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Difference and summing OP AMP

  1. Sep 9, 2006 #1
    hey guys, i have a HW problem of a combination of a summing and difference op amp. I got the first few steps of the problem, but i dont know how to continue from where i am. I am having a hard time solving for v_+. any help is greatly appreciatd, maybe if you guys can not solve the problem for me, bt direct me towards the right path. thanks in advance. i have attached a scanned image of my work.
     

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  2. jcsd
  3. Sep 10, 2006 #2

    berkeman

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    Staff: Mentor

    I'm having a little trouble undstanding the circuit diagram. What are the "1" and "Z"? Are they meant to represent static quantities? Or is Z a varying input? Or is Z really a 2, like in 2.0V?

    As for solving for V+ and V- and Vout, just write the sum of all currents into the + node is zero and do the same for the - node. Since the opamp is ideal, what simplification does that let you make?
     
  4. Sep 10, 2006 #3
    the "1",".25",".75",".67", and "2" are all voltages. are you saying that i should right nodal equations for the inputs? i didnt quite understand what you meant.....
     
  5. Sep 10, 2006 #4

    berkeman

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    Yeah, heck if they're static voltages, then you can solve for Vo directly. Just write the two equations for the V- and V+ inputs and solve for Vo. You didn't answer my quiz question yet about what properties does an ideal opamp have that makes this problem easier....
     
  6. Sep 10, 2006 #5

    berkeman

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    BTW, in order to avoid confusion, in engineering it is a convention to write those numbers like this:

    1.0, 0.25, 0.75, 0.67, 2.0

    That helps to identify them as numbers, rather than variables or other things. Also, the preceedinig 0 on the numbers less than 1.0 helps to be sure that people don't miss the decimal point. Especially when writing fast on paper, using this convention will help make your work clearer (to you and to others reading/correcting the work).
     
  7. Sep 10, 2006 #6
    IDEAL CHARACTERISTICS:
    A_ol=inf
    R_i=inf
    R_o=0
    i_+=0
    i_-=0
    v_in=v_+ - v_-
    v_out=A_ol(v_+ - v_-)
    v_+=v_-

    but i dont see how any of them would make this problem easier
     
  8. Sep 10, 2006 #7

    berkeman

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    I was referring to the first two that you listed. Very high gain, and zero input current. That lets you assume that Vin+ = Vin-, and just write the voltage divider equation for the + input. Set the - input voltage to the same, and write the voltage divider equation for the - input, including the Vo term. Then just solve for Vo.
     
  9. Sep 10, 2006 #8
    ok so from what i understand from what you said, here is what i have done:
    .067/2000 + 2/4000 + -2/4000 + ((2-v_o)/50000) = 1/2000 + 0.25/3000 + 0.75/5000

    then i solve for v_o?
    does that look correct?
     
  10. Sep 10, 2006 #9

    berkeman

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    No, or at least I'm not tracking what you are writing.

    Start with figuring out the voltage at the Vi+ terminal of the opamp. You should write the sum of the three curents into the Vi+ terminal (in terms like (1.0V - Vi+)/2000, and set that sum equal to zero. Then solve for Vi+. Then set Vi- = Vi+, and write the sum of the four currents into the Vi- node and set that sum to zero. That sum will have one term that looks like (Vo - Vi-)/50000, plus three other terms. The only unknown is Vo, so just solve for it then.

    EDIT -- Fixed a typo.
     
  11. Sep 10, 2006 #10
    heres another shot:
    ((1-v_i+)/2000) + ((.25-v_i+)/(3000)) + ((.75-v_i+)/5000)) = 0
    v_i+=.7097

    v_o-=v_i+

    ((v_o- - .7097)/(50000)) + ((1-.7097)/(2000)) + ((2-.7097)/(4000)) + ((0-.7097)/(4000))=0
    v_o-=-13.81
     
  12. Sep 10, 2006 #11
    what confuses me is that why cant the current i_1= 1/2k, i_2=.25/3k, i_3=.75/5k for the currents entering v_i+?

    also would the current into the node with the two 4k resistors be 2/4k + 0-v_i-/4k?
     
  13. Sep 10, 2006 #12

    berkeman

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    Vi+ is not ground, so to find the current through the resistor, you need to take the voltage *difference* across the resistor and divide it by the resistance.

    And on your second question, you need to sum all the currents into the node, not just two.

    This looks good to me. I didn't check the math in the final step, but the equation is set up correctly.
     
  14. Sep 10, 2006 #13
    so even though this is a part summer, part difference amplifier, the method we used will apply? anyway, thanks alot for your help berkeman, this problem was giving me nightmares last night..
     
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