32.png
    Would this be the node equation then? If I assume that the current is entering the node in all three directions. Since it would be Vs-Vo / (1/jwc) the the jwc comes to multiply on top ?
     
  7. Mar 13, 2017 #6

    gneill

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    Actually you've written it as all the currents flowing out of the node (the node is at potential Vs, so Vs - Vo would be for a current flowing from Vs to Vo). Looks good. Plug in the capacitor value and solve for Vo.
     
  8. Mar 13, 2017 #7
    Vo=0.2j+5Vs, now can I simply divide them to get the network function ?
     
  9. Mar 13, 2017 #8

    gneill

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    That doesn't look right. There should still be ω involved. Can you show more of your work?
     
  10. Mar 13, 2017 #9
    Sorry, ignore that it was completely wrong:

    Vo=200kwj+5Vs
     
  11. Mar 13, 2017 #10

    gneill

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    Nope. Show your work.
     
  12. Mar 13, 2017 #11
    upload_2017-3-13_20-45-55.png

    Tried to revise it
     
  13. Mar 13, 2017 #12

    gneill

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    Much better! You could clear the decimal by multiplying top and bottom by 5:

    ##V_o = V_s \left( \frac{25 + jω}{5 + jω} \right)##

    Now it just remains to form the transfer function H(ω) as described in the problem statement.


    Edit: Inadvertently wrote ##V_A## instead of ##V_s##. Fixed it.
     
  14. Mar 13, 2017 #13
    Ok, I do so by just dividing Vo/Vs. So in the end there is no more term with Vs ?
     
  15. Mar 13, 2017 #14

    gneill

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    Right. H is the ratio of the output to the input and is a function of ω.
     
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