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Checking the Gains of the Given Circuit

  1. Aug 21, 2011 #1
    Hi all,
    About a month ago I had thought I completed what was the final question of my EE summer assignment... I thought wrong. My professor just posted a final problem, an easier problem involving less work. But I decided to ask for some guidance before continuing. I have also attached an example solution of the problem he's provided.

    1. The problem statement, all variables and given/known data

    (Attachment below - EE P8)

    In each part of this problem, you will be given an expression for the gain of the above
    circuit. Assume that you solved the circuit and found the given expression (answer).
    For each given expression (answer), comment on whether the answer is either correct,
    plausible, or incorrect. Please state reasoning for the choice you picked.

    2. Relevant equations

    (Attachment below - EE P8)

    3. The attempt at a solution

    My question is, should I begin plugging in 0 for the different resistors as he did in the example solutions? I know it's a completely different circuit than the one that I have, but I thought that might be where I start. Any thoughts and input would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2011 #2

    gneill

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    Staff: Mentor

    You could start by solving for the gain and identifying any correct answers directly. Judging the result of setting this or that resistance to zero (or infinity) requires that you understand the circuit operation. This understanding can best be obtained by first solving the circuit!

    Also, on the solution sheet, for answer (d) I think that he meant to set G3 = 0, not R3 = 0. If G3 is zero then R3 goes to infinity (equivalent to removing the resistor from the circuit) and his argument stands. If R3 is set to zero then the gain must be zero, since the output would be shorted to ground.
     
  4. Aug 21, 2011 #3
    Gotcha, (I don't know if you remember helping me a month ago, I can help refresh your memory if needed) so I would solve as I did for the gains of the other circuits? Solving for [itex]\frac{V_o}{V_i}[/itex]
     
  5. Aug 21, 2011 #4

    gneill

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    I remember :smile:

    Yes, solve for [itex]\frac{V_o}{V_i}[/itex]. Use what you know about the characteristics of the ideal op-amp.
     
  6. Aug 21, 2011 #5
    Great :biggrin: out of curiosity, is R3 in parallel with R2 and R4?
     
  7. Aug 21, 2011 #6

    gneill

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    No. Although a certain property of the op-amp's input terminals will allow you to consider R2 and R3 to be in parallel. Not that this would be of tremendous help to you...

    Consider writing a KCL equation for the node formed by the confluence of R2, R3, and R4. Assign an arbitrary voltage to that node; Maybe call it V2. You should also be able to write KCL for the "-" input terminal node.
     
  8. Aug 21, 2011 #7
    Makes sense. I did this and these are my two equations. Now my thoughts would be to solve for [itex]\frac{V_z}{R_2}[/itex] on the "-" nodal equation and substitute into the Vz nodal equation..

    EDIT: Apologies on the small upload, if it's difficult to read just let me know! :smile:
     

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  9. Aug 21, 2011 #8

    gneill

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    If you solve the first equation for V2 then you can use it to eliminate all the V2's in the second equation and solve it for Vo (and then Vo/Vi).

    EDIT: Note that you can repeat the operations using conductances (G1, G2, ...) and find the gain in terms of them, too.
     
  10. Aug 21, 2011 #9
    Ok I've done so, just asking for a check on the current work I've done. If correct I think I know where I'm going with it and will continue, just thought I'd see if I was correct so far.
     

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  11. Aug 21, 2011 #10

    gneill

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    From what I can make out from the Lilliputian writings :smile:, you're doing okay so far.
     
  12. Aug 21, 2011 #11
    Hahaha apologies for that, I took a different approach... to the scanning that is :tongue: How does this look so far? I know there's got to be something in my next step dealing with that R3 I feel
     

    Attached Files:

  13. Aug 21, 2011 #12

    gneill

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    Looks good indeed. Can you combine the terms and put them over a single denominator?
     
    Last edited: Aug 21, 2011
  14. Aug 21, 2011 #13
    Great!! :biggrin: This is what I was mentioning, my thoughts now would be to eliminate the R3 in the denominator and bring it to the numerator
     
  15. Aug 22, 2011 #14
    In doing so I would multiply the right side of the equation with R3 over 1 which would eliminate it in the denominator and leave R1 in the denominator. Is that a legal algebraic move though?
     
  16. Aug 22, 2011 #15

    gneill

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    Nope. Anything you do to one side that changes its value you must do to the other. Multiplying the numerator or denominator by a constant that is not equal to 1 will change the value of the expression.

    There's nothing wrong with having a denominator that's R1*R3. :wink:
    Take a look at some of the proposed gain expressions in the question statement for confirmation.
     
  17. Aug 22, 2011 #16
    I see I see, I remember doing this algebraic move in one of my previous problems (only thing is all my completed homework is at school, I moved in on Saturday and thinking I wouldn't get another problem I brought all my school work/books and put them in my bedroom :frown:). I do see a provided answer with a denominator of R1*R3 - Isn't the required move to multiply the two terms without R3 by R3/R3?
     
  18. Aug 22, 2011 #17

    gneill

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    Yup. You can always multiply a term by one and leave the value unchanged. R3/R3 = 1.
     
  19. Aug 22, 2011 #18
    Got it :smile: - so with this gain, that proves that letter D is correct :biggrin: Now, dealing with conductance, I know a little about this and have briefly researched a little bit about it. From what I remember, and correct me if I'm wrong, it is the inverse of resistance?
     

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  20. Aug 22, 2011 #19

    gneill

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    Yup. G = 1/R.

    Now, you could stick in 1/Gi for each Ri in the gain formula you've derived, and then clean up the algebra. Or, which might be preferable, try the gain derivation from scratch using the conductances rather than the resistances.

    For conductance G = 1/R, with voltage across it V, the current is I = V/R = V*G. You'll find that your KCL equations look pretty clean when conductances are used!
     
  21. Aug 22, 2011 #20

    uart

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    Yes that's the correct answer. Also one of the other answers is correct too.

    To find an expression using conductances just follow these two simple steps.

    1. Replace all occurrences of resistance in the gain formula by 1/G, eg replace R1 with 1/G1 etc.

    2. Multiply top and bottom by the product (G1 G2 G3 G4)
     
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