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Weird looking circuit -- Can someone check my attempt?

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data
    This is kind of what the circuit looks like:
    upload_2016-11-28_0-4-31.png

    2. Relevant equations
    KCL

    3. The attempt at a solution
    This my attempt at "normalizing" it:
    upload_2016-11-28_0-12-40.png
    upload_2016-11-28_0-21-43.png

    Solving Attempt:
    ##\frac{V_{in}-V_o}{R_1}+C_1\dot{V_{in}}=0##
    ##\frac{V_o-V_{in}}{R_1}+\frac{V_o}{R_3+R_2}=0##

    Are my equations correct?
     

    Attached Files:

    Last edited: Nov 28, 2016
  2. jcsd
  3. Nov 28, 2016 #2

    cnh1995

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    Homework Helper

    No. The output voltage does not involve ground (-ve terminal of Vin). Hence, Vo=Vο+-Vo-. You need to use both Vo+ and Vo- in the KCL equation.
    Also, voltage across the capacitor is not Vin as you have taken in the first equation.

    What exactly is the question asking for?
     
  4. Nov 28, 2016 #3
    Sorry, I thought I wrote it out. The prompt is "Find the differential equation for Vo"
    node vo+ (Vo+ = Vop, Vo- = Vom)
    ##\frac{V_{op}-V_{in}}{R_1} + \frac{V_{op}-V_{om}}{R_3+R_2}=0##

    ##C_1(\dot{V_{om}}-\dot{V_{in}}) + \frac{V_{om}-V_{op}}{R_2+R_3}=0##
     
  5. Nov 28, 2016 #4

    cnh1995

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    Homework Helper

    I didn't follow your equations. Basically, the currents do not add to zero. You are taking only two currents out of three. I think KCL method is not useful here.
     
  6. Nov 28, 2016 #5
    Is this better?
    Vo+ = V1
    Vo- = V2
    upload_2016-11-28_11-21-45.png

    upload_2016-11-28_11-22-41.png
    ##\frac{V_{in}-V_1}{R_1} + C_1(\dot{V_{in}}-\dot{V_{2}})=0##

    upload_2016-11-28_11-24-56.png
    ##\frac{V_1}{R_3}+\frac{V_1-V_{in}}{R_1}=0##

    upload_2016-11-28_11-25-55.png
    ##\frac{V_2}{R_2}+C_1(\dot{V_2}-\dot{V_{in}})=0##
     
  7. Nov 28, 2016 #6

    cnh1995

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    Homework Helper

    The currents do not add to zero. I believe you need to use a different method. I'll get back to you on this later.
     
  8. Nov 28, 2016 #7
    Is nodal analysis doable or do I need to use meshcurrent?
     
  9. Nov 28, 2016 #8

    gneill

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    Staff: Mentor

    Nodal analysis is doable, you just have to recognize how to apply it. It might help if you were to forget +Vo and -Vo labels for a moment and replace them with V1 and V2 (Hence Vo = V1 - V2). Find V1 and V2 separately.
     
  10. Nov 28, 2016 #9
    upload_2016-11-28_20-40-33.png

    ##\frac{V_{in}-V_1}{R_1} + C_1(\dot{V_{in}}-\dot{V_2}) = 0##
    ##\frac{V_1-V_{in}}{R_1}+\frac{V_1}{R_3}=0##
    ##\frac{V_2}{R_2}+C_1(\dot{V_2}-\dot{V_{in}})##
    What is wrong with my equations?
     
  11. Nov 28, 2016 #10

    gneill

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    The first one is not a node equation. Vin is not an essential node, it is part of the reference supernode since it has a fixed potential with respect to the reference node.

    The second equation is good. Solve it for V1.

    The third equation is also good. Solve it for V2.
     
  12. Nov 28, 2016 #11
    ##R_3V_1-R_3V_{in}+R_1V_1=0##
    ##V_2+R_2C_1\dot{V_2}-R_2C_1\dot{V_{in}}##

    ##V_1=\frac{R_3V_{in}}{R_3+R_1}##
    ##V_2=R_2C_1(\dot{V_{in}}-\dot{V_2})##

    ##V_o=V_1-V_2##
    ##V_o=\frac{R_3V_{in}}{R_3+R_1}-R_2C_1(\dot{V_{in}}-\dot{V_2})##
    How would I get rid of ##\dot{V_2}##?
     
    Last edited: Nov 28, 2016
  13. Nov 28, 2016 #12

    gneill

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    Staff: Mentor

    Please do not edit content out your previous posts. It makes followups discussing that content nonsensical. If you do edit a post, make sure that you leave a note in the post indicating what changes you made and why. It is best to have a policy of only adding to a previous post, not removing content. Fixing typos is fine.

    Solve for V2 before you combine with V1. The V2 equation is a simple differential equation. Hint: What is ##\dot{V_{in}}##?
     
  14. Nov 28, 2016 #13
    I edited that post on accident and didn't know how to undo. My bad.
    ##\frac{V_1-V_{in}}{R_1}+\frac{V_1}{R_3}=0##
    ##V_{in} = V_1(1+\frac{R_1}{R_3})##
    ##\dot{V_{in}}=\dot{V_1}(1+\frac{R_1}{R_3})##
    ##V_2=R_2C_1(\dot{V_1}(1+\frac{R_1}{R_3})-\dot{V_2})##
     
  15. Nov 28, 2016 #14

    gneill

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    Question: Is Vin a DC source or an AC source?
     
  16. Nov 28, 2016 #15
    The problem was given like this as practice for the exam. I think the professor said something about Vin being an AC sinusoid for the exam. Also the prompt for this problem is "Find the differential equation for Vo". I forgot to include that in the OP.
     
  17. Nov 29, 2016 #16

    gneill

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    Staff: Mentor

    I see. That makes things a bit trickier. I don't think you can eliminate V2' without using the actual function for Vin. I suppose you could write V2 in terms of Vo, since Vo = V1 - V2. That would at least lead to a differential equation for Vo.
     
  18. Nov 29, 2016 #17
    What if ##V_{in}## was sin(t)?
    How would I go about solving?
     
  19. Nov 29, 2016 #18

    gneill

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    Staff: Mentor

    If you have the actual function then you have its derivative, too. You can plug those into your V2 node equation and solve the differential equation. But that would lead to the overall solution, not the differential equation for Vo.
     
  20. Nov 29, 2016 #19
    What do you mean for overall solution? Is that not the differential equation for Vo?
    How would I get Vo in terms of Vin? I don't see how to eliminate V2, even with a function for Vin
     
  21. Nov 29, 2016 #20

    gneill

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    Staff: Mentor

    It would be the time domain solution for Vo. Some combination of sine and cosine functions.
    You make the substitution for V2 that I suggested using Vo = V1 - V2. Then V2 becomes V1 - Vo, and you already have V1 in terms of Vin.
     
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