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What kind of op-amp is that?

  1. Feb 27, 2012 #1

    Femme_physics

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

    [​IMG]

    At first I thought a comparator, then I noticed a line connecting the Vout and the lines that connect to V- and V+, so it can't be a comparator, can it? I don't see any other option....

    As far as the question, I need to find out the unknown resistors, but I'll start working on it once I figure that basic fact.
     
  2. jcsd
  3. Feb 27, 2012 #2

    rude man

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    As a start, you can either ignore Rx or the circuit is in saturated mode ... this is a consideration since the output current is defined to be 3 mA. The rest is just the usual KVL equations of an active op amp circuit (input voltages equal etc.).
     
  4. Feb 27, 2012 #3

    gneill

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    Looks like an amplifier with feedback, although the feedback mechanism is made a bit trickier due to the presence of Rx. What effect do you suppose the zener diode is going to have?

    Rb and R1 form a voltage divider. What voltage are they dividing? What's the op-amp going to try to do with that voltage?
     
  5. Feb 27, 2012 #4

    I like Serena

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    I'd say it looks a bit like an inverting amplifier, but it's not quite it.
    KVL and KCL should tell what it does.
     
  6. Feb 27, 2012 #5

    rude man

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    I hadn't noticed the output voltage is given. Since R1 is given, all components are defined implicitly.

    You can solve this problem without writing any KVL equations. Hints: what must Ra be to satisfy Iz_min? What must Rb be to give a 12V output? And finally, what must Rx be to give 3 mA of current flowing into the op amp?
     
  7. Feb 27, 2012 #6
    There are a few opamps with open collector outputs. This could be one of them.
     
  8. Feb 27, 2012 #7
    I would say it is a non-inverting amplifier with voltage of 5V on the + input and voltage gain of R1/(R1+Rb).
    Vout =+12V so Rb can be calculated.... and so on
     
  9. Feb 27, 2012 #8

    rude man

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    Nope on your gain expression ... but you're warm ...
     
  10. Feb 27, 2012 #9
    gain of (R1+Rb)/R1 = 12/5 = 2.4
     
  11. Feb 29, 2012 #10

    Femme_physics

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    Well, this is the solution my classmate offered

    [​IMG]

    I agree on Ra. It's just KVL.

    But on Rb-- I'm confused as far as how to use a voltage divider.

    I CAN indeed apply KVL

    Since I know that the 3 mA split at this point marked in red:

    [​IMG]

    I know that there's only 1 mA going through Rb and R1, and I know they have a 12V potential difference to the ground. So,

    Sum of all V = 0 ; 12 - 1ma x Rb -1ma x R1 = 0

    I get that Rb 2000 ohms. Makes sense?


    As far as Rx -- well, I don't really understand something fundemental about the circuit. Is Vs some type of another Vout? Or does it just define the limits of the Op-Amp like we see in those Vcc+ Vcc- sort of thing? I really don't know how to approach Vs. How can 25 Volt comes out of an op-amp who only produces a Vout of 12v?!? And how does any of that helps me with Rx?


    Sorry-- A lot of questions, I know. Just a confusing circuit!
     
  12. Feb 29, 2012 #11

    NascentOxygen

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    +Vs is the power supply (you've previously seen it as +Vcc) to power the OP-AMP. There is no -Vcc here, the negative power supply here is ground (you've previously seen OP-AMP's negative supply as -Vcc, but not with this arrangement here).
     
  13. Feb 29, 2012 #12

    NascentOxygen

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    Proof please! :tongue:
     
  14. Feb 29, 2012 #13

    NascentOxygen

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    Are you referring to the words he's written to the right of the resistor string? :frown:
     
  15. Feb 29, 2012 #14

    Femme_physics

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    Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?


    Well, sum of all I entering that crossection I marked in red

    3 -2 -I1 = 0
    I1 = 1 mA

    There you go...

    LOL I didn't c that..sorry..ignore that part.
     
  16. Feb 29, 2012 #15

    I like Serena

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    It does.
    The arrow indicates that the wire goes to a power supply.
    It does not indicate the direction of the current.
    The actual current flows away from the 25V power supply.



    What happened to the current coming from the 25V power supply that is coming in through Rx?


    How can we now that it's out there! ;)
     
  17. Feb 29, 2012 #16

    Femme_physics

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    OOOOOhhhhhhhhh!!!! AHHHHHHA!!!

    OH! OH!

    Now I get it :)


    It's kinda late now but i'll sit with it tomorrow trying t finalize my results based on this new evidence!

    Well, it just says "Or take it"

    Or being my name. Telling me that I should take this exercise and try to solve it. He just put the "e" in front of the "k" by accident :)
     
  18. Feb 29, 2012 #17

    I like Serena

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    OOOOOhhhhhhhhh!!!! AHHHHHHA!!!
    Or take it!

    Yes, I think I understand now what he meant. ;)
     
  19. Feb 29, 2012 #18

    NascentOxygen

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    :yuck: I believe you realise this is so not right that you need to make a fresh start.

    Hint: you know V+ so determine V_.
     
  20. Feb 29, 2012 #19

    Femme_physics

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    Taking your criticisms into consideration, here is my new idea...

    [​IMG]


    ;)
     
  21. Mar 1, 2012 #20

    I like Serena

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